Introduction to Power Series: Learn It 4

Giselle Miller

Introduction to Power Series: Learn It 4

Representing Functions as Power Series

Why should we care about representing functions as power series? The answer lies in the incredible usefulness of polynomials.

Polynomials are the simplest functions to work with — they only involve basic arithmetic operations like addition, subtraction, multiplication, and division. When we represent a complicated function as an “infinite polynomial” (a power series), we gain several powerful advantages:

Easy differentiation and integration: We can differentiate or integrate term by term
Function approximation: We can use partial sums to approximate function values
Simplified analysis: Complex functions become more manageable

The key question becomes: when can we actually represent a function using a power series?

Let’s revisit the geometric series we’ve seen before:

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

Geometric Series Formula

The geometric series [latex]a + ar + ar^2 + ar^3 + \cdots[/latex] converges if and only if [latex]|r| < 1[/latex]. When it converges, the sum equals [latex]\frac{a}{1-r}[/latex].

For our series with [latex]a = 1[/latex] and [latex]r = x[/latex], we get convergence when [latex]|x| < 1[/latex], and the sum is [latex]\frac{1}{1-x}[/latex]. Therefore:

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

We’ve successfully represented the function [latex]f(x) = \frac{1}{1-x}[/latex] as a power series. This representation is valid whenever [latex]|x| < 1[/latex].

We can see how well this power series represents the original function by comparing the graph of [latex]f(x) = \frac{1}{1-x}[/latex] with the graphs of several partial sums of this infinite 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.

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.

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]

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].