Question

Find third-order Maclaurin

or Taylor polynomial for the given function about the indicated

point.

${\tan }^{-1}x,x_0=0$

Short answer

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

Step-by-step solution

Given information

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

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

$P_3\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)x+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{x}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{x}^3$

First, determine the function's value as well as $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right)atx=0$

Find the derivatives of the function

The value of the function $x=0$is,

localid="1653480203360" $$\begin{gathered} f\left(0\right)=ta{n}^{-1}0 \\ =0 \end{gathered}$$

The derivatives of the function $f\left(x\right)={\tan }^{-1}x$are,

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

When $x=0$

$f^{\text{'}}\left(0\right)=\frac{1}{1+0^2}$

then also, localid="1653118971406" $$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}{\left(1+x^2\right)}^{-1} \\ =-{\left(1+x^2\right)}^{-2}·2x \\ =-\frac{2x}{1+x^2} \end{gathered}$$

So,

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

Again,

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

At $x=0$

$$\begin{gathered} f^{\text{'}\text{'}}\left(0\right)=-2\left[\frac{4{\left(0\right)}^2}{{(1+0^2)}^3}+\frac{1}{{(1+0^2)}^2}\right] \\ =-2 \end{gathered}$$

Find the third-order Maclaurin series for the function.

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

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