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: Understanding the Question:
The question asks to apply the Newton-Raphson method twice to find the root of the given function \(f(x) = e^{-x} - x\), starting with an initial guess \(x_0 = 0.5\). The "first improved approximation" is \(x_1\) and the "second approximation" is \(x_2\). We need to find the value of \(x_2\).
Step 2: Key Formula or Approach:
The Newton-Raphson iteration formula is: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] First, we need to find the derivative of \(f(x)\). \[ f(x) = e^{-x} - x \] \[ f'(x) = -e^{-x} - 1 \] Step 3: Detailed Calculation:
Iteration 1 (to find \(x_1\)): We start with \(x_0 = 0.5\). \[ f(x_0) = f(0.5) = e^{-0.5} - 0.5 \approx 0.60653 - 0.5 = 0.10653 \] \[ f'(x_0) = f'(0.5) = -e^{-0.5} - 1 \approx -0.60653 - 1 = -1.60653 \] Now, calculate the first improved approximation, \(x_1\): \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.5 - \frac{0.10653}{-1.60653} = 0.5 + 0.06631 \approx 0.56631 \] Iteration 2 (to find \(x_2\)): Now we use \(x_1 = 0.56631\) as the input for the next iteration. \[ f(x_1) = f(0.56631) = e^{-0.56631} - 0.56631 \approx 0.56762 - 0.56631 = 0.00131 \] \[ f'(x_1) = f'(0.56631) = -e^{-0.56631} - 1 \approx -0.56762 - 1 = -1.56762 \] Now, calculate the second approximation, \(x_2\): \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.56631 - \frac{0.00131}{-1.56762} = 0.56631 + 0.000835 \approx 0.567145 \] Step 4: Final Answer:
The value of the second approximation (\(x_2\)) is approximately 0.567145. Rounding to two decimal places, we get 0.57.