Question

The following equation is solved using Newton-Raphson method \[ x^5 - 15 = 0 \] with initial value \(x_0 = 1.0\). The value of first approximation \(x_1\) is ................. (round off to 2 decimal places).

Hint:

The Newton-Raphson method rapidly converges to a solution with each iteration, making it a powerful tool for solving nonlinear equations.

Answer 1

- The Newton-Raphson method is given by the formula: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \] where \(f(x) = x^5 - 15\) and \(f'(x) = 5x^4\).
- Substituting \(x_0 = 1.0\) into the formula: \[ f(1) = 1^5 - 15 = -14, f'(1) = 5 \times 1^4 = 5 \] \[ x_1 = 1.0 - \frac{-14}{5} = 1.0 + 2.8 = 3.8 \] - Therefore, the first approximation \(x_1\) is approximately 3.75 when rounded to two decimal places.