Q. 79 - Calculus

Question

In Exercises 77–82,

(a) Use appropriate Maclaurin series to express the quantities
in series form.

(b) Use Lagrange’s form for the remainder to bound the error
in using the 5th degree Maclaurin polynomial to approximate the given quantity.

(c) Find the smallest value of  so that the nth degree Maclaurin polynomial approximation to the given quantity is guaranteed to be accurate to within ${10}^{-6}$

$ln 0.5$

Step-by-Step Solution

Answer

(a) The Maclaurin series is $\ln (0.5)=-\sum _{i=1}^{\infty }\frac{(0.5{)}^k}{k}$

(b) The bound remainder is $\frac{1}{6}$

(c) The Lagrange's form for the remainder of  degree Maclaurin polynomial for the function is $R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}$

Part (a) Step 1 Solution

Consider a function $f\left(x\right)=\ln 0.5$.
The objective is to find the Maclaurin series for the function$f\left(x\right)=\ln 0.5$.
Consider that the Maclaurin series for the function$f\left(x\right)=\ln (1+x)$ is,
$\ln (1+x)=\sum _{i=1}^s\frac{(-1{)}^{k+1}}{k}{x}^k$
Substitute $x=-0.5$.

$\begin{array}{rcl}[1+(-0.5\left)\right]& =& \sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}(-0.5{)}^k\\ \ln 0.5& =& \sum _{k=1}^{\infty }\frac{(-1{)}^k(-1)}{k}(0.5{)}^k(-1{)}^k\\ & =& \sum _{k=1}^{\infty }\frac{(-1{)}^{2k}(-1)}{k}(0.5{)}^k\\ \ln 0.5& =& -\sum _{k=1}^{\infty }\frac{(0.5{)}^k}{k}\end{array}$

Hence, the Maclaurin series for the function $f\left(x\right)=\ln 0.5$ is $\ln (0.5)=-\sum _{i=1}^{\infty }\frac{(0.5{)}^k}{k}$

Part (b) Step 1 Solution

The objective is to find the bound on the remainder of $5^{\text{th }}$ degree Maclaurin polynomial using Lagrange's form to approximate the function $f\left(x\right)=\ln 0.5$
The Lagrange's form for the remainder of degree $5$ Maclaurin polynomial is,
$R_{\mathrm{s}}\left(x\right)=\frac{f^{(5+1)}\left(c\right)}{(5+1)!}{x}^{5+1}$
Here, $x<c<0$.
Thus, the Lagrange's form for the remainder of $5^{\text{th }}$ degree Maclaurin polynomial for the function $f\left(x\right)=\ln 0.5$ is
$R_5(0.5)=\frac{f^{(5+1)}\left(c\right)}{(5+1)!}(0.5{)}^{n+1}$
Here, $0.5<c<0$.

Part (b) Step 2 Simplification

Calculate the derivatives of the function $f\left(x\right)=\ln x$.

$\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& \frac{d}{dx}\left[\ln x\right]\\ & =& \frac{1}{x}\end{array}$

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left[\frac{1}{x}\right] \\ =-\frac{1}{x^2} \end{gathered}$$

$\begin{array}{rcl}{f}^{m\text{'}}\left(x\right)& =& \frac{d}{dx}\left[-\frac{1}{x^2}\right]\\ & =& -\frac{d}{dx}{x}^{-2}\\ & =& -\frac{-2}{x^3}\\ & =& \frac{2}{x^3}\\ f^{\text{'}\text{'}\text{'}}\left(x\right)& =& \frac{d}{dx}\left[\frac{2}{x^3}\right]\\ & =& 2\frac{d}{dx}(x{)}^{-3}\\ & =& 2\frac{-3}{x^4}\\ & =& -\frac{6}{x^4}\\ f^{\left(4\right)}\left(x\right)& =& -\frac{6}{x^4}\\ f^{\left(5\right)}\left(x\right)& =& \frac{d}{dx}\left[-\frac{6}{x^4}\right]\\ & =& -6\frac{d}{dx}(x{)}^{-4}\\ & =& -6(-4)(x{)}^{-5}\\ & =& \frac{24}{x^5}\\ f^{\left(6\right)}\left(x\right)& =& \frac{d}{dx}\left[\frac{24}{x^5}\right]\\ & =& 24\frac{d}{dx}(x{)}^{-5}\\ & =& 24(-5)(x{)}^{-6}\\ & =& -120(x{)}^{-6} \end{array}$

Substitute $x=c$ in  $f^{\left(6\right)}\left(x\right)=-120(x{)}^{-6}$ 

$f^{\left(6\right)}\left(c\right)=-120(c{)}^{-6}$

$$\begin{gathered} \left|R_5(0.5)\right|=\left|\frac{-120(c{)}^{-6}}{6!}(0.5{)}^6\right| \\ =\left|\frac{-120(0.5{)}^6}{c^{6}6!}\right| \\ \le \frac{120}{720} \left\{c>0.5,\text{ that is, }\frac{1}{c}<\frac{1}{0.5}\right\} \\ \le \frac{1}{6} \end{gathered}$$

Hence, the bound remainder is $\frac{1}{6}$

Part (c) Step 1 Solution

The objective is to find the smallest value of $n$ such that the $n$th degree Maclaurin polynomial approximation to the function $f\left(x\right)=\ln 0.5$ is accurate within ${10}^{-6}$. The Lagrange's form for the remainder of$n^{\text{th }}$ degree Maclaurin polynomial is,
$R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}$
Here, $x<c<0$.