Question

Newton-Raphson iterative formula for finding \[ \frac{1}{\sqrt{N}}, (N>0) \] is

• A. \( x_{n+1} = x_n (2 - N x_n) \)
• B. \( x_{n+1} = \frac{1}{2} \left( x_n + \frac{N}{x_n} \right) \)
• C. \( x_{n+1} = \frac{1}{2} \left( x_n + \frac{1}{N x_n} \right) \)
• D. \( x_{n+1} = \frac{1}{2} \left( x_n + \frac{N}{x} \right) \)

Hint:

In Newton-Raphson, choose \( f(x) \) so that its root corresponds to your target expression.

Answer 1

The Correct Option is C

Using Newton-Raphson’s method for solving \( f(x) = 0 \), the iterative formula is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Let \[ f(x) = \frac{1}{x^2} - N \] Then, \[ f'(x) = -\frac{2}{x^3} \] Substituting in the formula: \[ x_{n+1} = x_n - \frac{\frac{1}{x_n^2} - N}{-2/x_n^3} \] Simplify: \[ = x_n + \frac{x_n (1 - N x_n^2)}{2} \] Simplify further: \[ = \frac{1}{2} \left( x_n + \frac{1}{N x_n} \right) \]