Use the newton method to approximate the roots of f(x) = x²+logx

To approximate the roots of \( f(x) = x^2 + \log(x) \) using Newton's method, follow these steps:

### Newton's Method Steps:

1. **Define the function and its derivative:**
   \[
   f(x) = x^2 + \log(x)
   \]
   \[
   f'(x) = 2x + \frac{1}{x}
   \]

2. **Choose an initial guess:** Let's denote the initial guess by \( x_0 \). It should be a value close to the expected root. Since \( f(x) \) is only defined for \( x > 0 \), choose a positive value for \( x_0 \).

3. **Iterate using the Newton-Raphson formula:**
   \[
   x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
   \]
   Substitute \( f(x) \) and \( f'(x) \):
   \[
   x_{n+1} = x_n - \frac{x_n^2 + \log(x_n)}{2x_n + \frac{1}{x_n}}
   \]

4. **Continue iterating until the change \( |x_{n+1} - x_n| \) is sufficiently small, indicating convergence to a root.**

### Example Calculation:

Let's choose an initial guess \( x_0 = 0.5 \):

1. **Initial guess:**
   \[
   x_0 = 0.5
   \]

2. **First iteration:**
   \[
   f(0.5) = 0.5^2 + \log(0.5) = 0.25 - 0.6931 = -0.4431
   \]
   \[
   f'(0.5) = 2(0.5) + \frac{1}{0.5} = 1 + 2 = 3
   \]
   \[
   x_1 = 0.5 - \frac{-0.4431}{3} = 0.5 + 0.1477 = 0.6477
   \]

3. **Second iteration:**
   \[
   f(0.6477) = 0.6477^2 + \log(0.6477) \approx 0.4196 - 0.4344 = -0.0148
   \]
   \[
   f'(0.6477) = 2(0.6477) + \frac{1}{0.6477} \approx 1.2954 + 1.543 = 2.8384
   \]
   \[
   x_2 = 0.6477 - \frac{-0.0148}{2.8384} = 0.6477 + 0.0052 = 0.6529
   \]

4. **Third iteration:**
   \[
   f(0.6529) = 0.6529^2 + \log(0.6529) \approx 0.4263 - 0.4281 = -0.0018
   \]
   \[
   f'(0.6529) = 2(0.6529) + \frac{1}{0.6529} \approx 1.3058 + 1.5317 = 2.8375
   \]
   \[
   x_3 = 0.6529 - \frac{-0.0018}{2.8375} = 0.6529 + 0.0006 = 0.6535
   \]

5. **Fourth iteration:**
   \[
   f(0.6535) = 0.6535^2 + \log(0.6535) \approx 0.4266 - 0.4268 = -0.0002
   \]
   \[
   f'(0.6535) = 2(0.6535) + \frac{1}{0.6535} \approx 1.307 + 1.5301 = 2.8371
   \]
   \[
   x_4 = 0.6535 - \frac{-0.0002}{2.8371} = 0.6535 + 0.0001 = 0.6536
   \]

The root approximates to \( x \approx 0.6536 \).

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

### Related Questions:

1. What are the convergence criteria for Newton's method?
2. How does the choice of initial guess affect the convergence of Newton's method?
3. Can Newton's method be applied to functions with multiple roots?
4. What are the potential pitfalls of using Newton's method?
5. How does Newton's method compare to other root-finding algorithms?
6. Can Newton's method be extended to solve systems of equations?
7. What are the implications of having a derivative that is zero or undefined at some points?
8. How can we improve the convergence rate of Newton's method?

### Tip:

Always verify that the derivative \( f'(x) \) does not approach zero at the initial guess or during iterations, as this can cause the method to fail or converge slowly.

Learn how to approximate roots of f(x) = x² + log(x) using Newton's method. Follow detailed steps and an example calculation to understand this root-finding method. Suitable for advanced high school students and beyond.

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