Calculus II 09.09 Using Power Series to Represent a Function

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9.9 Using Power Series to Represent a Function

• Find a geometric power series that represents a function.
• Construct a power series using series operations.

Geometric Power Series[1]

Figure 9.9.1

A Power Series can represent some function \(f(x)\), but only over a single interval. Consider the function $$ f(x)=\frac{1}{1-x}$$ This form for \(f\) closely resembles the sum for a geometric series $$ \sum_{n=0}^{\infty} ar^{n} = \frac{a}{1-r}, \: 0 < |r| < 1. $$ In other words, when \(a=1\) and \(r=x\), a power series representation for \(f\) centered at 0 looks like $$ \frac{a}{1-r} $$ $$= \sum_{n=0}^{\infty} ar^{n}$$ $$= \sum_{n=0}^{\infty} x^{n} $$ $$= 1+x+x^{2}+x^{3}+ \cdots,\:\:|x| <1.$$ This series represents \(f\) only on the interval (-1,1), whereas \(f\) is defined for all \(x \ne 1\), as shown in Figure 9.9.1. To represent \(f\) on another interval, a different series is needed. For instance, the power series centered at -1 could be written as $$ \frac{1}{1-x} = \frac{1}{2-(x+1)}=\frac{1/2}{1-[(x+1)/2]}= \frac{a}{1-r} $$ which implies that \(a=1/2\) and \(r=(x+1)/2\). Taking \(|x+1|<2\) yields $$ \frac{a}{1-x} $$ $$= \sum_{n=0}^{\infty} \left ( \frac{1}{2}\right ) \left ( \frac{x+1}{2} \right )^{n}$$ $$= \frac{1}{2} \left[ 1+ \frac{(x+1)}{2} + \frac{(x+1)^{2}}{4} + \frac{(x+1)^{3}}{8} + \cdots \right], \: |x+1|<2 $$ which converges on the interval (-3,1).

Example 9.9.1 Finding a Geometric Power Series Centered at Zero

Find a power series for

$$ f(x)=\frac{4}{x+2}$$

centered at 0.
Solution Writing \(f(x)\) as \(a/(1-r)\) produces

$$ \frac{4}{x+2} = \frac{2}{1-(-x/2)} = \frac{a}{1-r}$$

which implies that \(a=2\) and

$$r=-\frac{x}{2}. $$

The power series for \(f\) is

$$ \frac{4}{x+2} $$	$$= \sum_{n=0}^{\infty} ar^{n}$$
	$$= \sum_{n=0}^{\infty} 2 \left( - \frac{x}{2} \right)^{n} $$
	$$= 2 \left( 1 - \frac{x}{2} + \frac{x^{2}}{4} - \frac{x^{3}}{8} \right). $$

The power series converges when

$$\left | -\frac{x}{2} \right | < 1 $$

which produces a convergence interval at (-2,2).

Operations with Power Series

$$\text{Let }f(x)= \sum_{n=0}^{\infty} a_{n}x^{n} \text{ and } g(x)= \sum_{n=0}^{\infty} b_{n}x^{n}.$$

$$1.\:f(kx)= \sum_{n=0}^{\infty} a_{n}k^{n}x^{n} $$

$$2.\:f(x^{N})= \sum_{n=0}^{\infty} a_{n}x^{nN} $$

$$3.\:f(x)\pm g(x)= \sum_{n=0}^{\infty} (a_{n} \pm b_{n})x^{n} $$

The operations can change the convergence interval for the resulting series. For example, in the addition shown below, the convergence interval for the sum is the intersection between the two original series' convergence intervals. As shown in Figure 9.9.2 below.

Figure 9.9.2

Example 9.9.2 Adding Two Power Series

Find a power series for

$$ f(x)=\frac{3x-1}{x^{2}-1}$$

centered at 0.
Solution Using partial fractions \(f\) can be written as

$$ \frac{3x-1}{x^{2}-1} = \frac{2}{x+1} + \frac{1}{x-1}.$$

By adding the two geometric power series

$$ \frac{2}{x+1} = \frac{2}{1-(-x)} = \sum_{n=0}^{\infty} 2(-1)^{n}x^{n}, \: |x| < 1 $$

and

$$ \frac{1}{x-1} = \frac{-1}{1-x} = - \sum_{n=0}^{\infty} x^{n}, \: |x| < 1 $$

produces the power series below

$$ \frac{3x-1}{x^{2}-1} $$	$$= \sum_{n=0}^{\infty} [2(-1)^{n}-1]x^{n}$$
	$$= 1-3x+x^{2} - 3x^{3}+ x^{4}- \cdots $$

The convergence interval for the added power series is (-1,1).

Example 9.9.3 Adding Two Power Series

Find a power series for

$$ f(x)=\ln x$$

centered at 1.
Solution A power series centered at 1 produces

$$ \frac{1}{x} = \sum_{n=0}^{\infty} (-1)^{n}(x-1)^{n} \:\:\:\: \color{red}{\text{Convergence interval = (0,2) }}.$$

Integrating this series produces

$$ \ln x $$	$$= \int \frac{1}{x}dx+C$$
	$$= C+\sum_{n=0}^{\infty} (-1)^{n} \frac{(x-1)^{n-1}}{n+1}.$$

By letting \(x=1\), yields \(C=0\). Therefore,

$$ \ln x $$	$$= \int \frac{1}{x}dx+C$$
	$$= \frac{(x-1)}{1} - \frac{(x-1)^{2}}{2} + \frac{(x-1)^{3}}{3}-\frac{(x-1)^{4}}{4}+\cdots \:\:\:\: \color{red}{\text{Convergence interval = (0,2] }}$$

Note that the series converges at \(x=2\). This is consistent with the observation in Section 9.8 that power series integration may alter the convergence at the interval endpoints.

Example 9.9.4 Finding a Power Series by Integration

Find a power series for

$$ g(x)=\arctan x$$

centered at 0.
Solution Because

$$ D_{x}[\arctan x]=\frac{1}{(1+x^{2})}$$

the series below can be used.

$$ f(x) = \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^{n}x^{n} \:\:\:\: \color{red}{\text{Convergence interval = (-1,1) }}$$

Substituting \(x^{2}\) for \(x\) produces

$$ f(\color{red}{x^{2}}) = \frac{1}{1+\color{red}{x^{2}}} = \sum_{n=0}^{\infty} (-1)^{n}x^{2n}. $$

Integrating yields

$$ \arctan x $$	$$= \int \frac{1}{1+x^{2}}dx+C$$
	$$= C+\sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{2n+1}.$$
	$$= \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{2n+1} \:\:\:\: \color{red}{\text{Let }x=0\text{, then }C=0}$$
	$$= x - \frac{x^{3}}{3} + \frac{x^{5}}{5}- \frac{x^{7}}{7}+ \:\:\:\: \color{red}{\text{Convergence interval = (-1,1) }}$$

Example 9.9.5 Approximating \(\pi\) with Two Series

Use the trigonometric identity

$$ 4 \arctan \frac{1}{5} - \arctan \frac{1}{239} = \frac{\pi}{4} $$

to approximate the number \(\pi\).
Solution Using the first five terms from each series yields

$$ 4 \left ( 4 \arctan \frac{1}{5} - \arctan \frac{1}{239} \right ) \approx 3.1415926 $$

which agrees with the exact value for \(\pi\) with an error less than 0.0000001.

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Parent Article: Calculus II 09 Infinite Series