Question

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

$\sqrt{1-x}$

Short answer

The fourth Maclaurin polynomial is,

$1-\frac{1}{2}x-\frac{1}{8}{x}^2-\frac{1}{16}{x}^3-\frac{5}{128}{x}^4$

Step-by-step solution

Step 1. Given Information.

The function is,

$\sqrt{1-x}$

Step 2. Describing the polynomial.

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

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,

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

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

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

Also,

localid="1649479781950" $$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{d\left({\displaystyle \frac{-1}{2\sqrt{1-x}}}\right)}{dx} \\ =-\frac{d\left({\displaystyle \frac{1}{2\sqrt{1-x}}}\right)}{dx} \\ =-\frac{1}{2}.-\frac{1}{2}{(1-x)}^{-\frac{3}{2}}.-1 \\ =-\frac{1}{4}{(1-x)}^{-\frac{3}{2}} \\ f\text{'}\text{'}\left(0\right)=-\frac{1}{4}{(1-0)}^{-\frac{3}{2}} \\ =-\frac{1}{4}{\left(1\right)}^{-\frac{3}{2}} \\ =-\frac{1}{4} \end{gathered}$$

Also,

localid="1649503826988" $$\begin{gathered} f\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(-{\displaystyle \frac{1}{4}}{(1-x)}^{-{\displaystyle \frac{3}{2}}}\right)}{dx} \\ =-\frac{1}{4}\frac{d{(1-x)}^{-{\displaystyle \frac{3}{2}}}}{dx} \\ =-\frac{1}{4}.-\frac{3}{2}{(1-x)}^{-\frac{5}{2}}.-1 \\ =-\frac{3}{8}{(1-x)}^{-\frac{5}{2}} \\ f\text{'}\text{'}\text{'}\left(0\right)=-\frac{3}{8}{(1-0)}^{-\frac{5}{2}} \\ =-\frac{3}{8}{\left(1\right)}^{-\frac{5}{2}} \\ =-\frac{3}{8} \end{gathered}$$

Also,

localid="1649504103027" $$\begin{gathered} f\text{'}\text{'}\text{'}\text{'}\left(x\right)=\frac{d\left(-{\displaystyle \frac{3}{8}}{(1-x)}^{-{\displaystyle \frac{5}{2}}}\right)}{dx} \\ =-\frac{3}{8}\frac{d{(1-x)}^{-{\displaystyle \frac{5}{2}}}}{dx} \\ =-\frac{15}{16}{(1-x)}^{-\frac{7}{2}} \\ f\text{'}\text{'}\text{'}\text{'}\left(0\right)=-\frac{15}{16}{(1-0)}^{-\frac{7}{2}} \\ =-\frac{15}{16}{\left(1\right)}^{-\frac{7}{2}} \\ =-\frac{15}{16} \end{gathered}$$

The fourth Maclaurin polynomial is,

$$\begin{gathered} P_4\left(x\right)=1+(-\frac{1}{2})x+\frac{\left(-{\displaystyle \frac{1}{4}}\right)}{2!}{x}^2+\frac{\left(-{\displaystyle \frac{3}{8}}\right)}{3!}{x}^3+\frac{\left(-{\displaystyle \frac{15}{16}}\right)}{4!}{x}^4 \\ =1-\frac{1}{2}x-\frac{1}{8}{x}^2-\frac{1}{16}{x}^3-\frac{5}{128}{x}^4 \end{gathered}$$