SeriesCoefficient[x^m, {x, 0, p}]

This gives back the same thing. I tried Assumptions, FullSimplify etc. I keep getting back the same thing. The answer is clearly KroneckerDelta[p,m]

  • 1
    $\begingroup$ it seems to work only when you give it a numerical value for m. as in SeriesCoefficient[x^4,{x,0,p}] then it gives Piecewise[{{1, p == 4}}, 0] $\endgroup$
    – Nasser
    Feb 26, 2018 at 9:51

1 Answer 1


With Assumptions you get

SeriesCoefficient[x^m, {x, 0, p},Assumptions -> Element[{p, m}, Integers]]
(* Piecewise[{{1, m == p}}, 0] *)    
  • 2
    $\begingroup$ That is strange, because I tried this also when I looked at it. But It still returned unevaluated. I tried Reals also. which version are you using? I am using 11.2 on windows. !Mathematica graphics $\endgroup$
    – Nasser
    Feb 26, 2018 at 10:04
  • $\begingroup$ Thanks to Ulrich. Now it works. $\endgroup$ Feb 26, 2018 at 10:05
  • $\begingroup$ My version is 11.0.1/Windows $\endgroup$ Feb 26, 2018 at 10:05
  • 3
    $\begingroup$ may be then it is a bug? Why does it work on 11.0.1 and not on 11.2? May be if confirmed, this should be reported to WRI. $\endgroup$
    – Nasser
    Feb 26, 2018 at 10:11
  • $\begingroup$ I also see no evaluation on 11.2; I think it's a bug/regression. $\endgroup$
    – Pillsy
    Feb 26, 2018 at 16:53

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