Use the fixed point iteration method to approximate the root of f(x) = x(e^x) - 1

To approximate the root of \( f(x) = x e^x - 1 \) using the fixed-point iteration method, follow these steps:

### Step-by-Step Process:

1. **Rewrite the equation** in the form \( x = g(x) \):

   Given \( f(x) = x e^x - 1 \), we want to solve for \( x \) such that:
   \[ x e^x = 1 \]

   We can rewrite this as:
   \[ x = \frac{1}{e^x} \]

   Thus, our \( g(x) \) function is:
   \[ g(x) = \frac{1}{e^x} \]

2. **Choose an initial guess** \( x_0 \).

3. **Iterate using the recurrence relation**:
   \[ x_{n+1} = g(x_n) = \frac{1}{e^{x_n}} \]

4. **Continue iterating until the change between successive approximations** \( x_{n+1} \) and \( x_n \) is sufficiently small.

### Example Calculation:

Let's choose an initial guess \( x_0 = 0.5 \).

1. \( x_0 = 0.5 \)

2. Calculate:
   \[ x_1 = g(x_0) = \frac{1}{e^{0.5}} \approx \frac{1}{1.6487} \approx 0.6065 \]

3. Calculate:
   \[ x_2 = g(x_1) = \frac{1}{e^{0.6065}} \approx \frac{1}{1.8340} \approx 0.5452 \]

4. Calculate:
   \[ x_3 = g(x_2) = \frac{1}{e^{0.5452}} \approx \frac{1}{1.7249} \approx 0.5796 \]

5. Calculate:
   \[ x_4 = g(x_3) = \frac{1}{e^{0.5796}} \approx \frac{1}{1.7850} \approx 0.5602 \]

6. Calculate:
   \[ x_5 = g(x_4) = \frac{1}{e^{0.5602}} \approx \frac{1}{1.7509} \approx 0.5709 \]

7. Calculate:
   \[ x_6 = g(x_5) = \frac{1}{e^{0.5709}} \approx \frac{1}{1.7695} \approx 0.5653 \]

8. Calculate:
   \[ x_7 = g(x_6) = \frac{1}{e^{0.5653}} \approx \frac{1}{1.7580} \approx 0.5688 \]

### Iteration Steps:

- \( x_0 = 0.5 \)
- \( x_1 \approx 0.6065 \)
- \( x_2 \approx 0.5452 \)
- \( x_3 \approx 0.5796 \)
- \( x_4 \approx 0.5602 \)
- \( x_5 \approx 0.5709 \)
- \( x_6 \approx 0.5653 \)
- \( x_7 \approx 0.5688 \)

The values are starting to converge. Continue this process until \( x_{n+1} \) and \( x_n \) are very close.

### Summary:

The iteration process should continue until the values converge to a stable root. For practical purposes, we might stop when the difference between successive values is less than a small threshold like \( 10^{-6} \). The root of \( f(x) = x e^x - 1 \) is approximately \( x \approx 0.567143 \).

Would you like more detailed steps or explanations on any part of the process?

### Related Questions:

1. What are the convergence criteria for the fixed-point iteration method?
2. How do we determine the initial guess \( x_0 \) in practice?
3. What are the potential pitfalls of the fixed-point iteration method?
4. Can we use a different \( g(x) \) for the same equation \( f(x) = x e^x - 1 \)?
5. How does fixed-point iteration compare to other numerical methods like Newton's method?
6. What is the impact of the choice of \( g(x) \) on the convergence rate?
7. How can we modify the method to improve its convergence?
8. Can the fixed-point iteration method be applied to complex-valued functions?

### Tip:

Always check that the derivative \( |g'(x)| \) is less than 1 in the interval of interest to ensure convergence of the fixed-point iteration method.

Learn how to approximate the root of f(x) = x(e^x) - 1 using the fixed-point iteration method. Follow detailed steps and calculations to understand the iterative process. Suitable for advanced undergraduate students and those interested in numerical methods.

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