7.1 Power Series

Many functions can be written in terms of a power series

$\sum _{k=0}^{\infty} a_k(x-x_0)^k \nonumber$

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

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

(7.1.1)

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

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

denote the so-called partial sum. If for some $x$ , the limit

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

exists, then we say that the series (7.1.1) converges at $x$ . Note that for $x=x_0$ , the series always converges to $a_0$ . When (7.1.1) converges at any other point $x \neq x_0$ , we say that (7.1.1) is a convergent power series. In this case we write

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

If the series does not converge for any point $x \neq x_0$ , we say that the series is divergent.

Example 1

The series

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

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 (7.1.1) converges absolutely at $x$ whenever the limit

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

exists. That is, the series $\sum_{k=0}^\infty \lvert a_k \rvert \, {\lvert x-x_0 \rvert}^k$ 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

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

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 $x=1$ .

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 $\bigl\lvert a_k {(x-x_0)}^k \bigr\rvert \leq \bigl\lvert a_k {(x_1-x_0)}^k \bigr\rvert$ for all $k$ . As the numbers $\bigl\lvert a_k {(x_1-x_0)}^k \bigr\rvert$ sum to some finite limit, summing smaller positive numbers $\bigl\lvert a_k {(x-x_0)}^k \bigr\rvert$ 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 $\PageIndex{1}$ : Convergence of a power series.

See Figure $\PageIndex{1}$ . In Example $\PageIndex{1}$ the radius of convergence is $\rho = \infty$ as the series converges everywhere. In Example $\PageIndex{2}$ 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 ratio test. Suppose that

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

is a series such that the limit

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

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

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

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

Define $A$ by

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

Then if $1 > L = A \lvert x - x_0 \rvert$ the series (7.1.1) converges absolutely. If $A = 0$ , then the series always converges. If $A > 0$ , then the series converges absolutely if $\lvert x - x_0 \rvert < \frac{1}{A}$ , and diverges if $\lvert x - x_0 \rvert > \frac{1}{A}$ . That is, the radius of convergence is $\frac{1}{A}$ .

A similar test is the root test. Suppose

$L=\lim_{k\to\infty} \sqrt[3]{|c_{k}|} \nonumber$

exists. Then $\sum_{k=0}^{\infty}c_{k}$ converges absolutely if $L<1$ and diverges if $L>1$ . We can use the same calculation as above to find $A$ . Let us summarize.

Note

Ratio and Root Tests for Power Series

Let

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

be a power series such that

$A = \lim_{n\to\infty} \left \lvert \frac{a_{k+1}}{a_k} \right \rvert\quad\text{or}\quad A=\lim_{k\to\infty}\sqrt[3]{|a_{k}|} \nonumber$

exists. If $A = 0$ , 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

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

First we compute,

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

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

$A = \lim_{k\to\infty} \lim_{k\to\infty} \sqrt[k]{\lvert a_k \rvert} = \lim_{k\to\infty} \sqrt[k]{\lvert 2^{-k} \rvert} = \lim_{k\to\infty} 2^{-1} = \frac{1}{2}. \nonumber$

The root or the ratio test does not always apply. That is the limit of $\bigl \lvert \frac{a_{k+1}}{a_k} \bigr \rvert$ or $\sqrt[k]{\lvert a_k \rvert}$ might not exist. There exist more sophisticated ways of finding the radius of convergence, but those would be beyond the scope of this chapter. 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 series1 near a point $x_0$ . That is, for $x$ near $x_0$ we have

$\label{eq:21} f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!} {(x-x_0)}^k,$

(7.1.2)

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

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 $\PageIndex{2}$ : The sine function and its Taylor approximations around $x_o=0$ of and degree.

For example, sine is an analytic function and its Taylor series around $x_0 = 0$ is given by

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

In Figure $\PageIndex{2}$ 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 $x_0$ you need to take more and more terms of the Taylor series.

A wave runs the width of the frame, and two other curves cling to it over a stretch in the middle before peeling away and shooting off the top and bottom — one of them hugging the wave noticeably further before it breaks off than the other. Dragging the expansion point slides that whole window of agreement along the wave, carrying the break-away points with it: the accurate stretch always sits centred on the chosen point, and the longer-reaching curve always outlasts the shorter one. 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.

Interactive figureHow far out does degree n reach?Drag the degree n slider from 1 to 15.

XYZ Graph · viewer build 5edf91b · plotted curve with n = 9 checked against “Sine's Taylor polynomial through degree 9, computed” (mojocas 0.1.0): agrees

The figure above reproduces the book's Figure 2 — degrees 5 and 9, about a centre you can move. This one pins the centre at 0 and puts the DEGREE on the slider instead. Degree 1 is the line x, useful only at the origin. Each step adds one term and pushes the agreement outward: by degree 5 the rust curve holds the first hump, by degree 9 (the highest truncation drawn above) it tracks the wave to about x = 4, and at degree 15 it stays with it nearly to the frame's edge. What never changes is the shape of the failure — every truncation is a polynomial, so it must eventually run off to infinity. The degree only decides how far out it agrees first.

x - \frac{1}{6} \cdot x^{3} + \frac{1}{120} \cdot x^{5} - \frac{1}{5040} \cdot x^{7} + \frac{1}{362880} \cdot x^{9}

Sine's Taylor polynomial through degree 9, computed ✓ Computed · mojocas 0.1.0 ✓ Agrees with the textThe section states sine's Taylor series about 0 in closed form; written out to degree 9 it is exactly this polynomial, and the computer algebra system derives it from sin(x) itself rather than from the formula. These are the same coefficients the figure above plots — and on every rebuild the plotted polynomial, with its degree slider fixed at 9, is checked against this computed series, which is in turn checked against the text's formula. So the curve you drag and the series the text asserts really are checked against one another.

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

$\frac{d}{dx} \left[\sum_{k=0}^\infty a_k {(x-x_0)}^k\right]=\sum_{k=1}^\infty k a_k {(x-x_0)}^{k-1}. \nonumber$

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.

Example 5

Let us show that the exponential $y=e^x$ solves $y'=y$ . First write

$y = e^x = \sum_{k=0}^\infty \frac{1}{k!} x^k. \nonumber$

Now differentiate

$y' = \sum_{k=1}^\infty k \frac{1}{k!} x^{k-1} =\sum_{k=1}^\infty \frac{1}{(k-1)!} x^{k-1}. \nonumber$

We reindex 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.

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

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

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,

$\left(\sum_{k=0}^\infty a_k {(x-x_0)}^k\right)+\left(\sum_{k=0}^\infty b_k {(x-x_0)}^k\right)=\sum_{k=0}^\infty (a_k+b_k) {(x-x_0)}^k. \nonumber$

We can multiply by constants,

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

We can also multiply series together,

$\left(\sum_{k=0}^\infty a_k {(x-x_0)}^k\right) \, \left(\sum_{k=0}^\infty b_k {(x-x_0)}^k\right)=\sum_{k=0}^\infty c_k {(x-x_0)}^k, \nonumber$

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.

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 $x_0 = 1$ :

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

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 (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 $-1 < x < 1$ we have

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

This series is called the 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 $x \not= 1$ .

Two curves lie exactly on top of one another through a band around the origin and separate sharply outside it. To the right of that band one carries calmly on below the axis while the other climbs off the top of the frame; to the left the second flies apart as well. Adding more terms widens the agreement only up to the same fixed edges and never past them, and each extra term flips the direction of the left-hand blow-up from upward to downward and back. 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 $x_0$ . Note that as for polynomials, we could equivalently use the Taylor series expansion (7.1.2).

Example 6

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

$\begin{align}\begin{aligned} \frac{x}{1+2x+x^2} &=x {\left( \frac{1}{1-(-x)} \right)}^2 \\ &=x { \left( \sum_{k=0}^{\infty} {(-1)}^k x^k \right)}^2 \\ &=x \left(\sum_{k=0}^{\infty} c_k x^k \right) \\ &= \sum_{k=0}^{\infty} c_k x^{k+1},\end{aligned}\end{align} \nonumber$

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

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

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

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

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

$\frac{x^3+x}{x^2-1}=x + \frac{1}{1+x} - \frac{1}{1-x}=x + \sum_{k=0}^\infty {(-1)}^k x^k - \sum_{k=0}^\infty x^k= - x + \sum_{\substack{k=3 \\ k \text{ odd}}}^\infty (-2) x^k. \nonumber$

Footnotes

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

Adapted from Differential Equations for Engineers by Jiří Lebl (https://www.jirka.org/diffyqs/), © Jiří Lebl, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.