Chapter 8.7, Problem 56E

To determine

To Find: First three nonzero terms in the Maclaurin series

Answer to Problem 56E

First three nonzero terms

  $\sec x\approx 1+\frac{x^2}{2}+\frac{5x^4}{24}$

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

  $y=\sec x$

Concept Used:

Maclaurin series for $\cos x$

  $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}\mathrm{..}$

Calculation:

Maclaurin series of

  $y=\sec x$

UsingtheMaclaurin series for $\cos x$

  $\sec x=\frac{1}{\cos x}=\frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24}}$

Divide 1 by the series for $\cos x$

  $\begin{array}{l}1-\frac{x^2}{2}+\frac{x^4}{24}\mathrm{..}\begin{array}{c}\hfill 1+\frac{x^2}{2}+\frac{5x^4}{24}\\ \hfill \overline{)1}\end{array}\\ 1-\frac{x^2}{2}+\frac{x^4}{24}\mathrm{..}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \frac{x^2}{2}-\frac{x^4}{24}\mathrm{..}\\ \frac{x^2}{2}-\frac{x^4}{4}+\frac{x^6}{48}\mathrm{..}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \frac{5x^4}{24}-\frac{x^6}{48}\mathrm{..}\\ \frac{5x^4}{24}-\frac{5x^6}{48}\mathrm{..}\end{array}$

Hence,

  $\sec x\approx 1+\frac{x^2}{2}+\frac{5x^4}{24}$