Properties of $\frac {\theta_{1}''(z|\tau)}{\theta_{1}(z|\tau)}$?

Question

I'm looking for references concerning the properties of the function $\frac {\theta_{1}''(z|\tau)}{\theta_{1}(z|\tau)}$ where $\theta_{1}(z|\tau)$ is a Jacobi theta function defined here. I am trying to find special values of $z$ where I can rewrite this function in terms of modular forms (or maybe $E_2$) and other theta constants, or anything else that could appear there.

Answer 1

Notation: I write $$\begin{aligned} q_n &= \exp\frac{2\pi\mathrm{i}\tau}{n} & q &= q_1 & w &= \exp(\mathrm{i}z) \\ \dot{(\ )} &= \frac{1}{2\pi\mathrm{i}}\left.\frac{\partial(\ )}{\partial\tau}\right|_z = \frac{q_n}{n}\left.\frac{\partial(\ )}{\partial q_n}\right|_w &&& (\ )' &= \left.\frac{\partial(\ )}{\partial z}\right|_\tau = \mathrm{i} w \left.\frac{\partial(\ )}{\partial w}\right|_{q_n} \end{aligned}$$ Note that the $q$ used in the DLMF pages is written $q_2$ in this answer; their $q^2$ simplifies to $q$ here.

Expressions with $w$ and/or $q_n$ are implicitly regarded as functions of $z$ and/or $\tau$, respectively.

Answer: Given $$\begin{aligned} \theta_1(z|\tau) &= \sum_{k\in\mathbb{Z}} q_8^{(2k+1)^2} (-\mathrm{i}w)^{2k+1} = 2q_8\sum_{n=0}^\infty (-1)^n q^{n(n+1)/2}\sin\left((2n+1)z\right) \\ &= -\mathrm{i} w q_8\prod_{n=1}^\infty (1 - w^2 q^n)(1 - w^{-2} q^{n-1})(1 - q^n) \\ &= 2q_8\sin(z)\prod_{n=1}^\infty (1 - w^2 q^n)(1 - w^{-2} q^n)(1 - q^n) \end{aligned}$$ we can use the heat equation $$\theta_1''(z|\tau) = -8\dot{\theta}_1(z|\tau)$$ and thus arrive at $$\begin{aligned} \frac{\theta_1''(z|\tau)}{\theta_1(z|\tau)} &= -8\frac{\dot{\theta}_1(z|\tau)}{\theta_1(z|\tau)} = -8\left(\ln\theta_1(z|\tau)\right)^\cdot \\ &= -8q\left.\frac{\partial}{\partial q} \ln\left(2q_8\sin(z)\prod_{n=1}^\infty (1 - w^2 q^n)(1 - w^{-2} q^n)(1 - q^n)\right)\right|_w \\ &= -1 + 8\left(\sum_{n=1}^\infty\frac{n\,w^2 q^n}{1 - w^2 q^n} + \sum_{n=1}^\infty\frac{n\,w^{-2} q^n}{1 - w^{-2} q^n} + \sum_{n=1}^\infty\frac{n\,q^n}{1 - q^n}\right) \end{aligned}$$ In order to compress that expression, we apply the usual trick of expanding the fractions to series, swapping the series nesting orders, and simplifying the new inner series. Thus, for $|w^2 q| < 1$, $$\sum_{n=1}^\infty\frac{n\,w^2 q^n}{1 - w^2 q^n} = \sum_{n=1}^\infty n\sum_{m=1}^\infty w^{2m} q^{nm} = \sum_{m=1}^\infty w^{2m}\sum_{n=1}^\infty n\,q^{nm} = \sum_{m=1}^\infty w^{2m}\frac{q^m}{(1 - q^m)^2}$$ Doing the same with the other series, we obtain $$\frac{\theta_1''(z|\tau)}{\theta_1(z|\tau)} = -1 + 8\sum_{m=1}^\infty \frac{q^m}{(1 - q^m)^2}\left(1 + 2\cos(2mz)\right) \qquad (|\Im z| < \pi\,\Im\tau)$$ Note that the right-hand side has no singularity at $z=0$ and yields $$\lim_{z\to 0}\frac{\theta_1''(z|\tau)}{\theta_1(z|\tau)} = -1 + 24\sum_{m=1}^\infty \frac{q^m}{(1 - q^m)^2} = -\mathrm{E}_2(\tau)$$ which is quasimodular in $\tau$.