# Power series

Personal notes on power series in complex analysis.

Notes taken while taking a course [1].

## Power series

A power series centered at $$z_0 \in \Complex$$ is a series of the form

$\sum_{k=0}^\infty a_k (z-z_0)^k$

Power series are also known as Taylor series.

## Convergence

### Absolute convergence

A series $$\sum_{k=0}^\infty$$ converges absolutely if the series $$\sum_{k=0}^\infty |a_k|$$ converges.

If $$\sum_{k=0}^\infty a_k$$ converges absolutely, then it also converges, and $$\left|\sum_{k=0}^\infty a_k\right| \leq \sum_{k=0}^\infty |a_k|$$.

For a power series $$\sum_{k=0}^\infty a_k (z-z_0)^k$$, there exists a number $$0 \leq R \leq \infty$$ such that the series converges absolutely in $$\{|z-z_0| < R\}$$ and diverges in $$\{|z-z_0| > R\}$$. The convergence is uniform in $$|z-z_0| \leq r$$ for any $$0 \leq r < R$$.

#### Ratio test

If $$\left\{\left|\frac{a_k}{a_{k+1}}\right|\right\}$$ has a limit as $$k \rightarrow \infty$$, then the radius of convergence $$R$$ for $$\sum_{k=0}^\infty a_k (z-z_0)^k$$ is

$R = \lim_{k \rightarrow \infty} \left|\frac{a_k}{a_{k+1}}\right|$

$$\infty$$ is considered a valid limit value in this context.

#### Root test

If $$\left\{\sqrt[k]{|a_k|}\right\}$$ has a limit as $$k \rightarrow \infty$$, then the radius of convergence $$R$$ for $$\sum_{k=0}^\infty a_k (z-z_0)^k$$ is

$R = \frac{1}{\lim_{k \rightarrow \infty} \sqrt[k]{|a_k|}}$

If $$\lim_{k \rightarrow \infty} \sqrt[k]{|a_k|} = 0$$, then $$R=\infty$$, and if $$\lim_{k \rightarrow \infty} \sqrt[k]{|a_k|} = \infty$$, then $$R=0$$.

The radius of convergence $$R$$ for $$\sum_{k=0}^\infty a_k(z-z_0)^k$$ is

$R = \frac{1}{\limsup_{k \rightarrow \infty} \sqrt[k]{|a_k|}}$

## Analyticity of power series

If $$f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$$ has radius of convergence $$R>0$$, $$f(z)$$ is analytic in $$\{|z-z_0|<R\}$$.

The series can be differentiated term by term.

\begin{align*} \frac{df}{dz}(z) & = \sum_{k=1}^\infty a_k k(z-z_0)^{k-1} \\ \frac{d^2 f}{dz^2}(z) & = \sum_{k=2}^\infty a_k k(k-1)(z-z_0)^{k-2} \\ & \vdots \end{align*}

The coefficient $$a_k$$ can be derived using the $$k$$-order derivative.

$a_k = \frac{1}{k!} \frac{d^k f}{dz^k}(z_0)$

### Power series representation

If $$f : U \rightarrow \Complex$$ is analytic and $$r$$ is such that $$\{|z-z_0| < r\} \subset U$$, then $$f$$ has a power series representation in this disk.

$f(z) = \sum_{k=0}^\infty a_k (z-z_0)^k$

for $$|z-z_0|<r$$ and where

$a_k = \frac{1}{k!} \frac{d^k f}{dz^k}(z_0)$

The radius of convergence is at least $$r$$.

### Determination by derivatives

If $$f$$ and $$g$$ are analytic in $$D=\{|z-z_0|<r\}$$ and $$\frac{d^k f}{dz^k}(z_0) = \frac{d^k g}{dz^k}(z_0)$$ for all $$k \geq 0$$, then $$f(z)=g(z)$$ for all $$z \in D$$.

## Integration of power series

If $$\sum_{k=0}^\infty a_k(z-z_0)^k$$ has radius of convergence $$R$$, then for any $$w$$ such that $$|w-z_0| < R$$,

\begin{align*} \int_{z_0}^w \sum_{k=0}^\infty a_k(z-z_0)^k \, dz &= \sum_{k=0}^\infty a_k \int_{z_0}^w (z-z_0)^k \, dz \\ &= \sum_{k=0}^\infty \frac{a_k}{k+1}(w-z_0)^{k+1} \end{align*}

## References

[1]
Petra Bonfert-Taylor. Introduction to complex analysis. Course on Coursera.