Chapter 9, Problem 58RE

$\text{(a)}$

To determine

To find: The first four terms and the general term of the Taylor series.

Answer to Problem 58RE

The answer: $f(x)=1+2x+4x^2+8x^3+\dots +2^n{x}^n+\dots$

Explanation of Solution

Given information:

The equation is $f(x)=1/(1-2x)$

The Taylor series generated by $f(x)$ at $x=0$ .

Calculation:

The Maclaurin series, i.e. the Taylor series at $x=0$ for $1/(1-x)$ is:

Based on this, the Taylor series generated by $f$ at $x=0$ can be obtained by replacing $x$ to $2x$:

According to the previous step, the first four terms and the general term of the Taylor series generated by $x=0$ are:

  $f(x)=1+2x+4x^2+8x^3+\dots +2^n{x}^n+\dots$

$\text{(b)}$

To determine

To find: The interval of convergence for the series found in part $\text{(a)}$ .

Answer to Problem 58RE

The answer: $\left(-\frac{1}{2},\frac{1}{2}\right)$

Explanation of Solution

Given information:

The equation is $f(x)=1/(1-2x)$

Calculation:

Use the Ratio Test solely for series with non-negative terms; as a result, consider the series' absolute value of terms,

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\left|2^{n+1}{x}^{n+1}\right|}{\left|2^n{x}^n\right|}=\underset{n\to \infty }{\lim }\frac{2^{n+1}|x{|}^{n+1}}{2^n|x{|}^n}\\ =\underset{n\to \infty }{\lim }2\left|x\right|\\ =2\left|x\right|\end{array}$

Based on the previous step, the series converges absolutely if $2\left|x\right|<1$ i.e. $-1/2<x-<1/2$ but should test the convergence also at the endpoints.

Based on part $\text{(a)}$ , the series at the left endpoint is

  ${\displaystyle \sum _{n=0}^{\infty }{2}^n}{\left(-\frac{1}{2}\right)}^n={\displaystyle \sum _{n=0}^{\infty }{(-1)}^n}$

which diverges according to the $n$ th Term Test because the series' general term doesn't get closer to zero as $n$ gets closer to infinity.

Check to see if the series converges at $x=1/2$ , the appropriate endpoint of the interval. The series at the right endpoint is based on component $\text{(a)}$ , and it is

This diverges according to the $n$ th Term Test because the series' general term doesn't get.

According to the examination above, the series converges if $-1/2<x<1/2$ i.e. the interval of convergence is $(-1/2,1/2)$ .

Therefore, the required interval of convergence is $\left(-\frac{1}{2},\frac{1}{2}\right)$ .

$(c)$

To determine

To find: The value of $f(-1/4)$ and number terms of the series are adequate for approximating $f(-1/4)$ with an error not exceeding one percent in magnitude.

Answer to Problem 58RE

The answer: six terms of the series to approximate $f(-1/4)$ with an error not exceeding one percent in magnitude.

Explanation of Solution

Given information:

The equation is $f(x)=1/(1-2x)$

Calculation:

By direct substitution, obtain that,

  $\begin{array}{c}f\left(-\frac{1}{4}\right)=\frac{1}{1-(-2/4)}\\ =\frac{1}{3/2}\\ =\frac{2}{3}\end{array}$

Approximate the value

  $f(-1/4)$

To use,

  $P_n(-1/4)$

Hence,

Since the truncation error is less than $1/2^{n+1}$ , when $P_n$ is used, determine the value of $n$ for which the error does not exceed one percent, i.e. $0.01$ .

  $\begin{array}{c}\text{error }<\frac{1}{2^{n+1}}\le 0.01\\ 2^{n+1}\ge 100\\ n\ge 6\end{array}$

In order for the approximation to be accurate to within one percent, the series must have at least six terms.