Chapter 4.7, Problem 20E

To determine

To calculate: The roots of the equation,

  $3\sin (x^2)=2x$ .

Answer to Problem 20E

The roots are

  $x_2=0.69299995$

  $x_3=0.69299996$ .

Explanation of Solution

Given information:

The equation is given as:

  $3\sin (x^2)=2x$ .

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,

  $3\sin (x^2)=2x$

The initial approximation $x_1$ can be guessed by sketching both the graphs

  $y=3\sin (x^2)$ and $y=2x$ .

  

Both the graph intersect at two points whose x-coordinate can be seen in graph.

Let $x_1=0.693$ where

  $\begin{array}{l}f(x)=3\sin (x^2)-2x\\ f\text{'}(x)=3\left[2x\cos (x^2)\right]-2\end{array}$

  $=6x\cos (x^2)-2$

Now, at $n=1$

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

  $x_2=x_1-\left\{\frac{3\sin (x_1{}^2)-2x_1}{6x_1\cos (x_1^2)-2}\right\}$

  $\begin{array}{l}{x}_2=(0.693)-\left(\frac{3\sin \left[{\left(0.693\right)}^2\right]-2(0.693)}{6(0.693)\cos \left[{\left(0.693\right)}^2\right]-2}\right)\\ x_2=(0.693)-\left(\frac{1.386000069-1.386}{1.687646675}\right)\\ x_2=(0.693)-\left(\frac{0.000000069}{1.687646675}\right)\end{array}$

  $\begin{array}{l}{x}_2=0.693-(0.0000000408853352)\\ x_2=0.693-0.0000000408853352\\ x_2=0.692999959\end{array}$

The second approximation is $x_2=0.69299995$ .

Let $n=2$ ,

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

  $x_3=x_2-\left\{\frac{3\sin (x_2{}^2)-2x_2}{6x_2\cos (x_2^2)-2}\right\}$

  $\begin{array}{l}{x}_3=(0.69299995)-\left(\frac{3\sin \left[{\left(0.69299995\right)}^2\right]-2(0.69299995)}{6(0.69299995)\cos \left[{\left(0.69299995\right)}^2\right]-2}\right)\\ x_3=(0.69299995)-\left(\frac{1.385999883-1.3859999}{1.687646544}\right)\\ x_3=(0.69299995)-\left(\frac{-0.000000017}{1.687646544}\right)\end{array}$

  $\begin{array}{l}{x}_3=0.69299995-(-0.00000001007319931)\\ x_3=0.69299995+0.00000001007319931\\ x_3=0.69299996\end{array}$

The third approximation with six decimal places is $x_3=0.69299996$ .

Therefore, the roots with eight decimal places are :-

  $x_2=0.69299995$

  $x_3=0.69299996$