9
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Bug introduced in 8.0 or earlier and fixed in 9.0


I have:

Assuming[-nn <= m <= nn && m ∈ Integers, 
         Simplify[Sum[KroneckerDelta[m, n] f[n], {n, -nn, nn}]]]

And it gets me

$$\begin{cases} \mathtt{2\,f[0]}&\mathtt{m}=0\\ \mathtt{f[m]}&\mathtt{True} \end{cases}$$

Do you know why there's a factor of two if $m=0$?

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7
  • $\begingroup$ Huh... returned unevaluated in Mathematica 10.3.0. $\endgroup$
    – user58955
    Oct 19, 2015 at 17:11
  • $\begingroup$ Looks like a bug; what version are you on? It's unevaluated in 10.2. $\endgroup$ Oct 19, 2015 at 17:11
  • $\begingroup$ Did you try it with a fresh kernel? $\endgroup$
    – Karsten7
    Oct 19, 2015 at 17:12
  • $\begingroup$ In Mathematica 10.3 and 10.2 it doesn't work. I had to re-install the version 8 to get the result. $\endgroup$
    – Daniela
    Oct 19, 2015 at 17:19
  • $\begingroup$ Returns unevaluated in 10.0. $\endgroup$
    – march
    Oct 19, 2015 at 17:20

2 Answers 2

2
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There's been some progress on this input. In M11, the expression evaluates to the expected result:

Assuming[
    -nn<=m<=nn && m ∈ Integers,
    Sum[KroneckerDelta[m,n] f[n],{n,-nn,nn}]
]

f[m]

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1
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This bug was present in Mathematica version 8.0:

$Version
Assuming[
    -nn <= m <= nn && m \[Element] Integers, 
    Simplify[Sum[KroneckerDelta[m, n] f[n], {n, -nn, nn}]]
]

print screen v8.0

Since version 9.0 this Sum remains unevaluated:

print screen v9.0

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