Chapter 8.7, Problem 5E

To determine

To find: We have to find the maclaurin series of $f\left(x\right)={\left(1-x\right)}^{-2}$ by using it’s definition. And also we have to find associated the radius of convergent.

Answer to Problem 5E

The Radius of convergences of is $R=1$

Explanation of Solution

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

By definition, the maclaurin series of a function is

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

Calculation:

Now, we have to find the derivatives of $f\left(x\right)={\left(1-x\right)}^{-2}$

  $\begin{array}{l}f\left(x\right)={\left(1-x\right)}^{-2}\text{ , }f\left(0\right)=1\\ f^{\prime }\left(x\right)=\frac{2}{{\left(1-x\right)}^3}\text{ , }{f}^{\prime }\left(0\right)=2\\ f^{″}\left(x\right)=\frac{6}{{\left(1-x\right)}^4}\text{ , }{f}^{″}\left(0\right)=2\times 3=6\\ f^{‴}\left(x\right)=\frac{24}{{\left(1-x\right)}^5}\text{ , }{f}^{‴}\left(0\right)=2\times 3\times 4=24\\ \mathrm{.......}so\text{ on}\end{array}$

Then from $-\left(i\right)$ ,

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

.......so on.

Now from $\left(i\right)$

  $\begin{array}{l}f\left(x\right)=f\left(0\right)+f^{\prime }\left(0\right)x+\frac{f^{″}}{2!}\left(0\right)x^2+\frac{f^{‴}}{3!}\left(0\right)x^3+\mathrm{....}\\ \Rightarrow f\left(x\right)=1+2x+\frac{6}{2!}{x}^2+\frac{24}{3!}{x}^3+\mathrm{.......}\\ \Rightarrow f\left(x\right)=1+2x+3x^2+4x^3+\mathrm{.....}\\ \Rightarrow f\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\left(n+1\right)x^n}\end{array}$

Hence, the maclurin series of $f\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\left(n+1\right)x^n}----\left(ii\right)$

  $2^{nd}$Part: We try to find the radius of convergence of the series expansion of equation $-\left(ii\right)$

The coefficient of equation $-\left(ii\right)$ is

  $a_n=\left(n+1\right)$

The Radius of convergences

  $\begin{array}{l}\frac{1}{R}=\underset{n\to \infty }{\lim }\frac{a_n}{a_n+1}\\ \Rightarrow \frac{1}{R}=\underset{n\to \infty }{\lim }\frac{n+1}{n+1+1}\\ \Rightarrow \frac{1}{R}=\underset{n\to \infty }{\lim }\frac{n+1}{n+2}\\ \Rightarrow \frac{1}{R}=\underset{n\to \infty }{\lim }\frac{\left(1+\frac{1}{n}\right)}{\left(1+\frac{2}{n}\right)}\\ \Rightarrow \frac{1}{R}=\frac{1+0}{1+0}\\ \frac{1}{R}=1\end{array}$

Here, $R=1$

Hence, The Radius of convergences of is $R=1$