Consider the following even meromorphic doubly periodic function with poles at the gaussian integer lattice.

$H(z) = \prod_{n \in \mathbb{Z}} {1 \over{ 1 - {1 \over{\cosh\left(2\pi\left(z-n\right)\right)}}}}$

Computational evidence suggests special values for: $H({i\over4}) = H({3i\over4}) = 0$

and rather amazingly

$Real \left( H(z) \right) = H\left({{1+i}\over 2}\right) = 0.847201266746891$

to remain constant on the of the unit square diagonals.

How would one go proving this and/or finding an exact analytic formula for this constant?

Similar interrogations also arise in this post

  • 3
    $\begingroup$ inverse symbolic calculator suggests the exact value may be 1/2*(2^(1/4)+2^(3/4))^(1/2). To prove it, I'd probably try to calculate the exact behavior of H at the poles, and then use that information to express it in terms of the Weierstrauss P function. $\endgroup$
    – zeb
    Mar 21, 2012 at 20:01

2 Answers 2


Well, again you seem to have a constant multiple of the Weierstrass P-function, plus a constant. There is a formula in this case for P((1 + i)z), not just for P(2z) as there is for any period lattice. The tilting line with constant real part is clearly related. The fact that the real part is constant will be an aspect of the Schwarz reflection principle, which tells one the condition that a function is real on the real axis (http://en.wikipedia.org/wiki/Schwarz_reflection_principle).

  • 1
    $\begingroup$ Everything you are bringing up concerns, it seems, a vector space of two complex dimensions spanned by the P function for the lattice of Gaussian integers, and the constant function. A function in that space can be identified by the leading term of its Laurent expansion at 0, and a single value. Since P vanishes at (1 + i)/2 for this lattice, the value there tells you the constant part. Your question concerns the action of a form of a kind of "complex conjugation" of order 2 (linear over the real numbers) on what is a real vector space of dimension four. Linear algebra. $\endgroup$ Mar 22, 2012 at 10:59

Thanks to your indications and the study of Hancocks’ Lectures on the theory of elliptical functions I finally made some progresses on this question. I now realize there is probably no break-through here since the subject has been intensively studied throughout the nineteenth century. However, I found this theory fascinating and thought I very much wanted to share these findings with you, since you were able to point me towards the right direction.

First of all it is easy to see from the product expression above that the central term for which $n=0$, becomes zero when $z = \frac{i}{4}$. It follows that the entire product equals zero such that $H(\frac{i}{4})=0$. Similarly the same holds for $H(\frac{3i}{4})$, by periodicity and parity of $H$.

As you mentioned above, given the fact that the poles have order 2 on the grid of Gaussian integers, we should be looking for solutions of the form:

$\Large{H \left(z \right ) = A \left( \wp \left(z \right ) - \wp \left( \frac{i}{4} \right ) \right )}$

Due to known special values of the Weierstrass P function (see for instance Abramovitch and Stegun 18.14.12 pp. 658 ) we have:

$\Large{\wp \left( \frac{i}{4} \right) = -\left(1+\sqrt{2}\right) \frac{\Gamma^4\left(\frac{1}{4} \right )}{8 \pi} \simeq -16,59816685...}$

In addition, we can identify the value for A by computing the limit:

$\Large{A = \lim_{z\rightarrow 0} H\left(z \right ) \times z^2}$

which yields,

$\Large{A = \frac{\left ( -1 ; e^{-4\pi} \right )^2_{\infty}}{8 \pi^2\left ( e^{-2\pi} ; e^{-2\pi} \right) ^4_{\infty}} \simeq 0.051041857...}$

Consequently, because the Weierstrass P functions has a zero at $\frac{1+i}{2}$, the corresponding evaluation of H which is also the constant real part on the unit diagonal is given by:

$\Large{H \left( \frac{1+i}{2} \right) = -A \wp \left( \frac{i}{4} \right)}$

In addition, If the algebraic form suggested above by inverse symbolic calculator is true we also have

$\Large{H \left( \frac{1+i}{2} \right) =\frac{1}{2} \sqrt{2^\frac{1}{4}+2^\frac{3}{4}} \simeq 0.847201267...} (1)$

Which would lead to a rather nice corollary involving Q-Pochhamer symbols

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

Now of course the missing part consists in finding a proof for (1) if (2) is unknown, or if (2) is already documented somewhere then (1) would follow directly. Does anybody know if (2) is known (or not)?

I opened a dedicated question on the subject.


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