# 6.4 Working with taylor series  (Page 2/11)

 Page 2 / 11
$\sum _{n=0}^{\infty }\left(\begin{array}{c}r\hfill \\ n\hfill \end{array}\right){x}^{n}=1+rx+\frac{r\left(r-1\right)}{2\text{!}}{x}^{2}+\text{⋯}+\frac{r\left(r-1\right)\text{⋯}\left(r-n+1\right)}{n\text{!}}{x}^{n}+\text{⋯}.$

We now need to determine the interval of convergence for the binomial series [link] . We apply the ratio test. Consequently, we consider

$\begin{array}{cc}\hfill \frac{|{a}_{n+1}|}{|{a}_{n}|}& =\frac{{|r\left(r-1\right)\left(r-2\right)\text{⋯}\left(r-n\right)|x||}^{n+1}}{\left(n+1\right)\text{!}}·\frac{n}{|r\left(r-1\right)\left(r-2\right)\text{⋯}\left(r-n+1\right)|{|x|}^{n}}\hfill \\ & =\frac{|r-n||x|}{|n+1|}.\hfill \end{array}$

Since

$\underset{n\to \infty }{\text{lim}}\frac{|{a}_{n+1}|}{|{a}_{n}|}=|x|<1$

if and only if $|x|<1,$ we conclude that the interval of convergence for the binomial series is $\left(-1,1\right).$ The behavior at the endpoints depends on $r.$ It can be shown that for $r\ge 0$ the series converges at both endpoints; for $-1 the series converges at $x=1$ and diverges at $x=-1;$ and for $r<-1,$ the series diverges at both endpoints. The binomial series does converge to ${\left(1+x\right)}^{r}$ in $\left(-1,1\right)$ for all real numbers $r,$ but proving this fact by showing that the remainder ${R}_{n}\left(x\right)\to 0$ is difficult.

## Definition

For any real number $r,$ the Maclaurin series for $f\left(x\right)={\left(1+x\right)}^{r}$ is the binomial series. It converges to $f$ for $|x|<1,$ and we write

$\begin{array}{cc}\hfill {\left(1+x\right)}^{r}& =\sum _{n=0}^{\infty }\left(\begin{array}{c}r\hfill \\ n\hfill \end{array}\right){x}^{n}\hfill \\ & =1+rx+\frac{r\left(r-1\right)}{2\text{!}}{x}^{2}+\text{⋯}+\frac{r\left(r-1\right)\text{⋯}\left(r-n+1\right)}{n\text{!}}{x}^{n}+\text{⋯}\hfill \end{array}$

for $|x|<1.$

We can use this definition to find the binomial series for $f\left(x\right)=\sqrt{1+x}$ and use the series to approximate $\sqrt{1.5}.$

## Finding binomial series

1. Find the binomial series for $f\left(x\right)=\sqrt{1+x}.$
2. Use the third-order Maclaurin polynomial ${p}_{3}\left(x\right)$ to estimate $\sqrt{1.5}.$ Use Taylor’s theorem to bound the error. Use a graphing utility to compare the graphs of $f$ and ${p}_{3}.$
1. Here $r=\frac{1}{2}.$ Using the definition for the binomial series, we obtain
$\begin{array}{cc}\hfill \sqrt{1+x}& =1+\frac{1}{2}x+\frac{\left(1\text{/}2\right)\left(\text{−}1\text{/}2\right)}{2\text{!}}{x}^{2}+\frac{\left(1\text{/}2\right)\left(\text{−}1\text{/}2\right)\left(\text{−}3\text{/}2\right)}{3\text{!}}{x}^{3}+\text{⋯}\hfill \\ & =1+\frac{1}{2}x-\frac{1}{2\text{!}}\phantom{\rule{0.2em}{0ex}}\frac{1}{{2}^{2}}{x}^{2}+\frac{1}{3\text{!}}\phantom{\rule{0.2em}{0ex}}\frac{1·3}{{2}^{3}}{x}^{3}-\text{⋯}+\frac{{\left(-1\right)}^{n+1}}{n\text{!}}\phantom{\rule{0.2em}{0ex}}\frac{1·3·5\text{⋯}\left(2n-3\right)}{{2}^{n}}{x}^{n}+\text{⋯}\hfill \\ & =1+\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n\text{!}}\phantom{\rule{0.2em}{0ex}}\frac{1·3·5\text{⋯}\left(2n-3\right)}{{2}^{n}}{x}^{n}.\hfill \end{array}$
2. From the result in part a. the third-order Maclaurin polynomial is
${p}_{3}\left(x\right)=1+\frac{1}{2}x-\frac{1}{8}{x}^{2}+\frac{1}{16}{x}^{3}.$

Therefore,
$\begin{array}{cc}\hfill \sqrt{1.5}& =\sqrt{1+0.5}\hfill \\ & \approx 1+\frac{1}{2}\left(0.5\right)-\frac{1}{8}{\left(0.5\right)}^{2}+\frac{1}{16}{\left(0.5\right)}^{3}\hfill \\ & \approx 1.2266.\hfill \end{array}$

From Taylor’s theorem, the error satisfies
${R}_{3}\left(0.5\right)=\frac{{f}^{\left(4\right)}\left(c\right)}{4\text{!}}{\left(0.5\right)}^{4}$

for some $c$ between $0$ and $0.5.$ Since ${f}^{\left(4\right)}\left(x\right)=-\frac{15}{{2}^{4}{\left(1+x\right)}^{7\text{/}2}},$ and the maximum value of $|{f}^{\left(4\right)}\left(x\right)|$ on the interval $\left(0,0.5\right)$ occurs at $x=0,$ we have
$|{R}_{3}\left(0.5\right)|\le \frac{15}{4\text{!}{2}^{4}}{\left(0.5\right)}^{4}\approx 0.00244.$

The function and the Maclaurin polynomial ${p}_{3}$ are graphed in [link] .

Find the binomial series for $f\left(x\right)=\frac{1}{{\left(1+x\right)}^{2}}.$

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

## Common functions expressed as taylor series

At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form $f\left(x\right)={\left(1+x\right)}^{r}.$ In [link] , we summarize the results of these series. We remark that the convergence of the Maclaurin series for $f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(1+x\right)$ at the endpoint $x=1$ and the Maclaurin series for $f\left(x\right)={\text{tan}}^{-1}x$ at the endpoints $x=1$ and $x=-1$ relies on a more advanced theorem than we present here. (Refer to Abel’s theorem for a discussion of this more technical point.)

Maclaurin series for common functions
Function Maclaurin Series Interval of Convergence
$f\left(x\right)=\frac{1}{1-x}$ $\sum _{n=0}^{\infty }{x}^{n}$ $-1
$f\left(x\right)={e}^{x}$ $\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}$ $\text{−}\infty
$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$ $\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)\text{!}}$ $\text{−}\infty
$f\left(x\right)=\text{cos}\phantom{\rule{0.1em}{0ex}}x$ $\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)\text{!}}$ $\text{−}\infty
$f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(1+x\right)$ $\sum _{n=0}^{\infty }{\left(-1\right)}^{n+1}\frac{{x}^{n}}{n}$ $-1
$f\left(x\right)={\text{tan}}^{-1}x$ $\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{2n+1}$ $-1
$f\left(x\right)={\left(1+x\right)}^{r}$ $\sum _{n=0}^{\infty }\left(\begin{array}{c}r\hfill \\ n\hfill \end{array}\right){x}^{n}$ $-1

Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series in [link] , to create Maclaurin series for other functions.

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