Sum[DiscreteDelta[a - 1]^2, {a, -∞, ∞}]

evaluates to 1, as expected, but

Sum[DiscreteDelta[a - b]^2, {a, -∞, ∞}]

does not converge, according to Mathematica. Is there a way to cleanly circumvent this?

  • $\begingroup$ DiscreteDelta[] evaluates to 1 only if the argument is zero. MMA cannot decide, wether a-b equals zero, because properties of b aren't known! $\endgroup$ – Ulrich Neumann Jul 11 '18 at 13:59
  • $\begingroup$ @UlrichNeumann However, Sum[DiscreteDelta[a - b], {a, -[Infinity], [Infinity]}] does give zero. I think this specifically a problem with power. $\endgroup$ – Weather Report Jul 11 '18 at 14:09
  • $\begingroup$ I believe the result Sum[DiscreteDelta[a - b], {a, -[Infinity], [Infinity]}]==1 cannot be true for arbitrary b (If I understand the definition DiscreteDelta right) $\endgroup$ – Ulrich Neumann Jul 11 '18 at 16:31
  • $\begingroup$ OK, I see. Carl Woil also suggested that this is a bug. $\endgroup$ – Weather Report Jul 11 '18 at 20:48

You could use Assuming:

    b ∈ Integers,


  • $\begingroup$ Any ideas why I don't need this when the DiscreteDelta is not squared? $\endgroup$ – Weather Report Jul 11 '18 at 14:41
  • $\begingroup$ @WeatherReport My guess is that the unsquared version is a bug. For instance, if b=.5 the answer is incorrect. $\endgroup$ – Carl Woll Jul 11 '18 at 14:44
  • $\begingroup$ Hmm. I was really looking for an abstract notion of delta-symbol. which would evaluate Sum[AbstractDelta[a-b]f[a],{a,allvalues}] to f[b]. Like a DiracDelta but which should not diverge when squared. I thought that DiscreteDelta serves that purpose. But you are saying its probably a bug:) Is there an abstract notion that I'm looking for? $\endgroup$ – Weather Report Jul 11 '18 at 14:54

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