4
$\begingroup$

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,

$\endgroup$
2
  • 1
    $\begingroup$ This identities always look like shameless lies to me. :) $\endgroup$ May 7, 2012 at 16:41
  • $\begingroup$ @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 $$ (-e^{-4\pi};e^{-4\pi})^2_\infty =2^{\frac{1}{8}}e^\frac{\pi}{3}\sqrt{\sqrt{2}-1}.$$ $\endgroup$ May 7, 2012 at 18:49

1 Answer 1

10
$\begingroup$

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

$$\frac{\left(-e^{-4\pi};e^{-4\pi}\right)_{\infty}^{2}}{\left(e^{-2\pi};e^{-2\pi}\right)_{\infty}^{4}}=\frac{16\pi^{3}\left(\sqrt{2}-1\right)\sqrt{2^{\frac{1}{4}}+2^{\frac{3}{4}}}}{\Gamma^{4}\left(\frac{1}{4}\right)}.$$

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

$$\frac{32\pi^{3}2^{\frac{1}{8}}\sqrt{\sqrt{2}-1}}{\Gamma^{4}\left(\frac{1}{4}\right)}.$$

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

$$\left(e^{-2\pi}\right)_{\infty}=\frac{\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}e^{\frac{\pi}{12}}$$

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

$$\left(e^{-8\pi}\right)_{\infty}=\frac{2^{-\frac{13}{16}}\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}\left(\sqrt{2}-1\right)^{\frac{1}{4}}e^{\frac{\pi}{3}}.$$

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

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

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.

$\endgroup$
1
  • 2
    $\begingroup$ It's great to see this site working in a way that is close to actual research. $\endgroup$ May 10, 2012 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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