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 $R_4\left(x\right)$.

$\tan x,\frac{\pi }{4}$

Short answer

The required Lagrange's form is $R_4\left(x\right)=\frac{16{\left(1+{\tan }^{2}x\right)}^3+88{\left(1+{\tan }^{2}x\right)}^2{\tan }^{2}x+16\left(1+{\tan }^{2}x\right){\tan }^{4}x}{120}{\left(x-\frac{\pi }{4}\right)}^5$

Step-by-step solution

Given Information

Let us consider the function$f\left(x\right)=\tan x$

Derivatives

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

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

Also,

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

Again,

role="math" localid="1650292326399" $\begin{array}{rcl}{f}^{\text{'}\text{'}\text{'}}\left(x\right)& =& \frac{d}{dx}\left[2{\sec }^{2}x\tan x\right]\\ & =& 2\frac{d}{dx}\left[{\sec }^{2}x\tan x\right]\\ & =& 2\left\{{\sec }^{2}x\frac{d}{dx}\left[\tan x\right]+\tan x\frac{d}{dx}\left[{\sec }^{2}x\right]\right\}\\ & =& 2\left({\sec }^{2}x·{\sec }^{2}x+\tan x·2\sec x·\sec x\tan x\right)\end{array}$

Implies that,

role="math" localid="1650292365914" $\begin{array}{rcl}{f}^{\text{'}\text{'}\text{'}}\left(x\right)& =& 2\left({\sec }^{4}x+2{\sec }^{2}x{\tan }^{2}x\right)\\ & =& 2{\sec }^{4}x+4{\sec }^{2}x{\tan }^{2}x\end{array}$

Also,

role="math" localid="1650292381313" $\begin{array}{rcl}{f}^{\text{'}\text{'}\text{'}\text{'}}\left(x\right)& =& \frac{d}{dx}\left[2{\sec }^{4}x+4{\sec }^{2}x{\tan }^{2}x\right]\\ & =& 2\frac{d}{dx}\left[{\sec }^{4}x\right]+4{\sec }^{2}x\frac{d}{dx}\left[{\tan }^{2}x\right]+4{\tan }^{2}x\frac{d}{dx}\left[{\sec }^{2}x\right]\\ & =& 2·4{\sec }^{3}x·\sec x\tan x+4{\sec }^{2}x·2\tan x·{\sec }^{2}x+4{\tan }^{2}x·2\sec x·\sec x\tan x\\ & =& 8{\sec }^{4}x\tan x+8{\sec }^{4}x\tan x+8{\tan }^{3}x{\sec }^{2}x\end{array}$

Implies that,

$f^{\left(4\right)}\left(x\right)=16{\sec }^{4}x\tan x+8{\tan }^{3}x{\sec }^{2}x$

Finally,

$\begin{array}{rcl}{f}^{\left(5\right)}\left(x\right)& =& \frac{d}{dx}\left[16{\sec }^{4}x\tan x+8{\tan }^{3}x{\sec }^{2}x\right]\\ & =& 16{\sec }^{4}x\frac{d}{dx}\left[\tan x\right]+16\tan x\frac{d}{dx}\left[{\sec }^{4}x\right]+8{\tan }^{3}x\frac{d}{dx}\left[{\sec }^{2}x\right]+8{\sec }^{2}x\frac{d}{dx}\left[{\tan }^{3}x\right]\\ & =& 16{\sec }^{4}x·{\sec }^{2}x+16\tan x·4{\sec }^{3}x·\sec x\tan x+8{\tan }^{3}x·2\sec x·\sec x\tan x+8{\sec }^{2}x·3{\tan }^{2}x·{\sec }^{2}x\\ & =& 16{\sec }^{6}x+64{\sec }^{4}x{\tan }^{2}x+16{\sec }^{2}x{\tan }^{4}x+24{\sec }^{4}x{\tan }^{2}x\\ & & \end{array}$

Implies that,

$f^{\left(5\right)}\left(x\right)=16{\sec }^{6}x+88{\sec }^{4}x{\tan }^{2}x+16{\sec }^{2}x{\tan }^{4}x$

Or,

$f^{\left(5\right)}\left(x\right)=16{\left(1+{\tan }^{2}x\right)}^3+88{\left(1+{\tan }^{2}x\right)}^2{\tan }^{2}x+16\left(1+{\tan }^{2}x\right){\tan }^{4}x$

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 $I$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

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

Since $f^{\left(5\right)}\left(x\right)=16{\left(1+{\tan }^{2}x\right)}^3+88{\left(1+{\tan }^{2}x\right)}^2{\tan }^{2}x+16\left(1+{\tan }^{2}x\right){\tan }^{4}x$and $x_0=\frac{\pi }{4}$then

$R_4\left(x\right)=\frac{f^5\left(c\right)}{5!}{\left(x-\frac{\pi }{4}\right)}^5$

That is,

$R_4\left(x\right)=\frac{16{\left(1+{\tan }^{2}x\right)}^3+88{\left(1+{\tan }^{2}x\right)}^2{\tan }^{2}x+16\left(1+{\tan }^{2}x\right){\tan }^{4}x}{120}{\left(x-\frac{\pi }{4}\right)}^5$