Question

Consider the function \(f(x) = e^{-x} - x\). Using the Newton-Raphson method, obtain the first improved approximation starting from the initial guess \(x_0 = 0.5\). Enter the value of the second approximation, correct to two decimal places.}

Hint:

For Newton-Raphson, be systematic. For each iteration, calculate \(f(x_n)\), then \(f'(x_n)\), and then plug them into the formula \(x_{n+1} = x_n - f(x_n)/f'(x_n)\). Keep sufficient precision during intermediate steps.

Answer 1

Step 1: Formulas.
$f(x) = e^{-x} - x \implies f'(x) = -e^{-x} - 1$. Newton-Raphson: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.
Step 2: First Iteration ($x_1$).
$f(0.5) = e^{-0.5} - 0.5 = 0.6065 - 0.5 = 0.1065$. $f'(0.5) = -e^{-0.5} - 1 = -1.6065$. $x_1 = 0.5 - (0.1065 / -1.6065) = 0.5 + 0.0663 = 0.5663$.
Step 3: Second Iteration ($x_2$).
$f(0.5663) = e^{-0.5663} - 0.5663 = 0.5676 - 0.5663 = 0.0013$. $f'(0.5663) = -e^{-0.5663} - 1 = -1.5676$. $x_2 = 0.5663 - (0.0013 / -1.5676) = 0.5663 + 0.0008 = 0.5671$. Rounded to two decimal places: 0.57.