Q. 41 - Calculus

Question

In Exercises 41–48 in Section 8.2, you were asked to find the

fourth Taylor polynomial $P_{4}x$ for the specified function and

the given value of $x_0$ . In Exercises 37–44 give Lagrange’s form

for the remainder $R_4\left(x\right)$.

$\ln x, 3$

Step-by-Step Solution

Answer

We find the required Lagrange's form as $R_4\left(x\right)=\frac{1}{5c^5}(x-3{)}^5$

Step 1 Given Information

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

Step 2 Finding Derivatives

The derivatives of the function $f\left(x\right)=\ln x$ are

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

Also, 

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

Again, 

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

Also, 

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

Implies that, 

$f^{\left(4\right)}\left(x\right)=-\frac{6}{x^4}$

Finally, 

$\begin{array}{rcl}{f}^{\left(5\right)}\left(x\right)& =& \frac{d}{dx}\left[-\frac{6}{x^4}\right]\\ & =& -6\frac{d}{dx}\left[x^{-4}\right]\\ & =& -6·-4x^{-5}\\ & =& 24x^{-5}\end{array}$

Step 3 Calculation

Now, by the Lagrange's form for the remainder, if $f$ is a function that can be differentiated $n+1$ times in some open interval/containing the point $x_0$ and $R_n\left(x\right)$ be the $n$th remainder for $f$ at $x=x_0$. Then there exists at least one $c$ between $x_0$ and $x$ such that

$R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}{\left(x-x_0\right)}^{n+1}$

Since $f^{\left(3\right)}\left(x\right)=24x^{-5}$ and $x_0=3$ then

$R_4\left(x\right)=\frac{24c^{-5}}{5!}(x-3{)}^5$

That is,

$R_4\left(x\right)=\frac{1}{5c^5}(x-3{)}^5$