Chapter 4.7, Problem 24E

(a) Apply Newton’s method to the equation 1/x − a = 0 to derive the following reciprocal algorithm:

$x_{n+1}=2x_n-a{x}_n^2$

(This algorithm enables a computer to find reciprocals without actually dividing.)

(b) Use part (a) to compute 1/1.6984 correct to six decimal places.

To determine

To Derive: The square root of the given reciprocal equation by Newton’s method

Explanation of Solution

Given:

The equation is $\frac{1}{x}-a=0$.

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 $f\left(x\right)=\frac{1}{x}-a$ and obtain the derivative of $f\left(x\right)$.

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

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d\left(\frac{1}{x}-a\right)}{dx}\\ =\frac{d\left(\frac{1}{x}\right)}{dx}-\frac{d\left(a\right)}{dx}\\ =-\frac{1}{x^2}-0\\ =-\frac{1}{x^2}\end{array}$

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

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

Hence, the derivation.

To determine

To Compute: The root of $\frac{1}{1.6984}$ correct to six decimal places by using part (a).

Answer to Problem 24E

The root correct to six decimal places is $\underset{\_}{0.588788}$.

Explanation of Solution

Given:

The equation is $\frac{1}{x}-a=0$.

Choose the initial point $x_1=0.5$.

Calculation:

Calculate $f\left(x_1\right)$ at the point $x_1=0.5$ and $a=1.6984$.

$\begin{array}{c}f\left(0.5754\right)=\frac{1}{0.5}-1.6984\\ =2-1.6984\\ =0.301600\end{array}$

Calculate $f^{\prime }\left(x_1\right)$ at the point $x_1=0.5$.

$\begin{array}{c}{f}^{\prime }\left(0.5\right)=-\frac{1}{{\left(0.5\right)}^2}\\ =-4\end{array}$

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

Substitute $n=1$ to obtain $x_2=2x_1-a{x}_1{}^2$.

Substitute $x_1=0.5,f\left(x_1\right)=0.301600$ and $f^{\prime }\left(x_1\right)=-4$.

$\begin{array}{c}{x}_2=2\times 0.5-1.6984\times {\left(0.5\right)}^2\\ =1-1.6984\times 0.25\\ =1-0.424600\\ \approx 0.5754\end{array}$

Calculate $f\left(x_2\right)$ at the point $x_2=0.5754$ and $a=1.6984$.

$\begin{array}{c}f\left(0.5754\right)=\frac{1}{0.5754}-1.6984\\ =1.737921-1.6984\\ =0.039521\end{array}$

Calculate $f^{\prime }\left(x_2\right)$ at the point $x_2=0.5754$.

$\begin{array}{c}{f}^{\prime }\left(0.5754\right)=-\frac{1}{{\left(0.5754\right)}^2}\\ =-3.020370\end{array}$

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

$x_3=2x_2-a{x}_2^2$

Substitute $x_2=0.5754,f\left(x_2\right)=0.039521$ and $f^{\prime }\left(x_2\right)=-3.020370$.

$\begin{array}{c}{x}_3=0.5754-\frac{0.039521}{\left(-3.020370\right)}\\ =31.666666+\frac{0.039521}{\left(3.020370\right)}\\ =0.5754+0.013084\\ \approx 0.588484\end{array}$

Calculate $f\left(x_3\right)$ at the point $x_3=0.588484$ and $a=1.6984$.

$\begin{array}{c}f\left(0.588484\right)=\frac{1}{0.588484}-1.6984\\ =1.699279-1.6984\\ =0.000879\end{array}$

Calculate $f^{\prime }\left(x_3\right)$ at the point $x_3=0.588484$.

$\begin{array}{c}{f}^{\prime }\left(0.588484\right)=-\frac{1}{{\left(0.588484\right)}^2}\\ =-2.887549\end{array}$

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

$x_4=2x_3-a{x}_3^2$

Substitute $x_3=0.588484,f\left(x_3\right)=0.000879$ and $f^{\prime }\left(x_3\right)=-2.887549$.

$\begin{array}{c}{x}_4=0.588484-\frac{0.000879}{\left(-2.887549\right)}\\ =0.588484+\frac{0.000879}{\left(2.887549\right)}\\ =0.588484+0.000304\\ \approx 0.588788\end{array}$

Thus, the root corrected to six decimal places is $\underset{\_}{0.588788}$.