6
You can use Sum to iterate twice through the set of lattice points.
lattice = Catenate@Table[{x, y}, {x, 0, 3}, {y, 0, 3}]
(* {{0, 0}, {0, 1}, {0, 2}, {0, 3}, {1, 0}, {1, 1}, {1, 2}, {1, 3}, {2,
0}, {2, 1}, {2, 2}, {2, 3}, {3, 0}, {3, 1}, {3, 2}, {3, 3}} *)
Sum[F[p1, p2], {p1, lattice}, {p2, lattice}]]
(* F[{0, 0}, {0, 0}] + F[{0, 0}, {0, 1}] + F[{0, 0}...
3
Inactivate has attribute HoldFirst, that is why Inactivate[ft, Sum] by itself doesn't inactivate your sum. You have to first bypass this HoldFirst with Evaluate.
ft = easyFourierTrigSeries[f, {x, -\[Pi], \[Pi]}, \[Infinity]]
D[Inactivate[Evaluate[ft], Sum] // ReleaseHold, x]
(* Inactive[Sum][-((
4 (-1)^\[FormalK] Sin[\[FormalK] x])/\[FormalK]), {\[FormalK]...
3
sum1 = Total[F @@@ Tuples[Range[0, 3], {2, 2}]]
sum1 // Short
F[{0, 0}, {0, 0}] + F[{0, 0}, {0, 1}] + F[{0, 0}, {0, 2}] + << 250 >> +
F[{3, 3}, {3, 1}] + F[{3, 3}, {3, 2}] + F[{3, 3}, {3, 3}]
sum2 = Array[F[{#, #2}, {##3}] &, {4, 4, 4, 4}, 0, Plus];
sum2 == sum1
True
1
You can try permuting the variables and then sorting and taking the first element of the result. For instance:
canon[expr_, vars_List] := First @ Sort @ ReplaceAll[
expr,
Thread[vars -> Permutations @ vars]
]
Then:
c1 = canon[n1 + 2 n2, {n1, n2}]
c2 = canon[2 n1 + n2, {n1, n2}]
c1 === c2
2 n1 + n2
2 n1 + n2
True
1
Try with the following code:
RowsSum[nmax_Integer?Positive, length_Integer?Positive, vector_List] := Module[
{matrix, matrixrows, s},
matrix = Table[i, {i, nmax}, {length}];
matrixrows[vector2_List, matrix_] := matrix[[vector]];
s = Flatten@Map[Total, List[matrixrows[vector, matrix]]];
Return[s];
];
Test:
RowsSum[12, 4, {1, 5, 9}]
(*{15, 15, 15, 15}*)
...
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