How to sum over $\mathbb{Z} \times \mathbb{Z} \setminus \{(0,0)\}$? [closed]

Question

I want to sum over $\mathbb{Z} \times \mathbb{ Z} \setminus \{(0,0)\}$ . So, something like

Sum[f[m,n],{m,-Infinity,Infinity},{n,-Infinity,Infinity}]

but I want to exclude f[0,0] as infinities might appear.

Any suggestions?

Answer 1

I suggest using a function

sumZZ0[f_, M_:Infinity] := (
      Sum[f[n, k], {n, M}, {k, 0, M}]
    + Sum[f[k, -n], {n, M}, {k, 0, M}]
    + Sum[f[-n, -k], {n, M}, {k, 0, M}]
    + Sum[f[-k, n], {n, M}, {k, 0, M}]);

This code may or may not help you depending on which f[n,k] you want to sum. If the original sum is not absolutely convergent, then splitting the sum into four sums may not give correct results.

As a simple example, using sumZZ0[(#1 + #2 I)^-4 &] to evaluate the Eisenstein series $G_4(i)$ does not produce a useful result. However, the code

Sum[Piecewise[{{(n + k I)^-4, n^2 + k^2 != 0}, {0, True}}],
   {n, -Infinity, Infinity}, {k, -Infinity, Infinity}]

does produce a useful result after a few seconds.

Still, an example that works is sumZZ0[x^(#1^2 + #2^2) &] which returns -1 + EllipticTheta[3, 0, x]^2 as it should in under a second. The code

 Sum[If[n == 0 && k == 0, 0, x^(n^2 + k^2)],
    {n, -Infinity, Infinity}, {k, -Infinity, Infinity}]

takes much longer to produce the same result. Also, using the Piecewise method takes even longer.

Answer 2

Instead of dealing with indices, try

Sum[Piecewise[{{f[m, n], m^2 + n^2 != 0}, {0, True}}], {m, -Infinity,  Infinity}, {n, -Infinity, Infinity}]

It should be noticed the notation n,k is preferable over n,m which are similar.

Answer 3

You can do

Sum[If[n == 0 && m == 0,0, f[m,n]],{m,-Infinity,Infinity},{n,-Infinity,Infinity}]