# Summations with power of DiscreteDelta

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
• OK, I see. Carl Woil also suggested that this is a bug. – Weather Report Jul 11 '18 at 20:48

You could use Assuming:

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


1

• Any ideas why I don't need this when the DiscreteDelta is not squared? – Weather Report Jul 11 '18 at 14:41
• @WeatherReport My guess is that the unsquared version is a bug. For instance, if b=.5 the answer is incorrect. – Carl Woll Jul 11 '18 at 14:44
• 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? – Weather Report Jul 11 '18 at 14:54