Q. 64 - 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.

$\left(\sin 2x\right)\left({\tan }^{-1}{x}^3\right)$

Step-by-Step Solution

Answer

The required Maclaurin series is $2x^4-\frac{4}{3}{x}^6-\frac{2}{3}{x}^{10}+\frac{4}{9}{x}^{12}$

The interval of convergence for the series is $[-1,1]$.

Step 1. Given Information

Consider the function as follows $f\left(x\right)=\left(\sin 2x\right)\left({\tan }^{-1}{x}^3\right)$

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

Step 2. Calculation

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

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

Expand the above series in the following way:

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

Therefore, the Maclaurin series for the function $\sin 2x$ is,

$\sin 2x=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}(2x{)}^{2k+1}$

That is,

$\sin 2x=\left(2x\right)-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^3}{5!}-\frac{(2x{)}^7}{7!}+\cdots$

Step 3 . Calculation

The Maclaurin series for the function ${\tan }^{-1}x$ is,

${\tan }^{-1}x=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{2k+1}{x}^{2k+1}$

Expand the above series in the following way:

${\tan }^{-1}x=x-\frac{1}{3}{x}^3+\frac{1}{5}{x}^5-\frac{1}{7}{x}^7\cdots$

So, the Maclaurin series for the function ${\tan }^{-1}{x}^3$,

${\tan }^{-1}{x}^3=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{2k+1}{\left(x^3\right)}^{2k+1}$

Expand the above series in the following way,

$\begin{array}{rcl}{\tan }^{-1}{x}^3& =& x^3-\frac{1}{3}{\left(x^3\right)}^3+\frac{1}{5}{\left(x^3\right)}^5-\frac{1}{7}{\left(x^3\right)}^7+\cdots \\ & =& x^3-\frac{1}{3}{x}^9+\frac{1}{5}{x}^{15}-\frac{1}{7}{x}^{21}+\cdots \end{array}$

Step 4 . Find the value of coefficient

Multiply the preceding two series together term by term to get first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=\left(\sin 2x\right)\left({\tan }^{-1}{x}^3\right)$

There will be no constant term, since the series for ${\tan }^{-1}{x}^3$ does not contains any constant terms, so after multiplying the series for $\left(\sin 2x\right)$ and $\left({\tan }^{-1}{x}^3\right)$, the series having the smallest degree of $x$ is $4$ .

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

$2·1=2$

The coefficient of $x^6$ term is,

$\begin{array}{rcl}-\frac{8}{3!}·1& =& -\frac{8}{6}\\ & =& -\frac{4}{3}\end{array}$

The coefficient of $x^{10}$ term is,

$2·\left(-\frac{1}{3}\right)=-\frac{2}{3}$

Step 5 . Find the interval of convergence for the series

Also, the coefficient of $x^{12}$ term is,

$\begin{array}{rcl}\left(-\frac{8}{3!}\right)\left(-\frac{1}{3}\right)& =& \frac{8}{18}\\ & =& \frac{4}{9}\end{array}$

Therefore, the first four nonzero terms in the Maclaurin series for the function $f\left(x\right)=\left(\sin 2x\right)\left({\tan }^{-1}{x}^3\right)$are as follows:

$2x^4-\frac{4}{3}{x}^6-\frac{2}{3}{x}^{10}+\frac{4}{9}{x}^{12}$

The interval of convergence for the Maclaurin series of the given function is, $[-1,1]$