Chapter 10.2, Problem 7E

To determine

To find: First three nonzero term, general term and interval of convergence of the given function.

Answer to Problem 7E

The first three nonzero terms are, $T_0=x^2,T_1=-\frac{x^6}{3}\text{and}{T}_2=\frac{x^{10}}{5}$

The general term is, $T_n={\left(-1\right)}^n\frac{x^{2\left(2n+1\right)}}{2n+1}$

And the interval of convergence is $[-1,1]$

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)={\tan }^{-1}{x}^2$

Calculation:

As Maclaurin series of ${\tan }^{-1}x$ is

  ${\tan }^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\mathrm{........}+{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)}+\mathrm{......}$

So for Maclaurin series of $f\left(x\right)={\tan }^{-1}{x}^2$ replace x by $x^2$ .

So, the series is

  $\begin{array}{l}{\tan }^{-1}{x}^2=x^2-\frac{{\left(x^2\right)}^3}{3}+\frac{{\left(x^2\right)}^5}{5}-\mathrm{........}+{\left(-1\right)}^n\frac{{\left(x^2\right)}^{2n+1}}{\left(2n+1\right)}+\mathrm{......}\\ \Rightarrow {\tan }^{-1}{x}^2=x^2-\frac{x^6}{3}+\frac{x^{10}}{5}-\mathrm{........}+{\left(-1\right)}^n\frac{{\left(x\right)}^{-\left(2n+1\right)}}{\left(2n+1\right)}+\mathrm{......}\end{array}$

O, the first three nonzero terms are,

  $T_0=x^2,T_1=-\frac{x^6}{3}\text{and}{T}_2=\frac{x^{10}}{5}$

And the general term of the series is

  $T_n={\left(-1\right)}^n\frac{x^{2\left(2n+1\right)}}{2n+1}$

Now, for determining the interval of convergence use ration test. So,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{{\left(-1\right)}^n{x}^{2\left(2n+1\right)}}{2n+1}\times \frac{2n-1}{{\left(-1\right)}^n{x}^{2\left(2n-1\right)}}\right|\\ =\underset{n\to \infty }{\lim }\left|-\frac{x^4\left(2n-1\right)}{\left(2n+1\right)}\right|\\ =\left|-x^4\right|\underset{n\to \infty }{\lim }\frac{\left(2n-1\right)}{\left(2n+1\right)}\\ =\left|x^4\right|\times 1\\ =\left|x^4\right|\end{array}$

So, the series will converge if $\left|x^4\right|<1\text{i}\text{.e}\text{. }-1<x<1$

Now checking the convergence at the end points.

Put $x=-1$ in the original series.

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

It is alternating harmonic series so it converges.

Put $x=1$ in the original series.

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

It is also alternating harmonic series so it converges.

So, the series converges when $-1\le x\le 1$

Hence, the interval of convergence is $[-1,1]$