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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.
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  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented 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
    Commented 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
    Commented Sep 14, 2022 at 11:10

1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Sep 13, 2022 at 15:35

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