Chapter 4.7, Problem 23E

(a) Apply Newton’s method to the equation x² − a = 0 to derive the following square-root algorithm (used by the ancient Babylonians to compute $\sqrt{a}$):

$x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)$

(b) Use part (a) to compute $\sqrt{1000}$ correct to six decimal places.

To determine

To Derive: The algorithm for the square root of the given equation by Newton’s method.

Answer to Problem 23E

The derivation is $x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)$

Explanation of Solution

Given:

The equation is $x^2-a=0$.

Formula used:

The Newton’s method is $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$.

Calculation:

Let $f\left(x\right)=x^2-a$ and obtain the derivative of $f\left(x\right)$.

Let the derivative of $f^{\prime }\left(x\right)=\frac{df\left(x\right)}{dx}$.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d\left(x^2-a\right)}{dx}\\ =\frac{d\left(x^2\right)}{dx}-\frac{d\left(a\right)}{dx}\\ =2x-0\\ =2x\end{array}$

To derive the $x_{n+1}$ use the above mentioned formula.

$\begin{array}{c}{x}_{n+1}=x_n-\frac{x^2{}_n-a}{2x_n}\\ =\frac{2x^2{}_n-x^2{}_n+a}{2x_n}\\ =\frac{x^2{}_n+a}{2x_n}\\ =\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)\end{array}$

Hence, the derivation.

To determine

To Compute: The root of $\sqrt{1000}$ correct to six decimal places by using part (a).

Answer to Problem 23E

The root correct to six decimal places is $31.6227762$.

Explanation of Solution

Given:

The equation is $x^2-a=0$.

Choose the initial point $x_1=30$.

Calculation:

Calculate $f\left(x_1\right)$ at the point $x_1=30$ and $a=1000$.

$\begin{array}{c}f\left(30\right)={\left(30\right)}^2-1000\\ =900-1000\\ =-100\end{array}$

Calculate $f^{\prime }\left(x_1\right)$ at the point $x_1=30$.

$\begin{array}{c}{f}^{\prime }\left(30\right)=2\times 30\\ =60\end{array}$

Substitute $n=1$ in the above mentioned result in part (a) and obtain $x_2$

$x_2=\frac{1}{2}\left(x_1+\frac{a}{x_1}\right)$

Substitute $x_1=30,f\left(x_1\right)=-100$ and $f^{\prime }\left(x_1\right)=60$.

$\begin{array}{c}{x}_2=30-\frac{\left(-100\right)}{60}\\ =30+\frac{100}{60}\\ =30+1.666666\\ \approx 31.666666\end{array}$

Calculate $f\left(x_2\right)$ at the point $x_2=31.666666$ and $a=1000$.

$\begin{array}{c}f\left(31.666666\right)={\left(31.666666\right)}^2-1000\\ =1002.777735-1000\\ =2.777735\end{array}$

Calculate $f^{\prime }\left(x_2\right)$ at the point $x_2=31.666666$.

$\begin{array}{c}{f}^{\prime }\left(31.666666\right)=2\times 31.666666\\ =63.333332\end{array}$

Substitute $n=2$ in the above mentioned result in part (a) and obtain $x_3$.

$x_3=\frac{1}{2}\left(x_2+\frac{a}{x_2}\right)$

Substitute $x_2=31.666666,f\left(x_2\right)=2.777735$ and $f^{\prime }\left(x_2\right)=63.333332$.

$\begin{array}{c}{x}_3=31.666666-\frac{2.777735}{63.333332}\\ =31.666666-0.043859\\ \approx 31.622807\end{array}$

Calculate $f\left(x_3\right)$ at the point $x_3=31.622807$ and $a=1000$.

$\begin{array}{c}f\left(31.622807\right)={\left(31.666666\right)}^2-1000\\ =1000.001924-1000\\ =0.001924\end{array}$

Calculate $f^{\prime }\left(x_3\right)$ at the point $x_3=31.622807$.

$\begin{array}{c}{f}^{\prime }\left(31.622807\right)=2\times 31.622807\\ =62.666664\end{array}$

Substitute $n=3$ in the above mentioned result in part (a) and obtain $x_4$.

$x_4=\frac{1}{2}\left(x_3+\frac{a}{x_3}\right)$

Substitute $x_3=31.622807,f\left(x_3\right)=0.001924$ and $f^{\prime }\left(x_3\right)=62.666664$.

$\begin{array}{c}{x}_4=31.622807-\frac{0.001924}{62.666664}\\ =31.622807-0.000030\\ \approx 31.6227762\end{array}$

Thus, the root corrected to six decimal places is $\underset{\_}{31.6227762}$.