series expansion of the q-Pochhammer symbol - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:18:52Z http://mathoverflow.net/feeds/question/111648 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111648/series-expansion-of-the-q-pochhammer-symbol series expansion of the q-Pochhammer symbol Carlo Beenakker 2012-11-06T14:36:40Z 2012-11-20T01:50:43Z <p>The following identity arose while I was working on a recent <A HREF="http://mathoverflow.net/questions/108927/a-question-about-a-formal-power-series-manipulation" rel="nofollow">MO question</A>:</p> <p>$-\sum_{n=1}^{\infty}\frac{1}{n}\frac{(-x)^n}{1-x^n}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^n}{1-x^{2n}}.$</p> <p>I have no doubt that the identity is true, but I am not able to prove it. Can anyone help?</p> <p>It is easy to prove by Taylor expansion that the left-hand-side of the identity can equivalently be written as $\sum_{n=1}^{\infty}\ln(1+x^n)$, which is the logarithm of the <A HREF="http://en.wikipedia.org/wiki/Q-Pochhammer_symbol" rel="nofollow">q-Pochhammer symbol</A> $(-x,x)_{\infty}$, so an alternative way to pose my question is to ask for a proof of the series expansion</p> <p>$\ln(-x,x)_{\infty}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^{n}}{1-x^{2n}}.$</p> http://mathoverflow.net/questions/111648/series-expansion-of-the-q-pochhammer-symbol/111651#111651 Answer by Gjergji Zaimi for series expansion of the q-Pochhammer symbol Gjergji Zaimi 2012-11-06T15:10:13Z 2012-11-20T01:50:43Z <p>First notice that $$\sum _{n=1} ^{\infty} \frac{x^n}{n(1-x^{2n})} = \sum _{r=0} ^{\infty} \sum _{m=1} ^{\infty}\left(\frac{1}{2^r}\sum _{k|2m-1} \frac{1}{k}\right)x^{2^r(2m-1)}.$$ And similarly $$-\sum _{n=1}^{\infty}\frac{(-x)^n}{n(1-x^n)} = \sum _{s=1}^{\infty} \left(\sum _{k|s}\frac{(-1)^{k+1}}{k}\right)x^s.$$ So we need to show that the respective coefficients match, i.e.: $$\frac{1}{2^r}\sum _{k|2m-1} \frac{1}{k}=\sum _{k|s}\frac{(-1)^{k+1}}{k},$$ for $s=2^r(2m-1)$. But this is a simple corollary of $\frac{1}{2^r}=1-(\frac{1}{2}+\cdots+\frac{1}{2^r})$.</p> http://mathoverflow.net/questions/111648/series-expansion-of-the-q-pochhammer-symbol/111671#111671 Answer by Garth Payne for series expansion of the q-Pochhammer symbol Garth Payne 2012-11-06T19:06:09Z 2012-11-06T19:40:25Z <p>I would make a mere comment since Gjergji has already answered, but I am not allowed to make comments.</p> <blockquote> <p>... so an alternative way to pose my question is to ask for a proof of the series expansion <code>$\ln(-x,x)_{\infty}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^{n}}{1-x^{2n}}$</code>.</p> </blockquote> <p>This is a corollary of Euler's theorem that the number of partitions of $n$ into distinct parts is equal to the number of partitions of $n$ into odd parts. In terms of generating functions, <a href="http://en.wikipedia.org/wiki/Partition_%28number_theory%29#Generating_function" rel="nofollow">Euler's theorem is just</a> <code>$(-x,x)_{\infty}=\frac{1}{(x,x^2)_\infty}$</code>, which can be <em>easily</em> proved by replacing the term <code>$(1+x^i)$</code> in <code>$(-x,x)_\infty$</code> by <code>$\frac{1-x^{2i}}{1-x^i}$</code> and cancelling all the terms in the numerator against the corresponding terms in the denominator. By Euler's theorem, <code>$\ln \left( (-x,x)_{\infty}\right) =\ln\left( \frac{1}{(x,x^2)_\infty}\right) =\sum_{i=1}^\infty \ln \left( \frac{1}{1-x^{2i-1}} \right) =\sum_{i=1}^\infty \sum_{n=1}^\infty \frac 1 n x^{n (2i-1)}$</code> </p> <p><code>$=\sum_{n=1}^\infty \sum_{i=1}^\infty \frac 1 n x^{n (2i-1)}$</code> <code>$=\sum_{n=1}^\infty \frac 1 n \frac{x^n}{1-x^{2n}}$</code>.</p>