Question

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

$\ln \left(1+x\right)$.

Short answer

The fourth Maclaurin polynomial is,

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

Step-by-step solution

Step 1. Given Information.

The function is,

$\ln \left(1+x\right)$.

Step 2. Describing the polynomial.

Let $f\left(x\right)=\ln \left(1+x\right)$.

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 fourth Maclaurin polynomial

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

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

Finding the derivatives of the function $f\left(x\right)=\ln \left(1+x\right)$.

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

Also,

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

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left({\displaystyle \frac{-1}{{(1+x)}^2}}\right)}{dx} \\ =-\frac{d\left({\displaystyle \frac{1}{{(1+x)}^2}}\right)}{dx} \\ =\frac{2}{{(1+x)}^3} \\ f\text{'}\text{'}\text{'}\left(0\right)=\frac{2}{{(1+0)}^3} \\ =2 \end{gathered}$$

Also,

$$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left({\displaystyle \frac{2}{{(1+x)}^3}}\right)}{dx} \\ =\frac{-6}{{(1+x)}^4} \\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)=\frac{-6}{{(1+0)}^4} \\ =-6 \end{gathered}$$

Thus the fourth Maclaurin polynomial is,

$$\begin{gathered} P_4\left(x\right)=0+1.x+\frac{(-1)}{2!}{x}^2+\frac{2}{3!}{x}^3+\frac{(-6)}{4!}{x}^4 \\ =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} \end{gathered}$$