Chapter 8, Problem 38RQ

To determine

To Find: The McLaurin series and the radius of convergence.

Answer to Problem 38RQ

The series is ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n{x}^{n+2}}$ and the series convergence at $\left|x^2\right|<1\Rightarrow \left|x\right|<1$ and the radius of the convergence is $R=1$ .

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)={\tan }^{-1}\left(x^2\right)$

Calculation:

Consider the given function is $f\left(x\right)={\tan }^{-1}\left(x^2\right)$ .

Consider the equation as ${\tan }^{-1}\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{x^{2n+1}}{2n+1}}$ then the series is,

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

Then, the series is,

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

Consider the above series converges at $\left|x^2\right|<1\Rightarrow \left|x\right|<1$ .

The radius of convergence of the above series is $R=1$ .