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Math Help - Theorem involving delta function: proof by induction

  1. #1
    Senior Member bugatti79's Avatar
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    Theorem involving delta function: proof by induction

    Hi Folks,

    I am working through a theorem involving the delta function. I have some queries in the early stages of derivation. See attached.

    Thanks
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  2. #2
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    1. There is k -th derivative of delta function and differencial of k-th derivative
    gives k+1 derivative.
    2. The k-th derivative of delta function at \pm \; \infty =0.
    You may think delta function being proportional to e^{-x^2}.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by zzzoak View Post
    1. There is k -th derivative of delta function and differencial of k-th derivative
    gives k+1 derivative.
    2. The k-th derivative of delta function at \pm \; \infty =0.
    You may think delta function being proportional to e^{-x^2}.
    I am confused now because in the first line of attachment
    d  x^{k+1} = x^k but then for the delta function you state that

     d \delta ^k (x) = \delta^{k+1}
    Are both treated differently? Thanks
    Last edited by bugatti79; December 9th 2010 at 12:02 PM. Reason: press send by accident
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  4. #4
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    Yes, they are treated differently.

    Please look here
    Delta Function -- from Wolfram MathWorld
    (17) equation.

    For k-th derivative we write
    <br />
\delta^{(k)}<br />

    and k-th degree
    <br />
x^{k}.<br />
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by zzzoak View Post
    Yes, they are treated differently.

    Please look here
    Delta Function -- from Wolfram MathWorld
    (17) equation.

    For k-th derivative we write
    <br />
\delta^{(k)}<br />

    and k-th degree
    <br />
x^{k}.<br />
    Thanks,

    I have written out the problem fully, just have 2 queries highlighted in red. Hopefully it can be read!!

    1) At the Start, integration by parts query
    2) how we got from (k+1)(-1)(-1)^kK!=(-1)^(k+1)!
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  6. #6
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    I am not sure you have some problems with this line

    <br />
(-1)(-1)^k \; (k+1)k! \; = \; (-1)^{k+1} \; (k+1)! \; \; ?<br />
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  7. #7
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by zzzoak View Post
    I am not sure you have some problems with this line

    <br />
(-1)(-1)^k \; (k+1)k! \; = \; (-1)^{k+1} \; (k+1)! \; \; ?<br />
    Yes, Im wondering how the LHS equates to the RHS? Ie the k -th derivative goes to k+1 etc

    Thanks
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  8. #8
    Senior Member bugatti79's Avatar
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    (k+1)!=(k+1)k! and (-1)(-1)^k=(-1)^(k+1)

    Cheers
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