Summations with power of DiscreteDelta

Question

Expression

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?

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! — Ulrich Neumann, Jul 11 '18 at 13:59
@UlrichNeumann However, Sum[DiscreteDelta[a - b], {a, -[Infinity], [Infinity]}] does give zero. I think this specifically a problem with power. — Weather Report, Jul 11 '18 at 14:09
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) — Ulrich Neumann, Jul 11 '18 at 16:31

Carl Woll · Accepted Answer · 2018-07-11 14:10:27Z

3

You could use Assuming:

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

1

answered Jul 11 '18 at 14:10

Carl Woll

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$\begingroup$ 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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