Chapter 8.6, Problem 14E

To determine

To find: The power series representation of $f(x)=x^2{\tan }^{-1}(x^3)$ .

Answer to Problem 14E

Interval of convergence $(-1,1)$ .

Explanation of Solution

Given information: We know that $\frac{1}{1-x}$is a geometric series and its expression is $\frac{1}{1-x}=1+x+x^2+x^3+\mathrm{...}={\displaystyle \sum _{n=0}^{\infty }{x}^n}\left|x\right|<1.\mathrm{.....}\text{(1)}$

Solution: Now let’s work from $f(x)=x^2{\tan }^{-1}(x^3)$ to get $\frac{1}{1-x}$ .

Now ,

  $\begin{array}{l}\frac{1}{1+x^2}={\displaystyle \sum _{n=0}^{\infty }(-x^2}{)}^n\\ \text{ =}{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}·x^{2n}\end{array}$

Integrating to get ${\tan }^{-1}(x)$

  ${\tan }^{-1}(x)={\displaystyle \int \frac{dx}{1+x^2}}$

  $={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}·{\displaystyle \int x^{2n}dx}$

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

Now, put $x=0$ for finding integral constant c .

  $\therefore c=0$

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

Now replacing x by $x^3$

  ${\tan }^{-1}(x^3)={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{6n+3}}{2n+1}$

Multiplying by $x^2$

We get, $x^2{\tan }^{-1}(x^3)$

  $=x^2{\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{6n+3}}{2n+1}$

  ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{6n+3}}{2n+1}$

Hence , the series expression of

  $x^2{\tan }^{-1}(x^3)={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{x^{6n+3}}{2n+1}$

To find integral of convergence ,

By ratio test $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$ where $a_n=\frac{{(-1)}^n}{2n+1}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\left|\frac{{(-1)}^{n+1}}{2n+3}·\frac{2n+1}{{(-1)}^n}\right|\\ =1\end{array}$

Interval of convergence is $\left|x\right|<1$

i.e, $-1<x<1$

Hence, the interval of convergence is $(-1,1)$ .