# 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 $$z_0$$.

Created: 2023-06-25 Sun 02:26

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