On the growth of the Jacobi theta function So, I ran into this exercise from Stein & Shakarchi. CA, Chapter 5:
Show that if $\tau$ is fixed with positive imaginary part, then the Jacobi theta function
$$\theta(z | r) = \sum_{n=-\infty}^{\infty}{ e^{\pi i n^{2} \tau} e^{2\pi i n z} }$$
is of (growth) order $2$ as a function of z
They give this hint: $-n^{2}t + 2n|z| \leq - n^{2}t /2$ when $t > 0$ and $n \geq 4 |z|/t$, but I don't understand how to use it.
Any help is to be well received //or any reference of course.
Thanks!
 A: Ok, you can, after some calculations see that
the absolute value of your sum is less than or equal to
$$\sum_{n=-\infty}^\infty e^{\pi(-tn^2 + 2n|z|)}$$
where $t = im(\tau) >0.$
The part of the sum where $n<0$ will be less than some finite constant
depending only on $\tau,$ so we only need to worry about $n>0.$
(This you need to prove.)
Ok, so, divide that sum into two parts:
Let  $N$ be the least integer greater than $4|z|/t$
(where $t = im(\tau) >0.$)
If $n\geq N$ we have $nt/4 > |z|.$
This will imply that $-tn^2 + 2n|z| \leq -n^2t/2.$
Use this on the second sum, $N...\infty$ and use some other estimations on the first part.
Edit: Continuation:
First, you should see that
$$\sum_{n=-\infty}^0 e^{\pi(-tn^2 + 2n|z|)} \leq 
\sum_{n=-\infty}^0 e^{-\pi t n^2}
$$
and this clearly converges to something that only depends on $\tau.$
(The exponent will diverge to $-\infty$ rapidly, so the sum must converge.)
Now, if $n\geq N$ then
$$\sum_{n=N}^\infty e^{\pi(-tn^2 + 2n|z|)}\leq \sum_{n=N}^\infty e^{\pi(-n^2 t/2)} 
\leq \sum_{n=0}^\infty e^{\pi(-n^2 t/2)} 
$$
which converges to a finite number independent of $z$.
Now, you only need to do something about the last part:
$$\sum_{n=0}^{N-1} e^{\pi(-tn^2 + 2n|z|)}$$
(Hint: use $N-1< 4|z|/t$ which follows from the conditions above).
Edit: Continuation:
Ok, choose $r \in \mathbb{R}$ so that $\theta(r,\tau) \neq 0.$
By quasi-periodicity, we have for integers $m$ that
$$\theta(r+m\tau,\tau) = e^{-i\pi m^2\tau - 2im \pi r}\theta(r,\tau)$$
so by taking absolute values on each side, we see that
$$|\theta(r+m\tau,\tau)| = e^{t\pi m^2}|\theta(r,\tau)|$$
where $t = im(\tau)>0.$
Hence,
$$\lim_{m \rightarrow \infty} \frac{|\theta(r+m\tau,\tau)|}{e^{t\pi m^2}}
= |\theta(r,\tau)| > 0.$$
Thus, the order is at least 2.
