Use the fixed point iteration method to find the roots of f(x) = x - xcosx.

To find the root of \( f(x) = x - x \cos(x) \) using the fixed-point iteration method, we follow these steps:

### Step-by-Step Process:

1. **Rewrite the equation** in the form \( x = g(x) \):
   
   Given \( f(x) = x - x \cos(x) \), we can consider:
   \[ x = x \cos(x) \]

   This gives us:
   \[ g(x) = x \cos(x) \]

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

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

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 \( x_0 = 1 \).

1. \( x_0 = 1 \)

2. Calculate:
   \[ x_1 = g(x_0) = 1 \cos(1) \approx 1 \cdot 0.5403 = 0.5403 \]

3. Calculate:
   \[ x_2 = g(x_1) = 0.5403 \cos(0.5403) \approx 0.5403 \cdot 0.8576 = 0.4633 \]

4. Calculate:
   \[ x_3 = g(x_2) = 0.4633 \cos(0.4633) \approx 0.4633 \cdot 0.8950 = 0.4146 \]

5. Calculate:
   \[ x_4 = g(x_3) = 0.4146 \cos(0.4146) \approx 0.4146 \cdot 0.9141 = 0.3791 \]

6. Calculate:
   \[ x_5 = g(x_4) = 0.3791 \cos(0.3791) \approx 0.3791 \cdot 0.9303 = 0.3527 \]

7. Calculate:
   \[ x_6 = g(x_5) = 0.3527 \cos(0.3527) \approx 0.3527 \cdot 0.9382 = 0.3310 \]

8. Calculate:
   \[ x_7 = g(x_6) = 0.3310 \cos(0.3310) \approx 0.3310 \cdot 0.9454 = 0.3131 \]

9. Calculate:
   \[ x_8 = g(x_7) = 0.3131 \cos(0.3131) \approx 0.3131 \cdot 0.9513 = 0.2978 \]

Continue iterating until the difference between \( x_{n+1} \) and \( x_n \) is very small. For practical purposes, you might stop when the difference is less than a small threshold like \( 10^{-6} \).

### Summary:

The iteration process should continue until the values converge. For example, if the values stabilize around \( x \approx 0.317 \), this would be our approximate root for \( f(x) = x - x \cos(x) \).

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

### Related Questions:

1. What is the theoretical basis for the convergence of 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 \( f(x) = x - x \cos(x) \)?
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 verify that the function \( g(x) \) derived for the iteration has a derivative less than 1 in magnitude in the interval of interest to ensure convergence.

Learn how to use the fixed-point iteration method to find roots of the function f(x) = x - xcos(x). Follow a step-by-step process and calculation example to understand the iterative method's application. Suitable for advanced undergraduate students and those interested in numerical methods in mathematics.

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