0

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 $f$.

Dr Potato
  • 1,013
  • 4
    Do you happen to have a minus sign in $c_j$, like $c_j=e^{-j^2/s}$? – Andrei Sep 06 '19 at 18:14
  • I might be off base since this is the continuous fourier transform,but shouldn't it be something like: http://mathworld.wolfram.com/FourierTransformGaussian.html – Kitter Catter Sep 06 '19 at 18:18
  • If you replace $e^{it}\to z$ and let $a=e^{-1/s}$, then WolframAlpha provides an answer in terms of the q-Pochammer functions: https://www.wolframalpha.com/input/?i=sum+a%5E%28j%5E2%29+z%5Ej+from+j%3D-infinty+to+infinty – Semiclassical Sep 06 '19 at 18:25
  • By the convolution theorem, it is the inverse Fourier transform of the impulse train multiplied (in the frequency domain) by a gaussian, so in time domain it must be the superposition of the same gaussian, equally separated one from the next. But I'm looking for a closed form of this periodic function. – Dr Potato Sep 06 '19 at 18:26

2 Answers2

1

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, $$\sum_{n}e^{-\pi (t-nT)^2}$$

Mostafa Ayaz
  • 33,056
  • That is the inverse Fourier transform, not the same as Fourier Series. – Dr Potato Sep 06 '19 at 18:37
  • Sorry I missed that. I added more details.... – Mostafa Ayaz Sep 06 '19 at 18:41
  • So $\sum_n e^{-\pi x(t-n)^2} = x^{-1/2} \sum_k e^{-\pi k^2/x} e^{2i \pi kt}$. Also to find the Fourier transform of the Gaussian we need some sort of analytic continuation. – reuns Sep 06 '19 at 18:45
  • Which is easy. You may want to check http://mathworld.wolfram.com/FourierTransformGaussian.html – Mostafa Ayaz Sep 06 '19 at 18:48
  • I don't see what you mean. For $a$ real then $F(a) = \int_{-\infty}^\infty e^{-\pi (t-a)^2}dt-1=0$, the function of $a$ complex is analytic thus $\forall a \in \Bbb{C},F(a) = 0$ – reuns Sep 06 '19 at 19:00
  • Based on the link I sent to you, you can split the integral to real and imaginary parts and solve them separately.' – Mostafa Ayaz Sep 06 '19 at 19:28
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.

Chappers
  • 69,099
  • The Fourier series are of course a good way to understand such functions, it's how $\sum_n e^{-\pi x(t-n)^2} = x^{-1/2} \sum_k e^{-\pi k^2/x} e^{2i \pi kt}$ is obtained – reuns Sep 06 '19 at 19:23
  • @reuns It's not necessary to do it that way, though: one can do some reasonably simple calculations by integrating particular elliptic functions to derive it. The simplest version I've seen is in Walkers Elliptic functions: a constructive approach, although there is also a version in Weil's famous text Elliptic functions according to Eisenstein and Kronecker. – Chappers Sep 06 '19 at 20:28
  • There's also a contour integration proof due to Landsberg, according to Whittaker and Watson (pp. 124 and 474) – Chappers Sep 06 '19 at 20:38
  • But yes, there are uses for the Fourier series as well. – Chappers Sep 06 '19 at 20:39