Use the Newton-Raphson method to obtain the root of the equation x³-x²-x-1 = 0 within the interval 1.6 < x < 2.1

The Newton-Raphson method is an iterative numerical technique used to find the roots of a real-valued function. The formula for the Newton-Raphson iteration is:

\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]

Given the equation \( f(x) = x^3 - x^2 - x - 1 \), we first need to determine its derivative:

\[ f'(x) = 3x^2 - 2x - 1 \]

Let's start by selecting an initial guess, \( x_0 \), within the interval \( 1.6 < x < 2.1 \). For this example, we'll choose \( x_0 = 2.0 \).

Now, we will apply the Newton-Raphson iteration formula.

1. **First Iteration:**
   \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \]
   \[ x_1 = 2.0 - \frac{2^3 - 2^2 - 2 - 1}{3(2)^2 - 2(2) - 1} \]
   \[ x_1 = 2.0 - \frac{8 - 4 - 2 - 1}{12 - 4 - 1} \]
   \[ x_1 = 2.0 - \frac{1}{7} \]
   \[ x_1 = 2.0 - 0.142857 \]
   \[ x_1 \approx 1.857143 \]

2. **Second Iteration:**
   \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \]
   \[ f(1.857143) = (1.857143)^3 - (1.857143)^2 - 1.857143 - 1 \approx 0.482 \]
   \[ f'(1.857143) = 3(1.857143)^2 - 2(1.857143) - 1 \approx 5.143 \]
   \[ x_2 = 1.857143 - \frac{0.482}{5.143} \]
   \[ x_2 \approx 1.763157 \]

3. **Third Iteration:**
   \[ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} \]
   \[ f(1.763157) = (1.763157)^3 - (1.763157)^2 - 1.763157 - 1 \approx 0.045 \]
   \[ f'(1.763157) = 3(1.763157)^2 - 2(1.763157) - 1 \approx 4.088 \]
   \[ x_3 = 1.763157 - \frac{0.045}{4.088} \]
   \[ x_3 \approx 1.752 \]

4. **Fourth Iteration:**
   \[ x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} \]
   \[ f(1.752) = (1.752)^3 - (1.752)^2 - 1.752 - 1 \approx 0.002 \]
   \[ f'(1.752) = 3(1.752)^2 - 2(1.752) - 1 \approx 4.008 \]
   \[ x_4 = 1.752 - \frac{0.002}{4.008} \]
   \[ x_4 \approx 1.7514 \]

5. **Fifth Iteration:**
   \[ x_5 = x_4 - \frac{f(x_4)}{f'(x_4)} \]
   \[ f(1.7514) = (1.7514)^3 - (1.7514)^2 - 1.7514 - 1 \approx 0 \]
   \[ f'(1.7514) = 3(1.7514)^2 - 2(1.7514) - 1 \approx 4.004 \]
   \[ x_5 = 1.7514 - \frac{0}{4.004} \]
   \[ x_5 \approx 1.7514 \]

After five iterations, the root of the equation \( x^3 - x^2 - x - 1 = 0 \) is approximately \( x \approx 1.7514 \).

Would you like further details or have any questions?

### Further Questions:
1. How does the Newton-Raphson method compare to other root-finding methods like the bisection method?
2. Can you explain the convergence criteria for the Newton-Raphson method?
3. What happens if the initial guess is far from the actual root?
4. Can the Newton-Raphson method fail? If so, under what conditions?
5. How can we improve the accuracy of the Newton-Raphson method?
6. How is the derivative calculated if the function is not given in a simple algebraic form?
7. Can the Newton-Raphson method be used for complex roots? If so, how?
8. How does the choice of the initial guess affect the efficiency of the Newton-Raphson method?

### Tip:
Always check the derivative \( f'(x) \) at your initial guess to ensure it is not zero, as this would lead to division by zero and make the method inapplicable at that point.

Learn how to apply the Newton-Raphson method to find the root of the equation x^3 - x^2 - x - 1 = 0 within the interval 1.6 < x < 2.1. This problem involves iterative numerical techniques and covers the Newton-Raphson iteration formula. Suitable for undergraduate level study.

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