Chapter 9.3, Problem 6E

To determine

To find: The first four nonzero terms of the series and the general term of the Maclaurin series for the function.

Answer to Problem 6E

The first four nonzero terms of the series are $x^5/5!,x^7/7!,x^9/9\text{ !, }{x}^{11}/11!\text{ !}$ and the general term is ${(-1)}^n\frac{x^{2n+1}}{(2n+1)!}$ .

Explanation of Solution

Given information:

The function is $\sin x-x+x^3/3!$

Calculation:

To use the Maclaurin series of $\sin x$ obtain the Maclaurin series for the given function by subtracting $x$ from it, and adding $x^3/3!$ to it,

Hence,

  $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots +{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}+\dots$

The required series is:

  $\begin{array}{}$ \sin x-x+\frac{x^3}{3!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots +{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}+\dots $$ =\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{11!}\dots +{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}+\dots $\end{array}$

Therefore, the required first four nonzero terms of the series are $x^5/5!,x^7/7!,x^9/9\text{ !, }{x}^{11}/11!\text{ !}$ and the general term is ${(-1)}^n\frac{x^{2n+1}}{(2n+1)!}$ .