3
$\begingroup$

I'm looking for the Fourier coefficient of a "periodic Gaussian", which writes $$ f(x) = e^{-\frac{1}{2s}(\cos(\pi x) - m)^2} $$ It is a real even 2-periodic function, so its Fourier coefficients are real and even. I don't necessarily want a closed-form solution, anything which is numerically tractable to compute is fine!

I tried using the Taylor series of exponential, using the binomial coefficient then writing $\cos (\pi x)$ with Euler's formula, but the resulting formula is not very inspiring. I get $$ c_{2p} = \sum_{n \geq p} \frac{(-2s)^{-n}}{n!} \sum_{k=p}^n \binom{2n}{2k} \binom{2k}{k+p} (-m)^{2(n-k)} 4^{-k} $$ which gets midely simplified with Mathematica. Besides, the integral relation $$ c_p = \frac{1}{2} \int_{-1}^1 e^{-\frac{1}{2s}(\cos(\pi x) - m)^2} e^{-i\pi p x} {\rm d} x $$ did not work for me neither. I could simply use this last relation with integral quadrature solver but the cost is too high.

I found great tricks on this forum before, so I'm trying my luck now! Any help or pointer would be greatly appreciated.

Additional information

  • Something in the spirit of this related question would be incredible.
  • After comparing the formula of $c_{2p}$ in Mathematica with the integral formulation on random samples, it seems correct.
  • My use case focuses on $s \approx 1$ and $m \approx 0$, both real numbers.
Improve this question
$\endgroup$
6
  • $\begingroup$ how about first turning the squared cosine into cos(2x) and then maybe using summation formula to turn into to just cosine? Which has a neat formula mathoverflow.net/questions/272505/… $\endgroup$
    – Thomas Kojar
    Sep 12, 2022 at 21:17
  • $\begingroup$ the sum you wrote, assuming correct, should be ok numerically if you truncate it at some point, right? $\endgroup$
    – Thomas Kojar
    Sep 12, 2022 at 21:18
  • $\begingroup$ Indeed I was hoping to have the very neat formula of the post you linked (which I added to my question). Unfortunately I did not know how to deal with $\cos 2 \pi x + \cos \pi x$. Could you elaborate on the "summation formula" to deal with that? $\endgroup$
    – gaspardb
    Sep 13, 2022 at 11:45
  • $\begingroup$ My take will be to truncate the sum, but I need to find efficient ways to do so -- I also need to read about hypergeometric series. I was wondering if something simpler existed! $\endgroup$
    – gaspardb
    Sep 13, 2022 at 11:46
  • $\begingroup$ Regarding terminology: Is the term “periodic Gaussian” yours? Because what I would call a periodic Gaussian is (a shift of) the function whose Fourier coefficients are $c_j := \exp(-j^2)$ (up to scale), or equivalently a sum of periodically translated Gaussians: this of course should look somewhat like yours, but seems like a more natural object since it is, e.g., Green's function for the heat equation on the circle (which is how I define a Gaussian on Riemannian manifolds in general). $\endgroup$
    – Gro-Tsen
    Sep 14, 2022 at 11:10

1 Answer 1

Reset to default
4
$\begingroup$

For large $s$ a single sum in terms of a hypergeometric function may be useful, $$c_{p}=\frac{(-1)^p}{2^p p!}\sum_{n=0}^\infty\frac{(n)_p}{(2s)^{n}n!}(m+1)^{n-p} {}_2F_1\bigl(p+1/2,p-n,2p+1,2/(m+1)\bigr).$$

To test a numerical code, this closed-form answer for $m=0$ could help, $$c_{2p+1}(m=0)=0,\;\;c_{2p}(m=0)=(-1)^p e^{-1/4s}I_p(1/4s),$$ with $I_p$ a Bessel function.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you! I will investigate this. Mathematica simplifies the double sum in a similar fashion but with additional terms, yours seem simpler. I need to find a way to evaluate this incrementally or to find a stopping criterium, I'll let you know when I find but it does not seem trivial... $\endgroup$
    – gaspardb
    Sep 13, 2022 at 11:52
  • $\begingroup$ this series is helpful if $s$ is $\gg 1$; is that the case in your application? and by the way, is $m$ an integer, is it small, is it large? $\endgroup$
    – Carlo Beenakker
    Sep 13, 2022 at 13:15
  • $\begingroup$ That's not especially the case no... $s$ would be around $1$, potentially close to $0$. $m$ is real. I will add those information to my post, sorry for not providing it earlier. I'm trying to see how to compute these incrementally. $\endgroup$
    – gaspardb
    Sep 13, 2022 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.