Representing Functions as Power Series

Learning Outcomes

Use a power series to represent a function

Being able to represent a function by an “infinite polynomial” is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?

Consider again the geometric series

[latex]1+x+{x}^{2}+{x}^{3}+\cdots =\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex].

Recall that the geometric series

[latex]a+ar+a{r}^{2}+a{r}^{3}+\cdots[/latex]

converges if and only if [latex]|r|<1[/latex]. In that case, it converges to [latex]\frac{a}{1-r}[/latex]. Therefore, if [latex]|x|<1[/latex], the series in the example: Representing a Function with a Power Series converges to [latex]\frac{1}{1-x}[/latex] and we write

[latex]1+x+{x}^{2}+{x}^{3}+\cdots =\frac{1}{1-x}\text{for}|x|<1[/latex].

As a result, we are able to represent the function [latex]f\left(x\right)=\frac{1}{1-x}[/latex] by the power series

[latex]1+x+{x}^{2}+{x}^{3}+\cdots \text{when}|x|<1[/latex].

We now show graphically how this series provides a representation for the function [latex]f\left(x\right)=\frac{1}{1-x}[/latex] by comparing the graph of f with the graphs of several of the partial sums of this infinite series.

Example: Graphing a Function and Partial Sums of its Power Series

Sketch a graph of [latex]f\left(x\right)=\frac{1}{1-x}[/latex] and the graphs of the corresponding partial sums [latex]{S}_{N}\left(x\right)=\displaystyle\sum _{n=0}^{N}{x}^{n}[/latex] for [latex]N=2,4,6[/latex] on the interval [latex]\left(-1,1\right)[/latex]. Comment on the approximation [latex]{S}_{N}[/latex] as N increases.

Show Solution

try it

Sketch a graph of [latex]f\left(x\right)=\frac{1}{1-{x}^{2}}[/latex] and the corresponding partial sums [latex]{S}_{N}\left(x\right)=\displaystyle\sum _{n=0}^{N}{x}^{2n}[/latex] for [latex]N=2,4,6[/latex] on the interval [latex]\left(-1,1\right)[/latex].

Hint

Show Solution

Next we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.

Example: Representing a Function with a Power Series

Use a power series to represent each of the following functions [latex]f[/latex]. Find the interval of convergence.

[latex]f\left(x\right)=\frac{1}{1+{x}^{3}}[/latex]
[latex]f\left(x\right)=\frac{{x}^{2}}{4-{x}^{2}}[/latex]

Show Solution

You should recognize this function f as the sum of a geometric series, because[latex]\frac{1}{1+{x}^{3}}=\frac{1}{1-\left(\text{-}{x}^{3}\right)}[/latex]
Using the fact that, for [latex]|r|<1,\frac{a}{1-r}[/latex] is the sum of the geometric series
[latex]\displaystyle\sum _{n=0}^{\infty }a{r}^{n}=a+ar+a{r}^{2}+\cdots[/latex],

we see that, for [latex]|\text{-}{x}^{3}|<1[/latex],
[latex]\begin{array}{cc}\hfill \frac{1}{1+{x}^{3}}& =\frac{1}{1-\left(\text{-}{x}^{3}\right)}\hfill \\ & ={\displaystyle\sum _{n=0}^{\infty}}{\left(-{x}^{3}\right)}^{n}\hfill \\ & =1-{x}^{3}+{x}^{6}-{x}^{9}+\cdots .\hfill \end{array}[/latex]

Since this series converges if and only if [latex]|\text{-}{x}^{3}|<1[/latex], the interval of convergence is [latex]\left(-1,1\right)[/latex], and we have
[latex]\frac{1}{1+{x}^{3}}=1-{x}^{3}+{x}^{6}-{x}^{9}+\cdots \text{for}|x|<1[/latex].
This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate f to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain
[latex]\begin{array}{cc}\hfill \frac{{x}^{2}}{4-{x}^{2}}& =\frac{{x}^{2}}{4\left(\frac{1-{x}^{2}}{4}\right)}\hfill \\ & =\frac{{x}^{2}}{4\left(1-{\left(\frac{x}{2}\right)}^{2}\right)}.\hfill \end{array}[/latex]

Therefore, we have

[latex]\begin{array}{cc}\hfill \frac{{x}^{2}}{4-{x}^{2}}& =\frac{{x}^{2}}{4\left(1-{\left(\frac{x}{2}\right)}^{2}\right)}\hfill \\ & =\frac{\frac{{x}^{2}}{4}}{1-{\left(\frac{x}{2}\right)}^{2}}\hfill \\ & ={\displaystyle\sum _{n=0}^{\infty}} \frac{{x}^{2}}{4}{\left(\frac{x}{2}\right)}^{2n}.\hfill \end{array}[/latex]

The series converges as long as [latex]|{\left(\frac{x}{2}\right)}^{2}|<1[/latex] (note that when [latex]|{\left(\frac{x}{2}\right)}^{2}|=1[/latex] the series does not converge). Solving this inequality, we conclude that the interval of convergence is [latex]\left(-2,2\right)[/latex] and
[latex]\begin{array}{cc}\hfill \frac{{x}^{2}}{4-{x}^{2}}& ={\displaystyle\sum _{n=0}^{\infty}} \frac{{x}^{2n+2}}{{4}^{n+1}}\hfill \\ & =\frac{{x}^{2}}{4}+\frac{{x}^{4}}{{4}^{2}}+\frac{{x}^{6}}{{4}^{3}}+\cdots \hfill \end{array}[/latex]

for [latex]|x|<2[/latex].

try it

Represent the function [latex]f\left(x\right)=\frac{{x}^{3}}{2-x}[/latex] using a power series and find the interval of convergence.

Hint

Show Solution

Watch the following video to see the worked solution to the above Try It.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.1.3” here (opens in new window).

In the remaining sections of this chapter, we will show ways of deriving power series representations for many other functions, and how we can make use of these representations to evaluate, differentiate, and integrate various functions.

Module 6: Power Series