7
$\begingroup$

I am trying to find a closed form solution for the following series. The $\sqrt{i^2 + j^2}$ in the exponent comes from distances on the euclidean grid from the origin.

$x = \sum_{i,j} e^{-\sqrt{i^2 + j^2}}$

where $i,j$ range from $0$ to infinity.

It appears this expression is not a geometric series, so I have trouble analyzing it. I did some quick simulations to realize that the value converges quickly. For $i,j$ in range (0,40), and using double-precision floating point, the value converges to $2.95878712840391$. Altering the range of $i,j$ no longer changes the sum because the incremental values are beyond the precision of the floating point decimal.

I would greatly appreciate some help in approaching this series, and if there is a way to represent it in a closed form. Or if there is a way to approximate the answer to a desired precision.

$\endgroup$
5
$\begingroup$

By letting $$ r_2(n)=\left|\{(a,b)\in\mathbb{Z}^2:a^2+b^2=n\}\right|$$ we have $r_2(n) = 4(\chi_4 * 1)(n) = 4\sum_{d\mid n}\chi_4(d)$, with $\chi_4(n)=1$ if $n\equiv 1\pmod{4}$, $\chi_4(n)=-1$ if $n\equiv -1\pmod{4}$ and $\chi_4(n)=0$ if $n$ is even (i.e. $\chi_4$ is the non-principal Dirichlet character $\!\!\!\pmod{4}$).
It follows that $$ \sum_{i,j\geq 0}e^{-\sqrt{i^2+j^2}} = 1+4\sum_{n\geq 1}(\chi_4*1)(n) e^{-\sqrt{n}} $$ where $(\chi_4*1)(n)$ has a moderately erratic behaviour, but $e^{-\sqrt{n}}$ converges to zero really fast, such that in order to compute the LHS up to $N$ figures it is enough to compute $\sum_{n=1}^{N^2}(\chi_4*1)(n) e^{-\sqrt{n}}$. As an alternative, we have $$ e^{-\sqrt{n}}=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}e^{-s^2/4}e^{-n/s^2}\,ds=\frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}e^{-\frac{1}{4s}}e^{-ns}\,\frac{ds}{s^{3/2}} $$ hence by setting $$\Theta(x)=\sum_{n\geq 0} x^{n^2} $$ we have $$ \sum_{i,j\geq 0}e^{-\sqrt{i^2+j^2}}=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\Theta^2(e^{-s^2}) e^{-\frac{1}{4s^2}}\frac{ds}{s^{2}} $$ where the RHS can be approximated through numerical integration algorithms, especially if combined with the functional identity for the Jacobi $\Theta$ function, which is a consequence of the Poisson summation formula. My computations point towards an approximated value of $\color{green}{2.9587871284039}\color{red}{3}$.

$\endgroup$
  • $\begingroup$ This is very thorough, but it's not really an answer. The asker wants a closed form solution, they already have a numerical approximation. $\endgroup$ – Larry B. Oct 6 '18 at 0:05
  • 2
    $\begingroup$ @LarryB.: I am pretty confident there is no nice closed form solution. $\endgroup$ – Jack D'Aurizio Oct 6 '18 at 0:06
  • $\begingroup$ Are you sure? Maybe it can be sandwiched between a couple really nice integrals. $\endgroup$ – Larry B. Oct 6 '18 at 0:11
  • $\begingroup$ @LarryB.: even if that is the case, that only leads to a double bound, and we already have plenty of them. Do you have any actual reason for believing the last integral is related to some well-known set of constants? $\endgroup$ – Jack D'Aurizio Oct 6 '18 at 0:13
  • $\begingroup$ I re-read the question. This is an answer, since it "is a way to approximate the answer to a desired precision." $\endgroup$ – Larry B. Oct 6 '18 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.