Question

Consider a nonlinear algebraic equation, \( e^x - 2 = 0 \). Using the Newton-Raphson method, with the initial guess of \( x_0 = 1 \), the approximated value of the root of the equation after one iteration is ________.

Hint:

The Newton-Raphson method converges rapidly when the initial guess is close to the root. Ensure to use the correct function and its derivative for accurate results.

Answer 1

Step 1: Recall the Newton-Raphson formula.
The Newton-Raphson method is an iterative method used to find approximations to the roots of a real-valued function. The formula is given by: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}. \] Where:
\( f(x) \) is the function,
\( f'(x) \) is the derivative of the function,
\( x_n \) is the current approximation.
Step 2: Define the function and its derivative. We are given the equation \( e^x - 2 = 0 \), which can be rewritten as: \[ f(x) = e^x - 2. \] The derivative of \( f(x) \) is: \[ f'(x) = e^x. \] Step 3: Apply the Newton-Raphson method.
The initial guess is \( x_0 = 1 \). Now we apply the formula to find \( x_1 \) after one iteration: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}. \] Substitute the values of \( f(x_0) \) and \( f'(x_0) \) when \( x_0 = 1 \): \[ f(1) = e^1 - 2 = e - 2 \approx 0.7183, \] \[ f'(1) = e^1 = e \approx 2.7183. \] Now substitute into the Newton-Raphson formula: \[ x_1 = 1 - \frac{0.7183}{2.7183} \approx 1 - 0.2642 = 0.7358. \] Step 4: Conclusion.
After one iteration of the Newton-Raphson method, the approximated value of the root is \( \boxed{0.74} \) (rounded to two decimal places).