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Thread: Cauchy's Integral formula for Derivatives

  1. #1
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    Cauchy's Integral formula for Derivatives

    Higuys, I'm having a bit of trouble showing this.
    Any help is much appreciated

    $\displaystyle \displaystyle(\frac{1}{2i\pi})\oint_{C} (\frac{e^{tz}}{z^{n+1}}) dz = \frac{t^n}{n!} $

    where C is the unit circle |z|=1
    Last edited by Plato; Feb 17th 2011 at 12:39 PM. Reason: LaTeX fix
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  2. #2
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    $\displaystyle
    \displaystyle
    e^{tz}=\sum \ \frac{t^nz^n}{n!}
    $

    n term is only non zero:

    $\displaystyle \displaystyle
    (\frac{1}{2i\pi})\oint_{C} ( \ \frac{t^nz^n}{n!} \frac{1}{z^{n+1}}) dz =(\frac{1}{2i\pi}) \ \ \frac{t^n}{n!} \oint_{C} \frac{dz}{z}}= \frac{t^n}{n!}
    $
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  3. #3
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    Thank youu! That makes so much more sense, I had been using the wrong summation for $\displaystyle e^{tz}$, but got it now!
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