Chapter 8, Problem 44RQ

To determine

To Find: The McLaurin series and the radius of convergence.

Answer to Problem 44RQ

The series is ${\displaystyle \sum _{n=0}^{\infty }\frac{\left(n+4\right)}{4!n!}}{3}^n{x}^n$ and the radius converges at $R=\frac{1}{3}$ .

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)={\left(1-3x\right)}^{-5}$

Calculation:

Consider the given function $f\left(x\right)={\left(1-3x\right)}^{-5}$ then $f\left(0\right)=1$ .

Consider the successive derivate of $f\left(x\right)$ is,

  $\begin{array}{l}{f}^I\left(x\right)=\left(-5\right){\left(1-3x\right)}^{-6}\left(-3\right)\therefore f^I\left(0\right)=\left(-5\right)\left(-3\right)\\ f^{II}\left(x\right)=\left(-5\right)\left(-6\right){\left(1-3x\right)}^{-7}\left(-3\right)\therefore f^{II}\left(0\right)=\left(-5\right)\left(-6\right){\left(-3\right)}^2\\ f^{III}\left(x\right)={\left(1-3x\right)}^{-5}\\ f^{IV}\left(n\right)=\left(-5\right)\left(-6\right)\left(-7\right)\left(-8\right){\left(1-3x\right)}^{-9}{\left(-3\right)}^4\therefore f^{iv}=\left(-5\right)\left(-6\right)\left(-7\right)\left(-8\right){\left(-3\right)}^4\end{array}$

Consider the Maclaurin’s series for ${\displaystyle \int \frac{e^x}{x}}dx=\ln \left|x\right|+{\displaystyle \sum _{n=1}^{\infty }\ln \left|x\right|}+{\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n\cdot n!}}+c$ then,

  $\begin{array}{c}f\left(x\right)=1+\frac{\left(5\right)\left(-3\right)}{1!}x+\frac{\left(-5\right)\left(-6\right){\left(-3\right)}^2}{2!}{x}^2+\frac{\left(-5\right)\left(-6\right){\left(-3\right)}^3}{3!}{x}^2+\mathrm{..}\\ =1-\frac{5}{1!}\left(-3x\right)+\frac{5\cdot 6}{2!}{\left(-3x\right)}^2-\frac{5\cdot 6\cdot 7}{3!}\left(-3x^3\right)+\mathrm{..}\\ ={\displaystyle \sum _{n=0}^{\infty }\frac{\left(n+4\right)}{4!n!}}{3}^n{x}^n\end{array}$

Then, the series is convergent for $ $\left|3x\right|<1$ i.e. $\left|x\right|<\frac{1}{3}$ .

Then, the radius of the convergence is $R=\frac{1}{3}$ .