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DiracDelta[x] + Pi DiracDelta[x] == (1 + Pi) DiracDelta[x] is obviously true by inspection. And yet, FullSimplify[DiracDelta[x] + Pi DiracDelta[x] == (1 + Pi) DiracDelta[x]] simply returns the expression instead of simplifying to True. Why?

Mathematica v14.0.0.0.

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    $\begingroup$ DiracDelta[x] + Pi DiracDelta[x] // Factor $\endgroup$
    – Bob Hanlon
    Commented Mar 18 at 2:32

1 Answer 1

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FullSimplify[
 DiracDelta[x] + Pi DiracDelta[x] == (1 + Pi) DiracDelta[x], 
 TransformationFunctions -> Expand]

True

The documentation states:

DiracDelta[x] returns 0 for all real numeric x other than 0.

Both of the following evaluate to True automatically.

FullSimplify[
 DiracDelta[x] + Pi DiracDelta[x] == (1 + Pi) DiracDelta[x], 
 x ∈ PositiveReals]

FullSimplify[
 DiracDelta[x] + Pi DiracDelta[x] == (1 + Pi) DiracDelta[x], 
 x ∈ NegativeReals]

For the case where x = 0:

DiracDelta[x] + Pi DiracDelta[x] == (1 + Pi) DiracDelta[x] // Expand

returns True. Hence the use of TransformationFunctions above.

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  • $\begingroup$ It works if you include Factor in the TransformationFunctions as well. It's weird to me that Mathematica automatically evaluates DiracDelta[x] + 2 DiracDelta[x] to 3 DiracDelta[x], but not only does it not do an equivalent evaluation for DiracDelta[x] + Pi DiracDelta[x], but it doesn't even (by default) recognize that equivalence if you give it to it and then try to FullSimplify it. $\endgroup$
    – tparker
    Commented Mar 18 at 4:34
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    $\begingroup$ Someone familiar with the development aspect would know more. Please wait for an answer that explains it better. $\endgroup$
    – Syed
    Commented Mar 18 at 4:36
  • $\begingroup$ DiracDelta is not a usual function, but an attempt to implement the $\delta$-distribution in Mathematica. In the fifties of the 20-th century there were attempts to define the values of distributions at the reals and the definite integrals over distributions. These attempts are described, for example, in Mikusinskii Jan, Piotr Antosik, Roman Sikorski: Theory of distributions – the sequential approach. Elsevier 1973. It appeared that is not very successful and useful. For example, Walter Rudin, Functional analysis doesn't even mention it. $\endgroup$
    – user64494
    Commented Mar 18 at 5:02

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