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Dirac distribution

  1. Jun 3, 2006 #1

    naima

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    bonjour from france,

    I thought that sum of dirac(x - xi)/g'(xi), where the xi verify g(xi) = 0, was a definition for dirac(g(x)). It was proposed, as an exercise, to prove the equality of the 2 terms.
    can one help me
    thanks

    ps : can i write this in latex?
     
  2. jcsd
  3. Jun 3, 2006 #2

    arildno

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    Dearly Missed

    Yes, you may write this in LATEX. Please do so.
     
  4. Jun 3, 2006 #3

    naima

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    thanks for your help but it was only the second question.
     
  5. Jun 3, 2006 #4

    StatusX

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    The delta function is defined by its action under an integral. Try integrating both sides and verify that they behave the same in an integral expression.
     
  6. Jun 4, 2006 #5

    naima

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    Yes, you are true but if the equality I try to prove is not a definition, I do not know what means the first term. In this case I cannot integrate it with a function to compare!
     
  7. Jun 4, 2006 #6

    George Jones

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    I think StatusX means that what you want to show is true if and only if is true "under an integral", i.e., if

    [tex]\int^{\infty}_{-\infty} f(x) \delta \left( g \left( x \right) \right) dx = \int^{\infty}_{-\infty} f(x) \sum_{i} \frac{\delta \left( x - x_{i} \right)}{\left| g' \left( x_{i} \right) \right|} dx,[/tex]

    where [itex]g \left( x_{i} \right) = 0[/itex] for each [itex]x_i[/itex], and [itex]f[/itex] is an arbitrary test function.

    Note that I've inserted an absolute value and a summation.

    Hint:

    [tex]dx = \frac{1}{\frac{dg}{dx}} dg.[/tex]
     
    Last edited: Jun 4, 2006
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