Revision as of 17:18, 8 November 2009 by Adbohn (Talk | contribs)

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For problem 8.1, I set up the general formula:

$ z^k \equiv f(z) = \displaystyle\sum_{n=0}^{\infty}a_n(z-z_0)^n $

First off, the coefficients for n>k will be zero, as we do not need higher powers in z. You can differentiate both sides to obtain:

$ \frac{d^k}{dz^k}f(z) = \displaystyle\sum_{n=0}^{\infty} \frac{n!}{(n-k)!}a_n(z-z_0)^{n-k} $

From this, you should be able to set z = z_0 to get the required coefficients as factorials and derivatives of f(z_0)

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