Q. 25 - Calculus

Question

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

$\cos \left(2x\right)$.

Step-by-Step Solution

Answer

The fourth Maclaurin polynomial is,

$P_4\left(x\right)=1-2x^2+\frac{2}{3}{x}^4$.

Step 1. Given Information.

The function is,

$\cos \left(2x\right)$.

Step 2. Describing the polynomial.

Let $f\left(x\right)=\cos \left(2x\right)$.

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 fourth Maclaurin polynomial.

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

$$\begin{gathered} f\left(0\right)=\cos \left(2.0\right) \\ =\cos \left(0\right) \\ =1 \end{gathered}$$

Finding the derivatives of the function $f\left(x\right)=\cos \left(2x\right)$,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d\left(\cos \left(2x\right)\right)}{dx} \\ =-2\sin \left(2x\right) \\ f\text{'}\left(0\right)=-2\sin \left(2.0\right) \\ =-2\sin \left(0\right) \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d(-2\sin \left(2x\right))}{dx} \\ =-2\frac{d\left(\sin \left(2x\right)\right)}{dx} \\ =-4\cos \left(2x\right) \\ f\text{'}\text{'}\left(0\right)=-4\cos \left(2.0)\right) \\ =-4\cos \left(0\right) \\ =-4 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d(-4\cos \left(2x\right))}{dx} \\ =-4\frac{d\left(\cos \left(2x\right)\right)}{dx} \\ =8\sin \left(2x\right) \\ f\text{'}\text{'}\text{'}\left(0\right)=8\sin \left(2.0\right) \\ =8\sin \left(0\right) \\ =0 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(8\sin \left(2x\right)\right)}{dx} \\ =16\cos \left(2x\right) \\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)=16\cos \left(2.0\right) \\ =16\cos \left(0\right) \\ =16 \end{gathered}$$

Thus the fourth Maclaurin Polynomial is,

$$\begin{gathered} P_4\left(x\right)=1+0.x+\frac{(-4)}{2!}{x}^2+\frac{0}{3!}{x}^3+\frac{16}{4!}{x}^4 \\ =1-2x^2+\frac{2}{3}{x}^4 \end{gathered}$$