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Kronecker delta and Dirac delta

  1. Jun 29, 2007 #1
    I do not know if it is true but is this identity true

    [tex] \frac{\delta _{n}^{x} }{h} \rightarrow \delta (x-n) [/tex]

    as h tends to 0 ?, the first is Kronecker delta the second Dirac delta.

    i suspect that the above it is true but can not give a proof
  2. jcsd
  3. Jun 29, 2007 #2

    matt grime

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    It isn't true in any meaningful sense - of course, since you've not defined what you mean by convergent, then we're going to have to guess. What does convergence of generalized distributions even mean? The most meaningful, and traditional notion of a sequence of functions converging to the Dirac delta is as follows:

    we have functions f_r(x) for r in the natural numbers, normally, we require the integral of f_r over the real line to be 1 for all r, and for the sequence of numbers

    [tex]y_r:=\int f_r(x)g(x)dx[/tex]

    to converge to g(0).

    These are all trivially false in your case. It is true for suitably normalized Gaussians, for instance.
  4. Jun 29, 2007 #3
    ok, thanks i thought it was true by the similarity of the results

    [tex] \sum_{n=0}^{\infty} f(n) \delta _{n}^{2} =f(2) [/tex]

    (due to Kronecker delta only the value f(2) is obtained in the end)

    and [tex] \int_{0}^{\infty} f(x) \delta (x-2)dx =f(2) [/tex]

    in the first case f(n) takes only discrete values, whereas in the second f(x) involves a sum over the interval (0, infinity) although the sum and the series give the same result f(2)
    Last edited: Jun 29, 2007
  5. Dec 11, 2011 #4
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