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I Square of Dirac delta function

  1. Jun 8, 2017 #1
    Is the square of a Dirac delta function, ##(\delta(x))^2##, still a Dirac delta function, ##\delta(x)##?

    A Dirac delta function peaks at one value of ##x##, say 0. If it is squared, it still peaks at the same value, so it seems like the squared Dirac delta function is still a Dirac delta function, ##\delta(x)##, or some multiple of it, ##k\delta(x)##, where ##k>1##, since the area under graph seems larger.

    How about the square root of a Dirac delta function?
  2. jcsd
  3. Jun 8, 2017 #2


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    It is something totally different than just multiplying by a constant.One thing to think about. The Del function is sometimes described as a rectangle of width d, and height (1/d) then take the limit as d->0. (so height approaches infinity)
    At all values of d, you get an area of (d/d) = 1. But if you have (Del)^2, the width is essentially the same as Del, but the height is infinity^2 ?

    WolframAlpha produced a surprising result for this. http://www.wolframalpha.com/input/?i=(DiracDelta[t])*(DiracDelta[t])
    Last edited: Jun 8, 2017
  4. Jun 8, 2017 #3


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    The Dirac delta "function" is defined by its behavior inside an integral: ∫f(x)δ(x)dx = f(0). It is a generalized function, not a function.
    I think that your comment about an increased area of δ2 is confusing the delta function with an approximation of the δ function.
    One approach to multiplying generalized functions is to separate them into their "smooth" and "singular" parts. In that, δ(x)2 = 0. (see https://en.wikipedia.org/wiki/Generalized_function#Algebras_of_generalized_functions)
  5. Jun 10, 2017 #4
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