Consider the Newton-Raphson method applied to approximate the square root of

a positive number  $$\alpha$$​. A recursion relation for the error $$e_n = x_n - \sqrt{\alpha}$$​ is given by:

Correct option is C

The Newton-Raphson iteration is given by:

​$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \\[10pt]\text{We have } f(x_n) = x_n^2 - a \text{ and } f'(x_n) = 2x_n. \\[10pt]\text{Substituting these into the formula, we get:} \\[10pt]x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n} \\[10pt]= \frac{1}{2} \left( x_n + \frac{a}{x_n} \right) \\[10pt]\text{Now, let } e_n = x_n - \sqrt{a}, \text{ so } x_n = e_n + \sqrt{a}. \\[10pt]\text{Then, } x_{n+1} = e_{n+1} + \sqrt{a}. \\[10pt]\text{Substituting } x_n = e_n + \sqrt{a} \text{ into the Newton-Raphson iteration:} \\[10pt]e_{n+1} + \sqrt{a} = \frac{1}{2} \left( e_n + \sqrt{a} + \frac{a}{e_n + \sqrt{a}} \right) \\[10pt]e_{n+1} = \frac{1}{2} \left( e_n + \frac{a}{e_n + \sqrt{a}} \right) - \sqrt{a} \\[10pt]e_{n+1} = \frac{1}{2} \left( e_n + \frac{a - \sqrt{a} (e_n + \sqrt{a})}{e_n + \sqrt{a}} \right) \\[10pt]e_{n+1} = \frac{1}{2} \left( e_n + \frac{a - \sqrt{a}e_n - a}{e_n + \sqrt{a}} \right) \\[10pt]e_{n+1} = \frac{1}{2} \left( e_n + \frac{-\sqrt{a} e_n}{e_n + \sqrt{a}} \right) \\[10pt]e_{n+1} = \frac{1}{2} \left( \frac{e_n^2}{e_n + \sqrt{a}} \right) \\[10pt]$$​​