Question

In Problems 19 and 20 the given function is analytic at. Use appropriate series in (2) and multiplication to find the first four nonzero terms of the Maclaurin series of the given function.

${\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$

Short answer

$1-x+\frac{x^3}{3}-\frac{x^4}{6}$

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 $e^{-x}\cos x$analytic at $a=0$.

The Maclaurin series for $\cos x$& $e^x$is,

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots$and $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$

Replace $x$by $-x$to obtain that,

$$\begin{gathered} e^{-x}=1+\frac{-x}{1!}+\frac{{(-x)}^2}{2!}+\frac{{(-x)}^3}{3!}+\dots \\ =1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots \end{gathered}$$

Find maclaurin series of e-xcosx

The Maclaurin series for both $e^{-x}$and $\cos x$and then multiply them together to get the Maclaurin series for $e^{-x}\cos x$.

$$\begin{gathered} e^{-x}\cos x=\left(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots \right)\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \right) \\ =1\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \right)-\frac{x}{1!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \right) \\ +\frac{x^2}{2!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \right)-\frac{x^3}{3!}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots \right)+\dots \\ =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots -x+\frac{x^3}{2!}-\frac{x^5}{4!}+\frac{x^7}{6!}-\dots +\frac{x^2}{2!}-\frac{x^4}{2!·2!}+\frac{x^6}{2!·4!}-\frac{x^8}{2!·6!}+\dots \\ -\frac{x^3}{3!}+\frac{x^5}{3!·2!}-\frac{x^7}{3!·4!}+\frac{x^9}{3!·6!}-\dots \end{gathered}$$

Simplifying we get;

$$\begin{gathered} e^{-x}\cos x=1-x+\left(\frac{3-1}{6}\right)x^3+\left(\frac{1-6+1}{24}\right)x^4+\dots \\ =1-x+\frac{x^3}{3}-\frac{x^4}{6}+\dots \end{gathered}$$