Define the theta function as $$ \theta(x) = \sum_{n=\infty}^\infty e^{\gamma(x+n)^2} $$ where $\gamma>0$. Clearly, $\theta$ is 1periodic, nonzero and smooth. Therefore, the reciprocal map $x \mapsto \frac{1}{\theta(x)}$ is 1periodic 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 nontrivial map $f$ (with a fast decay) such that $$ a_n \leq f(n). $$ Are there any results in this direction? Thank you!

3$\begingroup$ Your best bet would be to analytically or meromorphically continue $1/\theta(x)$ into a strip around the real axis and shift the contour of integration for the Fourier coefficients to either the upper halfplane or lower halfplane, depending on whether the frequency $n$ is positive or negative (i.e., use the saddle point method). $\endgroup$– Terry TaoMay 14, 2021 at 17:51

$\begingroup$ Please refer to A Characterization in the space of Convolution Operators B.R.Nagaraj,Proc.AMS,Vol.94,July(1985) $\endgroup$– Nagaraj IyengarMay 14, 2021 at 21:59
2 Answers
As @TerryTao 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 WeylHeisenberg frame bound calculations, Indagationes Mathematicae, Volume 7, Issue 2, 1996, Pages 165183, Janssen obtains on page 178 the Fourier expansion of $\frac 1 {\vartheta_3}$:
$$ \frac 1 {\vartheta_3(z,q)} = \frac 1 C \sum_k (1)^k a_k e^{2ikz} $$ $$ C = \sum_{n \in \mathbb Z} (1)^n (2n+1)q^{(n+\frac 1 2)^2} $$ $$ a_k = 2 \sum_{m=0}^\infty (1)^m q^{(m+\frac 1 2)(2k+m+\frac 1 2)} $$ 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.

1$\begingroup$ Great, that's it! It indeed agrees, taking into account the famous expansion $\prod_{n\geqslant1}(1x^n)^3=\sum_{n\geqslant0}(1)^n(2n+1)x^{n(n+1)}$. Except for one minor detail: $a_0$ must be replaced by $\frac12a_0$ (this is stated correctly in Whittaker & Watson, though, and my $c_0$ agrees with that too). $\endgroup$ May 20, 2021 at 16:35
By no means anything like a complete answer, but what I have is an empirical explicit formula for the coefficients $a_n$ which might be useful.
For $\theta$ itself, $$ \theta(x)=\sqrt{\,\frac\pi\gamma\,}\ \left(1+2q\cos(2\pi x)+2q^4\cos(4\pi x)+2q^9\cos(6\pi x)+...+2q^{n^2}\cos(2\pi nx)+...\right), $$ where $q=e^{\frac{\pi^2}\gamma}$.
It seems that the following holds (checked in Mathematica to about 100th power of $q$): $$ \frac1{\theta(x)}=\sqrt{\,\frac\gamma\pi\,}\ \frac1{\left((1q^2)(1q^4)(1q^6)\cdots(1q^{2n})\cdots\right)^3}\ \sum_{n=0}^\infty c_n\cos(2\pi n x), $$ where \begin{align*} c_0=&\ \,1q^2+q^6q^{12}+...+(1)^kq^{k(k+1)}+...,\\[.5ex] c_1=2&\left(qq^5+q^{11}q^{19}+...+(1)^kq^{k(k+3)+1}+...\right),\\ c_2=2&\left(q^2q^8+q^{16}q^{26}+...+(1)^kq^{k(k+5)+2}+...\right),\\ c_3=2&\left(q^3q^{11}+q^{21}q^{33}+...+(1)^kq^{k(k+7)+3}+...\right),\\ \cdots\\ c_n=(1)^n\times2&\sum_{k=0}^\infty(1)^kq^{k(k+2n+1)+n} \end{align*} Again empirically, these $c_n$ are of order $q^n$, but I do not know how to prove this. Here is the plot of their absolute values for $\gamma=100$:


$\begingroup$ @SylvainJULIEN Whittaker & Watson (Example 14 on page 489) refer to the Mathematical Tripos of 1903 (see another answer). $\endgroup$ May 20, 2021 at 19:02