Chapter 9.9, Problem 19ES

In this exercise we will use the Remainder Estimation Theorem to determine the number of terms that are required in Formula (14) to approximate $\ln 2$ to five decimal-place accuracy. For this purpose let

$f(x)=\ln \frac{1+x}{1-x}=\ln (1+x)-\ln (1-x)(-1<x<1)$

(a) Show that $f^{(n+1)}(x)=n!\left[\frac{{(-1)}^n}{{(1+x)}^{n+1}}+\frac{1}{{(1-x)}^{n+1}}\right]$

(b) Use the result in part (a) to show that $\left|f^{(n+1)}\left(x\right)\right|\le n!⌈\frac{1}{{(1+x)}^{n+1}}+\frac{1}{{(1-x)}^{n+1}}⌉$

(c) Since we want to achieve five decimal-place accuracy, our goal is to choose $n$ so that the absolute value of the $nth$ remainder at $x=\frac{1}{3}$ does not exceed the value $0.000005=0.5\times {10}^{-5};$ that is, $\left|R_n\left(\begin{array}{c}1\\ 3\end{array}\right)\right|\le \mathrm{0.000005.}$ Use the Remainder Estimation Theorem to show that this condition will be satisfied if $n$ is chosen so that $\frac{M}{\left(n+1\right)!}{\left(\begin{array}{c}1\\ 3\end{array}\right)}^{n+1}\le 0.000005$ where $\left|f^{(n+1)}(x)\right|\le M$ on the interval $\left[0,\begin{array}{c}1\\ 3\end{array}\right].$

(d) Use the result in part (b) to show that $M$ can be taken as $M=n!\left[1+\frac{1}{{\left(\begin{array}{c}2\\ 3\end{array}\right)}^{n+1}}\right]$

(e) Use the results in parts (c) and (d) to show that five decimal-place accuracy will be achieved if $n$ satisfies $\frac{1}{n+1}\left[{\left(\begin{array}{c}1\\ 3\end{array}\right)}^{n+1}+{\left(\begin{array}{c}1\\ 2\end{array}\right)}^{n+1}\right]\le 0.000005$ and then show that the smallest value of $n$ that satisfies this condition is $n=13.$