Define the theta function as
$$
\theta(x) = \sum\_{n=-\infty}^\infty e^{-\gamma(x+n)^2}
$$
where $\gamma>0$. Clearly, $\theta$ is 1-periodic, non-zero and smooth. Therefore, the reciprocal map $x \mapsto \frac{1}{\theta(x)}$ is 1-periodic and smooth as well and can be expanded as a Fourier series:
$$
\frac{1}{\theta(x)} = \sum\_{n=-\infty}^\infty a\_ne^{2\pi i nx}.
$$
I'm interested in upper bounding the Fourier coefficients, i.e. I want to find a non-trivial map $f$ (with a fast decay) such that
$$
|a\_n| \leq f(n).
$$
Are there any results in this direction? Thank you!



Does there exist an upper bound on the Fourier coefficients of the reciprocal theta function $\frac {1}{\theta}$?
As [@TerryTao](https://mathoverflow.net/users/766/terry-tao) stated correctly in the comments, a bound on the Fourier coefficients can be obtained by the saddle point method: The extension of $\frac 1 \theta$ to a meromorphic function on $\mathbb C$ has poles which are bounded away from the real axis. This implies that the Fourier coefficients of $\frac 1 \theta$ decay at least exponentially. In fact, the Fourier coefficients can be derived explicitly from which we can show that the decay is exactly exponentially:


First note, that your map $\theta$ is essentially the Jacobi theta function of third kind. Using Poisson's summation formula we have
$$
\theta(t) = \sqrt \frac \pi \gamma \sum\_n \left ( e^{-\frac{\pi^2}{\gamma}} \right )^{n^2} e^{2\pi i n t} = \sqrt \frac \pi \gamma \vartheta\_3(it, e^{-\frac{\pi^2}{\gamma}}).
$$
Hence, for $q = e^{-\frac{\pi^2}{\gamma}}$ and $z=it$ we have
$$
\frac 1 {\theta(t)} = \sqrt \frac \gamma \pi \frac 1 {\vartheta\_3(z,q)}.
$$
In the paper *A.J.E.M. Janssen, Some Weyl-Heisenberg frame bound calculations,
Indagationes Mathematicae,
Volume 7, Issue 2,
1996,
Pages 165-183*, Janssen obtains on page 178 the Fourier expansion of $\frac 1 {\vartheta\_3}$:


$$
\frac 1 {\vartheta\_3(z,q)} = \frac 1 C \sum\_k (-1)^k a\_k e^{2ikz}
$$
$$
C = \sum\_{n \in \mathbb Z} (-1)^n (2n+1)q^{(n+\frac 1 2)^2}
$$
$$
a\_k = 2 \sum\_{m=0}^\infty (-1)^m q^{(m+\frac 1 2)(2|k|+m+\frac 1 2)}
$$
From the value $|k|$ in the exponent in the definition of $a\_k$ it follows directly that the $a\_k$'s decay exponentially assuming that $|q|<1$. Applying some trivial bounds on $a\_k$ gives you the desired map $f$ so that $|a\_k| \leq f(k)$.


To derive the above formula, Jenssen refers to an exercise in the book *Whittaker, E., & Watson, G. (1927). A Course of Modern Analysis (4th ed., Cambridge Mathematical Library)*. He essentially solves this exercise which leads to the above expression.


The above formula should agree with the one derived in the answer of [@მამუკა ჯიბლაძე](https://mathoverflow.net/users/41291/%E1%83%9B%E1%83%90%E1%83%9B%E1%83%A3%E1%83%99%E1%83%90-%E1%83%AF%E1%83%98%E1%83%91%E1%83%9A%E1%83%90%E1%83%AB%E1%83%94).


