# Reciprocals of theta functions

I've spent the last few months with partial fraction expansions, and thought to create a function with simple poles over a lattice of zeros, like that of any of the Jacobi theta functions... but I can't find anything online analyzing or even discussing the reciprocals of Jacobi theta functions (or other functions with equivalent zeros). Can anyone explain the lack of available information (or discussion) online?

(Edit) I've managed to develop one type of reciprocal theta function into a series: \begin{align} 1\over\sum_{k \in \mathbb{Z}}e^{-\pi q(k+z)^2}&=e^{\pi qz^2}\prod_{k \ge 1}\frac1{(1-e^{-2k\pi q})^3} \sum_{n \ge 1}{(-1)^{n-1}e^{-n\pi q (n-1)}\sinh(2n-1)\pi q\over\cosh{(2n-1)\pi q}+\cosh{2\pi qz}} \\ &=\sqrt q \prod_{k \ge 1}\frac1{(1-e^{-2k\pi /q})^3} \sum_{n \ge 1}{(-1)^{n-1}e^{-n(n-1)\pi/q}\sinh{\frac{(2n-1)\pi}q}\over \cosh{\frac{(2n-1)\pi}q}+\cos{2\pi z}} \end{align}

The sum in the second line admits an interesting Fourier series (it contains partial theta sums).

• Weierstrass elliptic functions have second-order poles on a lattice generated by $\omega_1,\omega_2$ Jul 31, 2015 at 2:53
• Reciprocal theta functions have single-order (or simple) poles though, and do not vanish unlike the Weierstrass elliptic functions. I want to know if anyone's studied just the denominators of $\wp(z)$. Jul 31, 2015 at 3:41
• Antonio DJC:Have you verified your calculations yet? Sep 12, 2015 at 4:05
• Yes, but I've been caught up with writing calculations for one of my posts, and it's proved to be very tortuous. Afterwards I'll post my calculations. Sep 16, 2015 at 17:16
• What you might be looking for is the Weierstrass or Jacobi zeta functions. They have simple poles over a lattice of zeros. However, there does not exist doubly periodic functions with only a simple pole in period parallelogram. Only double or higher order poles like $\wp(z)$ and its derivatives. Jan 6, 2018 at 15:17

I got the above formula by expanding the product $\prod_{k=1}^\infty{(1+e^{-(2k-1)\pi q+2\pi qz})(1+e^{-(2k-1)\pi q-2\pi qz})}^{-1}$ into the sum $\sum_{k\ge1}\frac{A_ke^{-(4k-2)\pi q}}{(1+e^{-(2k-1)\pi q+2\pi qz})(1+e^{-(2k-1)\pi q-2\pi qz})}$ and used the Heaviside cover-up method to calculate $A_k$.

$A_ke^{-(4k-2)\pi q}$ evaluates to\begin{align} \lim_{z\to \frac i{2q}+k-\frac12}&(1+e^{-(4k-2)\pi q}+2e^{-(2k-1)\pi q}\cosh{2\pi qz})\prod_{n\ge1}\frac1{1+e^{-(4n-2)\pi q}+2e^{-(2n-1)\pi q}\cosh{2\pi qz}}\\ &=\prod_{n=1}^{k-1}\frac1{\left(1+e^{-(2n-1)\pi q+2\pi q(\frac i{2q}+k-\frac12)}\right)\left(1+e^{-(2n-1)\pi q-2\pi q(\frac i{2q}+k-\frac12)}\right)}\cdot\prod_{n\ge{k+1}}\frac1{\left(1+e^{-(2n-1)\pi q+2\pi q(\frac i{2q}+k-\frac12)}\right)\left(1+e^{-(2n-1)\pi q-2\pi q(\frac i{2q}+k-\frac12)}\right)}\\ &=\prod_{n=1}^{k-1}\frac1{\left(1-e^{-2n\pi q+2k\pi q}\right)\left(1-e^{-2n\pi q-2k\pi q+2\pi q}\right)}\cdot\prod_{n\ge{k+1}}\frac1{\left(1-e^{-2n\pi q+2k\pi q}\right)\left(1-e^{-2n\pi q-2k\pi q+2\pi q}\right)}\\ &=\prod_{n=1}^{k-1}\frac1{1-e^{2n\pi q}}\cdot\prod_{n=k}^{2k-2}\frac1{1-e^{-2n\pi q}}\prod_{n\ge1}\frac1{1-e^{-2n\pi q}}\prod_{n\ge{2k}}\frac1{1-e^{-2n\pi q}}\\ &=\frac{(-1)^{k-1}}{e^{\pi qk(k-1)}}(1-e^{-(4k-2)\pi q})\prod_{n\ge 1}\frac1{(1-e^{-2n\pi q})^2} \end{align}

Then I plug this into the product $$\frac1{\sum_{k\in\mathbb{Z}}e^{-\pi q(k+z)^2}}=e^{\pi qz^2}\prod_{k\ge1}\frac1{1-e^{-2k\pi q}}\frac1{1+e^{-(2k-1)\pi q+2\pi qz}}\frac1{1+e^{-(2k-1)\pi q-2\pi qz}}$$ to get\begin{align} e^{\pi qz^2}&\prod_{k\ge1}\frac1{(1-e^{-2k\pi q})^3}\sum_{k\ge1} \frac{(-1)^{k-1}}{e^{\pi qk(k-1)}}\frac{e^{(4k-2)\pi q}-1}{1+e^{(4k-2)\pi q}+2e^{(2k-1)\pi q}\cosh{2\pi qz}}\\ &=e^{\pi qz^2}\prod_{k\ge 1}\frac1{(1-e^{-2k\pi q})^3}\sum_{k\ge1}\frac{(-1)^{k-1}e^{-\pi qk(k-1)}\sinh{(2k-1)\pi q}}{\cosh{(2k-1)\pi q}+\cosh{2\pi qz}} \end{align} The second part of my formula follows from the imaginary transformation $$\sum_{k\in\mathbb{Z}}e^{-\pi q(k+z)^2}=\frac1{\sqrt{q}}e^{-\pi qz^2}\sum_{k\in\mathbb{Z}}e^{-\pi(k+iqz)^2/q}$$

To justify the calculations I only need to employ the identity $$\tfrac1{1+e^{-(4k-2)\pi q}+2e^{-(2k-1)\pi q}\cosh{2\pi qz}}\cdot\tfrac1{1+e^{-(4n-2)\pi q}+2e^{-(2n-1)\pi q}\cosh{2\pi qz}}=\tfrac1{e^{(2n-2k)\pi q}+e^{-(2n+2k-2)\pi q}-1-e^{-(4k-2)\pi q}}\left(\tfrac{e^{(2n-2k)\pi q}}{1+e^{-(4k-2)\pi q}+2e^{-(2k-1)\pi q}\cosh{2\pi qz}}-\tfrac1{1+e^{-(4n-2)\pi q}+2e^{-(2n-1)\pi q}\cosh{2\pi qz}}\right)$$ to separate the factors in the product $\prod_{k\ge1}\frac1{(1+e^{-(2k-1)\pi q-2\pi qz})(1+e^{-(2k-1)\pi q+2\pi qz})}$

I'm peeved that my phone sometimes makes $\LaTeX$ look too big and thus mixes it with my posts' plaintext. I can't say for sure it's a formatting issue or a bug.

(Edit) I've extended my approach to cover powers of theta functions:

For simplicity, I'll denote $\vartheta(z,q)=\sum_{k\in\mathbb{Z}}e^{-\pi q(k+z)^2}$, $A_k=(-1)^{k-1}e^{-\pi qk(k-1)}\sinh(2k-1)\pi q$, and $c_k=\cosh(2k-1)\pi q$. Now when I square $\vartheta(z,q)$, I multiply each term in $\sum_{k\ge1}\frac{A_k}{c_k+\cosh2\pi qz}$ by a term of the form $\frac{A_n}{c_n+\cosh2\pi qz}$ and so by partial fractions I'll get a series that looks like $$e^{2\pi qz^2}\sum_{k\ge1} \left(\frac{A'_k}{c_k+\cosh2\pi qz}+\frac{B_k}{(c_k+\cosh2\pi qz)^2}\right)$$ Now if I consider $u_n=c_n+\cosh2\pi qz$ I can then multiply it out to get $$\frac{u_n^2}{\vartheta^2(z,q)}=e^{2\pi qz^2}\left[B_n+u_n^2\sum_{k\neq n}\left(\frac{A'_k}{u_k}+\frac{B_k}{u_k^2}\right)+u_nA'_n\right]$$ and so if I divide out the exponential I'll have a polynomial in $u_n$ on the right hand side (as the sum does not have singularities at $u_n=0$) and thus can differentiate it at $u_n=0$ enough times to determine $A'_n$ and $B_n$. Higher powers will result in a series of the form $$e^{m\pi qz^2}\sum_{k\ge1}\left(\frac{D_k}{u_k}+\frac{E_k}{u_k^2}+\frac{F_k}{u_k^3}+\cdots+\frac{Z_k}{u_k^m}\right)$$ and similarly multiplying by $e^{-m\pi qz^2}u_n^m$ will produce a polynomial that can be differentiated to determine its coefficients. I'll post the calculations in a separate answer to not clutter this one.