Question

If \( f(x) = x^2 - 153 = 0 \), then the iterative formula for Newton-Raphson method is _____ .

• A. \( x(n + 1) = 0.5 \left[ x(n) + \frac{153}{x(n)} \right] \)
• B. \( x(n + 1) = 0.5 \left[ x(n) - \frac{153}{x(n)} \right] \)
• C. \( x(n + 1) = \left[ x(n) + \frac{153}{x(n)} \right] \)
• D. \( x(n + 1) = \left[ x(n) - \frac{153}{x(n)} \right] \)

Hint:

The Newton-Raphson method converges quickly if the initial guess is close to the root. It is particularly effective for finding square roots.

Answer 1

The Correct Option is A

The Newton-Raphson method provides an iterative approach to finding roots of a function. The formula for the Newton-Raphson method is: \[ x(n + 1) = x(n) - \frac{f(x(n))}{f'(x(n))} \] For the function \( f(x) = x^2 - 153 \), the derivative is: \[ f'(x) = 2x \] Substituting into the Newton-Raphson formula: \[ x(n + 1) = x(n) - \frac{x(n)^2 - 153}{2x(n)} \] Simplifying the formula: \[ x(n + 1) = \frac{2x(n)^2 - (x(n)^2 - 153)}{2x(n)} = \frac{x(n)^2 + 153}{2x(n)} \] Thus, the iterative formula is **\( x(n + 1) = 0.5 \left[ x(n) + \frac{153}{x(n)} \right] \)**.