## Q. 12.7

A root of   $3 \sin x=x$  is near to x = 2.5. Use two iterations of the Newton–Raphson technique to find a more accurate approximation.

## Verified Solution

The equation must first be written in the form f (x) = 0, that is

$f(x)=3 \sin x-x=0$

Then

$\begin{array}{ll}x_1=2.5 & \\f(x)=3 \sin x-x & f\left(x_1\right)=-0.705 \\f^{\prime}(x)=3 \cos x-1 & f^{\prime}\left(x_1\right)=-3.403\end{array}$

Then

$x_2=2.5-\frac{(-0.705)}{(-3.403)}=2.293$

The process is repeated with  $x_1$  = 2.293 as the initial approximation:

$x_1=2.293 \quad f\left(x_1\right)=-0.042 \quad f^{\prime}\left(x_1\right)=-2.983$

Then

$x_2=2.293-\frac{(-0.042)}{(-2.983)}=2.279$

Using two iterations of the Newton–Raphson technique, we obtain x = 2.28 as an improved estimate of the root.