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

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- Feb 17th 2011, 12:22 PMNoodleCauchy'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 - Feb 17th 2011, 02:38 PMzzzoak
$\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!}

$ - Feb 17th 2011, 03:18 PMNoodle
Thank youu! That makes so much more sense, I had been using the wrong summation for $\displaystyle e^{tz}$, but got it now!