Prove this exponential/cosine relation?

Question

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.

Answer 1

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.