Question

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

$\frac{1+x}{1-x}$.

Short answer

The fourth Maclaurin polynomial is,

$1+2(x+x^2+x^3+x^4)$.

Step-by-step solution

Step 1. Given Information.

The function is,

$\frac{1+x}{1-x}$.

Step 2. Describing the polynomial.

Let $f\left(x\right)=\frac{1+x}{1-x}$.

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

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

Step 3.Finding the fourth Maclaurin polynomial.

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

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

The derivatives of the function $f\left(x\right)=\frac{1+x}{1-x}$are,

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

Also,

localid="1649703664746" $$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d\left(2{(1-x)}^{-2}\right)}{dx} \\ =2\frac{d\left({\left(1-x\right)}^{-2}\right)}{dx} \\ =4{(1-x)}^{-3} \\ f\text{'}\text{'}\left(0\right)=4{(1-0)}^{-3} \\ =4 \end{gathered}$$

Also,

localid="1649703850327" $$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(4{(1-x)}^{-3}\right)}{dx} \\ =4\frac{d\left({\left(1-x\right)}^{-3}\right)}{dx} \\ =12{(1-x)}^{-4} \\ f\text{'}\text{'}\text{'}\left(0\right)=12{(1-0)}^{-4} \\ =12 \end{gathered}$$

Also,

localid="1649703970448" $$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(12{(1-x)}^{-4}\right)}{dx} \\ =12\frac{d\left({\left(1-x\right)}^{-4}\right)}{dx} \\ =48{(1-x)}^{-5} \\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)=48{(1-0)}^{-5} \\ =48 \end{gathered}$$

The fourth Maclaurin polynomial is,

$$\begin{gathered} P_4\left(x\right)=1+2x+\frac{4}{2!}{x}^2+\frac{12}{3!}{x}^3+\frac{48}{4!}{x}^4 \\ =1+2(x+x^2+x^3+x^4) \end{gathered}$$