Asymptotics of Jacobi's third theta function.

Question

For $z\in \mathbb{C}$ and $\tau \in i\mathbb{R}_{+}$ consider the function $$ \theta_3(z;\tau)=\sum_{n\in\mathbb{Z}}\exp\left(2\pi i n z +\pi i \tau n^2\right) $$ This function satisfies the well known quasiperiodicity properties $$\theta_3(z+k;\tau)=\theta_3(z;\tau)$$ and $$\theta_3(z+k\tau;\tau)=\exp\left(-\pi i \tau k^2-2\pi i k z \right)\theta_3(z;\tau) $$ Asymptotics in the parameter $\tau$ are well known ( and easily found through a search engine ). Namely

$$\theta_3(z;\tau) \to 1 \text{, } \Im(\tau)\to +\infty$$ and $$ \theta_3(z;\tau) \sim \sqrt{\frac{1}{-i\tau}}\exp\left( \frac{z^2}{i\tau}\right) \text{, } \Im(\tau)\to 0^{+} $$

Asymptotics in $z$ are a little less straightforward.

Since for any $k\in \mathbb{Z}$, $$\theta_3(z+k;\tau)=\theta_3(z;\tau)$$ holds, if $z=x$ is real there is no growth as $x\to \pm \infty$, just oscillations.

The question is how to apply the second identity to compute an asymptotic formula for $\theta_3(\pm ix;\tau)$ as $x\to \infty$.

Writing $ix=\epsilon_x \tau +k_x\tau$ with $k_x=\lfloor \frac{ix}{\tau} \rfloor$ and $\epsilon_x=\frac{ix}{\tau}-k$, we immediately have that $$\theta_3(\pm ix;\tau)=\exp\left( -\pi i k_x^2 \pm 2\pi k_x \epsilon_x \tau \right)\theta_3(\epsilon_x \tau;\tau)$$

The right most term is bounded and well defined. My calculations on the exponential term have been somewhat lengthy and did not give a nice formula.

I would like to, first, invite suggestions for references, this seems to be too simple to not be published somewhere!

If not, then, any suggestions on how to deal with the floor terms would be much appreciated.

Answer 1

The Poisson summation formula gives $$\sum_n \exp(-\pi (n+z)^2 y) =\sum_n y^{-1/2}\exp\left(2\pi i n z -\pi n^2/y \right)$$ so as $y \to 0^+$ with $f(z) = z+1/2-\lfloor \Re(z) +1/2\rfloor$ for $\Re(z) \not\in \mathbb{Z}+1/2$ $$\sum_n \exp\left(2\pi i n z -\pi n^2 y \right) \sim y^{-1/2} \exp(-\pi f(z)^2/y)$$

For the asymptotic as $z=ix,x \to + \infty$ $$\sum_n \exp\left(2i\pi n (ix) -\pi n^2 y \right)= \exp(2\pi x^2/y)\sum_n \exp\left(-\pi ( n+x/y)^2y\right)$$ where the latter series is $y$-periodic in $x$. Together with the $1$-periodicity in $z=ix$ of the LHS it gives the asymptotic in $z$ in every direction.

The intermediate cases, the asymptotic in $y$ depending on $z$, are less obvious, they require splitting one of the the series in two parts depending on the magnitude and growth. Which one are you interested in exactly and for what application ?

Answer 2

Yes, it is $e^{c x^2}$. The constant is obtained from the expression in $k$ above. The problem is that the remaining function is oscillatory.

Answer 3

Let $x \in \mathbb R, \,\tau' = i/\tau, \, \tau' > 0$. The Poisson summation formula gives $$\theta_3(i x; \tau) = \sqrt{\tau'} \,e^{\pi \tau' x^2} \left( 1 + 2 \sum_{n > 0} e^{-\pi \tau' n^2} \cos(2 \pi \tau' n x) \right).$$ Setting the cosine to $\pm 1$ gives upper and lower bounds between which $\theta_3(i x; \tau)$ will oscillate as $x \to \pm \infty$ for fixed $\tau$.

It seems fairly obvious that setting the cosine to $1$ and to $(-1)^n$ gives exact upper and lower bounds. But I'm blanking on how to prove that the derivative has exactly two zeroes in a cell.