4.1 Differentiating of Power Series

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We defined many interesting functions via power series. Our goal in this section is to verify that power series are holomorphic on their open ball of convergence, and that their derivatives are easy to calculate.

Fix a power series $$\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ with radius of convergence $R>0$. Define $f:\mathsf{B}(b,R)\to \mathbb{C}$ by $$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n=a_0+a_1(z-b)+a_2(z-b{)}^2+a_3(z-b{)}^3+\cdots$$ for all $z\in \mathsf{B}(b,R)$. We do not know yet whether $f$ is holomorphic, but if it were, a reasonable guess for its derivative would be $$\sum _{n=0}^{\mathrm{\infty }}n{a}_n(z-b{)}^{n-1}=a_1+2a_2(z-b)+3a_3(z-b{)}^2+4a_3(z-b{)}^3+\cdots$$ which is itself a power series with $n$th coefficient $(n+1)a_{n+1}$. To verify this is the case we need to do three things.

1. Prove the purported derivative has radius of convergence $R$.
2. Prove $f$ is holomorphic on $\mathsf{B}(b,R)$.
3. Prove $f^{\prime }$ equals the purported derivative on $\mathsf{B}(b,R)$.

The first step will make use of the following result, which will be on the exercise sheet.

Lemma

For every $z\in \mathsf{B}(0,1)$ the series $$\sum _{n=0}^{\mathrm{\infty }}(n+1)z^n=1+2z+3z^2+4z^3+\cdots$$ converges.

Proof:

This follows from the ratio test: $$\underset{n\to \mathrm{\infty }}{lim}\left|{\displaystyle \frac{a_{n+1}}{a_n}}\right|=|z|\underset{n\to \mathrm{\infty }}{lim}{\displaystyle \frac{n+1}{n}}=|z|$$ so the series converges absolutely whenever $|z|<1$. $◻$

In fact $$\sum _{n=0}^{\mathrm{\infty }}(n+1)z^n={\displaystyle \frac{1}{(1-z{)}^2}}$$ which will be on the problem sheet.

Lemma

If a power series $${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n}$$ has radius of convergence $R>0$ then the power series $${\displaystyle \sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}(z-b{)}^n}$$ converges absolutely on the open ball $\mathsf{B}(b,R)$.

Proof:

Fix $|z-b|<R$ and choose $r$ such that $|z-b|<|r-b|<R$. Then $$\sum _{n=0}^{\mathrm{\infty }}{a}_n(r-b{)}^n$$ converges absolutely. Hence the summands must be bounded: there is $K>0$ such that $|a_n(r-b{)}^n|<K$ for all $n\ge 0$.

Let $q=(z-b)/(r-b)$ and note that $0<|q|<1$. Then $$\begin{array}{rl}|(n+1)a_{n+1}(z-b{)}^n|=& (n+1)|a_{n+1}(r-b{)}^n{q}^n|\\ =& \frac{n+1}{|r-b|}|a_{n+1}(r-b{)}^{n+1}||q{|}^n\\ <& \frac{K}{|z-b|}(n+1)|q{|}^n\end{array}$$ but $$\sum _{n=0}^{\mathrm{\infty }}(n+1)q^n$$ converges by the previous lemma. Hence $$\sum _{n=0}^{\mathrm{\infty }}|(n+1)a_{n+1}(z-b{)}^n|$$ converges by the comparison test. We conclude that $$\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}(z-b{)}^n$$ converges absolutely. $◻$

Remark

The Cauchy-Hadamard theorem can be used to prove that the two power series have the same radius of convergence.

The second and third steps will be proved together in the following theorem.

Theorem (Differentiating Power Series)

If a power series $$\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ has radius of convergence $R>0$ then it is holomorphic on the open ball $\mathsf{B}(b,R)$ and its derivative there is $$\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}(z-b{)}^n$$ as hoped.

Proof:

Write $$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ and $$g(z)=\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}(z-b{)}^n$$ for all $z\in \mathsf{B}(b,R)$. We need to compare the quantities $$\begin{array}{rl}& \frac{f(z+h)-f(z)}{h}\\ =& \frac{1}{h}(\sum _{n=0}^{\mathrm{\infty }}{a}_n(z+h-b{)}^n-\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-b{)}^n)\\ =& \sum _{n=1}^{\mathrm{\infty }}{a}_n\frac{(z+h-b{)}^n-(z-b{)}^n}{(z+h-b)-(z-b)}\\ =& \sum _{n=1}^{\mathrm{\infty }}{a}_n((z+h-b{)}^{n-1}(z-b{)}^0+(z+h-b{)}^{n-2}(z-b{)}^1+(z+h-b{)}^{n-3}(z-b{)}^2+\\ & \cdots +(z+h-b{)}^1(z-b{)}^{n-2}+(z+h-b{)}^0(z-b{)}^{n-1})\end{array}$$ and $$g(z)=\sum _{n=0}^{\mathrm{\infty }}(n+1)a_{n+1}(z-b{)}^n=a_1+2a_2(z-b)+3a_3(z-b{)}^2+\cdots$$ as we need to prove their difference tends to zero as $h\to 0$.

The first terms in both expressions agree. The second terms are $$a_2\frac{2h(z-b)+h^2}{h}\text{ and }2a_2(z-b)$$ and these are close for $h$ small. We can make similar arguments for later terms in the series, but cannot carry out the argument for all terms simultaneously as the errors would conspire against us. What we need, therefore, is to bound the tails of the two series separately, and then choose $h$ small enough to control the remaining terms manually. Formally, this goes as follows.

Fix $z\in \mathbb{C}$ with $|z-b|<R$. Fix $ϵ>0$. Choose $r>0$ such that $|z-b|<r<R$. If $|h|<r-|z-b|$ then we also $|z+h-b|<r$. For such $h$ we can estimate $$\begin{array}{rl}& |a_n((z+h-b{)}^{n-1}(z-b{)}^0+(z+h-b{)}^{n-2}(z-b{)}^1\\ & +(z+h-b{)}^{n-3}(z-b{)}^2+\cdots +(z+h-b{)}^1(z-b{)}^{n-2}\\ & +(z+h-b{)}^0(z-b{)}^{n-1})|\\ \le & |a_n|(r^{n-1}{r}^0+r^{n-2}{r}^1+\cdots +r^1{r}^{n-2}+r^0{r}^{n-1})\\ =& n|a_n|r^{n-1}\end{array}$$ and, as in the proof of the previous lemma, find $N\in \mathbb{N}$ such that $$\begin{array}{c}|\frac{1}{h}\sum _{n=N}^{\mathrm{\infty }}{a}_n(z+h-b{)}^n-a_n(z-b{)}^n|<\frac{ϵ}{4}\\ |\sum _{n=N}^{\mathrm{\infty }}n{a}_n(z-b{)}^{n-1}|<\frac{ϵ}{4}\end{array}$$ both hold. Now $$\sum _{n=0}^{N-1}{a}_n\frac{(z+h-b{)}^n-(z-b{)}^n}{h}\text{ and }\sum _{n=0}^{N-1}n{a}_n(z-b{)}^{n-1}$$ are both polynomials in $h$ taking the same value at $h=0$. Therefore, continuity of polynomials implies there is $\delta >0$ such that $|h|<\delta$ guarantees $$|\sum _{n=0}^{N-1}{a}_n(z+h-b{)}^n-a_n(z-b{)}^n-\sum _{n=0}^{N-1}n{a}_n(z-b{)}^{n-1}|<\frac{ϵ}{2}$$ giving the result when combined with the previous estimate. $◻$