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Dirac function

  1. Apr 13, 2014 #1
    Sorry if the question seems naive but if we have the Dirac delta function delta(x-y) is it the same as delta(y-x)?? Or there are opposite in sign? And why ?
    Thank you for your time
     
  2. jcsd
  3. Apr 13, 2014 #2

    strangerep

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    Yes.

    Because it is only nonzero when x = y.
     
  4. Apr 13, 2014 #3
    Thank you very much just want to make sure
     
  5. Apr 13, 2014 #4
    Hm. I find this argument somewhat misleading. You can as well argue that its weak derivative [itex]\delta'(x)[/itex] is zero for [itex]x\neq y[/itex]. But there we have [itex]\delta'(-x)=-\delta'(x)[/itex].

    I think if you want to see why the Dirac delta is an "even" distribution is to go back to the definition:

    [itex] \int_{I(0)} \delta(x) f(x) dx = f(0) [/itex]

    where [itex]I(0)[/itex] is a neighbourhood of 0. Now let's see what happens if we use [itex]\delta(-x)[/itex] instead:

    [itex] \int_{I(0)} \delta(-x) f(x) dx = - \int_{I(0)} \delta(+x) f(-x) (-dx) = \int_{I(0)} \delta(x) f(-x) dx = f(-0) = f(0) [/itex]

    The second step involves the substitution of [itex]x\to-x[/itex] and as we can see we do get the same result. That means according to the definition of the Dirac delta as the generator of the linear functional that extracts f(0) both [itex]\delta(x)[/itex] and [itex]\delta(-x)[/itex] are identical.

    Cheers,

    Jazz
     
  6. Apr 13, 2014 #5
    Wooow Jazz amazing explanation I didn't know that it could be like that
    Thank you
     
  7. Apr 13, 2014 #6

    strangerep

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    (Sigh)

    Well, I was trying to find a simple explanation since I wasn't sure whether the OP had studied distribution theory.

    Silly me. I should have remembered: "simple explanations = trouble". :frown:
     
    Last edited: Apr 13, 2014
  8. Apr 14, 2014 #7
    I'm not really sure what this is about. It was obvious to me that you tried to give a simple answer, and I don't think anyone ever doubted that you know the exact answer. I wasn't sure either if the OP understands distribution theory. Nevertheless I felt that your answer should not be the only one, because it can be problematic if generalised.

    So please don't take this personal, there's really no reason for it.

    Cheers,

    Jazz
     
  9. Apr 14, 2014 #8

    strangerep

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    Agreed. I would have deleted my answer, but the editing time window had expired.

    I didn't take it personally. I was just annoyed at myself for not realizing my answer could indeed be misleading in the way you pointed out.

    Thanks for contributing.
     
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