suppose $x, y\in\mathbb{R}^2$ are given such that $\vert x \vert=\vert y \vert$ (where $\vert x \vert$ denote the Euclidean norm). I want to investigate, if the following limits are equal or not:
$$\sum_{\ell\in\mathbb{Z^2}} e^{-\vert x-\ell \vert^2}=\sum_{\ell\in\mathbb{Z^2}} e^{-\vert y-\ell \vert^2}$$
If $y=-x$ then both limits are equal. But what happens, if the angle between $x,y$ is between $0$ and $\pi$? I tried to use $\vert x-\ell\vert^2=\vert x \vert^2 - 2\vert x \vert\vert y \vert\cos(\angle(x,\ell))+ \vert \ell\vert ^2$, but it did not help me.
Can someone tell me how to approach this problem?
Best regards
For any $z=(z_1,z_2)\in\mathbb{R}^2$, we have $$ g(z_1,z_2)=\sum_{(x,y)\in\mathbb{Z}^2}e^{-(x-z_1)^2-(y-z_2)^2}=\sum_{s\in\mathbb{Z}}e^{-(s-z_1)^2}\sum_{t\in\mathbb{Z}}e^{-(t-z_2)^2}=f(\{z_1\})\cdot f(\{z_2\})$$ where $$ f(z) = \sum_{n\in\mathbb{Z}} e^{-(z-n)^2} = \sqrt{\pi}\sum_{k\in\mathbb{Z}}e^{-k^2 \pi^2}e^{-2\pi k i z}=\sqrt{\pi}+2\sqrt{\pi}\sum_{k\geq 1}e^{-k^2\pi^2}\cos(2\pi k z)$$ follows from the Poisson summation formula. In particular $g(z_1,z_2)$ is always pretty close to $\pi$, since the coefficients of the last Fourier cosine series decay to zero really fast, and if $(\pm u_1,\pm u_2)$ (for some choice of signs) and $(v_1,v_2)$ are the same element of $\mathbb{R}^2/\mathbb{Z}^2$ we have the exact equality $g(u_1,u_2)=g(v_1,v_2)$.
Corollary: $$ \frac{1}{4}\left[\sum_{n\in\mathbb{Z}}e^{-n^2}+\sum_{n\in\mathbb{Z}}e^{-\left(n-\frac{1}{2}\right)^2}\right]^2 $$ is an excellent approximation of $\pi$.