A function whose Fourier series are a gaussian?

Question

What function $f$ is such that its Fourier coefficients are a normal distribution? that is to say: given some number $s$, write $$\forall j\in\mathbb{Z}, \ c_j=e^{-j^2/s}$$ and if $$\lim_{N\to\infty} \sum_{j=-N}^{N} c_je^{ijt} \to_{a.e.} f(t) $$ then find $f$.

Answer 1

Yes, such functions are well-known, and have been studied since Jacobi. They are called theta functions. The simplest one is Jacobi's function $$ \theta(z;\tau) = \sum_{n=-\infty}^{\infty} \exp{(\pi i n^2 \tau + 2\pi i nz)} , $$ which is periodic, $\theta(z+1;\tau) = \theta(z;\tau)$, and has several other important properties, including the pseudoperiodicity $\theta(z+\tau;\tau) = e^{-\pi i \tau - 2\pi i z} \theta(z;\tau)$, and Jacobi's imaginary transformation $ \theta(z/\tau;-1/\tau) = \sqrt{-i\tau} e^{i\pi z^2/\tau} \theta(z;\tau) $. These are used in the theory of doubly-periodic functions, among other places, and have an appalling number of further identities.

The Fourier series is not the best way to understand such functions: generally, more useful are the product expansions, which are found using Weierstrass's product expansion. See also What is a Theta Function?

A related function is Weierstrass's $\sigma$-function, although this generally is pseudoperiodic in two directions, rather than just $1$, and so does not have a Fourier series.

Answer 2

Hint

An interesting property of the Gaussian function $f(x)=e^{-\pi t^2}$ is that $$F(f)=e^{-\pi f^2}=f(f)$$Try to prove it by expanding the Fourier integral and using $$\int_{\Bbb R}e^{-\pi t^2}dt=1$$ Additional Remark

For the discrete case, it is like we have sampled the Fourier transform by a sampling period of $1$, which is equivalent to periodizing the function, that is, $$\sum_{n}e^{-\pi (t-nT)^2}$$