Chapter 10.2, Problem 5E

To determine

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

Answer to Problem 5E

The first three nonzero terms are, $T_0=2x,T_1=-\frac{4x^3}{4}\text{and}{T}_2=\frac{4x^5}{15}$

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

And the series converges for all real values of x .

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=\sin 2x$

Calculation:

As Maclaurin series of $\sin x$ is

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

So, Maclaurin series of $f\left(x\right)=\sin 2x$ will be

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

So the first three term are,

  $T_0=2x,T_1=-\frac{4x^3}{4}\text{and}{T}_2=\frac{4x^5}{15}$

And the general term of the series is

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

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

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{{\left(-1\right)}^n{\left(2x\right)}^{2n+1}}{\left(2n+1\right)!}\times \frac{\left(2n-1\right)!}{{\left(-1\right)}^n{\left(2x\right)}^{2n-1}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{-{\left(2x\right)}^2}{\left(2n\right)\left(2n+1\right)}\right|\\ =\left|4x^2\right|\underset{n\to \infty }{\lim }\frac{1}{\left(2n\right)\left(2n+1\right)}\\ =0\end{array}$

As it is very less than 1 so the series converges for all real values of x .