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How to find the second derivitive of delta function?

  1. Feb 10, 2009 #1

    KFC

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    how to find the second derivitive of delta function?
     
  2. jcsd
  3. Feb 10, 2009 #2

    lurflurf

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    let f=the dirac delta
    D=differentiation
    [D^n]f=[n!(-1/x)^n]f
    or
    {[D^n]-[n!(-1/x)^n]}f=0
    if n=2
    [D^2]f=[n!(-1/x)^2]f
     
  4. Feb 10, 2009 #3

    KFC

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    Thanks. It helps. What about if x is complex?
     
  5. Feb 11, 2009 #4

    lurflurf

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    It is not possible to formulate a consistent general theory for that. What specifically did you want to do?
     
  6. Feb 11, 2009 #5

    KFC

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    I want to know how to carry out first derivitative of a delta function with parameter is complex. For example, [tex]\delta(z)[/tex], where [tex]z=x+iy[/tex]

    how to carray out [tex]\frac{d\delta(z)}{dz^*}[/tex] where [tex]z^*[/tex] mean the conjugate of z
     
  7. Feb 13, 2009 #6

    Mute

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    The definition of the derivative of the delta function is

    [tex]\int dx~\delta^{(n)}(x-y)f(x) = (-1)^n f^{(n)}(y)[/tex]

    Formally, you integrate by parts to get

    [tex]\int dx \delta^{(n)}(x-y)f(x) = (-1)^{n}\int dx \delta(x-y) f^{(n)}(x)[/tex].

    If it's at all possible to generalize this to the case you want, you should start with a method like this. I haven't run into a complex delta function, though, so I don't know if such as thing is well defined, let alone it's derivative (and especially its derivative with respect to the complex congugate).

    This book seems to discuss the complex delta function:

    http://books.google.com/books?id=Z2...X&oi=book_result&resnum=6&ct=result#PPA131,M1
     
    Last edited: Feb 13, 2009
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