Find the root of the equation 𝑥 = 𝑒
−𝑥
, using the Newton Raphson Method up to 
4
th iteration.

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.

Find the root of the equation 𝑥 = 𝑒^{-𝑥}, using the Newton Raphson Method up to 4th iteration.

Solve the equation x = e^(-x) using the Newton-Raphson method up to 4 iterations. This step-by-step solution demonstrates the use of calculus and numerical methods, including the derivation of the Newton-Raphson formula and the convergence process. Suitable for students in calculus or numerical analysis courses.

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