Chapter 9.5, Problem 60E

a.

To determine

To state: The Taylor series for $f\left(x\right)=\ln \left(1+x\right)\text{ at }x=0$ . Also include an expression for the general term.

Answer to Problem 60E

The resultant answer is $\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots +{\left(-1\right)}^{n+1}\frac{x^n}{n}+\cdots$ .

Explanation of Solution

Given information:

It is given that $f\left(x\right)=\ln \left(1+x\right)\text{ at }x=0$ .

Consider the given information:

  $f\left(x\right)=\ln \left(1+x\right)\text{ at }x=0$

Form the list of Maclaurin series:

  $\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots +{\left(-1\right)}^{n+1}\frac{x^n}{n}+\cdots$

b.

To determine

To state: The value of x at which the series in part (a) converge.

Answer to Problem 60E

The series converges for $-1<x\le 1$ .

Explanation of Solution

Given information:

The given statement says to find the value of x at which the series in part (a).

The series of absolute values is:

  ${\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|=}{\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{\left|x\right|}^n}{n}}$

The ratio Test:

Let ${\displaystyle \sum a_n}$ be a series with positive terms, and with ${}_n{\underrightarrow{\lim }}_{\infty }\frac{a_{n+1}}{a_n}=L$ .

Then series converges if $L<1$ , the series diverges if $L>1$ ,and the test is inconclusive if $L=1$ .

Using the ratio test, the absolute convergence can be checked as:

  $\begin{array}{c}{}_n{\underrightarrow{\lim }}_{\infty }\left|\frac{a_{n+1}}{a_n}\right|={}_n{\underrightarrow{\lim }}_{\infty }\frac{{\left|x\right|}^{n+1}}{n+1}\times \frac{n}{{\left|x\right|}^n}\\ ={\left|x\right|}_n{\underrightarrow{\lim }}_{\infty }\frac{n}{n+1}\\ =\left|x\right|\end{array}$

The series converges absolutely for $\left|x\right|<1$ and diverges for $\left|x\right|>1$ .

Now check at $x=1$ : ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}}$ converges by the alternating series test.

Check at $x=-1$ : ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$ diverges by the p- test.

Therefore, the series converges for $-1<x\le 1$ .

c.

To determine

To state: A bound for the error in evaluating $\ln \left(\frac{3}{2}\right)$ by using only the first five nonzero terms of the series in part (a). Use Theorem 16.

Answer to Problem 60E

The resultant answer is $0.002605$ .

Explanation of Solution

Given information:

The given information is $\ln \left(\frac{3}{2}\right)$ .

Let $x-\frac{1}{2}$ to estimate $\ln \left(\frac{3}{2}\right)$ .

The truncation error is less than the magnitude of the sixth nonzero term, that is

  $\begin{array}{c}\left|-\frac{x^6}{6}\right|=\frac{1}{2^6\times 6}\\ =\frac{1}{384}<0.002605\end{array}$

Thus, the bound for the absolute truncation error is $0.002605$ .

d.

To determine

To state: The logarithmic function whose Taylor series is ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}{x}^{2n}}{2n}}$ use the result found in part (a).

Answer to Problem 60E

The resultant answer is $\frac{1}{2}\ln \left(1+x^2\right)$ .

Explanation of Solution

Given information:

The given series is ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}{x}^{2n}}{2n}}$ .

Consider the given series:

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

Now solve this to determine the logarithmic function as:

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}{x}^{2n}}{2n}}=\frac{1}{2}{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}{\left(x^2\right)}^n}{n}}\\ =\frac{1}{2}\ln \left(1+x^2\right)\end{array}$

Therefore, the resultant answer is $\frac{1}{2}\ln \left(1+x^2\right)$ .