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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,

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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

1 Answer 1

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

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

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.

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Splendid! Thanks a thousand ;) –  handelskai May 7 '12 at 18:49
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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

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