Chapter 4.7, Problem 16E

To determine

To calculate: The roots of the equation,

  $\frac{1}{x}=1+x^2$ .

Answer to Problem 16E

The roots are

  $x_2=0.666666$

  $x_3=0.682170$

  $x_4=0.682328$

Explanation of Solution

Given information:

The equation is given as:

  $\frac{1}{x}=1+x^2$ .

Formula used:

Newton’s Method:

We seek a solution of $f(x)=0$ , starting from an initial estimate $x=x_1$ .

For $x=x_n$ , compute the next approximation $x_{n+1}$ by

  $x_{n+1}=x_n-\frac{f(x_n)}{f\text{'}(x_n)}$ and so on.

Calculation:

Consider the equation,

  $\frac{1}{x}=1+x^2$

Now,

  $\begin{array}{l}\frac{1}{x}=1+x^2\\ f(x)=x^2-\frac{1}{x}+1\\ f\text{'}(x)=2x+\frac{1}{x^2}\end{array}$

Let the initial approximation $x_1=1$

Now, at $n=1$

  $x_2=x_1-\frac{f(x_1)}{f\text{'}(x_1)}$

  $x_2=x_1-\frac{x_1{}^2-\frac{1}{x_1}+1}{2x_1+\frac{1}{x_1{}^2}}$

  $\begin{array}{l}{x}_2=(1)-\left(\frac{{(1)}^2-\frac{1}{1}+1}{2(1)+\frac{1}{1}}\right)\\ x_2=(1)-\left(\frac{1-1+1}{2+1}\right)\\ x_2=(1)-\left(\frac{1}{3}\right)\end{array}$

  $\begin{array}{l}{x}_2=1-\frac{1}{3}\\ x_2=\frac{2}{3}\\ x_2=0.666666666\end{array}$

The second approximation is $x_2=0.666666$ .

Let $n=2$ ,

  $x_3=x_2-\frac{f(x_2)}{f\text{'}(x_2)}$

  $x_3=x_2-\frac{x_2{}^2-\frac{1}{x_2}+1}{2x_2+\frac{1}{x_2{}^2}}$

  $\begin{array}{l}{x}_3=(0.666666)-\left(\frac{{(0.666666)}^2-\frac{1}{0.666666}+1}{2(0.666666)+\frac{1}{{(0.666666)}^2}}\right)\\ x_3=(0.666666)-\left(\frac{0.444443555-1.5000015+1}{1.333332+2.250004503}\right)\\ x_3=0.666666-\left(\frac{-0.05555795}{3.583336503}\right)\end{array}$

  $\begin{array}{l}{x}_3=0.666666+0.01550453\\ x_3=0.68217053\end{array}$

The third approximation with six decimal places is $x_3=0.682170$ .

Let $n=3$ ,

  $x_4=x_3-\frac{f(x_3)}{f\text{'}(x_3)}$

  $x_4=x_3-\frac{x_3{}^2-\frac{1}{x_3}+1}{2x_3+\frac{1}{x_3{}^2}}$

  $\begin{array}{l}{x}_4=(0.682170)-\left(\frac{{(0.682170)}^2-\frac{1}{0.682170}+1}{2(0.682170)+\frac{1}{{(0.682170)}^2}}\right)\\ x_4=(0.682170)-\left(\frac{0.465355908-1.465910257+1}{1.36434+2.148892886}\right)\\ x_4=0.682170-\left(\frac{-0.0005554349}{3.513232886}\right)\end{array}$

  $\begin{array}{l}{x}_4=0.682170+0.0001580979451\\ x_4=0.682328097\end{array}$

The fourth approximation with six decimal places is $x_4=0.682328$ .

Therefore, the roots with six decimal places are :-

  $x_2=0.666666$

  $x_3=0.682170$

  $x_4=0.682328$