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Distribution question

  1. Aug 24, 2008 #1
    let be the distribution

    [tex] D^{m} \delta (x-a) D^{k} \delta (x) [/tex]

    my questions are , what happens whenever x=a or x=a ??

    is this identity correct

    [tex] \delta (x-a) = e^{-a D} \delta (x)= \sum_{n=0}^{\infty}(-a)^{n} \frac{D^{n}}{n!}\delta (x) [/tex]
  2. jcsd
  3. Aug 24, 2008 #2
    I don't understand the first question, but this

    is correct if you using such test functions that the Taylor series work for them. It can be checked with integration by parts. If the Taylor series don't converge for the test functions, then this distribution thing stops working as well.
  4. Aug 24, 2008 #3


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    This doesn't make sense to me. What meaning did you intend?
  5. Aug 24, 2008 #4


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    Probably meant to say x=a or x=0, since the expression contains the terms δ(x-a) and δ(x-0).
  6. Aug 24, 2008 #5


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    I meant that the idea of "plugging in" a value for x doesn't appear to make sense in this context.
  7. Aug 24, 2008 #6
    the idea is

    [tex] D^{m} \delta (x-a) D^{k} \delta (x) [/tex] however the value

    [tex] D^{m} \delta (-a) D^{k} \delta (0) [/tex] and [tex] D^{m} \delta (0) D^{k} \delta (a) [/tex]

    is not defined since delta functions are just oo

    another question , how would we define [tex] \int_{-\infty}^{\infty}dx D^{m} \delta (x-a) D^{k} \delta (x) [/tex]

    also , under suitable test function f , then

    [tex] < f | \delta (x-a) > = \sum_{n=0}^{\infty} (-a)^{n} \frac{ < \delta | D^{n} f>}{n!} [/tex]

    although it would make no sense , i think we could say

    [tex] (2\pi ) i^{m}D^{m}\delta (0) = \int_{-\infty}^{\infty}dx x^{m} [/tex] which is divergent... although in Cauchy's principal value the integral should be 0 for m Odd
    Last edited: Aug 24, 2008
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