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how to find the second derivitive of delta function?

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- #1

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how to find the second derivitive of delta function?

- #2

lurflurf

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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

- #3

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Thanks. It helps. What about if x is complex?

- #4

- #5

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how to carray out [tex]\frac{d\delta(z)}{dz^*}[/tex] where [tex]z^*[/tex] mean the conjugate of z

- #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

[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

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