Q. 80 - 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 

${10}^{-6}$

$ln 0.7$

Step-by-Step Solution

Answer

(a) The function in series form is $\ln (0.7)=-\sum _{k=1}^{\infty }\frac{(0.3{)}^k}{k}$

(b) $\left|R_5\left(x\right)\right|\le \frac{1}{6}$

(c)  by trial and error, the smallest value of  so that the th degree Maclaurin polynomial approximation to the given quantity is guaranteed to be accurate to within  is .

Step 1 Given Information

Consider the function $f\left(x\right)=\ln (0.7)$

Part (a) Step 1 Solution

The objective is to use the appropriate Maclaurin series to express the term in series form. The Maclaurin series for the function $f\left(x\right)=\ln (1+x)$ is,
$\ln (1+x)=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}{x}^k$
Therefore, the function $\ln (1+x)$ for $x=-0.3$ in series form is,
$\ln [1+(-0.3\left)\right]=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}(-0.3{)}^k$
Therefore, the function $f\left(x\right)=\ln (0.7)$ in the series form is $\ln (0.7)=-\sum _{k=1}^{\infty }\frac{(0.3{)}^k}{k}$.

f ( x ) = ln ( 0 . 7 ) Part (b) Step 1 Solution

Consider the function .
The objective is to use the Lagrange's form for the remainder to bound the error in using the $5^{\text{th }}$ degree Maclaurin polynomial to approximate the function $f\left(x\right)=\ln (0.7)$.
The Lagrange's form for the remainder is $R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{x}^{n+1}$.
Therefore, the Lagrange's form for the remainder to bound the error in using the $5^{\text{th }}$ degree Maclaurin polynomial to approximate the given quantity is to find $R_5\left(x\right)$, that is
$\left|R_{\mathrm{s}}\left(x\right)\right|\le \left|\frac{f^{\left(6\right)}\left(c\right)}{6!}{x}^6\right|$

Part (b) Step 2 Simplification

The first derivative of the function $f\left(x\right)=\ln (1+x)$ is calculated as.
$\begin{array}{rcl}{f}^{\text{'}}\left(x\right)& =& \frac{d}{dx}\left[\ln \right(1+x\left)\right]\\ & =& \frac{{\displaystyle 1}}{{\displaystyle 1+x}}\end{array}$

The second derivative of the function $f\left(x\right)=\ln (1+x)$ is calculated as,
$\begin{array}{rcl}{f}^{\text{'}\text{'}}\left(x\right)& =& \frac{d}{dx}\left[\frac{1}{1+x}\right]\\ & =& -\frac{1}{(1+x{)}^2}\end{array}$

The third derivative of the function $f\left(x\right)=\ln (1+x)$ is calculated as,
$\begin{array}{rcl}{f}^{\text{'}\text{'}}\left(x\right)& =& \frac{d}{dx}\left[-\frac{1}{(1+x{)}^2}\right]\\ & =& -\frac{d}{dx}[1+x{]}^{-2}\\ & =& -\frac{-2}{(1+x{)}^3}\\ & =& \frac{2}{(1+x{)}^3}\end{array}$

Part (b) Step 3

The fourth derivative of the function $f\left(x\right)=\ln (1+x)$ is calculated as,
$\begin{array}{rcl}{f}^{m\text{'}\text{'}}\left(x\right)& =& \frac{d}{dx}\left[\frac{2}{(1+x{)}^3}\right]\\ & =& 2\frac{{\displaystyle d}}{{\displaystyle dx}}(1+x{)}^{-3}\\ & =& 2\frac{{\displaystyle -3}}{{\displaystyle (1+x{)}^4}}\\ & =& -\frac{{\displaystyle 6}}{{\displaystyle (1+x{)}^4}}\end{array}$

The fifth derlvatlve of the functlon $f\left(x\right)=\ln (1+x)$ is calculated as,
$\begin{array}{rcl}{f}^{\left(5\right)}\left(x\right)& =& \frac{d}{dx}\left[-\frac{6}{(1+x{)}^4}\right]\\ & =& -6\frac{{\displaystyle d}}{{\displaystyle dx}}(1+x{)}^{-4}\\ & =& -6·-4(1+x{)}^{-5}\\ & =& 24(1+x{)}^{-5}\end{array}$

Part (b) Step 4

The sixth derivative of the function $f\left(x\right)=\ln (1+x)$  is calculated as,

$\begin{array}{rcl}{f}^{\left(6\right)}\left(x\right)& =& \frac{d}{dx}\left[\frac{24}{(1+x{)}^5}\right] \\ & =& 24\frac{d}{dx}(1+x{)}^{-5}\\ & =& 24·-5(1+x{)}^{-6}\\ & =& -120(1+x{)}^{-6}\end{array}$

Substitute$x=-0.3$ into the sixth derivative of the function $f\left(x\right)=\ln (1+x)$ to calculate the bound for $\left|R_5\left(x\right)\right|$ as follows:
$f^{\left(6\right)}(0.7)=-120(0.3{)}^{-6}$
Substitute the value of $f^{\left(6\right)}(0.7)=-120(0.3{)}^{-6}$ into $\left|R_5\left(x\right)\right|\le \left|\frac{f^{\left(6\right)}\left(c\right)}{6!}{x}^6\right|$.
$\left|R_5\left(x\right)\right|\le \frac{-120(0.3{)}^{-6}(-0.3{)}^6}{720}$
Therefore, $\left|R_5\left(x\right)\right|\le \frac{1}{6}$.

Part (c) Step 1 Solution

Consider the function $f\left(x\right)=\ln (0.7)$.
The objective is to find the smallest value $n$ so that the approximation by Maclaurin series is accurate within ${10}^{-6}$.
As $\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$, therefore, $\ln (0.7)=-0.3-\frac{(0.3{)}^2}{2}-\frac{(0.3{)}^3}{3}-\cdots$ Hence, by trial and error, the smallest value of $n$ so that the $n$th degree Maclaurin polynomial approximation to the given quantity is guaranteed to be accurate to within ${10}^{-6}$ is$999,999$ .