02 - Power Series

Radius of Convergence

Let ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ be a power series
Let $S$ be the set of $z\in \mathbb{C}$ at which it converges

Radius of Convergence of the Power Series is

$$\sup \{|z|:z\in \mathcal{S}\}$$

or $\infty$ is the set $S$ is unbounded

Property of the Radius of Convergence

Let ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ be a power series
Let $S$ be the set of $z\in \mathbb{C}$ at which it converges and $R$ be it’s radius of convergence then

$$B(0,R)\subseteq S\subseteq \overline{B}(0,R)$$

With series converging absolutely on $B(0,R)$ and if $0\le r\le R$ then it converges uniformly on $\overline{B}(0,r)$ with

$$\frac{1}{R}=\underset{n}{\mathrm{limsup}}|a_n{|}^{1/n}$$

Proof

Containment $S\subseteq \overline{B}(0,R)$ is immediate from definition of the radius of convergence

Other containment $B(0,R)\subseteq S$ with series converging absolutely on $B(0,R)$
is a consequence of the series converging uniformly on $\overline{B}(0,r)$ when $0\le r<R$ as

$$B(0,R)=\bigcup _{r<R}\overline{B}(0,r)$$

By definition of $R$ there is some $w$ with $|w|>r$ such that

$$\sum _{n=0}^{\infty }{a}_n{w}^n\text{ converges}$$

Hence the terms of the sums are bounded

$$|a_n{w}^n|\le M\text{ for some }M$$

But then if $|z|\le r$ then

$$|a_n{z}^n|=|a_n{w}^n|{\left|\frac{z}{w}\right|}^n\le M{\left|\frac{r}{w}\right|}^n$$

As the geometric series ${\sum }_{n=0}^{\infty }{\left|\frac{r}{w}\right|}^n$ converges since $|w|>r$
Therefore by Weierstrass Test

$$\sum _{n=0}^{\infty }{a}_n{z}^n\text{ converges uniformly for }|z|<r$$

Suppose the radius of convergence is $R$
Let $0\le r<R$ then by the above there is $w$ with $|w|>r$ such that

$$|a_n{w}^n|\le M\text{ for all }n$$

Assume that $M\ne 0$ then by taking $n$th roots

$$|a_n{|}^{1/n}|w|\le M^{1/n}$$

Since $M^{1/n}\to 1$ as $n\to \infty$ then

$$\underset{n}{\mathrm{limsup}}|a_n{|}^{1/n}\le \frac{1}{|w|}<\frac{1}{r}$$

Since $r<R$ was arbitrary then it follows that

$$\underset{n}{\mathrm{limsup}}|a_n{|}^{1/n}\le \frac{1}{R}$$

Note that it when $R=\infty$ with $\frac{1}{R}=0$ in this case

For the other direction, suppose that ${\mathrm{limsup}}_n|a_n{|}^{1/n}=L$ and that $L\in (0,\infty )$
If $L^{\prime }>L$ then

$$|a_n{|}^{1/n}\le L^{\prime }\text{ for all sufficiently large }n$$

Therefore

$$|a_n{z}^n|\le |L^{\prime }z{|}^n\text{ for sufficiently large }n$$

By the geometric series formula then

$$\sum _{n=0}^{\infty }{a}_n{z}^n\text{ converges provided }|z|<\frac{1}{L^{\prime }}$$

Therefore

$$R\ge \frac{1}{L^{\prime }}$$

Since $L^{\prime }>L$ was arbitrary then

$$R\ge \frac{1}{L}$$

Hence

$$\underset{n}{\mathrm{limsup}}|a_n{|}^{1/n}\ge \frac{1}{R}$$

Minimal changes to argument for when $L=0$ or $L=\infty$

Combining Power Series lemma

Let ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ and ${\sum }_{n=0}^{\infty }{b}_n{z}^n$ be power series of radii of convergence $R_1$ and $R_2$ respectively

For $|z|<\min \{R_1,R_2\}$ let $s(z),t(z)$ be the function to which these series converge to then

1. ${\sum }_{n=0}^{\infty }(a_n+b_n)z^n$ converges for $|z|<\min \{R_1,R_2\}$ to

$$s(z)+t(z)$$

2. ${\sum }_{n=0}^{\infty }\left({\sum }_{k+l=n}{a}_k{b}_l\right)z^n$ converges for $|z|<\min \{R_1,R_2\}$ to

$$s(z)t(z)$$

Note that $\min \{R_1,R_2\}$ is only a lower bound for the radii of convergence in each case0

Differentiation of Power Series

Let ${\sum }_{n=0}^{\infty }{a}_n{z}^n$ be a power series, with radius of convergence $R$
Let $s(t)$ be the function to which this series converges on $B(0,R)$ then

Power Series

$$t(z)=\sum _{n=1}^{\infty }n{a}_n{z}^{n-1}$$

also has radius of convergence $R$

Additionally on $B(0,R)$ then power series $s$ is complex differentiable with

$$s^{\prime }(z)=t(z)$$

In particular, a power series is infinitely differentiable within its radius of convergence

Power Series about other Points

Power Series about points $z_0$ are functions given by an expression of form

$$f(z)=\sum _{n=0}^{\infty }{a}_n(z-z_0{)}^n$$

with

$$f^{\prime }(z)=\sum _{n=1}^{\infty }n{a}_n(z-z_0{)}^{n-1}$$

Taylor Series of Power Series

Let $f(z)={\sum }_{n=0}^{\infty }{a}_n(z-z_0{)}^n$ be a function given by power series with radius of convergence $R$

As radius of convergence for the derivative $f^{\prime }(z)={\sum }_{n=1}^{\infty }n{a}_n(z-z_0{)}^{n-1}$ is at least $R>0$ then

$$f^{\prime }\text{ is also holomorphic in }B(z_0,R)$$

By induction then all derivatives of $f$ as well

Moreover as $f(z_0)=a_0,f^{\prime }(z_0)=a_1$ etc then

$$f(z)=\sum _{n=0}^{\infty }\frac{f^{(n)}(z_0)}{n!}(z-z_0{)}^n$$

so $f$ is given by it’s Taylor Series and is called analytic