Chapter 8, Problem 47RQ

(a)

To determine

To Find: The approximation of $f$ by the Taylor polynomial with the degree.

Answer to Problem 47RQ

The required expression is $\sqrt{x}=1+\frac{1}{2}\left(x-1\right)-\left(\frac{1}{8}\right){\left(x-1\right)}^2+\frac{1}{16}\left(x-1\right)$ .

Explanation of Solution

Given:

The given expression is $f\left(x\right)=\sqrt{x}$ , $a=1$ , $n=3$ for the range $0.9\le x\le 1.1$ .

Calculation:

Consider the expression as,

  $f\left(x\right)=\sqrt{x}$

Then,

  $\begin{array}{l}{f}^{\prime }\left(1\right)=-\frac{1}{2}\\ f^{″}\left(1\right)=-\frac{1}{4}\\ f^{‴}\left(1\right)=\frac{3}{8}\\ f^{iv}\left(1\right)=-\frac{15}{16}\end{array}$

Then, the value of $\sqrt{x}=T_2\left(x\right)$ that equals,

  $\begin{array}{c}\sqrt{x}=T_3\left(x\right)\\ =1+\frac{\left(\frac{1}{2}\right)}{1!}\left(x-1\right)+\frac{\left(-\frac{1}{4}\right)}{2!}{\left(x-1\right)}^2+\frac{\left(\frac{3}{8}\right)}{3!}{\left(x-1\right)}^3\\ =1+\frac{1}{2}\left(x-1\right)-\left(\frac{1}{8}\right){\left(x-1\right)}^2+\frac{1}{16}\left(x-1\right)\end{array}$