Q 11. - Calculus

Question

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

$x \sin x, x_0 = 0$

Step-by-Step Solution

Answer

$P_3\left(x\right)=x^2$

Step 1: Given information

$x_0=0$

Step 2: Concept

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

Step 3: Calculation

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

Since the third Maclaurin polynomial is given by for every function $f$ with a derivative of order $3$ at $x=0$

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

As a result, first, determine the function's value as well as$f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right)$ and $f^{\text{'}\text{'}}\left(x\right)$ at $x=0$

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

$$\begin{gathered} f\left(0\right)=0·\sin 0 \\ =0 \end{gathered}$$

Step 4: Calculation

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

 $$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[x\sin x\right] \\ =x\frac{d}{dx}\left[\sin x\right]+\sin x\frac{d}{dx}\left[x\right] \\ =x\cos x+\sin x·1 \\ =x\cos x+\sin x \\ \text{ So, at }x=0 \\ {f}^{\text{'}}\left(0\right)=0·\cos 0+\sin 0 \\ =0·1+0 \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}[x\cos x+\sin x] \\ =x\frac{d}{dx}\left[\cos x\right]+\cos x\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[\sin x\right] \\ =-x(-\sin x)+\cos x·1+\cos x \\ =x\sin x+2\cos x \end{gathered}$$
So, at $x=0$

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

Again

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}[x\sin x+2\cos x] \\ =x\frac{d}{dx}\left[\sin x\right]+\sin x\frac{d}{dx}\left[x\right]+2\frac{d}{dx}\left[\cos x\right] \\ =x·\cos x+\sin x·1+2(-\sin x) \\ =x\cos x-\sin x \\ \text{ So, at }x=0 \\ {f}^{\text{'}\text{'}}\left(0\right)=0·\cos 0-\sin 0 \\ =0·1-0 \\ =0 \end{gathered}$$

As a result, the third-order Maclaurin series for the function $f\left(x\right)=x\sin x$ is

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

Implies that

$P_3\left(x\right)=x^2$