Chapter 8.7, Problem 57E

To determine

To Find: the first three nonzero terms in the Maclaurin series

Answer to Problem 57E

First three nonzero terms

  $\frac{x}{\sin x}\approx 1+\frac{x^2}{6}+\frac{7x^4}{360}$

Explanation of Solution

Given information:

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for function

  $y=\frac{x}{\sin x}$

Concept Used:

Maclaurin series for $\sin x$

  $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\mathrm{..}$

Calculation:

Maclaurin series of

  $y=\frac{x}{\sin x}$

UsingtheMaclaurin series for $\sin x$

  $\begin{array}{l}\frac{x}{\sin x}=\frac{x}{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\mathrm{..}}\\ \frac{x}{\sin x}=\frac{x}{x-\frac{x^3}{6}+\frac{x^5}{120}\mathrm{..}}\end{array}$

Divide x by the series for $\sin x$

  $\begin{array}{l}x-\frac{x^3}{6}+\frac{x^5}{120}\mathrm{..}\begin{array}{c}\hfill 1+\frac{x^2}{6}+\frac{7x^4}{360}\\ \hfill \overline{)x}\end{array}\\ x-\frac{x^3}{6}+\frac{x^5}{120}\mathrm{..}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \frac{x^3}{6}-\frac{x^5}{120}\mathrm{..}\\ \frac{x^3}{6}-\frac{x^5}{36}\mathrm{..}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \frac{7x^5}{360}\mathrm{..}\\ \frac{7x^5}{360}\mathrm{..}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \end{array}$

Hence,

  $\frac{x}{\sin x}\approx 1+\frac{x^2}{6}+\frac{7x^4}{360}$