Calculus of power series

Functions defined by power series

If the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ has the radius of convergence $r>0$, then it defines a function $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ on open interval $(-r,r)$.

Theorem (Continuous power series)

Let $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ be a power series with its radius of convergence $r>0$. Then, the function $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ is continuous on $(-r,r)$.

Proof. It suffices to show that $f(x)$ is continuous on open interval $I=(-s,s)$ where $s$ is an arbitrary real number such that $0<s<r$. Let $t=\frac{r+s}{2}$. Then $s<t<r$ and the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges absolutely at $x=t$.

Step 1. For the partial sum $f_n(x)=\sum _{k=0}^n{a}_k{x}^k$, we show the following:

• (*) For any $\epsilon >0$, there exists an $N\in {\mathbb{N}}_0$ such that, for all $n\in {\mathbb{N}}_0$ and all $x\in I$, if $n\ge N$, then $|f(x)-f_n(x)|<\epsilon$.

(In other words, $\{f_n(x)\}$ uniformly converges to $f(x)$.) Take an arbitrary $\epsilon >0$. Since the series $\sum _{n=0}^{\mathrm{\infty }}|a_n|t^n$ has a sum, $\underset{n\to \mathrm{\infty }}{lim}|a_n|t^n=0$. In particular, the sequence $\{|a_n|t^n\}$ is bounded so that there exists $M>0$ such that $|a_n|t^n<M$ for all $n$. Also, since $0<\frac{s}{t}<1$, the geometric series $\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{s}{t}\right)}^n$ converges. Therefore, we can find a sufficiently large $N$ such that $\sum _{k=n+1}^{\mathrm{\infty }}{\left(\frac{s}{t}\right)}^k<\frac{\epsilon }{M}$ for all $n\ge N$.

Noting that, for all $n\ge N$ and $x\in (-s,s)$, $|x|<s$, we have

$$\begin{array}{rcl}|f(x)-f_n(x)|& =& \left|\sum _{k=n+1}^{\mathrm{\infty }}{a}_k{x}^k\right|\le \sum _{k=n+1}^{\mathrm{\infty }}|a_k||x{|}^k\\ & =& \sum _{k=n+1}^{\mathrm{\infty }}|a_k|t^k{\left(\frac{|x|}{t}\right)}^k<\sum _{k=n+1}^{\mathrm{\infty }}|a_k|t^k{\left(\frac{s}{t}\right)}^k\\ & <& \sum _{k=n+1}^{\mathrm{\infty }}M{\left(\frac{s}{t}\right)}^k<\epsilon .\end{array}$$

Thus, (*) holds. This means that the function $f(x)$ can be approximated by a polynomial function $f_n(x)$ to arbitrary precision. In other words, $f(x)=\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)$.

Step 2. Next, consider a sequence of functions $\{f_n(x)\}$ that are continuous on interval $I$ and satisfy the property (*) in Step 1. Then, we show that its limit $f(x)=\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)$ is also continuous on $I$.

Take an arbitrary $a\in I$. Let $\epsilon >0$. By (*), for all $x\in I$, there exists some $n$ such that $|f(x)-f_n(x)|<\frac{\epsilon }{3}$; in particular, $|f(a)-f_n(a)|<\frac{\epsilon }{3}$. Furthermore, since $f_n(x)$ is continuous, there exists some $\delta >0$ such that, if $|x-a|<\delta$, then $|f_n(x)-f_n(a)|<\frac{\epsilon }{3}$. Combining these together, we have

$$\begin{array}{rcl}|f(x)-f(a)|& \le & |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|\\ & <& \frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon .\end{array}$$

Now,

$$\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^n{a}_k{x}^k$$

and polynomial functions (i.e., partial sums) are continuous (everywhere). Therefore, $f(x)$ is continuous on $I$. ■

Term-wise integration and term-wise differentiation

Given the power series $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$, consider the following power series:

$$\begin{array}{rcl}F(x)& =& \sum _{n=0}^{\mathrm{\infty }}\frac{a_n}{n+1}{x}^{n+1},\\ g(x)& =& \sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}.\end{array}$$

If $f(x)$ is a finite sum (i.e., polynomial function; $\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\ge N(a_n=0)$), then $F(x)$ and $g(x)$ are the primitive function and derivative of $f(x)$, respectively. As we show below, this holds true even when $f(x)$ is not a polynomial. That is, if $f(x)$ has the radius of convergence $r>0$, then $F(x)$ is the primitive function of $f(x)$ and $g(x)$ is the derivative of $f(x)$ on the open interval $(-r,r)$. This means that we can obtain the primitive function and derivative of a function defined by a power series by term-wise (term-by-term) integration and term-wise (term-by-term) differentiation of $f(x)$, respectively.

Theorem (Term-wise integration of power series)

Suppose the power series $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ has a radius of convergence $r>0$. Then, the following holds on $(-r,r)$:

$${\int }_0^{x}f(t)dt=\sum _{n=0}^{\mathrm{\infty }}\frac{a_n}{n+1}{x}^{n+1}.$$

Proof. Consider the sequence $\{f_n(x)\}$ of polynomial functions given by the partial sums $f_n(x)=\sum _{k=0}^n{a}_k{x}^k$. For $x\in (-r,r)$, take an $s\in \mathbb{R}$ such that $|x|<s<r$. Then, on the interval $I=(-s,s)$, we have $f(x)=\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)$ (c.f., the property (*) in the proof of the above Theorem (Continuous power series)). For any $\epsilon >0$, there exists an $N\in {\mathbb{N}}_0$ such that, for all $n\ge N$, $|f(x)-f_n(x)|<\epsilon$. Thus, if $n\ge N$, then

$$|{\int }_0^{x}f(t)dt-{\int }_0^x{f}_n(t)dt|\le {\int }_0^x|f(t)-f_n(t)|dt<\epsilon |x|.$$

Therefore,

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^x{f}_n(t)dt={\int }_0^{x}f(t)dt.$$

Since

$${\int }_0^x\left(\sum _{k=0}^n{a}_k{t}^k\right)dt=\sum _{k=0}^n\left({\int }_0^x{a}_k{t}^{k}dt\right)=\sum _{k=0}^n\frac{a_k}{k+1}{x}^{k+1},$$

we have

$${\int }_0^{x}f(t)dt=\sum _{n=0}^{\mathrm{\infty }}\frac{a_n}{n+1}{x}^{n+1}.$$

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Theorem (Term-wise differentiation of power series)

Given the power series $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$, consider the power series $g(x)=\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$. The following hold:

1. The radius of convergence of $g(x)$ is equal to that of $f(x)$.
2. If $f(x)$ has a radius of convergence $r>0$, then $f(x)$ is differentiable on the open interval $(-r,r)$ and $f^{\prime }(x)=g(x)$.

Proof. 

1. Let $r$ and $r^{\prime }$ be the radii of convergence of $f(x)$ and $g(x)$, respectively. First, we show that $r\le r^{\prime }$. This is trivial if $r=0$. Suppose $r>0$. It suffices to show that $g(x)$ converges absolutely for all $x$ such that $|x|<r$. Let $s$ be such that $|x|<s<r$. Since the series $\sum _{n=0}^{\mathrm{\infty }}|a_n|s^n$ has a sum, there exists $M$ such that $|a_n|s^n<M$ for all $n$. Let $t=\frac{|x|}{s}$, then $0<t<1$ and $$\begin{array}{}\text{(Eq:gbound)}& \sum _{n=1}^{\mathrm{\infty }}|n{a}_n{x}^{n-1}|=\sum _{n=1}^{\mathrm{\infty }}\frac{n}{s}|a_n{s}^n|t^{n-1}<M\sum _{n=1}^{\mathrm{\infty }}\frac{n}{s}{t}^{n-1}.\end{array}$$ But $$\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{n+1}{s}{t}^n}{\frac{n}{s}{t}^{n-1}}=\underset{n\to \mathrm{\infty }}{lim}\frac{(n+1)t}{n}=t<1,$$ so by D'Alembert's criterion, the right-hand side of (Eq:gbound) has a sum, and hence $g(x)$ converges absolutely. Next, we show that $r^{\prime }\le r$. This is trivial if $r^{\prime }=0$. Suppose $r^{\prime }>0$. It suffices to show that $f(x)$ converges absolutely for all $x\in (-r^{\prime },r)$. But this is true because for all $n\ge 1$, $$|a_n{x}^n|\le |x||n{a}_n{x}^{n-1}|<r^{\prime }|n{a}_n{x}^{n-1}|.$$ That is, $r^{\prime }g(x)$ is a dominating series of $f(x)$. Therefore, we have shown that $r=r^{\prime }$.
2. By Part 1, $g(x)$ is a power series with the radius of convergence $r>0$. By term-wise integration, for all $x\in (-r,r)$, we have $${\int }_0^{x}g(t)dt=\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}^n=f(x)-a_0.$$ The left-hand side is differentiable with respect to $x$. Thus, $f(x)$ is also differentiable. By differentiating both sides, we have $f^{\prime }(x)=g(x)$ (c.f. the Fundamental Theorem of Calculus).

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Corollary 1

Suppose the power series $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ has a radius of convergence $r>0$. Then, the function $f(x)$ is of class $C^{\mathrm{\infty }}$ on the open interval $(-r,r)$, and its derivative $f^{\prime }(x)$ is the power series $\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$ with the radius of convergence $r$.

Proof. Exercise. ■

Corollary 2

Suppose the power series $f(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ has a radius of convergence $r>0$. For each $k=0,1,2,\cdots$, we have

$$\begin{array}{}\text{(Eq:maccof)}& a_k=\frac{f^{(k)}(0)}{k!}\end{array}$$

where $f^{(k)}(x)$ is the $k$-th derivative of $f(x)$.

Proof. For $k=0$, (Eq:maccof) is trivial. By repeatedly term-wise differentiating $f(x)$, we can show by mathematical induction (exercise!) that

$$f^{(k)}(x)=\sum _{n=k}^{\mathrm{\infty }}n(n-1)\cdots (n-k+1)a_n{x}^{n-k}.$$

Substituting $x=0$ to this, we have

$$f^{(k)}(0)=k!a_k.$$

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