# 6.1 Power series and functions  (Page 4/8)

 Page 4 / 8

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

$\sum _{n=0}^{\infty }\frac{{x}^{n+3}}{{2}^{n+1}}$ with interval of convergence $\left(-2,2\right)$

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.

## Key concepts

• For a power series centered at $x=a,$ one of the following three properties hold:
1. The power series converges only at $x=a.$ In this case, we say that the radius of convergence is $R=0.$
2. The power series converges for all real numbers x . In this case, we say that the radius of convergence is $R=\infty .$
3. There is a real number R such that the series converges for $|x-a| and diverges for $|x-a|>R.$ In this case, the radius of convergence is R .
• If a power series converges on a finite interval, the series may or may not converge at the endpoints.
• The ratio test may often be used to determine the radius of convergence.
• The geometric series $\sum _{n=0}^{\infty }{x}^{n}=\frac{1}{1-x}$ for $|x|<1$ allows us to represent certain functions using geometric series.

## Key equations

• Power series centered at $x=0$
$\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\text{⋯}$
• Power series centered at $x=a$
$\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}={c}_{0}+{c}_{1}\left(x-a\right)+{c}_{2}{\left(x-a\right)}^{2}+\text{⋯}$

In the following exercises, state whether each statement is true, or give an example to show that it is false.

If $\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ converges, then ${a}_{n}{x}^{n}\to 0$ as $n\to \infty .$

True. If a series converges then its terms tend to zero.

$\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ converges at $x=0$ for any real numbers ${a}_{n}.$

Given any sequence ${a}_{n},$ there is always some $R>0,$ possibly very small, such that $\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ converges on $\left(\text{−}R,R\right).$

False. It would imply that ${a}_{n}{x}^{n}\to 0$ for $|x| If ${a}_{n}={n}^{n},$ then ${a}_{n}{x}^{n}={\left(nx\right)}^{n}$ does not tend to zero for any $x\ne 0.$

If $\sum _{n=1}^{\infty }{a}_{n}{x}^{n}$ has radius of convergence $R>0$ and if $|{b}_{n}|\le |{a}_{n}|$ for all n , then the radius of convergence of $\sum _{n=1}^{\infty }{b}_{n}{x}^{n}$ is greater than or equal to R .

Suppose that $\sum _{n=0}^{\infty }{a}_{n}{\left(x-3\right)}^{n}$ converges at $x=6.$ At which of the following points must the series also converge? Use the fact that if $\sum {a}_{n}{\left(x-c\right)}^{n}$ converges at x , then it converges at any point closer to c than x .

1. $x=1$
2. $x=2$
3. $x=3$
4. $x=0$
5. $x=5.99$
6. $x=0.000001$

It must converge on $\left(0,6\right]$ and hence at: a. $x=1;$ b. $x=2;$ c. $x=3;$ d. $x=0;$ e. $x=5.99;$ and f. $x=0.000001.$

Suppose that $\sum _{n=0}^{\infty }{a}_{n}{\left(x+1\right)}^{n}$ converges at $x=-2.$ At which of the following points must the series also converge? Use the fact that if $\sum {a}_{n}{\left(x-c\right)}^{n}$ converges at x , then it converges at any point closer to c than x .

1. $x=2$
2. $x=-1$
3. $x=-3$
4. $x=0$
5. $x=0.99$
6. $x=0.000001$

In the following exercises, suppose that $|\frac{{a}_{n+1}}{{a}_{n}}|\to 1$ as $n\to \infty .$ Find the radius of convergence for each series.

$\sum _{n=0}^{\infty }{a}_{n}{2}^{n}{x}^{n}$

$|\frac{{a}_{n+1}{2}^{n+1}{x}^{n+1}}{{a}_{n}{2}^{n}{x}^{n}}|=2|x||\frac{{a}_{n+1}}{{a}_{n}}|\to 2|x|$ so $R=\frac{1}{2}$

$\sum _{n=0}^{\infty }\frac{{a}_{n}{x}^{n}}{{2}^{n}}$

$\sum _{n=0}^{\infty }\frac{{a}_{n}{\pi }^{n}{x}^{n}}{{e}^{n}}$

$|\frac{{a}_{n+1}{\left(\frac{\pi }{e}\right)}^{n+1}{x}^{n+1}}{{a}_{n}{\left(\frac{\pi }{e}\right)}^{n}{x}^{n}}|=\frac{\pi |x|}{e}|\frac{{a}_{n+1}}{{a}_{n}}|\to \frac{\pi |x|}{e}$ so $R=\frac{e}{\pi }$

$\sum _{n=0}^{\infty }\frac{{a}_{n}{\left(-1\right)}^{n}{x}^{n}}{{10}^{n}}$

$\sum _{n=0}^{\infty }{a}_{n}{\left(-1\right)}^{n}{x}^{2n}$

$|\frac{{a}_{n+1}{\left(-1\right)}^{n+1}{x}^{2n+2}}{{a}_{n}{\left(-1\right)}^{n}{x}^{2n}}|=|{x}^{2}||\frac{{a}_{n+1}}{{a}_{n}}|\to |{x}^{2}|$ so $R=1$

$\sum _{n=0}^{\infty }{a}_{n}{\left(-4\right)}^{n}{x}^{2n}$

In the following exercises, find the radius of convergence R and interval of convergence for $\sum {a}_{n}{x}^{n}$ with the given coefficients ${a}_{n}.$

$\sum _{n=1}^{\infty }\frac{{\left(2x\right)}^{n}}{n}$

${a}_{n}=\frac{{2}^{n}}{n}$ so $\frac{{a}_{n+1}x}{{a}_{n}}\to 2x.$ so $R=\frac{1}{2}.$ When $x=\frac{1}{2}$ the series is harmonic and diverges. When $x=-\frac{1}{2}$ the series is alternating harmonic and converges. The interval of convergence is $I=\left[-\frac{1}{2},\frac{1}{2}\right).$

$\sum _{n=1}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{n}}{\sqrt{n}}$

$\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{2}^{n}}$

${a}_{n}=\frac{n}{{2}^{n}}$ so $\frac{{a}_{n+1}x}{{a}_{n}}\to \frac{x}{2}$ so $R=2.$ When $x=\text{±}2$ the series diverges by the divergence test. The interval of convergence is $I=\left(-2,2\right).$

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