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Mathematics LibreTexts

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

\begin{equation} \label{ps:tayloreq}

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

\end{equation}

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.