Chapter 8.7, Problem 3E

To determine

To find: We have to find the maclaurin series for $f$ and it’s radius of convergence.

Answer to Problem 3E

The Radius of convergences of is $R=1$

Explanation of Solution

Given information:

  $\begin{array}{l}{f}^{\left(n\right)}=\left(n+1\right)!\\ \text{for, }n=0,1,2,3----\left(i\right)\end{array}$

Calculation:

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

Then from $\left(i\right)$ we have

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

.......so on.

Hence, the maclurin series of $f$ is $f\left(x\right)=1+2x+3x^2+\mathrm{....}$

  $2^{nd}$Part: Now we have to find the radius of convergence.

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 $R=1$