Series Expansions

Gives you calculable solutions, approximations, prove things, give arbitrary functions as polynomials.

Taylor Series

If $f (z)$ is holomorphic everywhere within and on a circular contour $C$ centered at $z_{0}$ . Then every point inside $C$ , $f (z) = \sum_{i} \frac{f^{(i)} (z_{0})}{n!} (z - z_{0})^{n}$ converges uniformly to $f (z)$ . So in complex analysis, $C^{\infty}$ has all of its derivatives $\equiv C^{ω}$ which is analytic, i.e. the function converges to its Taylor series.

Binomial Series

$1 + α + α^{2} + \dots + α^{n} = \frac{1 - α^{n + 1}}{1 - α}$ . For $| α | < 1$ , $1 + α + \dots = \frac{1}{1 - α}$ .

Recalling Cauchy’s Integral Representation

$f (z) = \frac{1}{2 π i} \oint d z^{'} \frac{f (z^{'})}{z^{'} - z}$ . For $| \frac{z - z_{0}}{z^{'} - z_{0}} | < 1$ . Then, $\frac{1}{z^{'} - z} = \frac{1}{z^{'} - z_{0}} \frac{1}{1 - \frac{z - z_{0}}{z^{'} - z_{0}}} = \frac{1}{z^{'} - z} \sum_{n = 0}^{\infty} {(\frac{z - z_{0}}{z^{'} - z_{0}})}^{n}$ . Thus, $f (z) = \sum_{n} \frac{(z - z_{0})^{n}}{2 π i} \oint d z^{'} \frac{f (z^{'})}{(z^{'} - z_{0})^{n + 1}} = \sum - n \frac{(z - z_{0})^{n}}{2 π i} 2 π i a_{n} = \sum_{n} a_{n} (z - z_{0})^{n}$ . So, $a_{n} = \frac{f^{(n)} (z_{0})}{n!}$ .

This Taylor series expansion’s radius of convergence is as large as the distance to the nearest singularity from $z_{0}$ .