Q 14. - Calculus

Question

Maclaurin and Taylor polynomials: Find third-order Maclaurin or Taylor polynomial for the given function about the indicated point. 

$\sqrt{x}, x_0= 1$

Step-by-Step Solution

Answer

$P_3\left(x\right)=1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1{)}^2+\frac{1}{16}(x-1{)}^3$

Step 1: Given information

$x_0= 1$

Step 2: Calculation

Consider the function $f\left(x\right)=\sqrt{x}$

Since for any function $f$ with a derivative of order 3 at $x=0$ the third-order Taylor polynomial at $x_0=1$ is given by

$P_3\left(x\right)=f\left(\frac{\pi }{2}\right)+f^{\text{'}}\left(1\right)(x-1)+\frac{f^{\text{'}\text{'}}\left(1\right)}{2!}(x-1{)}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(1\right)}{3!}(x-1{)}^3$

Therefore, first find the value of the function along with $f^{\text{'}}\left(x\right),f^{\text{'}\text{'}}\left(x\right)$ and $f^{\text{'}\text{'}}\left(x\right)$ at $x_0=1$

Step 3: Calculation

Thus, the value of the function at $x=1$ is

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

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

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}\left[\sqrt{x}\right] \\ =\frac{1}{2\sqrt{x}} \end{gathered}$$

So, at $x=1$

$$\begin{gathered} f^{\text{'}}\left(1\right)=\frac{1}{2\sqrt{1}} \\ =\frac{1}{2} \end{gathered}$$

Also

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

So, at $x=1$

$$\begin{gathered} f^{\text{'}\text{'}}\left(1\right)=-\frac{1}{4}(1{)}^{-\frac{3}{2}} \\ =-\frac{1}{4} \end{gathered}$$

Again

$$\begin{gathered} f^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left[-\frac{1}{4}{x}^{-\frac{3}{2}}\right] \\ =-\frac{1}{4}\frac{d}{dx}(x{)}^{-\frac{3}{2}} \\ =-\frac{1}{4}·-\frac{3}{2}{x}^{\frac{5}{2}} \\ =\frac{3}{8}{x}^{-\frac{5}{2}} \end{gathered}$$

So, at $x=1$

$$\begin{gathered} f^{\text{'}\text{'}\text{'}}\left(1\right)=\frac{3}{8}(1{)}^{-\frac{5}{2}} \\ =\frac{3}{8} \end{gathered}$$

Therefore, the third-order Taylor polynomial for the function $f\left(x\right)=\sqrt{x}$ at $x=1$ is

$P_3\left(x\right)=1+\frac{1}{2}·(x-1)+\frac{\left(-\frac{1}{4}\right)}{2!}(x-1{)}^2+\frac{\frac{3}{8}}{3!}(x-1{)}^3$

Implies that

$P_3\left(x\right)=1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1{)}^2+\frac{1}{16}(x-1{)}^3$