I want to evaluate this simple expression $$ \sum_n f(n)\delta_{mn} = f(m) $$ using this:
Sum[KroneckerDelta[m, n] f[n], {n, Infinity}]
However Mathematica didn't reduce this expression. Could someone tell me how to do this correctly?
1 Answer
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Edit: The following solution works in Mathemtica 8.0.4, but not in 9.0.1:
This requires an assumption about the parameter m:
Assuming[m > 0 && m ∈ Integers,
Sum[KroneckerDelta[m, n] f[n], {n, Infinity}]]
(* ==> f[m] *)
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$\begingroup$ I have tried your code, but mathematica didn't reduce the expression neither. I am using Mathematica 9. $\endgroup$– entronJun 25, 2014 at 12:57
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$\begingroup$ @entron You're right, it works in version 8 and not in version 9! I'll think about it. Seems like a bug to me. $\endgroup$– JensJun 25, 2014 at 17:14
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$\begingroup$ Does this work in 8?
Assuming[m > 0 && m ∈ Integers && nn > m, Sum[KroneckerDelta[m, n] f[n], {n, nn}]]$\endgroup$ Jun 25, 2014 at 17:57 -
$\begingroup$ @george2079 yes - that works fine. It's the infinite case that causes trouble. $\endgroup$– JensJun 25, 2014 at 18:07
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$\begingroup$ that then suggests another approach to the linked question $\endgroup$ Jun 25, 2014 at 18:28
Infinity. But in version 8, the simplification works even when the upper limit isInfinity. $\endgroup$