Chapter 8.7, Problem 7E

To determine

To find: The Maclaurin series of $f\left(x\right)=\sin \pi x$

Answer to Problem 7E

Radius of convergences of $\sin 3x$ is $R=0$

Explanation of Solution

Given information: A maclaurin series is a Taylor series expansion of a function about $0.$

Maclaurin series expansion by definition,

  $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(0\right)}{n!}}{\left(x\right)}^n$

Calculation:

Now, we have to find the derivatives

  $f\left(0\right),f^1\left(0\right),f^{11}\left(0\right),f^{111}\left(0\right),\mathrm{.....},$ so on

  $\begin{array}{l}f\left(x\right)=\sin \pi x\text{ , }f\left(0\right)=0\\ f^{\prime }\left(x\right)=\pi \cdot \cos \pi x\text{ , }{f}^{\prime }\left(0\right)=\pi \\ f^{″}\left(x\right)=-{\pi }^2\cdot \sin \pi x\text{ , }{f}^{″}\left(0\right)=0\\ f^{‴}\left(x\right)=-{\pi }^3\cdot \cos \pi x\text{ , }{f}^{‴}\left(0\right)=-{\pi }^3\\ {f^{″}}^{″}\left(x\right)={\pi }^4\cdot \sin \pi x\text{ , }{f^{″}}^{″}\left(0\right)=0\\ {f^{‴}}^{″}\left(x\right)={\pi }^5\cdot \cos \pi x\text{ , }{f^{‴}}^{″}\left(0\right)={\pi }^5\end{array}$

Hence,

The expansion is,

  $\begin{array}{l}\sin 3x=0+\pi x+0+\frac{1}{3!}\cdot \left(-{\pi }^3\right)x^3+0+\frac{1}{5!}\cdot {\pi }^5{x}^5\mathrm{.......}\\ \Rightarrow \sin 3x=\pi x-\frac{{\pi }^3}{3!}{x}^3+\frac{1}{5!}{\pi }^5{x}^5\\ \Rightarrow \sin 3x={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{{\left(\pi x\right)}^{2n+1}}{\left(2n+1\right)!}}\end{array}$

Hence, maclurin expansion of $\sin \pi x$ is ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{{\left(\pi x\right)}^{2n+1}}{\left(2n+1\right)!}}$

  $2^{nd}$Part: We try to find the radius of convergence of the series expansion of $f\left(x\right)=\sin 3x$

The Radius of convergences

  $\begin{array}{l}\frac{1}{R}=\underset{n\to \infty }{\lim }\frac{a_n}{a_n+1}\\ \Rightarrow \frac{1}{R}=\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^n}{\left(2n+1\right)!}\times \frac{\left(2n+3\right)!}{{\left(-1\right)}^{n+1}}\\ \Rightarrow \frac{1}{R}=\underset{n\to \infty }{\lim }\frac{\left(2n+3\right)!}{\left(2n+1\right)!\cdot \left(-1\right)}\\ \Rightarrow \frac{1}{R}=-\underset{n\to \infty }{\lim }\left(2n+3\right)\cdot \left(2n+2\right)\cdot \frac{\left(2n+1\right)!}{\left(2n+1\right)!}\\ \Rightarrow \frac{1}{R}=-\underset{n\to \infty }{\lim }\left(2n+3\right)\cdot \left(2n+2\right)\\ \frac{1}{R}=-\infty \end{array}$

Here, $R=0$

Radius of convergences of $\sin \pi x$ is $R=0$