Chapter 10, Problem 59RE

a.

To determine

To find: Radius of convergence of series.

Answer to Problem 59RE

Radius of convergence is equal to $\frac{1}{e}$

Explanation of Solution

Given:

Function is given,

  $f\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\frac{x^n{n}^n}{n!}}$

Calculation:

First by using ratio test checking convergence test, we get

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{\left|x^{n+1}\right|{\left(n+1\right)}^{n+1}}{\left(n+1\right)!}\frac{n!}{\left|x^n\right|n^n}\\ =\underset{n\to \infty }{\lim }\frac{{\left(n+1\right)}^{n+1}}{{\left(n+1\right)}^{n+1}{n}^n}\left|x\right|\\ =\left|x\right|\underset{n\to \infty }{\lim }{\left(\frac{n+1}{n}\right)}^n\\ =\left|x\right|e\end{array}$

Since,

  $\begin{array}{l}\left|x\right|e<1\\ \left|x\right|<\frac{1}{e}\end{array}$

Therefore, series is convergent in nature and radius of convergence is $\frac{1}{e}$

b.

To determine

To Approximate: value of $f\left(\frac{-1}{3}\right)$ in first three terms of series

Answer to Problem 59RE

Value of $f\left(\frac{-1}{3}\right)$ is $\frac{-5}{18}$

Explanation of Solution

Given:

Function is given,

  $f\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\frac{x^n{n}^n}{n!}}$

First three terms of series are,

  $f\left(x\right)=x+\frac{2^2{x}^2}{2!}+\frac{3^3{x}^3}{3!}$

Now, substituting $x=\frac{-1}{3}$ in $f\left(x\right)$ .

  $\begin{array}{l}f\left(\frac{-1}{3}\right)=\frac{-1}{3}+\frac{2^2{\left(\frac{-1}{3}\right)}^2}{2!}+\frac{3^3{\left(\frac{-1}{3}\right)}^3}{3!}+\mathrm{.......}\\ =\frac{-1}{3}+\frac{2}{9}-\frac{1}{6}\\ =\frac{-5}{18}\end{array}$

c.

To determine

To estimate: The error involved in part |(b)

Answer to Problem 59RE

Error is equal to 0.132

Explanation of Solution

Given:

Function is given,

  $f\left(x\right)={\displaystyle \sum _{n=1}^{\infty }\frac{x^n{n}^n}{n!}}$

Calculation:

Now, by alternating series estimation theory error of the series cannot be more than fourth term of series.

Therefore, error is equal to.

  $\begin{array}{l}\text{error=}\left|\left(\frac{-1}{3}\right)\frac{4^4}{4!}\right|\\ =\frac{1}{3}\times \frac{4^4}{4\times 3\times 2\times 1}\\ =0.132\end{array}$

So error in estimation is 0.132