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 \(\mathcal{C}\) centered at \(z_0\). Then every point inside \(\mathcal{C}\), \(f(z) = \sum_i \frac{f^{(i)}(z_0)}{n!}(z-z_0)^n\) converges uniformly to \(f(z)\). So in complex analysis, \(C^\infty\) has all of its derivatives \(\equiv C^\omega\) which is analytic, i.e. the function converges to its Taylor series.

Binomial Series

\(1+\alpha+\alpha^2+\cdots+\alpha^n=\frac{1-\alpha^{n+1}}{1-\alpha}\). For \(|\alpha|\lt 1\), \(1+\alpha+\cdots = \frac{1}{1-\alpha}\).

Recalling Cauchy’s Integral Representation

\(f(z)= \frac{1}{2\pi i}\oint dz' \frac{f(z')}{z'-z}\). For \(\left|\frac{z-z_0}{z'-z_0}\right|\lt 1\). Then, \(\frac{1}{z'-z}=\frac{1}{z'-z_0}\frac{1}{1-\frac{z-z_0}{z'-z_0}} = \frac{1}{z'-z}\sum_{n=0}^\infty\left(\frac{z-z_0}{z'-z_0}\right)^n\). Thus, \(f(z)=\sum_n\frac{(z-z_0)^n}{2\pi i}\oint dz' \frac{f(z')}{(z'-z_0)^{n+1}} = \sum-n\frac{(z-z_0)^n}{2\pi i}2\pi i a_n = \sum_n a_n(z-z_0)^n\). 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 \(z_0\).

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:26