I'm relatively inexperienced with mathematica, so I apologize if this is a trivial question. I want to take a double sum over a function $f(i,j)$ of two indices, of the form $$ \sum_{i = -\infty}^\infty \sum_{j = 0\atop j\not = i}^m f(i,j). $$ That is, in the inner sum I want to sum over only those indices $j$ in my range of summation that satisfy the assumption $j \not = i$. How can I input such a sum to Mathematica?
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You can use Boole as follows:
Sum[f[i, j] Boole[i != j], {j, 0, m}, {i, -Infinity, Infinity}]
Here's an example where f[i, j] = Sin[i] Cos[j]
Sum[Sin[i] Cos[j] Boole[i != j], {j, 0, 3}, {i, 0, 3}]
Which gives
Sin[1] + Cos[2] Sin[1] + Cos[3] Sin[1] + Sin[2] + Cos[1] Sin[2] +
Cos[3] Sin[2] + Sin[3] + Cos[1] Sin[3] + Cos[2] Sin[3]
RunnyKine
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2What if f[i,j] diverges when i=j? Imagine f[i,j]=M[i,j]^(-a), where M[i,j] is a matrix with M[i,j]=0 if i=j, and a is a positive integer number. @RunnyKine – Ana S. H. May 05 '15 at 16:15
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@RunnyKine Do you know how to do double summation with conditions. https://mathematica.stackexchange.com/questions/250873/double-summation-with-condition – Jasmine Jul 12 '21 at 05:30
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You can use If to put the condition in the argument of the sum. It would be something like
Sum[If[i != j, f[i, j], 0], {i, -Infinity, Infinity}, {j, 0, m}]
where If[i != j, f[i,j], 0] tells Mathematica to use $f(i,j)$ in the sum if $i\neq{j}$ or 0 if $i=j$.
Pedro
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14I found that your If solution works better than the Boole by RunnyKine, because it allows f[i,j] to be undefined at i == j. – Peeter Joot Sep 19 '13 at 01:57
Boole[]orKroneckerDelta[]. See this as well. – J. M.'s missing motivation May 26 '13 at 09:14Sign[Abs[i - j]] f[i,j], not very elegant but it works. – Kuba May 26 '13 at 09:24