Consider the following function: $G(z) = \prod_{n \in \mathbb{Z}} {1 \over{\tanh^2\left(\pi\left(z-n\right)\right)}}$.

By constuction, it has poles at $z=m+in$ with $m,n \in \mathbb{Z}^2$.

Additionally, it is doubly periodic such that for any other value $G(z)=G(z+1)=G(z+i)$.

Now, strong computational evidence shows that it admits the following special values:

$G({i \over 2}) = 0, G({1 \over 2}+i{1 \over 4}) = 1$ and $G({1 \over 2}+i{1 \over 2}) = \frac{\sqrt{2}}{2}$

However, I have no idea how to find a proof for this.

Anyone can help?


2 Answers 2


As a meromorphic doubly-periodic function, it is an elliptic function in the classical sense. You didn't mention the fact that it is an even function, but that also helps. The poles are double, so that up to addition of a constant this must be a constant multiple of the Weierstrass P-function for the lattice of Gaussian integers. Since the P-function in this case is 0 at a known point, your third value gives the additive constant as (presumably) 0.5.

Everything about this function is standard in the theory of complex multiplication. However you came up with it, the formulae are probably written down somewhere.

'''Edit''': What I wrote before is probably fine, but starting from your formula it probably makes sense to use the Weierstrass sigma function to express the infinite product first. We're in the case where one of the roots of the cubic in the equation for the Weierstrass P-function is 0, and that means the P-function itself is easy to express in terms of the sigma function.


Thanks to replies in other posts I now have enough elements to completely answer this question. Looking at the above product formula, we may rewrite the product term:

$$T_n(z)={{\left(1-\tanh\left(\pi z\right) \tanh\left(\pi n\right)\right)^2}\over{\left(\tanh\left(\pi z\right) - \tanh\left(\pi n\right)\right)^2}}, \space \space (1)$$

Such that for $n=0$ it equals ${1}\over \tanh^2\left(\pi z\right) $, which becomes 0 when $z \rightarrow \frac{i}{2}$.

As a result, the entire product becomes zero since it is absolutely convergent for $z$ different from a Gaussian prime. This achieves to prove the first special value:

$$G\left( \frac{i}{2} \right) = 0$$

Now, if $z=\frac{1}{2}+\frac{i}{4}$, the product term expression (1) may be rewritten:

$$T_n\left(\frac{1}{2}+\frac{i}{4}\right)=-{{\left(i + \tanh\left(\frac{\pi}{2} \left(2n-1 \right) \right)\right)^2}\over{\left(i - \tanh\left(\frac{\pi}{2} \left(2n-1 \right) \right)\right)^2}},$$

It is then easy to verify that:

$$T_n\left(\frac{1}{2}+\frac{i}{4}\right)T_{1-n}\left(\frac{1}{2}+\frac{i}{4}\right) = 1$$

Such that the terms in the product may be rearranged by $(n, 1-n)$ pairs to yield:

$$G\left(\frac{1}{2}+\frac{i}{4}\right)=1, \space (2)$$

Similarly to a related post over an elliptic function (*), we are seeking solutions in a linear form involving Weierstrass P function such as to have the zeros at $z=\frac{i}{2}$:

$${G\left(z \right ) = A \left( \wp \left(z \right ) - \wp \left( \frac{i}{2} \right ) \right )}, \space (3)$$

Now, since the values of the Weierstrass Function for $z=\frac{i}{2}$ and $z=\frac{1}{2}+\frac{i}{4}$ are well known (see for instance Abramowitz and Stegun) we can readily infer the value of $A$.

Indeed, We have

$$\wp\left(\frac{i}{2}\right)=-\frac{\Gamma^4\left(\frac{1}{4}\right)}{8 \pi} \space \text{and, }\space \space\wp\left(\frac{1}{2}+\frac{i}{4}\right)=\left(\sqrt{2}-1\right)\frac{\Gamma^4\left(\frac{1}{4}\right)}{8 \pi}, \space (4)$$

So combining (2), (3) and (4) yields:

$$A =\frac{4 \sqrt{2}\pi}{\Gamma^4\left( \frac{1}{4} \right)}$$

In addition, we have:

$$A = \lim_{z\rightarrow 0} {G\left(z\right)z^2} = {{\left(-1;e^{-2\pi}\right)^4_{\infty}} \over {2 \pi^2 \left(e^{-2\pi};e^{-2\pi}\right)^4_{\infty}}}$$

As in a previous post this leads us to a Q-Pochhammer symbols identity:

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

But most interestingly, when combining with the obtained results in (*) we get a very amazing formula, with a flavour of Ramanujan's Cos/Cosh Identity!

$$2^{\frac{3}{8}}\sqrt{\sqrt{2}+1}\left(\prod_{n \in \mathbb{Z}} {1 \over{ 1 - {1 \over{\cosh\left(2\pi\left(z-n\right)\right)}}}}\right) -\left(\prod_{n \in \mathbb{Z}} {1 \over{\tanh^2\left(\pi\left(z-n\right)\right)}}\right)=1 $$

Where the linear combination of the two series is constant equal to $1$, independently of the variable $z$.


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