Question

What is the value of $\sqrt{12}$ by Newton-Raphson's method after first iteration?

Answer 1

$\sqrt{12}$

Let $x=\sqrt{12}$

$x^2=12$

$x^2-12=0$

Iterative eqn. for Newton-Raphson method is

$x_{n+1}=x_n-\frac{f\left(x_n\right)}{f\prime \left(x_n\right)}$

Substitute $f\left(x\right)=x^2-12$

$\therefore$ $x_{n+1}=x_n-\frac{{(x}_n^2-12)}{2x_n}$

First iteration :-

$x_1=x_0-\frac{{(x}_0^2-12)}{2x_0}$

$x_0$ is a no. closer to $\sqrt{12}$

$\sqrt{9}<\sqrt{12}<\sqrt{16}$

$\therefore$ $x_0=3.5$

$x_1=3.5-\frac{[{\left(3.5\right)}^2-12]}{2\ast 3.5}$

$x_1=3.5-0.0357$

$x_1=3.4643$