Question

The Newton-Raphson method is used to find the roots of the equation \(x = e^{-2x}\). If the initial guess \(x_0 = 0\), the estimate of the root after the first iteration, \(x_1 =\)

• A. 1
• B. 1/2
• C. 1/3
• D. 1/4

Hint:

Use \(x_1 = x_0 - f(x_0)/f'(x_0)\) with \(f(x)=x-e^{-2x}\); at \(x_0=0\), \(f=-1\) and \(f'=3\).

Answer 1

The Correct Option is C

Step 1: Set up the function to find the root of.
The equation is \(x = e^{-2x}\). To use Newton-Raphson we need it in the form \(f(x) = 0\), so write
\[ f(x) = x - e^{-2x} \]
The root of \(f(x)=0\) is the same as the solution of \(x = e^{-2x}\).

Step 2: Find the derivative.
Differentiate \(f(x)\) with respect to \(x\). The derivative of \(x\) is \(1\), and the derivative of \(-e^{-2x}\), using the chain rule, is \(-e^{-2x}\cdot(-2) = 2e^{-2x}\). So
\[ f'(x) = 1 + 2e^{-2x} \]

Step 3: Evaluate \(f\) and \(f'\) at the initial guess.
At \(x_0 = 0\):
\[ f(x_0) = 0 - e^{0} = 0 - 1 = -1 \]
\[ f'(x_0) = 1 + 2e^{0} = 1 + 2(1) = 3 \]

Step 4: Apply the Newton-Raphson update formula.
The Newton-Raphson iteration is
\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \]
Substituting the values found above:
\[ x_1 = 0 - \frac{-1}{3} = \frac{1}{3} \]

Final Answer:
The estimate of the root after the first iteration is \(x_1 = 1/3\), matching option (C), not the values 1, 1/2 or 1/4 given by (A), (B) and (D).
\[ \boxed{x_1 = \frac{1}{3}} \]