Chapter 8.7, Problem 19E

To determine

To Prove: The given series represents $\sin \pi x$ for all the value of $x$ .

Explanation of Solution

Given:

The given series is ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{{\pi }^{n+1}}{\left(2n+1\right)!}{x}^{2n+1}}$ for the value of $a=0$ and the radius is $R=\infty$ .

Calculation:

Consider that $f^{\left(n+1\right)}\left(x\right)$ is $\left(\pm {\pi }^{n+1}\sin \pi x\right)$ or $\left(\pm {\pi }^{n+1}\cos \pi x\right)$ as we know that $\left|f^{\left(n+1\right)}\left(x\right)\right|\le {\pi }^{n+1}$ for all $x$ .

Consider $M={\pi }^{n+1}$ is the Taylor’s inequality.

  $\left|R_n\left(\right)\right|\le \frac{M}{\left(n+1\right)!}\left|{\left(x-a\right)}^{n+1}\right|=\frac{\pi }{\left(n+1\right)!}\left|x^{n+1}\right|$

Since, $\underset{n\to \infty }{\lim }\frac{x^n}{n!}=0$ then the right side of the inequality is $0$ as $n\to \infty$

Then, by the squeeze theorem $R_n\left(x\right)\to 0$ as $n\to \infty$ and the series is equal to $\sin \pi x$ for all the value of $x$ .

Hence, Proved.