To find the root of the equation $x = e^{-x}$ using the Newton-Raphson method, we need to:

1. Rewrite the equation in the form $f(x) = 0$. We have: $f(x) = x - e^{-x}$

2. The Newton-Raphson formula is given by: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ where $f'(x)$ is the derivative of $f(x)$.

Step 1: Compute $f'(x)$

We need the derivative of $f(x) = x - e^{-x}$: $f'(x) = 1 + e^{-x}$

Step 2: Newton-Raphson Iterations

1. Initial Guess $x_0$: Let's assume an initial guess $x_0 = 0.5$.

2. Iteration Formula: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n - e^{-x_n}}{1 + e^{-x_n}}$

Now let's perform the iterations up to the 4th step.

First Iteration $x_1$:

$f(x_0) = 0.5 - e^{-0.5} \approx 0.5 - 0.60653 = -0.10653$ $f'(x_0) = 1 + e^{-0.5} \approx 1 + 0.60653 = 1.60653$ $x_1 = 0.5 - \frac{-0.10653}{1.60653} \approx 0.5 + 0.06633 = 0.56633$

Second Iteration $x_2$:

$f(x_1) = 0.56633 - e^{-0.56633} \approx 0.56633 - 0.56789 = -0.00156$ $f'(x_1) = 1 + e^{-0.56633} \approx 1 + 0.56789 = 1.56789$ $x_2 = 0.56633 - \frac{-0.00156}{1.56789} \approx 0.56633 + 0.00099 = 0.56732$

Third Iteration $x_3$:

$f(x_2) = 0.56732 - e^{-0.56732} \approx 0.56732 - 0.56732 = 0$ $f'(x_2) = 1 + e^{-0.56732} \approx 1 + 0.56732 = 1.56732$ $x_3 = 0.56732 - \frac{0}{1.56732} = 0.56732$

Fourth Iteration $x_4$:

Since $f(x_3) \approx 0$, we can stop here as $x_4 = x_3 = 0.56732$.

Result:

The root of the equation $x = e^{-x}$ after 4 iterations using the Newton-Raphson method is approximately $x \approx 0.56732$.

Would you like further details or have any questions?

Here are 5 questions to explore further:

1. What are the conditions for convergence in the Newton-Raphson method?
2. How can the initial guess affect the convergence of the method?
3. Can Newton-Raphson fail, and in what cases?
4. How would the method change for a system of nonlinear equations?
5. What are alternative methods for solving nonlinear equations?

Tip: In numerical methods, starting with a good initial guess can significantly speed up the convergence process.