Chapter 8.7, Problem 9E

To determine

To find: The McLaurin series for $f\left(x\right)$ and also find the associated radius.

Answer to Problem 9E

The required series is ${\displaystyle \sum _{n!}^{\infty }\frac{5^n{x}^n}{n!}}$ and the series converges for all the values of x by the ratio test as the radius of the convergence is $R=\infty$ .

Explanation of Solution

Given:

The given function $f\left(x\right)=e^{5x}$ .

Calculation:

Consider the given function is,

  $f\left(x\right)=e^{5x}$

Then,

  $\begin{array}{l}{f}^{\prime }\left(0\right)=5\\ f^{″}\left(0\right)=5^2\\ f^{‴}\left(0\right)=5^3\end{array}$

Consider the general form of the McLaurin series is,

  $\begin{array}{c}f\left(0\right)+\frac{f^{\prime }\left(0\right)x}{1!}+\frac{f^{″}\left(0\right)x}{2!}+\frac{f^{‴}\left(0\right)x^3}{3!}+\mathrm{...}=1+\frac{5x}{1!}+\frac{5^2{x}^2}{2!}+\frac{5^3{x}^3}{3!}+\mathrm{...}\\ ={\displaystyle \sum _{n!}^{\infty }\frac{5^n{x}^n}{n!}}\end{array}$

Consider the limit as,

  $\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|={\displaystyle \sum _{n!}^{\infty }\left|\frac{5^{n+1}{x}^{n+1}}{\left(n+1\right)!}\cdot \frac{n!}{5^n!x^n}\right|}\\ =\underset{n\to \infty }{\lim }\frac{5\left|x\right|}{n+1}=0<1\end{array}$

Thus, the given series ${\displaystyle \sum _{n!}^{\infty }\frac{5^n{x}^n}{n!}}$ converges for all the values of x by the ratio test as the radius of the convergence is $R=\infty$ .