Q. 62 - Calculus

Question

Use appropriate Maclaurin series to find the first four nonzero

terms in the Maclaurin series for the product functions in

Exercises 61–66. Also, give the interval of convergence for the

series.

$e^x\cos x$

Step-by-Step Solution

Answer

The required answer is $1+x-\frac{1}{3}{x}^3-\frac{1}{6}{x}^4$

Step 1. Given Information

Consider the function as follows:

$f\left(x\right)=e^x\cos x$

The objective is to find the first four nonzero terms of the Maclaurin series for the product of the functions mentioned in the above function and also the interval of convergence.

 

Step 2: Find the interval of convergence for the series.

The Maclaurin series for the function $e^x$ is,

$e^x=\sum _{k=0}^n\frac{x^k}{k!}$

Let us expand the above series in the following way:

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$

The Maclaurin series for the function $\cos x$ is,

$\cos x=\sum _{i=0}^{\infty }\frac{(-1{)}^k}{\left(2k\right)!}{x}^{2k}$

Let us expand the above series in the following way:

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

Step 3 : Calculation

Multiply the preceding two series together term by term to get the first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=e^x\cos x$

There are constant terms in the series of $e^x$ and $\cos x$

Those constant terms are respectively $1$ and $1$, So, the new series has $1$ as its constant term.

Therefore, the coefficient of $x$ term is,

$1.1=1$

The coefficient of $x^2$ term is,

$\begin{array}{rcl}1·-\frac{1}{2}+\frac{1}{2}·1& =& -\frac{1}{2}+\frac{1}{2}\\ & =& 0\end{array}$

The coefficient for $x^3$ term is,

$\begin{array}{rcl}1·\left(-\frac{1}{2}\right)+\frac{1}{6}·1& =& -\frac{1}{2}+\frac{1}{6}\\ & =& \frac{-3+1}{6}\\ & =& -\frac{1}{3}\end{array}$

Step 4: Simplify

Also, the coefficient of $x^4$ term is,

$\begin{array}{rcl}1·\frac{1}{24}+\frac{1}{24}·1-\frac{1}{4}& =& \frac{1}{24}+\frac{1}{24}-\frac{1}{4}\\ & =& \frac{1}{12}-\frac{1}{4}\\ & =& -\frac{1}{6}\end{array}$

Therefore, the first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=e^x\cos x$ are as follows:

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

The interval of convergence for the Maclaurin series of the given function is $\mathrm{ℝ}$