Question

In Exercises $41-48$in Section $8.2$, you were asked to find the fourth Taylor polynomial $P_4\left(x\right)$for the specified function and the given value of $x_0$. In Exercises $37-44$give Lagrange's form for the remainder $R_4\left(x\right)$.

Short answer

The required Lagrange's form for the remainder is$R_4\left(x\right)=\frac{-48c^4{\left(1+c^2\right)}^{-5}+12c^2{\left(1+c^2\right)}^{-4}+{\left(1+c^2\right)}^{-3}}{15}(x-1{)}^5$

Step-by-step solution

Given Information

Consider the function$f\left(x\right)={\tan }^{-1}x$

Finding Derivatives

We have,

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

Also,

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

Again,

$\begin{array}{rcl}f\text{'}\text{'}\text{'}\left(x\right)& =& \frac{d}{dx}\left[-2x{\left(1+x^2\right)}^{-2}\right]\\ & =& -2x\frac{d}{dx}{\left(1+x^2\right)}^{-2}-2{\left(1+x^2\right)}^{-2}\frac{d}{dx}\left[x\right]\\ & =& -2x·-2{\left(1+x^2\right)}^{-3}·2x-2{\left(1+x^2\right)}^{-2}·1\\ & =& 8x^2{\left(1+x^2\right)}^{-3}-2{\left(1+x^2\right)}^{-2}\end{array}$

Also,

$\begin{array}{rcl}{f}^{\text{'}\text{'}\text{'}\text{'}}\left(x\right)& =& \frac{d}{dx}\left[8x^2{\left(1+x^2\right)}^{-3}-2{\left(1+x^2\right)}^{-2}\right]\\ & =& 8x^2\frac{d}{dx}{\left(1+x^2\right)}^{-3}+8{\left(1+x^2\right)}^{-3}\frac{d}{dx}\left[x^2\right]-2\frac{d}{dx}{\left(1+x^2\right)}^{-2}\\ & =& 8x^2·-3{\left(1+x^2\right)}^{-4}·2x+8{\left(1+x^2\right)}^{-3}·2x-2·-2{\left(1+x^2\right)}^{-3}·2x\\ & =& 48x^3{\left(1+x^2\right)}^{-4}+16x{\left(1+x^2\right)}^{-3}-8x{\left(1+x^2\right)}^{-3}\end{array}$

Implies that,

$f^{\left(4\right)}\left(x\right)=48x^3{\left(1+x^2\right)}^{-4}+8x{\left(1+x^2\right)}^{-3}$

Finally,

$\begin{array}{rcl}{f}^{\left(5\right)}\left(x\right)& =& \frac{d}{dx}\left[48x^3{\left(1+x^2\right)}^{-4}+8x{\left(1+x^2\right)}^{-3}\right]\\ & =& 48x^3\frac{d}{dx}{\left(1+x^2\right)}^{-4}+48{\left(1+x^2\right)}^{-4}\frac{d}{dx}\left[x^3\right]+8x\frac{d}{dx}{\left(1+x^2\right)}^{-3}+8{\left(1+x^2\right)}^{-3}\frac{d}{dx}\left[x\right]\\ & =& 48x^3·-4{\left(1+x^2\right)}^{-5}·2x+48{\left(1+x^2\right)}^{-4}·3x^2+8x·-3{\left(1+x^2\right)}^{-4}·2x+8{\left(1+x^2\right)}^{-3}·1\end{array}$

Implies that,

$\begin{array}{rcl}{f}^{\left(5\right)}\left(x\right)& =& 48x^3·-4{\left(1+x^2\right)}^{-5}·2x+48{\left(1+x^2\right)}^{-4}·3x^2+8x·-3{\left(1+x^2\right)}^{-4}·2x+8{\left(1+x^2\right)}^{-3}·1\\ & =& -384x^4{\left(1+x^2\right)}^{-5}+144x^2{\left(1+x^2\right)}^{-4}-48x^2{\left(1+x^2\right)}^{-4}+8{\left(1+x^2\right)}^{-3}\end{array}$

Therefore,

$f^{\left(5\right)}\left(x\right)=-384x^4{\left(1+x^2\right)}^{-5}+96x^2{\left(1+x^2\right)}^{-4}+8{\left(1+x^2\right)}^{-3}$

Step 3

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 $I$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

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

So,

$R_4\left(x\right)=\frac{f^{\left(5\right)}\left(c\right)}{5!}{\left(x-x_0\right)}^5$

Since $f^{\left(5\right)}\left(x\right)=-384x^4{\left(1+x^2\right)}^{-5}+96x^2{\left(1+x^2\right)}^{-4}+8{\left(1+x^2\right)}^{-3}$and $x_0=1$then

$R_4\left(x\right)=\frac{-384c^4{\left(1+c^2\right)}^{-5}+96c^2{\left(1+c^2\right)}^{-4}+8{\left(1+c^2\right)}^{-3}}{5!}(x-1{)}^5$

That is,

$R_4\left(x\right)=\frac{-48c^4{\left(1+c^2\right)}^{-5}+12c^2{\left(1+c^2\right)}^{-4}+{\left(1+c^2\right)}^{-3}}{15}(x-1{)}^5$