Q. 21 - Calculus

Question

Find the fourth Maclaurin polynomial $P_4\left(x\right)$ for the specified function:

$\cos \left(x\right)$.

Step-by-Step Solution

Answer

The fourth Maclaurin polynomial is,

$P_4\left(x\right)=1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4$.

Step 1. Given Information.

The function is,

$\cos \left(x\right)$.

Step 2. Finding the fourth Maclaurin polynomial.

Let $f\left(x\right)=\cos \left(x\right)$.

Since for any function $f$ with a derivative of order $4$ at $x=0$, the fourth Maclaurin polynomial is,

$P_4\left(x\right)=f\left(0\right)+f\text{'}\left(0\right)x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{f\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3+\frac{f\text{'}\text{'}\text{'}\text{'}\left(0\right)}{4!}{x}^4$.

Step 3. Finding the derivatives.

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

$$\begin{gathered} f\left(0\right)=\cos \left(0\right) \\ =1 \end{gathered}$$

Finding the derivatives of the function $f\left(x\right)=\cos \left(x\right)$,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d\left(\cos \left(x\right)\right)}{dx} \\ =-\sin \left(x\right) \\ f\text{'}\left(0\right)=-\sin \left(0\right) \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d(-\sin \left(x\right))}{dx} \\ =-\frac{d\left(\sin \left(x\right)\right)}{dx} \\ =-\cos \left(x\right) \\ f\text{'}\text{'}\left(0\right)=-\cos \left(0\right) \\ =-1 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d(-\cos \left(x\right))}{dx} \\ =-\frac{d\left(\cos \left(x\right)\right)}{dx} \\ =-(-\sin \left(x\right)) \\ =\sin \left(x\right) \\ f\text{'}\text{'}\text{'}\left(0\right)=\sin \left(0\right) \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(\sin \left(x\right)\right)}{dx} \\ =\cos \left(x\right) \\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)=\cos \left(0\right) \\ =1 \end{gathered}$$

Thus, the fourth Maclaurin polynomial is,

$$\begin{gathered} P_4\left(x\right)=1+0.x+\frac{(-1)}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{1}{4!}{x}^4 \\ =1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4 \end{gathered}$$