3
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Suppose I have this expression

    Sum[F[i, j] KroneckerDelta[j, k], {j, 1, 3}]


(* F[i, 1] KroneckerDelta[1, k] + F[i, 2] KroneckerDelta[2, k] +
            F[i, 3] KroneckerDelta[3, k] *)

Is it possible for Mathematica to present the contracted result

F[i,k]

?

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1
  • $\begingroup$ note this is only true for i,k integers in {1,2,3}. That said even with appropriate assumptions it doesn't simplify as you'd like. $\endgroup$
    – george2079
    Jun 24, 2014 at 18:56

1 Answer 1

3
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While doing the Sum over j, you turn the second argument of F into a number. This concrete substitution has to be avoided.

To do this, we have to tell Mathematica that F[i,j] with a concrete number for j should be rewritten immediately as an expression that can be simplified to something involving F[i,k]. To do this, I add the following symbolic rule for F:

Clear[i, j, k];
F[i_, j_?NumericQ] := F[i, k] KroneckerDelta[k, j]

Assuming[k > 0 && k ∈ Integers && k < 4,
 Simplify[Sum[F[i, j] KroneckerDelta[j, k], {j, 1, 3}]
  ]
 ]

(* ==> F[i, k] *)
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2
  • $\begingroup$ interesting but it seems not all that useful in general since you can only handle the one specific symbol k that way. $\endgroup$
    – george2079
    Jun 25, 2014 at 15:58
  • $\begingroup$ @george2079 Well, that's what the question is about. $\endgroup$
    – Jens
    Jun 25, 2014 at 16:58

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