How do you pick initial guess for f(x) = x(e^x) - 1 using fixed point iteration method?.

Picking an initial guess for \( f(x) = x e^x - 1 \) using the fixed-point iteration method involves several steps to ensure that the guess is close enough to the actual root for efficient convergence. Here's a structured approach:

### 1. Graphical Inspection

Start by plotting the function \( f(x) = x e^x - 1 \) to get a visual sense of where the root might be.

- Plot \( f(x) \) and observe where it crosses the x-axis.
- Alternatively, plot \( g(x) = \frac{1}{e^x} \) and \( y = x \). The intersection points indicate potential fixed points.

### 2. Analyze the Function Behavior

Evaluate \( f(x) \) at several points to see where the function changes sign:

\[ f(0) = 0 \cdot e^0 - 1 = -1 \]
\[ f(1) = 1 \cdot e^1 - 1 = e - 1 \approx 1.718 - 1 = 0.718 \]

Since \( f(x) \) changes sign between \( x = 0 \) and \( x = 1 \), there is a root in this interval.

### 3. Derivative Analysis

Ensure the function derived for the iteration \( g(x) = \frac{1}{e^x} \) has a derivative less than 1 in magnitude around the initial guess:

\[ g'(x) = -\frac{e^{-x}}{e^{x}} = -\frac{1}{e^{2x}} \]

Since \( |g'(x)| = \frac{1}{e^{2x}} \), it is always less than 1 for all \( x \), ensuring the convergence of the fixed-point iteration method.

### 4. Select an Initial Guess

Based on the function's behavior and derivative analysis, we need an initial guess in the interval where \( f(x) \) changes sign, such as \( x = 0.5 \).

### Example Calculation with Initial Guess \( x_0 = 0.5 \)

1. **Choose an initial guess** \( x_0 = 0.5 \).

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

3. **Perform the Iterations**:

   \[
   \begin{align*}
   x_0 & = 0.5 \\
   x_1 & = \frac{1}{e^{0.5}} \approx \frac{1}{1.6487} \approx 0.6065 \\
   x_2 & = \frac{1}{e^{0.6065}} \approx \frac{1}{1.8340} \approx 0.5452 \\
   x_3 & = \frac{1}{e^{0.5452}} \approx \frac{1}{1.7249} \approx 0.5796 \\
   x_4 & = \frac{1}{e^{0.5796}} \approx \frac{1}{1.7850} \approx 0.5602 \\
   x_5 & = \frac{1}{e^{0.5602}} \approx \frac{1}{1.7509} \approx 0.5709 \\
   x_6 & = \frac{1}{e^{0.5709}} \approx \frac{1}{1.7695} \approx 0.5653 \\
   x_7 & = \frac{1}{e^{0.5653}} \approx \frac{1}{1.7580} \approx 0.5688 \\
   \end{align*}
   \]

### Convergence:

The values are starting to converge. Continue this process until the difference between successive values is less than a small threshold like \( 10^{-6} \).

### Summary:

Based on graphical inspection, function behavior, and derivative analysis, an initial guess of \( x_0 = 0.5 \) is reasonable. The root of \( f(x) = x e^x - 1 \) is approximately \( x \approx 0.567143 \).

### Related Questions:

1. What are the convergence criteria for the fixed-point iteration method?
2. How does the choice of initial guess affect the convergence rate?
3. Can fixed-point iteration fail to converge? If so, under what conditions?
4. How can we improve convergence speed if the initial guess is not optimal?
5. Are there specific functions for which fixed-point iteration is particularly well-suited?
6. How can we determine if \( |g'(x)| < 1 \) in the interval of interest?
7. What are some practical examples where fixed-point iteration is applied?
8. How does the convergence criterion (tolerance) affect the accuracy and performance of the fixed-point iteration method?

### Tip:

When selecting an initial guess, consider using a combination of graphical and analytical methods to ensure that the guess is as close to the actual root as possible. This can significantly enhance the efficiency and convergence of the iteration process.

Learn how to pick an initial guess for f(x) = x(e^x) - 1 using the fixed-point iteration method. This problem involves graphical inspection, function behavior analysis, and derivative considerations to ensure efficient convergence. Suitable for advanced undergraduate students and professionals.

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