# 6.1 Power series and functions  (Page 3/8)

 Page 3 / 8

Find the interval and radius of convergence for the series $\sum _{n=1}^{\infty }\frac{{x}^{n}}{\sqrt{n}}.$

The interval of convergence is $\left[-1,1\right).$ The radius of convergence is $R=1.$

## Representing functions as power series

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

$1+x+{x}^{2}+{x}^{3}+\text{⋯}=\sum _{n=0}^{\infty }{x}^{n}.$

Recall that the geometric series

$a+ar+a{r}^{2}+a{r}^{3}+\text{⋯}$

converges if and only if $|r|<1.$ In that case, it converges to $\frac{a}{1-r}.$ Therefore, if $|x|<1,$ the series in [link] converges to $\frac{1}{1-x}$ and we write

$1+x+{x}^{2}+{x}^{3}+\text{⋯}=\frac{1}{1-x}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}|x|<1.$

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

$1+x+{x}^{2}+{x}^{3}+\text{⋯}\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}|x|<1.$

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

## Graphing a function and partial sums of its power series

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

From the graph in [link] you see that as N increases, ${S}_{N}$ becomes a better approximation for $f\left(x\right)=\frac{1}{1-x}$ for x in the interval $\left(-1,1\right).$

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

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.

## Representing a function with a power series

Use a power series to represent each of the following functions $f.$ Find the interval of convergence.

1. $f\left(x\right)=\frac{1}{1+{x}^{3}}$
2. $f\left(x\right)=\frac{{x}^{2}}{4-{x}^{2}}$
1. You should recognize this function f as the sum of a geometric series, because
$\frac{1}{1+{x}^{3}}=\frac{1}{1-\left(\text{−}{x}^{3}\right)}.$

Using the fact that, for $|r|<1,\frac{a}{1-r}$ is the sum of the geometric series
$\sum _{n=0}^{\infty }a{r}^{n}=a+ar+a{r}^{2}+\text{⋯},$

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

Since this series converges if and only if $|\text{−}{x}^{3}|<1,$ the interval of convergence is $\left(-1,1\right),$ and we have
$\frac{1}{1+{x}^{3}}=1-{x}^{3}+{x}^{6}-{x}^{9}+\text{⋯}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}|x|<1.$
2. 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
$\begin{array}{cc}\hfill \frac{{x}^{2}}{4-{x}^{2}}& =\frac{{x}^{2}}{4\left(\frac{1-{x}^{2}}{4}\right)}\hfill \\ & =\frac{{x}^{2}}{4\left(1-{\left(\frac{x}{2}\right)}^{2}\right)}.\hfill \end{array}$

Therefore, we have
$\begin{array}{cc}\hfill \frac{{x}^{2}}{4-{x}^{2}}& =\frac{{x}^{2}}{4\left(1-{\left(\frac{x}{2}\right)}^{2}\right)}\hfill \\ & =\frac{\frac{{x}^{2}}{4}}{1-{\left(\frac{x}{2}\right)}^{2}}\hfill \\ & =\sum _{n=0}^{\infty }\frac{{x}^{2}}{4}{\left(\frac{x}{2}\right)}^{2n}.\hfill \end{array}$

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

for $|x|<2.$

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