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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?

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  • $\begingroup$ related mathematica.stackexchange.com/questions/51424/… $\endgroup$ – george2079 Jun 24 '14 at 21:25
  • $\begingroup$ In my opinion this should be tagged as a bug because it's a regression. In version 9, the sum works with any finite numerical upper limit, but not with Infinity. But in version 8, the simplification works even when the upper limit is Infinity. $\endgroup$ – Jens Jun 25 '14 at 17:24
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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. $\endgroup$ – entron Jun 25 '14 at 12:57
  • $\begingroup$ @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. $\endgroup$ – Jens Jun 25 '14 at 17:14
  • $\begingroup$ Does this work in 8? Assuming[m > 0 && m ∈ Integers && nn > m, Sum[KroneckerDelta[m, n] f[n], {n, nn}]] $\endgroup$ – george2079 Jun 25 '14 at 17:57
  • $\begingroup$ @george2079 yes - that works fine. It's the infinite case that causes trouble. $\endgroup$ – Jens Jun 25 '14 at 18:07
  • $\begingroup$ that then suggests another approach to the linked question $\endgroup$ – george2079 Jun 25 '14 at 18:28

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