Chapter 10.3, Problem 9E

To determine

To find:The Maclaurin series for the function ${\sin }^{2}x$

Answer to Problem 9E

  $-{\displaystyle \sum _{n=1}^{\infty }{(-1)}^n}\frac{2^{(2n-1)}}{(2n)!}{x}^{2n}$

Explanation of Solution

Given information: ${\sin }^{2}x$

Calculation:The Maclaurin series generated by $f$ is,

  $f(0)+f\text{'}(0)x+\frac{f\text{'}\text{'}(0)}{2!}{x}^2+\mathrm{...}+\frac{f^{(n)}(0)}{n!}{x}^n+\mathrm{...}$

Here, $f(x)={\sin }^{2}x=\frac{1-\cos 2x}{2}=\frac{1}{2}-\frac{\cos 2x}{2}$

  $f(0)={\left[\frac{1}{2}-\frac{\cos 2x}{2}\right]}_{x=0}=\frac{1}{2}-\frac{1}{2}=0$

  $f\text{'}(0)={\left[\sin 2x\right]}_{x=0}=0$

  $f\text{'}\text{'}(0)={\left[2\cos 2x\right]}_{x=0}=2$

  $f\text{'}\text{'}\text{'}(0)={\left[-2^2\sin 2x\right]}_{x=0}=0$

  $f^{(4)}(0)={\left[-2^3\cos 2x\right]}_{x=0}=-2^3$

  $f^{(5)}(0)={\left[2^4\sin 2x\right]}_{x=0}=0$

  $f^{(6)}(0)={\left[2^5\cos 2x\right]}_{x=0}=2^5$

Similarly, $f^{(7)}(0)=0,f^{(8)}=-2^7,f^{(9)}=0,f^{(10)}=2^9\mathrm{...}$

The Maclaurin series generated by $f(x)={\sin }^{2}x$ is given by,

  $\begin{array}{l}0+0x+\frac{2}{2!}{x}^2+\frac{0}{3!}{x}^3-\frac{2^3}{4!}{x}^4+\frac{0}{5!}{x}^5+\frac{2^5}{6!}{x}^6+\frac{0}{7!}{x}^7-\frac{2^7}{8!}{x}^8+\frac{0}{9!}{x}^9+\frac{2^9}{10!}{x}^{10}+\mathrm{...}\\ =\frac{2}{2!}{x}^2-\frac{2^3}{4!}{x}^4+\frac{2^5}{6!}{x}^6-\frac{2^7}{8!}{x}^8+\frac{2^9}{10!}{x}^{10}+\mathrm{...}\\ =-{\displaystyle \sum _{n=1}^{\infty }{(-1)}^n}\frac{2^{(2n-1)}}{(2n)!}{x}^{2n}\end{array}$