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.

Definition

As we said, a power series is an expression such as

$\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k} = a_{0} + a_{1} (x - x_{0}) + a_{2} {(x - x_{0})}^{2} + a_{3} {(x - x_{0})}^{3} + \dots,$

(7.1.1)

where $a sub 0, a sub 1, a sub 2, …, a sub k, …$ and $x sub 0$ are constants. Let

$S_{n} (x) = \sum_{k = 0}^{n} a_{k} {(x - x_{0})}^{k} = a_{0} + a_{1} (x - x_{0}) + a_{2} {(x - x_{0})}^{2} + a_{3} {(x - x_{0})}^{3} + \dots + a_{n} {(x - x_{0})}^{n},$

denote the so-called 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 (7.1.1) converges at $x$ . Note that for $x equals x sub 0$ , the series always converges to $a sub 0$ . When (7.1.1) converges at any other point $x not equal to x sub 0$ , we say that (7.1.1) is a 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 $x not equal to x sub 0$ , we say that the series is divergent.

Example 1

The series

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

is convergent for any $x$ . Recall that $k! = 1 \cdot 2 \cdot 3 \dots k$ is the factorial. By convention we define $0 ! equals 1$ . In fact, you may recall that this series converges to $e^{x}$ .

We say that (7.1.1) converges absolutely at $x$ whenever the limit

${lim}_{n \to \infty} \sum_{k = 0}^{n} | a_{k} | {| x - x_{0} |}^{k}$

exists. That is, the series $\sum_{k = 0}^{\infty} | a_{k} | {| x - x_{0} |}^{k}$ is convergent. If (7.1.1) converges absolutely at $x$ , then it converges at $x$ . However, the opposite implication is not true.

Example 2

The series

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

converges absolutely for all $x$ in the interval $open parenthesis minus 1, 1 close parenthesis$ .

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 $x equals 1$ .

Radius of Convergence

If a power series converges absolutely at some $x sub 1$ , then for all $x$ such that $| x - x_{0} | \leq | x_{1} - x_{0} |$ (that is, $x$ is closer than $x sub 1$ to $x sub 0$ ) we have $| a_{k} {(x - x_{0})}^{k} | \leq | a_{k} {(x_{1} - x_{0})}^{k} |$ for all $k$ . As the numbers $| a_{k} {(x_{1} - x_{0})}^{k} |$ sum to some finite limit, summing smaller positive numbers $| a_{k} {(x - x_{0})}^{k} |$ must also have a finite limit. Hence, the series must converge absolutely at $x$ .

Diagram of power series convergence on a number line: the shaded band from x sub 0 minus rho to x sub 0 plus rho is labelled converges absolutely, diverges beyond each endpoint. — Figure $1$ : Convergence of a power series.

See Figure $1$ . In Example $1$ the radius of convergence is $ρ = \infty$ as the series converges everywhere. In Example $2$ the radius of convergence is $ρ equals 1$ . We note that $ρ equals 0$ is another way of saying that the series is divergent. A useful test for convergence of a series is the ratio test. Suppose that

$∑ k equals 0 infinity c sub k$

is a series such that the limit

$L = {lim}_{n \to \infty} | \frac{c_{k + 1}}{c_{k}} |$

exists. Then the series converges absolutely if $L less than 1$ and diverges if $L greater than 1$ .

Let us apply this test to the series (7.1.1). That is we let $c_{k} = a_{k} {(x - x_{0})}^{k}$ in the test. Compute

$L = {lim}_{n \to \infty} | \frac{c_{k + 1}}{c_{k}} | = {lim}_{n \to \infty} | \frac{a_{k + 1} {(x - x_{0})}^{k + 1}}{a_{k} {(x - x_{0})}^{k}} | = {lim}_{n \to \infty} | \frac{a_{k + 1}}{a_{k}} | | x - x_{0} | .$

Define $A$ by

$A = {lim}_{n \to \infty} | \frac{a_{k + 1}}{a_{k}} | .$

Then if $1 > L = A | x - x_{0} |$ the series (7.1.1) converges absolutely. If $A equals 0$ , then the series always converges. If $A greater than 0$ , then the series converges absolutely if $| x - x_{0} | < \frac{1}{A}$ , and diverges if $| x - x_{0} | > \frac{1}{A}$ . That is, the radius of convergence is $\frac{1}{A}$ .

A similar test is the root test. Suppose

$L = {lim}_{k \to \infty} \sqrt[3]{| c_{k} |}$

exists. Then $∑ k equals 0 infinity c sub k$ converges absolutely if $L less than 1$ and diverges if $L greater than 1$ . We can use the same calculation as above to find $A$ . Let us summarize.

Note

Ratio and Root Tests for Power Series

Let

$\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}$

be a power series such that

$A = {lim}_{n \to \infty} | \frac{a_{k + 1}}{a_{k}} | or A = {lim}_{k \to \infty} \sqrt[3]{| a_{k} |}$

exists. If $A equals 0$ , then the radius of convergence of the series is $\infty$ . Otherwise the radius of convergence is $\frac{1}{A}$ .

Example 3

Suppose we have the series

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

First we compute,

$A = {lim}_{k \to \infty} | \frac{a_{k + 1}}{a_{k}} | = {lim}_{k \to \infty} | \frac{2^{- k - 1}}{2^{- k}} | = 2^{- 1} = \frac{1}{2} .$

Therefore the radius of convergence is $2$ , and the series converges absolutely on the interval $open parenthesis minus 1, 3 close parenthesis$ . And we could just as well have used the root test:

$A = {lim}_{k \to \infty} {lim}_{k \to \infty} \sqrt[k]{| a_{k} |} = {lim}_{k \to \infty} \sqrt[k]{| 2^{- k} |} = {lim}_{k \to \infty} 2^{- 1} = \frac{1}{2} .$

The root or the ratio test does not always apply. That is the limit of $| \frac{a_{k + 1}}{a_{k}} |$ or $the kth root of | a sub k |$ might not exist. There exist more sophisticated ways of finding the radius of convergence, but those would be beyond the scope of this chapter. The two methods above cover many of the series that arise in practice. Often if the root test applies, so does the ratio test, and vice versa, though the limit might be easier to compute in one way than the other.

Analytic Functions

Functions represented by power series are called 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 Taylor seriesto the power 1 near a point $x sub 0$ . That is, for $x$ near $x sub 0$ we have

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

(7.1.2)

where $f to the power open parenthesis k close parenthesis open parenthesis x sub 0 close parenthesis$ denotes the $k^{th}$ derivative of $f (x)$ at the point $x sub 0$ .

Plot of the sine curve with two Taylor approximations, x from minus 10 to 10. The green polynomial tracks the sine only to about x equals plus or minus 3 and the red to about plus or minus 4, then both run off the frame. — Figure $2$ : The sine function and its Taylor approximations around $x sub o equals 0$ of and degree.

For example, sine is an analytic function and its Taylor series around $x sub 0 equals 0$ is given by

$\sin (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1} .$

In Figure $2$ 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 $x sub 0$ you need to take more and more terms of the Taylor series.

Function graph showing y = sin(x), y = sin(a) + cos(a)*(x-a) - sin(a)*(x-a)^2/2 - cos(a)*(x-a)^3/6 + sin(a)*(x-a)^4/24 + cos(a)*(x-a)^5/120 and y = sin(a) + cos(a)*(x-a) - sin(a)*(x-a)^2/2 - cos(a)*(x-a)^3/6 + sin(a)*(x-a)^4/24 + cos(a)*(x-a)^5/120 - sin(a)*(x-a)^6/720 - cos(a)*(x-a)^7/5040 + sin(a)*(x-a)^8/40320 + cos(a)*(x-a)^9/362880. Adjustable parameter: Expansion point x0 (a) = 0. Viewing window: x from -8.44 to 8.44, y from -5.22 to 5.22. — The book's sine picture with the expansion point set free: sin x in blue, its degree-5 Taylor polynomial in red and its degree-9 in green, both taken about x₀ = a. It opens at a = 0, the book's own picture; drag a along the sine curve and watch the sharp accuracy window slide with it — degree 9 always reaching further than degree 5, both failing at a distance, and the failure region moving rigidly with the centre.

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

$\frac{d}{d x} [\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}] = \sum_{k = 1}^{\infty} k a_{k} {(x - x_{0})}^{k - 1} .$

Notice that the term corresponding to $k equals 0$ disappeared as it was constant. The radius of convergence of the differentiated series is the same as that of the original.

Example 5

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 reindex the series by simply replacing $k$ with $k plus 1$ . The series does not change, what changes is simply how we write it. After reindexing the series starts at $k equals 0$ again.

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

That was precisely the power series for $e^{x}$ that we started with, so we showed that $\frac{d}{d x} [e^{x}] = e^{x}$ .

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,

$(\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}) + (\sum_{k = 0}^{\infty} b_{k} {(x - x_{0})}^{k}) = \sum_{k = 0}^{\infty} (a_{k} + b_{k}) {(x - x_{0})}^{k} .$

We can multiply by constants,

$α (\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}) = \sum_{k = 0}^{\infty} α a_{k} {(x - x_{0})}^{k} .$

We can also multiply series together,

$(\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}) (\sum_{k = 0}^{\infty} b_{k} {(x - x_{0})}^{k}) = \sum_{k = 0}^{\infty} c_{k} {(x - x_{0})}^{k},$

where $c_{k} = a_{0} b_{k} + a_{1} b_{k - 1} + \dots + a_{k} b_{0}$ . 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.

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 $x sub 0$ by writing the polynomial as a polynomial in $open parenthesis x minus x sub 0 close parenthesis$ . For example, let us write $2 x^{2} - 3 x + 4$ as a power series around $x sub 0 equals 1$ :

$2 x^{2} - 3 x + 4 = 3 + (x - 1) + 2 {(x - 1)}^{2} .$

In other words $a sub 0 equals 3$ , $a sub 1 equals 1$ , $a sub 2 equals 2$ , and all other $a sub k equals 0$ . To do this, we know that $a sub k equals 0$ for all $k greater than or equal to 3$ 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 $a sub 0$ , $a sub 1$ , and $a sub 2$ . We could have also differentiated at $x equals 1$ and used the Taylor series formula (7.1.2).

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 $minus 1 less than x less than 1$ we have

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

This series is called the geometric series. The ratio test tells us that the radius of convergence is $1$ . The series diverges for $x less than or equal to minus 1$ and $x greater than or equal to 1$ , even though $\frac{1}{1 - x}$ is defined for all $x ⧸ equals 1$ .

Function graph showing y = 1/(1-x) and y = (1-x^(n+1))/(1-x). Adjustable parameter: Highest power kept (n) = 4. Viewing window: x from -4.87 to 4.87, y from -3.01 to 3.01. — 1/(1 − x) in blue, the partial sum Sₙ = 1 + x + ⋯ + xⁿ in red. Drag n: the red curve locks onto the blue one across (−1, 1) and nowhere else — exactly the radius of convergence 1 this section derives. Past x = 1 the partial sum climbs off the top of the frame while 1/(1 − x) runs calmly along the lower right, still perfectly well defined. Left of x = −1 the partial sum flies apart, and watch its direction: it flips from up to down at every single step of n, because the terms alternate. At x = −1 itself it neither converges nor blows up — it just hops 1, 0, 1, 0, never settling on the value 1/2 the function has there.

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 $x sub 0$ . Note that as for polynomials, we could equivalently use the Taylor series expansion (7.1.2).

Example 6

Expand $\frac{x}{1 + 2 x + x^{2}}$ as a power series around the origin ( $x sub 0 equals 0$ ) and find the radius of convergence. First, write $1 + 2 x + x^{2} = {(1 + x)}^{2} = {(1 - (- x))}^{2}$ . Now we compute

$\begin{matrix} \frac{x}{1 + 2 x + x^{2}} = x {(\frac{1}{1 - (- x)})}^{2} \\ = x {(\sum_{k = 0}^{\infty} {(- 1)}^{k} x^{k})}^{2} \\ = x (\sum_{k = 0}^{\infty} c_{k} x^{k}) \\ = \sum_{k = 0}^{\infty} c_{k} x^{k + 1}, \end{matrix}$

where using the formula for the product of series we obtain, $c sub 0 equals 1$ , $c_{1} = - 1 - 1 = - 2$ , $c_{2} = 1 + 1 + 1 = 3$ , etc $\dots$ .

Therefore

$\frac{x}{1 + 2 x + x^{2}} = \sum_{k = 1}^{\infty} {(- 1)}^{k + 1} k x^{k} = x - 2 x^{2} + 3 x^{3} - 4 x^{4} + \dots$

The radius of convergence is at least 1. We use the ratio test

${lim}_{k \to \infty} | \frac{a_{k + 1}}{a_{k}} | = {lim}_{k \to \infty} | \frac{{(- 1)}^{k + 2} (k + 1)}{{(- 1)}^{k + 1} k} | = {lim}_{k \to \infty} \frac{k + 1}{k} = 1 .$

So the radius of convergence is actually equal to 1.

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

$\frac{x^{3} + x}{x^{2} - 1} = x + \frac{1}{1 + x} - \frac{1}{1 - x} = x + \sum_{k = 0}^{\infty} {(- 1)}^{k} x^{k} - \sum_{k = 0}^{\infty} x^{k} = - x + \sum_{\begin{matrix} k = 3 \\ k odd \end{matrix}}^{\infty} (- 2) x^{k} .$

Footnotes

[1] Named after the English mathematician Sir Brook Taylor (1685–1731).

Adapted from Differential Equations for Engineers by Jiří Lebl (Oklahoma State University), hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.