Chapter 10.2, Problem 9E

To determine

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

Answer to Problem 9E

The first three nonzero terms are, $T_0=\cos 2,T_1=\cos \left(2+\frac{\pi }{2}\right)\text{and}{T}_2=\frac{\cos \left(2+\pi \right)}{2!}{x}^2$

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

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

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=\cos \left(x+2\right)$

Calculation:

As the function is,

  $\begin{array}{l}f\left(x\right)=\cos \left(x+2\right)\\ \Rightarrow f\left(x\right)=\cos 2\cos x-\sin 2\sin x\end{array}$

Now,

Maclaurin series of $\sin x$ is

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

And 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 Maclaurin series of $f\left(x\right)=\cos 2\cos x-\sin 2\sin x$ is

  $\begin{array}{l}\cos 2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\mathrm{...}+{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}+\mathrm{...}\right)-\sin 2\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\mathrm{....}+{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)!}+\mathrm{...}\right)\\ \Rightarrow f\left(x\right)=\cos 2-\left(\sin 2\right)x-\frac{\left(\cos 2\right)x^2}{2!}+\frac{\left(\sin 2\right)x^3}{3!}+\frac{\left(\cos 2\right)x^4}{4!}-\mathrm{...}\\ \Rightarrow f\left(x\right)=\cos 2+\cos \left(2+\frac{\pi }{2}\right)x+\frac{\cos \left(2+\frac{2\pi }{2}\right)}{2!}{x}^2+\mathrm{...}+\frac{\cos \left(2+\frac{n\pi }{2}\right)}{n!}{x}^n+\mathrm{...}\end{array}$

So, the first three nonzero terms are,

  $T_0=\cos 2,T_1=\cos \left(2+\frac{\pi }{2}\right)\text{and}{T}_2=\frac{\cos \left(2+\pi \right)}{2!}{x}^2$

And the general term of the series is

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

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

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{cos\left(2+\frac{\pi }{2}\right)}{n!}{x}^n\times \frac{\left(n-1\right)!}{cos\left(2+\frac{\left(n-1\right)\pi }{2}\right)x^{n-1}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{x}{n}k\right|\left(k=\pm \tan 2\text{or }\pm \text{cot2}\right)\\ =\left|xk\right|\underset{n\to \infty }{\lim }\frac{1}{n}\\ =0\end{array}$

As it is always less than 1.

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