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This seems correct numerically, but my efforts at a proof have not succeeded.

$$f(x)\equiv\sum_{n=-\infty}^{\infty}e^{-(x-n)^{2}}$$ Prove $$f(x)-\sqrt{\pi}=(f(0)-\sqrt{\pi})\cos(2\pi x)$$ Bonus points for the simplest proof.

Xander Henderson
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Jerry Guern
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  • It's a theta function; your formula will only be approximate, not exact, but to obtain it think Fourier series and Poisson summation. – Angina Seng Sep 27 '17 at 19:40
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    It's not true. $f(x)$ is an even periodic function with period $1$, so it has a Fourier series. Just the coefficients are $O(e^{-An^2})$, so numerically, only the first two, essentially $\sqrt{\pi}$ and the (small) $f(0)-\sqrt{\pi}$, play a role. –  Sep 27 '17 at 19:43
  • @LordSharktheUnknown Yes, this involves a theta function, but I deliberately posed the question to factor out the theta function and focus everyone's helpful efforts on the cosine. – Jerry Guern Sep 27 '17 at 20:04
  • @ProfessorVector Is it possible the higher Fourier terms vanish? – Jerry Guern Sep 27 '17 at 20:20
  • A related question https://math.stackexchange.com/q/891974/269624 – Yuriy S Sep 27 '17 at 20:37

1 Answers1

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According to this answer:

$$f(x)=\sum_{n=-\infty}^\infty e^{-(x-n)^2} =\sum_{n=-\infty}^\infty e^{-(x+n)^2} = \sqrt{\pi}+2\sqrt{\pi}\sum_{k=1}^\infty e^{-k^2 \pi^2}\cos 2\pi kx$$

Thus:

$$f(x)-\sqrt{\pi}=2\sqrt{\pi}\sum_{k=1}^\infty e^{-k^2 \pi^2}\cos 2\pi kx$$

$$f(0)-\sqrt{\pi}=2\sqrt{\pi}\sum_{k=1}^\infty e^{-k^2 \pi^2}$$

It seems your relation is not correct.

Yuriy S
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