Use the fixed iteration method to approximate the root of f(x) = x - xcosx

To use the fixed iteration method to approximate the root of \( f(x) = x - x \cos(x) \), we first need to rewrite the equation in the form \( x = g(x) \).

Given \( f(x) = x - x \cos(x) \), we can simplify this to:

\[ f(x) = x(1 - \cos(x)) \]

We want to find the root of this function, i.e., \( f(x) = 0 \). This implies:

\[ x(1 - \cos(x)) = 0 \]

One obvious solution is \( x = 0 \), but we are interested in non-trivial roots.

To apply the fixed-point iteration method, we rewrite the equation \( x(1 - \cos(x)) = 0 \) in the form \( x = g(x) \). 

One possible way is to consider:

\[ x = x \cos(x) \]

Thus, our \( g(x) \) becomes:

\[ g(x) = x \cos(x) \]

The fixed-point iteration method involves using the recurrence relation:

\[ x_{n+1} = g(x_n) = x_n \cos(x_n) \]

Let's choose an initial guess \( x_0 \) and iterate using the above formula. 

### Steps to apply fixed-point iteration:
1. **Choose an initial guess \( x_0 \)**
2. **Iterate using \( x_{n+1} = x_n \cos(x_n) \) until convergence**

### Example:
Let's use an initial guess \( x_0 = 1 \).

1. \( x_0 = 1 \)
2. Calculate \( x_1 = g(x_0) = 1 \cdot \cos(1) \approx 1 \cdot 0.5403 = 0.5403 \)
3. Calculate \( x_2 = g(x_1) = 0.5403 \cdot \cos(0.5403) \approx 0.5403 \cdot 0.8576 = 0.4638 \)
4. Calculate \( x_3 = g(x_2) = 0.4638 \cdot \cos(0.4638) \approx 0.4638 \cdot 0.8948 = 0.4149 \)

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

### Iteration Steps:
- \( x_0 = 1 \)
- \( x_1 \approx 0.5403 \)
- \( x_2 \approx 0.4638 \)
- \( x_3 \approx 0.4149 \)
- \( x_4 \approx 0.3842 \)
- \( x_5 \approx 0.3650 \)
- \( x_6 \approx 0.3523 \)
- \( x_7 \approx 0.3435 \)
- \( x_8 \approx 0.3372 \)

Notice that the values are converging. We continue this process until \( x_{n+1} \) and \( x_n \) are very close.

Would you like further details or have any questions about this method?

### Related Questions:
1. What are the convergence criteria for the fixed-point iteration method?
2. How do we choose a suitable initial guess \( x_0 \) for fixed-point iteration?
3. What are some potential issues with the fixed-point iteration method?
4. How can we ensure the function \( g(x) \) leads to convergence in fixed-point iteration?
5. What are the advantages of fixed-point iteration compared to other root-finding methods?
6. Can fixed-point iteration be applied to systems of equations?
7. How does the rate of convergence of fixed-point iteration compare to Newton's method?
8. What modifications can be made to improve the convergence of fixed-point iteration?

### Tip:
Always check the derivative \( g'(x) \) to ensure it is less than 1 in magnitude for all \( x \) in the interval of interest. This helps guarantee the convergence of the fixed-point iteration method.

Learn how to use the fixed-point iteration method to approximate the root of the function f(x) = x - xcos(x). This method involves iteratively applying a function g(x) = x cos(x) until convergence, starting from an initial guess. Suitable for advanced high school students and those interested in numerical methods for root-finding.

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