Question

To find the root of the equation $f(x) \equiv e^{-x} - x = 0$, the Newton-Raphson method is used. If the initial guess for the root is 1, then the estimate of the root after first iteration is _______}

• A. $\frac{1}{1+e}$
• B. $\frac{e}{1+e}$
• C. $\frac{2}{1+e}$
• D. $\frac{2e}{1+e}$

Hint:

Be careful with the signs during the derivative and division steps.
The negative derivative term $-e^{-x} - 1$ cancels the subtraction sign in the Newton-Raphson formula, converting the operation into simple addition.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the next approximation $x_1$ of the root of the non-linear equation $e^{-x} - x = 0$ using the Newton-Raphson numerical method, starting with an initial guess $x_0 = 1$.

Step 2: Key Formula or Approach:
The iterative formula for the Newton-Raphson method is:
\[ x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} \] For our function:
\[ f(x) = e^{-x} - x \] The derivative of the function is:
\[ f'(x) = -e^{-x} - 1 \]

Step 3: Detailed Explanation:

• Substitute $x_0 = 1$ into $f(x)$ and $f'(x)$:
\[ f(1) = e^{-1} - 1 = \frac{1}{e} - 1 = \frac{1 - e}{e} \] \[ f'(1) = -e^{-1} - 1 = -\frac{1}{e} - 1 = -\frac{1 + e}{e} \]
• Apply the Newton-Raphson iteration formula for $k = 0$:
\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \] \[ x_1 = 1 - \frac{\frac{1-e}{e}}{-\frac{1+e}{e}} \]
• Cancel the $e$ in the denominators and resolve the double negative:
\[ x_1 = 1 + \frac{1 - e}{1 + e} \]
• Find a common denominator to simplify:
\[ x_1 = \frac{(1 + e) + (1 - e)}{1 + e} \] \[ x_1 = \frac{2}{1 + e} \]

Step 4: Final Answer:
The estimated root after the first iteration is $\frac{2}{1+e}$.