
# 7.1: Power series

Many functions can be written in terms of a power series

$$\sum_{k=0}^\infty a_k {(x-x_0)}^k .$$

If we assume that a solution of a differential equation is written as a power series, then perhaps we can use a method reminiscent of undetermined coefficients.  That is, we will try to solve for the numbers $$a_k$$. Before we can carry out this process, let us review some results and concepts about power series.

As we said, a \emph{\myindex{power series}} is an expression such as

$\sum_{k=0}^\infty a_k {(x-x_0)}^k =a_0 + a_1 (x-x_0) +a_2 {(x-x_0)}^2 +a_3 {(x-x_0)}^3 + \cdots,$

where $$a_0,a_1,a_2,\ldots,a_k,\ldots$$ and $$x_0$$ are constants.  Let

$$S_n(x) = \sum_{k=0}^n a_k {(x-x_0)}^k =a_0 + a_1 (x-x_0) + a_2 {(x-x_0)}^2 + a_3 {(x-x_0)}^3 + \cdots + a_n {(x-x_0)}^n$$

denote the so-called \emph{\myindex{partial sum}}.  If for some $$x$$,the limit

$$\lim_{n\to \infty} S_n(x) = \lim_{n\to\infty} \sum_{k=0}^n a_k {(x-x_0)}^k$$

exists, then we say that the series \eqref{ps:sereq1} \emph{converges}\index{convergence of a power series} at $$x$$.

Note that for $$x=x_0$$,the series always converges to $$a_0$$.

When \eqref{ps:sereq1} converges at any other point $$x \not= x_0$$,we say that \eqref{ps:sereq1} is a \emph{\myindex{convergent power series}}.  In this case we write

$$\sum_{k=0}^\infty a_k {(x-x_0)}^k = \lim_{n\to\infty} \sum_{k=0}^n a_k {(x-x_0)}^k.$$

If the series does not converge for any point $$x \not= x_0$$,we say that the series is \emph{divergent}\index{divergent power series}.

\begin{example} \label{ps:expex}

The series

$$\sum_{k=0}^\infty \frac{1}{k!} x^k = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$

is convergent for any $$x$$.

Recall that $$k! = 1\cdot 2\cdot 3 \cdots k$$ is the

factorial.  By convention we define $$0! = 1$$.

In fact, you may recall that this series converges to $$e^x$$.

We say that \eqref{ps:sereq1}\emph{\myindex{converges absolutely}}\index{absolute convergence} at $$x$$ whenever the limit

$$\lim_{n\to\infty} \sum_{k=0}^n \lvert a_k \rvert \, {\lvert x-x_0 \rvert}^k$$

exists.  That is, the series

$\sum_{k=0}^\infty \lvert a_k \rvert \, {\lvert x-x_0 \rvert}^k$

is convergent.

If \eqref{ps:sereq1} converges absolutely at $$x$$,then it

converges at $$x$$.  However, the opposite implication is not true.

\begin{example} \label{ps:1kex}

The series

$$\sum_{k=1}^\infty \frac{1}{k} x^k$$

converges absolutely for all $$x$$ in the interval $$(-1,1)$$.

It converges at $$x=-1, as \sum_{k=1}^\infty \frac{{(-1)}^k}{k}$$ converges (conditionally)

by the alternating series

test.

But the power series does not converge absolutely at $$x=-1$$,because

$\sum_{k=1}^\infty \frac{1}{k}\) does not converge. The series diverges at $$x=1$$. \end{example} \subsection{Radius of convergence} If a power series converges absolutely at some $$x_1$$,then for all $$x$$ such that$\lvert x - x_0  \rvert \leq \lvert x_1 - x_0 \vert\) (that is, $$x$$ is

closer than $$x_1$$ to $$x_0) we have \left\lvert a_k {(x-x_0)}^k \right\rvert \leq \left\lvert a_k {(x_1-x_0)}^k \right\rvert for all \(k$$.

As the numbers $$\left\lvert a_k {(x_1-x_0)}^k \right\rvert$$ sum to some finite

limit, summing smaller positive numbers

$\left\lvert a_k {(x-x_0)}^k \right\rvert\) must also have a finite limit. Therefore, the series must converge absolutely at $$x$$. We have the following result. \begin{theorem} For a power series \eqref{ps:sereq1}, there exists a number$\rho\) (we allow $$\rho=\infty) called the \emph{\myindex{radius of convergence}} such that the series converges absolutely on the interval (x_0-\rho,x_0+\rho)$$ and diverges for $$x < x_0-\rho$$ and $$x > x_0+\rho$$.

We write $$\rho=\infty$$ if

the series converges for all $$x$$.

\end{theorem}

\begin{figure}[h!t]

\capstart

\begin{center}

\inputpdft{ps-conv}

\caption{Convergence of a power series.\label{ps:convfig}}

\end{center}

\end{figure}

See \figurevref{ps:convfig}.

In \exampleref{ps:expex} the radius of convergence is $$\rho = \infty as the series converges everywhere. In \exampleref{ps:1kex} the radius of convergence is \(\rho=1$$.

We note that $$\rho = 0$$ is another way of saying that the series is

divergent.

A useful test for convergence of a series is the

\emph{ratio test}\index{ratio test for series}.  Suppose that

$$\sum_{k=0}^\infty c_k$$

is a series such that the limit

$$L = \lim_{n\to\infty} \left \lvert \frac{c_{k+1}}{c_k} \right \rvert$$

exists.  Then the series converges absolutely if $$L < 1$$ and diverges

if $$L > 1$$.

Let us apply this test to the series \eqref{ps:sereq1}.  That is

we let $$c_k = a_k {(x - x_0)}^k$$ in the test.  Compute

$$L = \lim_{n\to\infty} \left \lvert \frac{c_{k+1}}{c_k} \right \rvert = \lim_{n\to\infty} \left \lvert \frac{a_{k+1} {(x - x_0)}^{k+1}}{a_k {(x - x_0)}^k} \right \rvert = \lim_{n\to\infty} \left \lvert \frac{a_{k+1}}{a_k} \right \rvert \lvert x - x_0 \rvert .$$

Define $$A$$ by

$$A = \lim_{n\to\infty} \left \lvert \frac{a_{k+1}}{a_k} \right \rvert .$$

Then if $$1 > L = A \lvert x - x_0 \rvert$$ the series \eqref{ps:sereq1}

converges absolutely.

If $$A = 0$$,then the series always converges.  If $$A > 0$$,then

the series converges absolutely

if $$\lvert x - x_0 \rvert < \nicefrac{1}{A}, and diverges if \(\lvert x - x_0 \rvert > \nicefrac{1}{A}$$.  That is,

the radius of convergence is $$\nicefrac{1}{A}$$.  Let us summarize.

\begin{theorem}

Let

$$\sum_{k=0}^\infty a_k {(x-x_0)}^k$$

be a power series such that

$$A = \lim_{n\to\infty} \left \lvert \frac{a_{k+1}}{a_k} \right \rvert$$

exists.  If $$A = 0$$,then the radius of convergence of the series

is $$\infty$$.  Otherwise the radius of convergence is $$\nicefrac{1}{A}$$.

\end{theorem}

\begin{example}

Suppose we have the series

$$\sum_{k=0}^\infty 2^{-k} {(x-1)}^k .$$

First we compute,

$$A = \lim_{k\to\infty} \left \lvert \frac{a_{k+1}}{a_k} \right \rvert = \lim_{k\to\infty} \left \lvert \frac{2^{-k-1}}{2^{-k}} \right \rvert = 2^{-1} = \nicefrac{1}{2}.$$

Therefore the radius of convergence is $$2$$,and the series

converges absolutely on the interval $$(-1,3)$$.

\end{example}

The ratio test does not always apply.  That is

the limit

of $$\bigl \lvert \frac{a_{k+1}}{a_k} \bigr \rvert$$ might not exist.

There exist more sophisticated ways of finding the radius of convergence,

but those would be beyond the scope of this chapter.

\subsection{Analytic functions}

Functions represented by power series are called

\emph{\myindex{analytic functions}}.  Not every function is analytic,

although the majority of the functions you have seen in calculus are.

An analytic function $$f(x)$$ is equal to its \emph{\myindex{Taylor series}}%

\footnote{Named after the English mathematician

\href{http://en.wikipedia.org/wiki/Brook_Taylor}{Sir Brook Taylor}

(1685--1731).}

near a point $$x_0$$.

That is, for $$x$$ near $$x_0$$ we have

\label{ps:tayloreq}

f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!} {(x-x_0)}^k ,

where $$f^{(k)}(x_0)$$ denotes the $$k^{\text{th}}$$ derivative of $$f(x) at the point \(x_0$$.

\begin{figure}[h!t]

\capstart

\begin{center}

\diffyincludegraphics{width=3in}{width=4.5in}{ps-sin}

\caption{The sine function and its Taylor approximations

around $$x_0=0 of \(5^{\text{th}}$$ and $$9^{\text{th}}$$ degree.\label{ps:sin}}

\end{center}

\end{figure}

For example, sine is an analytic function and its Taylor series

around $$x_0 = 0 is given by  \sin(x) = \sum_{n=0}^\infty \frac{{(-1)}^n}{(2n+1)!} x^{2n+1} .  In \figurevref{ps:sin} we plot \(\sin(x)$$ and the truncations of the

series up to degree 5 and 9.  You can see that the approximation is very

good for $$x$$ near 0, but gets worse for $$x$$ further away from 0.  This is

what happens in general.

To get a good approximation far away from $$x_0$$ you

need to take more and more terms of the Taylor series.

\subsection{Manipulating power series}

One of the main properties of power series that we will use is

that we can differentiate them term by term.  That is,

suppose that

$\sum a_k {(x-x_0)}^k\) is a convergent power series. Then for $$x$$ in the radius of convergence we have $$\frac{d}{dx} \left[\sum_{k=0}^\infty a_k {(x-x_0)}^k\right] = \sum_{k=1}^\infty k a_k {(x-x_0)}^{k-1} .$$ Notice that the term corresponding to $$k=0$$ disappeared as it was constant. The radius of convergence of the differentiated series is the same as that of the original. \begin{example} Let us show that the exponential $$y=e^x$$ solves $$y'=y$$. First write $$y = e^x = \sum_{k=0}^\infty \frac{1}{k!} x^k .$$ Now differentiate $$y' = \sum_{k=1}^\infty k \frac{1}{k!} x^{k-1} = \sum_{k=1}^\infty \frac{1}{(k-1)!} x^{k-1} .$$ We \emph{reindex}\index{reindexing the series} the series by simply replacing $$k$$ with $$k+1$$. The series does not change, what changes is simply how we write it. After reindexing the series starts at $$k=0$$ again. $$\sum_{k=1}^\infty \frac{1}{(k-1)!} x^{k-1} = \sum_{k+1=1}^\infty \frac{1}{\bigl((k+1)-1\bigr)!} x^{(k+1)-1} = \sum_{k=0}^\infty \frac{1}{k!} x^k .$$ That was precisely the power series for $$e^x$$ that we started with, so we showed that $$\frac{d}{dx} [ e^x ] = e^x$$. \end{example} Convergent power series can be added and multiplied together, and multiplied by constants using the following rules. First, we can add series by adding term by term, $$\left(\sum_{k=0}^\infty a_k {(x-x_0)}^k\right) + \left(\sum_{k=0}^\infty b_k {(x-x_0)}^k\right) = \sum_{k=0}^\infty (a_k+b_k) {(x-x_0)}^k .$$ We can multiply by constants, $$\alpha \left(\sum_{k=0}^\infty a_k {(x-x_0)}^k\right) = \sum_{k=0}^\infty \alpha a_k {(x-x_0)}^k .$$ We can also multiply series together, $$\left(\sum_{k=0}^\infty a_k {(x-x_0)}^k\right) \, \left(\sum_{k=0}^\infty b_k {(x-x_0)}^k\right) = \sum_{k=0}^\infty c_k {(x-x_0)}^k ,$$ where$c_k = a_0b_k + a_1 b_{k-1} + \cdots + a_k b_0\).

The radius of convergence of the sum or the product

is at least the minimum of the radii of convergence of

the two series involved.

\subsection{Power series for rational functions}

Polynomials are simply finite power series.  That is, a polynomial

is a power series where

the $$a_k$$ are zero for all $$k$$ large enough.  We can always expand

a polynomial as a power series about any point $$x_0$$ by writing

the polynomial as a polynomial in $$(x-x_0)$$.  For example,

let us write

$2x^2-3x+4\) as a power series around $$x_0 = 1: 2x^2-3x+4 = 3 + (x-1) + 2{(x-1)}^2 . In other words \(a_0 = 3$$,$$a_1 = 1$$,$$a_2 = 2$$,and all other$a_k = 0\).  To do this, we know that $$a_k = 0$$ for all $$k \geq 3$$ as the

polynomial is of degree 2.

We write $$a_0 + a_1(x-1) + a_2{(x-1)}^2$$,we expand, and we solve

for $$a_0$$,$$a_1$$,and $$a_2$$.  We could have also differentiated at $$x=1 and used the Taylor series formula \eqref{ps:tayloreq}. Let us look at rational functions, that is, ratios of polynomials. An important fact is that a series for a function only defines the function on an interval even if the function is defined elsewhere. For example, for -1 < x < 1$$ we have

$$\frac{1}{1-x} = \sum_{k=0}^\infty x^k = 1 + x + x^2 + \cdots$$

This series is called the \emph{\myindex{geometric series}}.  The ratio

test tells us that the radius of convergence is $$1$$.  The series

diverges for $$x \leq -1$$ and $$x \geq 1$$,even though

$\frac{1}{1-x}\) is defined for all $$x \not= 1$$. We can use the geometric series together with rules for addition and multiplication of power series to expand rational functions around a point, as long as the denominator is not zero at $$x_0$$. Note that as for polynomials, we could equivalently use the Taylor series expansion \eqref{ps:tayloreq}. \begin{example} Expand $$\frac{x}{1+2x+x^2}$$ as a power series around the origin ($x_0 = 0$) and find the radius of convergence. First, write $$1+2x+x^2 = {(1+x)}^2 = {\bigl(1-(-x)\bigr)}^2$$. Now we compute $$\begin{split} \frac{x}{1+2x+x^2} &= x \, {\left( \frac{1}{1-(-x)} \right)}^2 \\ &= x \, { \left( \sum_{k=0}^\infty {(-1)}^k x^k \right)}^2 \\ &= x \, \left( \sum_{k=0}^\infty c_k x^k \right) \\ &= \sum_{k=0}^\infty c_k x^{k+1} , \end{split}$$ where using the formula for the product of series we obtain, $$c_0 = 1$$,$$c_1 = -1 -1 = -2$$,$$c_2 = 1+1+1 = 3$$,etc\ldots. Therefore $$\frac{x}{1+2x+x^2} = \sum_{k=1}^\infty {(-1)}^{k+1} k x^k = x-2x^2+3x^3-4x^4+\cdots$$ The radius of convergence is at least 1. We use the ratio test $$\lim_{k\to\infty} \left\lvert \frac{a_{k+1}}{a_k} \right\rvert = \lim_{k\to\infty} \left\lvert \frac{{(-1)}^{k+2} (k+1)}{{(-1)}^{k+1}k} \right\rvert = \lim_{k\to\infty} \frac{k+1}{k} = 1 .$$ So the radius of convergence is actually equal to 1. \end{example} When the rational function is more complicated, it is also possible to use method of partial fractions. For example, to find the Taylor series for $$\frac{x^3+x}{x^2-1}$$,we write $$\frac{x^3+x}{x^2-1} = x + \frac{1}{1+x} - \frac{1}{1-x} = x + \sum_{k=0}^\infty {(-1)}^k x^k - \sum_{k=0}^\infty x^k = - x + \sum_{\substack{k=3 \\ k \text{ odd}}}^\infty (-2) x^k .$$ \subsection{Exercises} \begin{exercise} Is the power series $$\displaystyle \sum_{k=0}^\infty e^k x^k$$ convergent? If so, what is the radius of convergence? \end{exercise} \begin{exercise} Is the power series $$\displaystyle \sum_{k=0}^\infty k x^k$$ convergent? If so, what is the radius of convergence? \end{exercise} \begin{exercise} Is the power series $$\displaystyle \sum_{k=0}^\infty k! x^k$$ convergent? If so, what is the radius of convergence? \end{exercise} \begin{exercise} Is the power series $$\displaystyle \sum_{k=0}^\infty \frac{1}{(2k)!} {(x-10)}^k convergent? If so, what is the radius of convergence? \end{exercise} \begin{exercise} Determine the Taylor series for \(\sin x$$ around the point $$x_0 = \pi$$. \end{exercise} \begin{exercise} Determine the Taylor series for $$\ln x$$ around the point $$x_0 = 1, and find the radius of convergence. \end{exercise} \begin{exercise} Determine the Taylor series and its radius of convergence of \(\dfrac{1}{1+x} around \(x_0 = 0$$. \end{exercise} \begin{exercise} Determine the Taylor series and its radius of convergence of$\dfrac{x}{4-x^2}\) around $$x_0 = 0$$.  Hint: You will not be able to

use the ratio test.

\end{exercise}

\begin{exercise}

Expand $$x^5+5x+1$$ as a power series around $$x_0 = 5$$.

\end{exercise}

\begin{exercise}

Suppose that the ratio test applies to a series

$\displaystyle \sum_{k=0}^\infty a_k x^k\). Show, using the ratio test, that the radius of convergence of the differentiated series is the same as that of the original series. \end{exercise} \begin{exercise} Suppose that $$f$$ is an analytic function such that$f^{(n)}(0) = n\).  Find $$f(1)$$.

\end{exercise}

\setcounter{exercise}{100}

\begin{exercise}

Is the power series

$\displaystyle \sum_{n=1}^\infty {(0.1)}^n x^n$

convergent? If so, what is the radius of convergence?

\end{exercise}

\exsol{%

Yes.  Radius of convergence is $$10$$.

}

\begin{exercise}[challenging]

Is the power series

$\displaystyle \sum_{n=1}^\infty \frac{n!}{n^n} x^n$

convergent? If so, what is the radius of convergence?

\end{exercise}

\exsol{%

Yes.  Radius of convergence is $$e$$.

}

\begin{exercise}

Using the geometric series, expand $$\frac{1}{1-x}$$ around $$x_0=2$$.

For what $$x$$ does the series converge?

\end{exercise}

\exsol{%

$\frac{1}{1-x} = -\frac{1}{1-(2-x)}\) so$\frac{1}{1-x} =

\sum\limits_{n=0}^\infty {(-1)}^{n+1} {(x-2)}^n$, which converges for $$1 < x < 3$$. } \begin{exercise}[challenging] Find the Taylor series for $$x^7 e^x$$ around $$x_0 = 0$$. \end{exercise} \exsol{%$\sum\limits_{n=7}^\infty

\frac{1}{(n-7)!} x^n$} \begin{exercise}[challenging] Imagine $$f$$ and $$g$$ are analytic functions such that$f^{(k)}(0) = g^{(k)}(0)\) for all large enough $$k$$.  What can you

say about $$f(x)-g(x)? \end{exercise} \exsol{% f(x)-g(x)$$ is a polynomial.  Hint: Use Taylor series.