Q. 66 - Calculus

Question

Use appropriate Maclaurin series to find the first four nonzero terms in the Maclaurin series for the product functions in $\frac{{\tan }^{-1}x}{1-x^3}$. Also, give the interval of convergence for the series.

Step-by-Step Solution

Answer

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, .

step 1: given information

consider the function as follows 

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

step 2: To find the Maclaurin series for the function

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:

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

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

$\frac{1}{1-x}=\sum _{k=0}^{\infty }{x}^k$

Expand the above series in the following way:

$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$

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

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

step 3: To get the first four nonzero terms and the interval of convergence of Maclaurin series in the given function

Expand the above series in the following way:

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

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).