Chapter 10, Problem 27RE

To determine

To find:The Maclaurin series for the function $\sin \pi x$

Answer to Problem 27RE

  ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}\frac{{\pi }^{(2n+1)}}{(2n+1)!}{x}^{(2n+1)}$

Explanation of Solution

Given information: $\sin \pi x$

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, $f(x)=\sin \pi x$ . Then,

  $\begin{array}{l}f(0)={\left[\sin \pi x\right]}_{x=0}=0\\ f\text{'}(0)={\left[\pi \cos \pi x\right]}_{x=0}=\pi \\ f\text{'}\text{'}(0)={\left[-{\pi }^2\sin \pi x\right]}_{x=0}=0\\ f\text{'}\text{'}\text{'}(0)={\left[-{\pi }^3\cos \pi x\right]}_{x=0}=-{\pi }^3\\ f^{(4)}(0)={\left[{\pi }^4\sin \pi x\right]}_{x=0}=0\\ f^{(5)}(0)={\left[{\pi }^5\cos \pi x\right]}_{x=0}={\pi }^5\\ f^{(6)}(0)={\left[-{\pi }^6\sin \pi x\right]}_{x=0}=0\end{array}$

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

The Maclaurin series generated by $f(x)=\sin \pi x$ is given by,

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