Rapidly decaying Fourier coefficients for $\sum e^{-(x-2n)^2}$

Question

Consider the function $$f(x) = \sum \limits_{n \in \mathbb{Z}}e^{-(x-2n)^2}$$

This is an even periodic function with period 2, so its Fourier series has only cosines and a constant. Calculating this numerically, we find that $$f(x) \approx 0.8862269 + 0.150312900032 \cos(\pi x) + 0.00009167696 \cos(2\pi x) + 4*10^{-10} \cos(3\pi x) + \dots$$ In fact, numerics seem to suggest that the magnitude of the coefficients is roughly $10^{-n^2}$. Is there any underlying reason why this function is so sinusoidal? This really surprised me, but I can't find any closed form for the coefficients because of the Gaussian integrand.

Answer 1

It's easier to use the complex form $f(x)=\sum_{n\in\mathbb{N}}c_n e^{in\pi x}$, where \begin{align*} 2c_m&=\int_{-1}^1 f(x)e^{-im\pi x}\,dx \\&=\sum_{n\in\mathbb{Z}}\int_{-1}^1 e^{-(x-2n)^2-im\pi x}\,dx \\&=\sum_{n\in\mathbb{Z}}\int_{-2n-1}^{-2n+1}e^{-t^2-im\pi t}\,dt \\&=\int_{-\infty}^\infty e^{-t^2-im\pi t}\,dt=\sqrt\pi\ e^{-(m\pi)^2/4} \end{align*} (the last integral is well-known; see e.g. here).

Answer 2

Without Fourier analysis.

It is precisely that we know that $$\int e^{-(x-2n)^2}\, dn=\frac{\sqrt{\pi }}{4}\, \text{erf}(2 n-x)$$ and, then, the simplest form of Euler-MacLaurin summation is very useful.

$$\sum_{n=0}^\infty e^{-(x-2n)^2}=\frac{\sqrt{\pi }}{4}\, (1+\text{erf}(x))+e^{-x^2}\, P(x)$$ where $P(x)$ is a polynomial in odd powers of $x$; then the oscillations. Its degree depends on the number of terms you use in the summation.

I let you the work of finding what is $P(x)$.

Notice that in youe expansion, the constant term is just $\frac{\sqrt{\pi }}{2}$

Answer 3

On another forum I used Poisson's summation formula to obtain:

\begin{align*} \sum_{n=-\infty}^\infty e^{-(x-2n)^2} = \frac{\sqrt{\pi}}{2} + \sqrt{\pi} \sum_{m=1}^\infty e^{-m^2 \pi^2 / 4} \cos \pi m x . \end{align*}

Details:

First write:

\begin{align*} \sum_{n=-\infty}^\infty e^{-(x-2n)^2} = \sum_{n=-\infty}^\infty e^{-(n-x/2)^2 4} \end{align*}

which is periodic with period 1 in $x/2$. Poisson's summation formula:

\begin{align*} \sum_{n=-\infty}^\infty F(n+t) & = \sum_{m=-\infty}^\infty e^{2 \pi i m t} \int_0^1 e^{-2 \pi i m s} \sum_{n=-\infty}^\infty F(n+s) ds \nonumber \\ & = \sum_{m=-\infty}^\infty e^{2 \pi i m t} \sum_{n=-\infty}^\infty \int_0^1 e^{-2 \pi i m s} F(n+s) ds \nonumber \\ & = \sum_{m=-\infty}^\infty e^{2 \pi i m t} \sum_{n=-\infty}^\infty \int_n^{n+1} F(s) e^{-2 \pi i m s} ds \nonumber \\ & = \sum_{m=-\infty}^\infty e^{2 \pi i m t} \int_{-\infty}^\infty F(s) e^{-2 \pi i m s} ds \nonumber \\ & = \sum_{m=-\infty}^\infty \tilde{F} (2 \pi m) e^{2 \pi i m t} \end{align*}

where

\begin{align*} \tilde{F} (y) = \int_{-\infty}^\infty F(s) e^{-i y s} ds \end{align*}

Put $F(s)=e^{-s^2 4}$, then

\begin{align*} \tilde{F} (y) & = \int_{-\infty}^\infty e^{-s^2 4} e^{-i y s} ds = e^{-y^2 / 16} \frac{1}{2} \sqrt{\pi} . \end{align*}

So that

\begin{align*} \sum_{n=-\infty}^\infty e^{-(x-2n)^2} & = \frac{\sqrt{\pi}}{2} \sum_{m=-\infty}^\infty e^{-(2 \pi m)^2 / 16}e^{2 \pi i m (-x/2)} \nonumber \\ & = \frac{\sqrt{\pi}}{2} + \sqrt{\pi} \sum_{m=1}^\infty e^{-m^2 \pi^2 / 4} \cos \pi m x . \end{align*}