Chapter 10.2, Problem 10E

To determine

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

Answer to Problem 10E

The first three nonzero terms are, $T_0=x^2,T_1=-\frac{x^4}{2!}\text{and}{T}_2=\frac{x^6}{4!}$

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

And the series will converge for all real values of x .

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=x^2\cos x$

Calculation:

Maclaurin series of $\cos x$ is

  $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\mathrm{...}+{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}+\mathrm{...}$

So for Maclaurin series of $f\left(x\right)=x^2\cos x$ multiply by $x^2$

  $\begin{array}{l}f\left(x\right)=x^2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\mathrm{...}+{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}+\mathrm{...}\right)\\ \Rightarrow f\left(x\right)=x^2-\frac{x^4}{2!}+\frac{x^6}{4!}-\mathrm{...}+{\left(-1\right)}^n\frac{x^{2n+2}}{\left(2n\right)!}+\mathrm{...}\end{array}$

So, the first three nonzero terms are,

  $T_0=x^2,T_1=-\frac{x^4}{2!}\text{and}{T}_2=\frac{x^6}{4!}$

And the general term of the series is

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

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

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=\underset{n\to \infty }{\lim }\left|{\left(-1\right)}^n\frac{x^{2n+2}}{\left(2n\right)!}\times \frac{2\left(n-1\right)!}{{\left(-1\right)}^{n-1}{x}^{2n}}\right|\\ =\underset{n\to \infty }{\lim }\left|-\frac{x^2}{2n}\right|\\ =\left|-x^2\right|\underset{n\to \infty }{\lim }\frac{1}{2n}\\ =0\end{array}$

As it is always less than 1.

Hence, the series will converge for all real values of x .