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I conjecture that:

$$ \sum_{n=-\infty}^{\infty}(-1)^n e^{-\left(x - \frac{n}{2}\right)^2} = K\cos(2\pi q x)$$

for some $K \in \mathbb{R}, q \in \mathbb{Z}$. Let $G(x)$ denote the left side.

It's easy to show that $G(x + 1) = G(x)$ (by a shifting sum argument).

Furthermore it can be shown that $G(-x) = G(x)$ by noting that for each summand

$$(-1)^n e^{-\left(x- \frac{n}{2}\right)^2} $$ If we invert it $$(-1)^n e^{-\left(-x- \frac{n}{2}\right)^2}$$ Thats the same as $$(-1)^n e^{-\left(x+ \frac{n}{2}\right)^2}$$ And since the sum is over all natural numbers that implies this mapping is a bijection of terms in the sum to terms in the sum.

But is the set of functions $\mathbb{C} \rightarrow \mathbb{C}$ that satisfy $ F(x+1) = F(x) , F(x) = F(-x)$ really equal to

$K\cos(2\pi q x)$?

I worry that some pathological examples might arise that satisfy both equations but are excluded from this list. What other conditions should I chase? Trying to show that the second derivative is a multiple of the original function has not been very fruitful. I end up with the expression

$$ G'' = 2\sum_{n=-\infty}^{\infty}(-1)^{n+1} \left(1 - 2\left( x - \frac{n}{2} \right)^2 \right) e^{-\left(x - \frac{n}{2}\right)^2} $$

(if we let $u = e^{x^2} G$ then I think this yields)

$$ (4x^2 - 2)e^{-x^2} U -4xe^{-x^2}U' + e^{-x^2} U'' = (e^{x^2}U)'' $$

But involving matrix exponentials to solve this seems like a mess.

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  • $\begingroup$ Looks rather unlikely: not every even periodic function is a cosine. $\endgroup$ – Lord Shark the Unknown Oct 16 '17 at 5:20
  • $\begingroup$ Intuition wise: the graphs look exactly the same! but that just means there might be an error term some 5+ decimal places out $\endgroup$ – frogeyedpeas Oct 16 '17 at 5:21
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    $\begingroup$ It seems like any $K_1\cos(2\pi q_1 x)+K_2\cos(2\pi q_2 x)$ satisfies these conditions. $\endgroup$ – Thomas Andrews Oct 16 '17 at 5:50
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    $\begingroup$ Hmm ... presumably this thing would not satisfy $$y'' = - Q y $$ for any constant $Q$. So that would be a fruitful angle to chase $\endgroup$ – frogeyedpeas Oct 16 '17 at 5:51
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    $\begingroup$ Related: en.wikipedia.org/wiki/Poisson_summation_formula $\endgroup$ – Jack D'Aurizio Apr 11 '18 at 19:17
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We have from the Poisson summation formula that

$$f_p(x)=\sum_{n \in \mathbb{Z}} f(x+np) = \frac{1}{p}\sum_{n\in\mathbb{Z}}\hat{f}\left(\frac{n}{p}\right)e^{2i\pi\frac{n}{p}x}.$$

In our case $f(x)=e^{-x^2}-e^{-\left(x+\frac{1}{2}\right)^2}$ and $p=1$, which gives $\hat{f}(\xi)=\sqrt{\pi}e^{-\pi^2 \xi^2}(1-e^{\pi i \xi})$. Thus,

$$G(x)=f_{1}(x)=\sqrt{\pi}\sum_{n\in\mathbb{Z}}e^{-\pi^2 (2n+1)^2 }e^{(4n+2)i \pi x} = 2\sqrt{\pi}\sum_{n\geq 0} e^{-\pi^2 (2n+1)^2}cos((4n+2)\pi x).$$

Thus, $G(x)$ is not a multiple of cosine.

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  • $\begingroup$ Perhaps this explains why numerically the function $G$ is so close to a single cosine-- the series is very quickly converging, and summands $n \geq 1$ are on the order of $e^{- 9}$. $\endgroup$ – A Blumenthal Apr 14 '18 at 20:37
  • $\begingroup$ @ablumenthal I was almost certain that $G$ was sinusoidal, but then I noticed that replacing $e^{-\left(x+\frac{n}{2}\right)^2}$ in the sum with other functions such a as $\frac{1}{(x+n)^2 +1}$ yielded functions that also had striking resemblance to a cosine but upon closer inspection had a very small error term, which led me to realize the same was true for $G$ as well. $\endgroup$ – Jacob Apr 14 '18 at 20:48
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The conjecture is that the equation holds for $q=1$, which was supported in comments by numerical evidence.

If this is true, then the continuous Fourier spectrum should exhibit a single peak at frequency $f =1$, and be zero elsewhere.

We already know from OP that the LHS $G(x)$ is periodic with period $1$ in $x$.

Consider a rational $f = p/r$ with natural numbers $p$ and $r$.

The Fourier spectrum G(f) of the LHS can then be computed, for any $k \in \mathbb{Z}$, from (apart from a fixed normalization factor)
$$ G(f) = \int_{k r}^{(k+1)r} G(x) \cos(2\pi f x) dx $$ since $r$ is the GCD of $1$ and $f$.

Then it may also be computed as the average of several $(2 M)$ of these intervals: $$ G(f) = \frac{1}{2M} \sum_{k=-M}^{M-1} \int_{k r}^{(k+1)r} G(x) \cos(2\pi f x) dx = \frac{1}{2M} \int_{-Mr}^{Mr} G(x) \cos(2\pi f x) dx $$

Consider $f \ne 1$ (more generally, exclude all odd $f$ , (1,3,5,...)).

Compute $G(f)$, with some help by Wolfram alpha, as

$$ G(f) = \lim_{M \to \infty}\frac{1}{2M} \int_{-Mr}^{Mr} G(x) \cos(2\pi f x) dx \\ = \lim_{M \to \infty}\frac{1}{2M} \sum_{n=-\infty}^{\infty}(-1)^n \int_{-\infty}^{\infty} e^{-\left(x - \frac{n}{2}\right)^2} \cos(2\pi f x) d x\\ = \lim_{M \to \infty}\frac{1}{2M} \sum_{n=-\infty}^{\infty}(-1)^n \sqrt\pi e^{-\pi^2 f^2} \cos(\pi n f)\\ = \lim_{M \to \infty}\frac{1}{2M} \sqrt\pi e^{-\pi^2 f^2} (-1 + 2\lim_{m \to \infty} \sum_{n=0}^{m}(-1)^n \cos(\pi n f) )\\ = \lim_{M \to \infty}\frac{1}{2M} \sqrt\pi e^{-\pi^2 f^2} \lim_{m \to \infty} (-\cos(\pi (f + 1) (m + 1)) - \tan(\pi f/2) \sin(\pi (f +1) (m + 1)) ) $$

Now note that the inner limit $\lim_{m \to \infty} $ is bounded in absolute values by a finite $1 + |\tan(\pi f/2)|$. Hence

$$ |G(f)| \leq \lim_{M \to \infty}\frac{1}{2M} \sqrt\pi e^{-\pi^2 f^2} (1 + |\tan(\pi f/2)|) \to 0 $$

Hence all Fourier components other that for $f=1$ disappear, which proves that the LHS $G(x)$ has the only frequency $f=1$, with $K=G(0)$, which in turn shows that it equals the RHS. $\qquad \Box$

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