Chapter 9, Problem 59RE

$\text{(a)}$

To determine

To find: The radius of convergence of this series.

Answer to Problem 59RE

The answer: $R=\frac{1}{e}$

Explanation of Solution

Given information:

The equation is

Calculation:

Take the absolute value of the terms in the series since only use the Ratio Test for series with non-negative terms, and then look at the following limit:

  

The series converges absolutely if and only,

  $\begin{array}{c}e\left|x\right|<1\\ \left|x\right|<\frac{1}{e}\\ R=\frac{1}{e}\end{array}$

Therefore, the required radius of convergence of this series is $R=\frac{1}{e}$ .

$\text{(b)}$

To determine

To find: The first three terms of this series to approximate $f(-1/3)$ .

Answer to Problem 59RE

The answer: $f\left(-\frac{1}{3}\right)\approx -\frac{5}{18}$

Explanation of Solution

Given information:

The equation is

Calculation:

By use the given series and replace $x$ to $-1/3$ , and obtain,

  $\begin{array}{c}f\left(-\frac{1}{3}\right)\approx \frac{{(-1/3)}^1\cdot 1^1}{1!}+\frac{{(-1/3)}^2\cdot 2^2}{2!}+\frac{{(-1/3)}^3\cdot 3^3}{3!}\\ =-\frac{1}{3}+\frac{2}{9}-\frac{1}{6}\\ =-\frac{5}{18}\end{array}$

Therefore, the required first three terms of this series to approximate $f\left(-\frac{1}{3}\right)\approx -\frac{5}{18}$ .

$(c)$

To determine

To find: The error involved in the approximation in part $\text{(b)}$ .

Answer to Problem 59RE

The answer: The truncation error is less than $\frac{32}{243}\approx 0.1317$ .

Explanation of Solution

Given information:

The equation is

Calculation:

The given series at $x=-1/3$

Is:

  

This is an alternating series and meets the requirements for the Alternating Series Error Bound; as a result, provide an upper bound for the truncation error.

According to the preceding step, the truncation error is less than the absolute value of the series' fourth term, i.e.

  $\begin{array}{c}\text{error }<\frac{{(4/3)}^4}{4!}=\frac{32}{243}\\ \approx 0.1317\end{array}$

Therefore, the required truncation error is less than $\frac{32}{243}\approx 0.1317$ .