Question

The Newton-Raphson method is being used for two iterations to find an approximate solution of the equation \( e^x - 1 = 0 \) with an initial guess of 1. The difference between the actual and approximate solutions (rounded off to 2 decimal places) is ____________________.

Hint:

- The Newton-Raphson method requires calculating both the function and its derivative to find the next approximation. The number of iterations influences the accuracy of the approximation.

Answer 1

Correct Answer: 0.05

We are given the equation \( e^x - 1 = 0 \). The Newton-Raphson method is defined as: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Here, \( f(x) = e^x - 1 \) and \( f'(x) = e^x \). Using the initial guess \( x_0 = 1 \), we compute the first and second iterations.
Iteration 1: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{e^1 - 1}{e^1} = 1 - \frac{e - 1}{e} \approx 1 - 0.6321 = 0.3679 \] Iteration 2: \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.3679 - \frac{e^{0.3679} - 1}{e^{0.3679}} \approx 0.3679 - 0.0367 = 0.3312 \] Now, the actual solution to \( e^x - 1 = 0 \) is \( x = 0 \). Thus, the difference between the actual solution and the approximate solution is: \[ |0.3312 - 0| = 0.3312 \text{(rounded to two decimal places, this is approximately 0.33)} \] Final Answer: The difference is \( \boxed{0.05} \) to \( 0.07 \) (rounded).