Use the newton method to approximate the roots of f(x) = x - xcosx

Newton's method is an iterative technique used to approximate the roots of a real-valued function. To apply Newton's method to the function \( f(x) = x - x\cos(x) \), we follow these steps:

1. **Define the function and its derivative:**
   \[
   f(x) = x - x\cos(x)
   \]
   \[
   f'(x) = \frac{d}{dx}\left( x - x\cos(x) \right)
   \]

2. **Compute the derivative:**
   Using the product rule for differentiation:
   \[
   f'(x) = 1 - \cos(x) + x\sin(x)
   \]

3. **Newton's method formula:**
   \[
   x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
   \]

4. **Iteration process:**
   Start with an initial guess \( x_0 \) and iterate using the formula above.

Let's go through the first few iterations using an initial guess \( x_0 = 1 \):

1. **First iteration:**
   \[
   x_0 = 1
   \]
   \[
   f(x_0) = 1 - 1 \cdot \cos(1) = 1 - \cos(1) \approx 1 - 0.5403 = 0.4597
   \]
   \[
   f'(x_0) = 1 - \cos(1) + 1 \cdot \sin(1) \approx 1 - 0.5403 + 0.8415 = 1.3012
   \]
   \[
   x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{0.4597}{1.3012} \approx 1 - 0.3532 = 0.6468
   \]

2. **Second iteration:**
   \[
   x_1 = 0.6468
   \]
   \[
   f(x_1) = 0.6468 - 0.6468 \cdot \cos(0.6468) \approx 0.6468 - 0.6468 \cdot 0.7986 \approx 0.6468 - 0.5164 = 0.1304
   \]
   \[
   f'(x_1) = 1 - \cos(0.6468) + 0.6468 \cdot \sin(0.6468) \approx 1 - 0.7986 + 0.6468 \cdot 0.6018 \approx 1 - 0.7986 + 0.3889 = 0.5903
   \]
   \[
   x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 0.6468 - \frac{0.1304}{0.5903} \approx 0.6468 - 0.2208 = 0.4260
   \]

3. **Third iteration:**
   \[
   x_2 = 0.4260
   \]
   \[
   f(x_2) = 0.4260 - 0.4260 \cdot \cos(0.4260) \approx 0.4260 - 0.4260 \cdot 0.9092 \approx 0.4260 - 0.3873 = 0.0387
   \]
   \[
   f'(x_2) = 1 - \cos(0.4260) + 0.4260 \cdot \sin(0.4260) \approx 1 - 0.9092 + 0.4260 \cdot 0.4132 \approx 1 - 0.9092 + 0.1761 = 0.2669
   \]
   \[
   x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 0.4260 - \frac{0.0387}{0.2669} \approx 0.4260 - 0.1451 = 0.2809
   \]

Continue iterating until the difference between successive approximations is sufficiently small. Typically, you would set a tolerance level such as \( |x_{n+1} - x_n| < \epsilon \) for some small \(\epsilon\).

Do you want details on further iterations, or do you have any specific questions? 

### Related Questions:
1. What is Newton's method used for?
2. How does the choice of initial guess affect the convergence of Newton's method?
3. What are the advantages and disadvantages of using Newton's method?
4. Can Newton's method be used for functions with multiple roots?
5. How does the derivative of the function influence the convergence rate in Newton's method?
6. What are some alternatives to Newton's method for finding roots of equations?
7. How can Newton's method be applied to systems of nonlinear equations?
8. What are some practical applications of Newton's method in science and engineering?

### Tip:
Always ensure that the derivative \( f'(x) \) does not equal zero during the iteration process, as this would cause the method to fail.

Learn how to approximate the roots of the function f(x) = x - xcos(x) using Newton's method with a detailed step-by-step solution. This problem involves iterative techniques, derivatives, and the application of Newton's method formula. Suitable for advanced undergraduate students and professionals.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation