Chapter 10.5, Problem 60E

(a)

To determine

To find: The Taylor series.

Answer to Problem 60E

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

Explanation of Solution

Given information:

The function is f(x) = In (1 + x) at x = 0.

Formula used:

Maclaurin series is used.

Calculation:

Form the list of Maclaurin series

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

Conclusion:

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

(b)

To determine

To find: The value of x the series converges.

Answer to Problem 60E

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

Explanation of Solution

Given information:

The value of x does the series converges

Formula used:

The ratio Test is used.

Calculation:

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

  $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=L$

Then, the series converges if L < I, the series diverges if L >1 and the test is inconclusive if

L =1.

Using the ratio test, we check for absolute convergence as follows,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\frac{{\left|x\right|}^{\left[n+1\right]}}{n+1}\times \frac{n}{{\left|x\right|}^n}\\ =\left|x\right|\underset{n\to \infty }{\lim }\frac{n}{n+1}\\ \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\left|x\right|\end{array}$

The series converges absolutely for

  $\left|x\right|<1$

And diverges for

  $\left|x\right|>1$

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.

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

Conclusion:

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

(c)

To determine

The first five non-zero terms.

Answer to Problem 60E

A bound for the (absolute) truncation error is 0.002605

Explanation of Solution

Given information:

To bound for the error in evaluating In (3/2) by using only the first five nonzero terms of the series in part (a).

Formula used:

  $\left|-\frac{x^6}{6}\right|=\frac{1}{2^6\times 6}$

Calculation:

To estimate $\ln \left(\frac{3}{2}\right)$ , we would let $x=\frac{1}{2}$

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

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

Thus, a bound for the (absolute) truncation error is 0.002605

Conclusion:

A bound for the (absolute) truncation error is 0.002605

(d)

To determine

The logarithmic function

Answer to Problem 60E

The logarithmic function is $\frac{1}{2}\ln \left(1+x^2\right)$

Explanation of Solution

Given information:

To determine the logarithmic function whose Taylor series is ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}{x}^{2n}}{2n}}$

Formula used:

Taylor series is used.

Calculation:

The series is

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

To determine the logarithmic function as

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

Conclusion:

The logarithmic function is $\frac{1}{2}\ln \left(1+x^2\right)$