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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!

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...which "Stein and Shakarchi"? –  J. M. Oct 22 '11 at 10:23
    
it's Exercise 3 from Chapter 5 of Complex Analysis. –  Anna Oct 22 '11 at 14:11
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1 Answer

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

$$\sum_{n=N}^\infty e^{\pi(-tn^2 + 2n|z|)}\leq \sum_{n=N}^\infty e^{\pi(-n^2 t/2)} $$ 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.

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Hi, thanks. This is nice and might be indeed the right way to start. But I still don't understand how to get to the conclusion. How do you compute $\sum_{n=-\infty}^{\infty}{e^{\pi(-n^{2}t /2 )}}$ for example... –  Anna Oct 24 '11 at 14:45
    
See my edits... –  Paxinum Oct 25 '11 at 16:05
    
Thanks a lot. Can you please explain why does it have order precisely $2$? This solution only gives order $\leq 2$. –  Anna Nov 2 '11 at 1:04
    
Ok, see what I added. I did not prove that in my original solution, since (by definition) a function is of growth order 2 if one can prove $\lim_{|z| \rightarrow \infty} |f(z)|/e^{B|z|^2} < \infty.$ Thus, the question is to prove this, and does not ask you to find <i>the</i> order of the function, which is what I assume, what you're looking for. However, I added a sketch above, on how to prove that <i>the</i> order is 2. –  Paxinum Nov 6 '11 at 16:49
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