Why does this theta function value yield such a good Riemann sum approximation?

Question

Let $T(q)$ denote the third Jacobi theta function at $z=0$; i.e., $$T(q) := \sum_{n\in\mathbb{Z}} q^{n^2}.$$ Then for any $x>0$, $T(\exp(-1/x)) \approx \sqrt{\pi x}$. For example, as the entry for A195907 in the OEIS notes, when $x=1$, we can regard $T(\exp(-1))$ as a Riemann sum approximation to the integral $\int_{-\infty}^\infty \exp(-x^2)\, dx = \sqrt{\pi}$. Numerically, $$\eqalign{T(\exp(-1)) &\approx 1.772637 \cr \sqrt{\pi} &\approx 1.772454\cr}$$ What surprises me, however, is that the ratio $T(\exp(-1/x))/\sqrt{\pi x}$ converges to $1$ very rapidly, more rapidly than I would naïvely expect based solely on the Riemann sum idea. For example, if we take $x=10$, then $$\eqalign{T(\exp(-1/10)) &\approx 5.604991216397928699311282433868800893854325311 \cr \sqrt{10\pi} &\approx 5.604991216397928699311282433868800893854323775\cr}$$ Is there some theta-function magic that explains this rapid convergence? Or maybe my intuition is wrong, and the rate of convergence is not any faster than one would "expect"?

Answer 1

I would give a slightly more direct explanation than WhatsUp. The Poisson summation formula states that for $f$ a smooth rapidly decreasing function on $\mathbb R$, $\hat{f}$ the Fourier transform, and $y>0$ that

$$ y\sum_{n \in \mathbb Z} f(yn) = \sum_{n \in \mathbb Z} \hat{f} (y^{-1} n)$$

where $\hat{f}(0) = \int_{-\infty}^\infty f(t)dt$ is the integral to which $y\sum_{n \in \mathbb Z} f(yn)$ is a Riemann sum. Hence $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is the error in the Riemann sum approximation to this integral.

For $y$ small, $y^{-1}$ is large, so $y^{-1} n$ is large for all nonzero $n$. Thus the more rapidly decreasing $\hat{f}$ is, the smaller the error term $\sum_{n\neq 0} \hat{f} (y^{-1} n)$ is, and the more rapidly this Riemann sum converges.

The Fourier transform of $e^{-x^2}$ is another function of the form $C e^{ - B x^2}$ and thus is pretty rapidly decreasing.

The Poisson summation formula is exactly how the functional equation of the theta function is proved, showing the relation to WhatsUp's answer.

Answer 2

This is because the theta function has a functional equation.

Under the usual definition $\theta(t) = \sum_n e^{-\pi t n^2}$, the functional equation is $$\theta(t) = \frac 1{\sqrt t} \theta\left(\frac 1 t\right).$$

Translating to your $T$, it gives $$T\left(e^{-\frac 1 x}\right) = \theta\left(\frac 1 {\pi x}\right) = \sqrt{\pi x} \theta(\pi x).$$

So, what you are observing is the fact that $\theta(\pi x)$ converges very fast to $1$. This is nothing magical because $$\theta(\pi x) - 1 = \sum_{n \ne 0} e^{-x(\pi n)^2}$$ which is pretty small, even for $x = 1$.