Question

In Problems 21 and 22 the given function is analytic at $\mathit{a}\mathbf{=}\mathbf{0}$. Use appropriate series in (2) and long division to find the first four nonzero terms of the Maclaurin series of the given function.

$\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathit{x}$

Short answer

$\sec x=1+\frac{1}{2}{x}^2+\frac{5}{24}{x}^4+\frac{61}{720}{x}^6$

Step-by-step solution

Definition

Maclaurin seriesare a type of series expansion in which all terms are nonnegative integer powers of the variable.

Maclaurin series

Consider the function$secx$ analytic at $a=0$.

The Maclaurin series for$\cos x$ is,

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

The relation between$secx$ and$\cos x$ is,

$$\begin{gathered} \sec x=\frac{1}{\cos x} \\ \Rightarrow \sec x=\frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}} \end{gathered}$$

Find Maclaurin series

Consider the Maclaurin series for $\cos x$and then divide it by 1.

role="math" localid="1663922140515" $$\begin{gathered} 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}1+\frac{x^2}{2}+\frac{5x^4}{24}+\frac{61x^6}{720}11-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}----------0+ \frac{x^2}{2}-\frac{x^4}{24}+\frac{x^6}{720} \\ \frac{x^2}{2}-\frac{x^4}{4}+\frac{x^6}{48}-\frac{x^8}{1440}--------------0+\frac{5x^4}{24}-\frac{7x^6}{360}+\frac{x^8}{1440}\frac{5x^4}{24}-\frac{5x^6}{48}+\frac{5x^8}{576}-\frac{5x^{10}}{17280}------------------- 0+\frac{61x^6}{720}-\frac{23x^8}{2880}+\frac{5x^{10}}{17280}\frac{61x^6}{720}-\frac{61x^8}{1440}+......... \end{gathered}$$

Therefore, the Maclaurin series is,

$\sec x=1+\frac{1}{2}{x}^2+\frac{5}{24}{x}^4+\frac{61}{720}{x}^6$