I have following sum: $$ \sum_{\substack{n,j\\j<n}}^{3} x_nx_j$$. How can I give this $j<n$ condition in Sum?
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4 Answers
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For more complex condition:
Sum[Boole[j<n] x[n] x[j] ,{n,1,3},{j,1,3}]
For this simple condition:
@cvgmt give it as comment.
Just for fun:
Subsets[Array[x, 3], {2}] // Map[Apply@Times] // Total
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$\begingroup$ thank you. I have used Boole to provide equal or not equal condition in sum previously; I was confused about whether it can be useful in this case also, $\endgroup$ Nov 12, 2020 at 17:18
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The most direct solution is simply:
Sum[x[n] x[j], {n, 1, 3}, {j, 1, n}]
(* x[1]^2 + x[1]*x[2] + x[2]^2 + x[1]*x[3] + x[2]*x[3] + x[3]^2 *)
The same sort of thing works with NSum, Product and Product.
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$\begingroup$ I think you are almost right. Just the upper limit of j should be n-1. $\endgroup$ Nov 12, 2020 at 16:49
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SparseArray[{n_, j_} /; j < n -> Subscript[x, n] Subscript[x, j], {3,
3}] // Flatten // Total
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As usual, there are several ways to do it with Mathematica depending on the specific features that you want. Suppose you want to do some processing on all cases where $i<j$ where $i,j$ are in some range or list. The following code, in order of complexity, will each produce such a list of tuples:
r = Range[3];
Subsets[r, {2}]
Cases[Tuples[r, 2], {i_,j_}/;i<j]
Select[Tuples[r, 2], #[[1]]<#[[2]]&]
Flatten[Table[If[i<j, {i,j}, Nothing], {i,r},{j,r}]
Once you have such a list L of {i,j} tuples, then you can
find the sum of f[i,j] over all tuples with code such as:
Plus@@(L /. {i_,j_}->f[i,j])
Plus@@(# /. ij_->f@@ij& /@ L)
Same thing with the product using Times instead of Sum.
Table[x[n] x[j], {n, 1, 3}, {j, n, 3}] // Flatten // Total$\endgroup$Table[x[n] x[j], {n, 1, 3}, {j, 1, n - 1}] // Flatten // Total$\endgroup$