Question

Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point.

$\tan x,x_0=\frac{\pi }{3}$

Short answer

The third-order Maclaurin series for the function$f\left(x\right)=\tan x$is$P_3\left(x\right)=\sqrt{3}+\left(x-\frac{\pi }{3}\right)+\frac{1}{3}{\left(x-\frac{\pi }{3}\right)}^3$

Step-by-step solution

Given information

The function is$f\left(x\right)=\tan x$

The third Maclaurin polynomial is given for any function f with a derivative of order 3  at x=0

The third-order Taylor polynomial at $x_0=\frac{\pi }{3}$is

$P_3\left(x\right)=f\left(\frac{\pi }{3}\right)+f^{\text{'}}\left(\frac{\pi }{3}\right)\left(x-\frac{\pi }{3}\right)+\frac{f^{\text{'}\text{'}}\left(\frac{\pi }{3}\right)}{2!}{\left(x-\frac{\pi }{3}\right)}^2+\frac{f^{\text{'}\text{'}}\left(\frac{\pi }{3}\right)}{3!}{\left(x-\frac{\pi }{3}\right)}^3$

First, determine the function's value as well as$f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right)and{f}^{\text{'}\text{'}\text{'}}\left(x\right)at{x}_0=\frac{\pi }{3}$

Find the derivatives of the function 

The value of the function $x=\frac{\pi }{3}$is

$$\begin{gathered} f\left(\frac{\pi }{3}\right)=\tan \left(\frac{\mathrm{\pi }}{3}\right) \\ =\sqrt{3} \end{gathered}$$

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

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

At $x=0$

$$\begin{gathered} f^{\text{'}}\left(0\right)=se{c}^{2}0 \\ =1 \end{gathered}$$

Again

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left(se{c}^{2}x\right) \\ =2secx·secx\tan x \\ =2se{c}^{2}x\tan x \end{gathered}$$

At $x=0$

$$\begin{gathered} f^{\text{'}\text{'}}\left(0\right)=2se{c}^{2}0\tan 0 \\ =2·1·0 \\ =0 \end{gathered}$$

Again

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=2\frac{d}{dx}\left[{\sec }^{2}x\tan x\right] \\ =2\left[{\sec }^{2}x\frac{d}{dx}\left(\tan x\right)+\tan x\frac{d}{dx}\left({\sec }^{2}x\right)\right] \\ =2\left[{\sec }^{2}x·{\sec }^{2}x+\tan x·2\sec x·\sec x\tan x\right] \\ =2\left[{\sec }^{4}x+2{\sec }^{2}x{\tan }^{2}x\right] \end{gathered}$$

At $x=0$

$$\begin{gathered} f^{\text{'}\text{'}\text{'}}\left(0\right)=2\left({\sec }^{4}0+2{\sec }^{2}0{\tan }^{2}0\right) \\ =2(1+2·1·0) \\ =2 \end{gathered}$$

Find the third-order Maclaurin series for the function.

The third-order Maclaurin series for the function $f\left(x\right)=\tan x$at $x=\frac{\pi }{3}$is

$$\begin{gathered} P_3\left(x\right)=\sqrt{3}+1·\left(x-\frac{\pi }{3}\right)+\frac{0}{2!}{\left(x-\frac{\pi }{3}\right)}^2+\frac{2}{3!}{\left(x-\frac{\pi }{3}\right)}^3 \\ {P}_3\left(x\right)=\sqrt{3}+\left(x-\frac{\pi }{3}\right)+\frac{1}{3}{\left(x-\frac{\pi }{3}\right)}^3 \end{gathered}$$