Approximate root of the function f(x)=cosx-x using Newton Raphson method

To approximate the root of the function \( f(x) = \cos(x) - x \) using the Newton-Raphson method, we follow these steps:

### Steps in Newton-Raphson Method:
1. **Formula**: The iterative formula for Newton-Raphson is given by:
   \[
   x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
   \]
   Where \( f'(x) \) is the derivative of \( f(x) \).

2. **Function and Derivative**:
   - \( f(x) = \cos(x) - x \)
   - The derivative of \( f(x) \), \( f'(x) \), is:
     \[
     f'(x) = -\sin(x) - 1
     \]

3. **Initial Guess**: Choose an initial guess for \( x_0 \). Let's use \( x_0 = 0.5 \) as a reasonable starting point because the root is near \( x = 0.74 \) (known from prior analysis).

4. **Iterate** using the Newton-Raphson formula until the value converges to a satisfactory approximation.

Let me compute the root using these steps.The approximate root of the function \( f(x) = \cos(x) - x \) using the Newton-Raphson method, starting with an initial guess of \( x_0 = 0.5 \), is \( x \approx 0.739085 \). The method converged in 3 iterations.

Would you like further details or have any questions?

Here are five related questions to expand on this topic:
1. What happens if you choose a different initial guess in the Newton-Raphson method?
2. How does the Newton-Raphson method perform on functions with multiple roots?
3. Can you explain why the Newton-Raphson method sometimes fails to converge?
4. How does the convergence of Newton-Raphson compare to other root-finding methods like the bisection method?
5. What are some practical applications of root-finding methods in real-world problems?

**Tip**: Choosing a good initial guess is important in the Newton-Raphson method, as a poor choice may lead to divergence or slow convergence.

Learn how to approximate the root of the function f(x) = cos(x) - x using the Newton-Raphson method. This solution involves key numerical concepts, including iterative root-finding algorithms, and shows how to compute the solution with f'(x) = -sin(x) - 1. The final root approximation is x ≈ 0.739085 after 3 iterations, ideal for calculus students.

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