Question

In Exercises 41–48 find the fourth Taylor polynomial $P_4\left(x\right)$for the specified function and the given value of $x_0$.

$\sqrt{x},1$

Short answer

Ans: The fourth Taylor polynomial for the specified function is$=1+\frac{1}{2}·(x-1)-\frac{{\displaystyle 1}}{8}(x-1{)}^2+\frac{{\displaystyle 1}}{48}(x-1{)}^3-\frac{{\displaystyle 15}}{384}(x-1{)}^4$

Step-by-step solution

Step 1. Given information:

$\sqrt{x},1$

Step 2. The fourth Taylor polynomial:

Since for any function $f$with a derivative of order 4 at $x=1$, the fourth Taylor polynomial for $x=1$ is given by

$P_4\left(x\right)=f\left(1\right)+f^{\text{'}}\left(1\right)(x-1)+\frac{f^{\text{'}\text{'}}\left(1\right)}{2!}(x-1{)}^2+\frac{f^{\text{'}\text{'}}\left(1\right)}{3!}(x-1{)}^3+\frac{f^{\text{'}\text{'}\text{'}}\left(1\right)}{4!}(x-1{)}^4$

Therefore, first find the value of the function along with $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right),f^{\text{'}\text{'}\text{'}}\left(x\right)$and $f^{\text{'}\text{'}\text{'}}\left(x\right)$at $x=1$

Step 3. Finding the fourth Taylor polynomial through derivative of order 4:

$$\begin{gathered} Thus,thevalueofthethefunctionstx=1is \\ f\left(1\right)=\sqrt{1} \\ =1 \\ Thederivativesofthefunctionf\left(x\right)=\sin \left(x\right)are \\ f\text{'}\left(x\right)=\frac{d}{dx}\left[\sqrt{x}\right] \\ =\frac{1}{2\sqrt{x}} \\ So,atx=1 \\ f\text{'}\left(1\right)=\frac{1}{2\sqrt{1}} \\ =\frac{1}{2} \\ Also, \\ f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left[\frac{1}{2\sqrt{x}}\right] \\ =\frac{1}{2}\frac{d}{dx}\left[x^{\frac{1}{2}}\right] \\ =-\frac{1}{2}·\frac{1}{2}{\left(x\right)}^{\frac{3}{2}} \\ =-\frac{1}{4}{x}^{\frac{3}{2}} \\ So,atx=1 \\ f\text{'}\left(1\right)=-\frac{1}{4}{x}^{\frac{3}{2}} \\ =-\frac{1}{4}{1}^{\frac{3}{2}} \\ =-\frac{1}{4} \\ Again, \\ f\text{'}\text{'}\text{'}\left(x\right)=-\frac{d}{dx}\left[-\frac{1}{4}{x}^{\frac{3}{2}}\right] \\ =-\frac{1}{4}\frac{d}{dx}\left[x^{\frac{3}{2}}\right] \\ =-\frac{1}{4}·-\frac{3}{2}·x^{\frac{5}{2}} \\ =\frac{3}{8}·x^{\frac{5}{2}} \\ So,atx=1 \\ \mathrm{f}\text{'}\text{'}\text{'}\left(1\right)=-\frac{1}{4}·-\frac{3}{2}·x^{\frac{5}{2}} \\ =-\frac{1}{4}·-\frac{3}{2}·{\left(1\right)}^{\frac{5}{2}} \\ =\frac{3}{8} \\ f\text{'}\text{'}\text{'}\text{'}\left(x\right)=-\frac{d}{dx}\left[-\frac{1}{4}·-\frac{3}{2}·x^{\frac{5}{2}}\right] \\ =-\frac{1}{4}·-\frac{3}{2}·\frac{d}{dx}\left[x^{\frac{5}{2}}\right] \\ =-\frac{1}{4}·-\frac{3}{2}·-\frac{5}{2}·\left[x^{\frac{7}{2}}\right] \\ =-\frac{15}{16}·\left[x^{\frac{7}{2}}\right] \\ So,atx=1 \\ f\text{'}\left(1\right)=-\frac{1}{4}·-\frac{3}{2}·-\frac{5}{2}·\left[{\mathrm{x}}^{\frac{7}{2}}\right] \\ =-\frac{1}{4}·-\frac{3}{2}·-\frac{5}{2}·\left[1^{\frac{7}{2}}\right] \\ =-\frac{15}{16} \end{gathered}$$

Step 4. Substituting the derivative of order 4 in the fourth Taylor polynomial :

Therefore, the fourth Taylor polynomial for the function $f\left(x\right)=\sqrt{x}$is

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

$=1+\frac{1}{2}·(x-1)-\frac{{\displaystyle 1}}{8}(x-1{)}^2+\frac{{\displaystyle 1}}{48}(x-1{)}^3-\frac{{\displaystyle 15}}{384}(x-1{)}^4$