Unpopular "elementary" theorems/identities to impress an audience of mathematicians. - MathOverflow [closed]most recent 30 from http://mathoverflow.net2013-05-23T13:38:36Zhttp://mathoverflow.net/feeds/question/51013http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/51013/unpopular-elementary-theorems-identities-to-impress-an-audience-of-mathematiciUnpopular "elementary" theorems/identities to impress an audience of mathematicians.To be cont'd2011-01-03T13:31:00Z2011-01-03T20:05:06Z
<p>This question grew out of my recent job interview. Since the interviewers were math professors, I had a hard time searching for interesting elementary theorems in case I got asked for one.
I thought topics such as the Banach-Tarski paradox, Godel's theorems, the Mandelbrot set, the Brouwer Fixed Point Theorem, etc were well-known and wouldn't do the job. However, after a cursory search, I found some to my taste:</p>
<p>1.<s><a href="http://en.wikipedia.org/wiki/Marden%2527s_theorem" rel="nofollow">Marden's theorem</a> (or <a href="http://www.maa.org/joma/Volume8/Kalman/index.html" rel="nofollow"> here</a>)(It is not Marsden.)</s> <a href="http://en.wikipedia.org/wiki/Gauss-Lucas_theorem" rel="nofollow">Gauss–Lucas theorem</a></p>
<p>2.The identity $\int_{0}^1 \frac{x^4(1-x)^4}{1+ x^4} dx = \frac{22}{7}- \pi$</p>
<p>So, my question here is an invitation to expand the list (of theorems that would get an interviewee accepted).</p>
<p>To recap, my criteria for selection are </p>
<ol>
<li>Not widely known,</li>
<li>Elementary- understandable to a first year grad student, and</li>
<li>Interesting-i.e. MOtizens, assuming they are the audience, will be delighted to have come across it. </li>
</ol>
<p>Thank you.</p>
http://mathoverflow.net/questions/51013/unpopular-elementary-theorems-identities-to-impress-an-audience-of-mathematici/51015#51015Answer by Andrey Rekalo for Unpopular "elementary" theorems/identities to impress an audience of mathematicians.Andrey Rekalo2011-01-03T13:43:32Z2011-01-03T13:43:32Z<p>The AM-GM inequality is implied by an identity.</p>
<p>For a function $f=f(x_1,x_2,\dots,x_n):\mathbb R^n\to\mathbb R$ let $Pf(x_1,x_2,\dots,x_n)$ denote the sum of $f$ over the $n!$ quantities that result from all possible $n!$ permutations of the $x_i$. Then
$$\frac{x_1^n+x_2^n+\dots+x_n^n}{n}-x_1x_2\dots x_n=\frac{1}{2\ n!}(\phi_1+\phi_2+
\dots \phi_n),$$
where
$$\phi_k=P[(x_1^{n-k}-x_2^{n-k})(x_1-x_2)x_3x_4\dots x_{k+1}]$$
$$=P[(x_1-x_2)^2(x_1^{n-k-1}+\dots x_2^{n-k-1})x_3x_4\dots x_{k+1}]\geq0.$$
The proof dates back to Hurwitz. </p>