Q 12. - Calculus

Question

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

$\sin x, x_0= \frac{\pi }{2}$

Step-by-Step Solution

Answer

$P_3\left(x\right)=1-\frac{1}{2}{\left(x-\frac{\pi }{2}\right)}^2$

Step 1: Given information

$x_0= \frac{\pi }{2}$

Step 2: Concept

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

Step 3: Calculation

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

Since for any function $f$ with a derivative of order 3 at $x=0$ the third-order Taylor polynomial at $x_0=\frac{\pi }{2}$ is given by

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

Therefore, first find the value of the function along with $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right)$ and $f^{\text{'}\text{'}}\left(x\right)$ at $x_0=\frac{\pi }{2}$

Step 4: Calculation

Thus, the value of the function$x=\frac{\pi }{2}$ is

$$\begin{gathered} f\left(\frac{\pi }{2}\right)=\sin \frac{\pi }{2} \\ =1 \end{gathered}$$

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

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\sin x\right] \\ =\cos x \end{gathered}$$

So, at $x=\frac{\pi }{2}$

$$\begin{gathered} f^{\text{'}}\left(\frac{\pi }{2}\right)=\cos \frac{\pi }{2} \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left(\cos x\right) \\ =-\sin x \end{gathered}$$

So, at $x=\frac{\pi }{2}$

$$\begin{gathered} f^n\left(\frac{\pi }{2}\right)=-\sin \frac{\pi }{2} \\ =-1 \end{gathered}$$

Again

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}[-\sin x] \\ =-\frac{d}{dx}\left[\sin x\right] \\ =-\cos x \end{gathered}$$

So, at $x=\frac{\pi }{2}$

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

As a result, the third-order Taylor polynomial for the function $f\left(x\right)=\sin x$ at $x=\frac{\pi }{2}$ is

$P_3\left(x\right)=1+0·\left(x-\frac{\pi }{2}\right)+\frac{(-1)}{2!}{\left(x-\frac{\pi }{2}\right)}^2+\frac{0}{3!}{\left(x-\frac{\pi }{2}\right)}^3$

Implies that

$P_3\left(x\right)=1-\frac{1}{2}{\left(x-\frac{\pi }{2}\right)}^2$

$P_3\left(x\right)=1-\frac{1}{2}{\left(x-\frac{\pi }{2}\right)}^2$