Question

By Newton - Raphson's method the formula for finding the square root of any number $y$ is:

• A. $x_{n+1}=\frac{1}{2}[x_n+\frac{y}{x_n}]$
• B. $x_{n+1}=\frac{1}{2}[x_0+\frac{y}{x_0}]$
• C. $x_{n+1}=\frac{1}{3}[2x_n+\frac{y}{x_n^2}]$
• D. $x_{n+1}=\frac{1}{3}[2x_0+\frac{y}{x_0^2}]$

Answer 1

The correct option is A $x_{n+1}=\frac{1}{2}[x_n+\frac{y}{x_n}]$

Let $x=\sqrt{y}$

$⟹x^2=y$

$⟹x^2-y=0$

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

$⟹f^{\prime }\left(x\right)=2x$ ........... $[\because y$ is any number $\therefore f^{\prime }\left(y\right)=0]$

$x_{n+1}=x_n-\frac{x_n^2-y}{2x_n}$

$=x_n-\frac{x_n^2}{2x_n}+\frac{y}{2x_n}$

$=x_n-\frac{x_n}{2}+\frac{y}{2x_n}$

$x_{n+1}=\frac{x_n}{2}+\frac{y}{2x_n}$

$x_{n+1}=\frac{1}{2}[x_n+\frac{y}{x_n}]$