Question

The Newton-Raphson method is used to find the root of \[ f(x) \equiv x^2 - x - 1 = 0 \] Starting with an initial guess \( x_0 = 1 \), the second iterate \( x_2 \) is __________ (rounded off to 2 decimal places).

Hint:

The Newton-Raphson method converges quickly for well-behaved functions. Always check if the function and its derivative are well-defined and continuous.

Answer 1

Step 1: Apply the Newton-Raphson Formula.
The Newton-Raphson method formula is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Where: \[ f(x) = x^2 - x - 1, \quad f'(x) = 2x - 1 \] Step 2: Calculate the First Iterate \( x_1 \).
Starting with \( x_0 = 1 \): \[ f(x_0) = 1^2 - 1 - 1 = -1, \quad f'(x_0) = 2(1) - 1 = 1 \] \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{-1}{1} = 1 + 1 = 2 \] Step 3: Calculate the Second Iterate \( x_2 \).
Using \( x_1 = 2 \): \[ f(x_1) = 2^2 - 2 - 1 = 1, \quad f'(x_1) = 2(2) - 1 = 3 \] \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2 - \frac{1}{3} = 1.6667 \] Step 4: Final Answer.
The second iterate \( x_2 \) is approximately \( 1.67 \) (rounded off to 2 decimal places).