MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is this identity or an equivalent one already referenced in the litterature? In particular, is it even true?

${\frac{\left ( -1 ; e^{-4\pi} \right) ^2_{\infty}}{\left ( e^{-2\pi} ; e^{-2\pi} \right) ^4_{\infty}} = \frac{32 \pi^3 \left(\sqrt{2}-1 \right )\sqrt{2^\frac{1}{4}+2^\frac{3}{4}}}{\Gamma^4{\left(\frac{1}{4} \right )}} \simeq 4,030103529...}$

It appears in relation to a particular elliptic function.

Similar identities also arise in this post

Thanks in advance,

share|cite|improve this question
This identities always look like shameless lies to me. :) – Mariano Suárez-Alvarez May 7 '12 at 16:41
@Handelskai: That was my comment, sorry I removed it, I did so as I realized I could post a complete answer. The denominator is not exactly $\eta(i)^4$, there is a factor of $e^\frac{\pi}{3}$ missing. You instead have $$\frac{1}{2}(-1;e^{-4\pi})^2_\infty =2^{\frac{1}{8}}e^\frac{\pi}{3}\sqrt{\sqrt{2}-1}$$ or in other words $$ (-e^{-4\pi};e^{-4\pi})^2_\infty =2^{\frac{1}{8}}e^\frac{\pi}{3}\sqrt{\sqrt{2}-1}.$$ – Eric Naslund May 7 '12 at 18:49
up vote 9 down vote accepted

The equality is indeed correct. It follows from identities in Ramanujan's notebook.

First notice that $\left(-1;e^{-4\pi}\right)_{\infty}^{2}=2\left(-e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2},$ so we are trying to prove the identity


The right hand side many be cleaned up further, and written as


Now, since $1+q^{4}=\frac{1-q^{8}}{1-q^{4}},$ the left hand side is

$$\frac{\left(e^{-8\pi};e^{-8\pi}\right)_{\infty}^{2}}{\left(e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2}\left(e^{-2\pi};e^{-2\pi}\right)_{\infty}^{4}}=\frac{\left(e^{-8\pi}\right)_{\infty}^{2}}{\left(e^{-4\pi}\right)_{\infty}^{2}\left(e^{-2\pi}\right)_{\infty}^{4}}.\ \ \ \ \ \ \ \ \ \ (1)$$

On page 326 of Bruce C Brendts “Ramanujan's Notebook Part V” he shows that


$$\left(e^{-4\pi}\right)_{\infty}=\frac{2^{-\frac{3}{8}}\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}e^{\frac{\pi}{6}}$$ and


Combining these three together in equation (1) yields the desired result.

Remark: You could have proceeded in a different manner by noticing that the denominator is (almost) the Dedkind eta function. The product can be written as


Indeed there are many ways to write this product, another I stumbled across is $$\frac{\left(-e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2}}{\left(e^{-2\pi};e^{-2\pi}\right)_{\infty}^{4}}=\left(\frac{1}{\vartheta_{4}\left(e^{-4\pi}\right)\vartheta_{4}\left(e^{-2\pi}\right)}\right)^{2},$$ where $\vartheta_4(q)$ is a Jacobi theta function.

share|cite|improve this answer
Splendid! Thanks a thousand ;) – handelskai May 7 '12 at 18:49
It's great to see this site working in a way that is close to actual research. – Charles Matthews May 10 '12 at 15:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.