91Ó°ÊÓ

consider the function as follows

$f\left(x\right)=\frac{{\tan }^{-1}x}{1-x^3}$

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

${\tan }^{-1}x=\underset{k=0}{\overset{∞}{∑}}\frac{(-1{)}^k}{2k+1}{x}^{2k+1}$

Expand the above series in the following way:

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

The Maclaurin series for the function $\frac{1}{1-x}$is

$\frac{1}{1-x}=\underset{k=0}{\overset{∞}{∑}}{x}^k$

Expand the above series in the following way:

$\frac{1}{1-x}=1+x+x^2+x^3+⋯$

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

$\frac{1}{1-x^3}=\underset{k=0}{\overset{∞}{∑}}{\left(x^3\right)}^k$

Expand the above series in the following way:

$$\begin{gathered} \frac{1}{1-x^3}=1+\left(x^3\right)+{\left(x^3\right)}^2+{\left(x^3\right)}^3+⋯ \\ =1+x^3+x^6+x^4+⋯ \end{gathered}$$

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

There will be no constant term, since the series for ${\tan }^{-1}x$does not contains any constant terms, so after multiplying the series for ${\tan }^{-1}x$and $\frac{1}{1-x^3}$, we get the series having the smallest degree of $x$is $1$.

Therefore, the coefficient of $x$term is,

$1·1=1$

The coefficient of $x^3$term is,

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

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

$\left(1\right)·\left(1\right)=1$

The coefficient of $x^5$term is,

$\left(\frac{1}{5}\right)\left(1\right)=\frac{1}{5}$

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

$x-\frac{1}{3}{x}^3+x^4+\frac{1}{5}{x}^5$

The interval of convergence for the Maclaurin series of the given function is, (-1,1).