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?
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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?
You could use Assuming:
Assuming[
b ∈ Integers,
Sum[DiscreteDelta[a-b]^2,{a,-∞,∞}]
]
1
DiscreteDelta
is not squared?
$\endgroup$
– Weather Report
Jul 11 '18 at 14:41
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
DiscreteDelta[]
evaluates to1
only if the argument is zero. MMA cannot decide, wethera-b
equals zero, because properties of b aren't known! $\endgroup$ – Ulrich Neumann Jul 11 '18 at 13:59Sum[DiscreteDelta[a - b], {a, -[Infinity], [Infinity]}]
does give zero. I think this specifically a problem with power. $\endgroup$ – Weather Report Jul 11 '18 at 14:09Sum[DiscreteDelta[a - b], {a, -[Infinity], [Infinity]}]==1
cannot be true for arbitrary b (If I understand the definitionDiscreteDelta
right) $\endgroup$ – Ulrich Neumann Jul 11 '18 at 16:31