ដោះស្រាយ​បញ្ហា​គណិតវិទ្យា​ណាមួយ

ជំហាន​ដោយ​ជំហាន

1. f(x) = x^{3} - x - 2,\quad x_0 = 2

   Newton's method: x_{n+1} = x_n − f(x_n)/f′(x_n) — follow the tangent line down to the axis, repeat.

2. f'(x) = 3 x^{2} - 1

   Differentiate once; the same derivative serves every iteration.

3. x_{1} = 2.0 - \frac{4.0}{11.0} = 1.63636364

   Iteration 1.

4. x_{2} = 1.63636364 - \frac{0.74530423}{7.03305817} = 1.53039205

   Iteration 2.

5. x_{3} = 1.53039205 - \frac{0.05393892}{6.02629948} = 1.52144147

   Iteration 3.

6. x_{4} = 1.52144147 - \frac{0.0003671}{5.94435215} = 1.52137971

   Iteration 4.

7. x_{5} = 1.52137971 - \frac{2.0 \cdot 10^{-8}}{5.94378853} = 1.52137971

   Iteration 5.

8. x_{6} = 1.52137971 - \frac{0.0}{5.94378853} = 1.52137971

   Iteration 6.

9. \text{converged}

   The iterates have stopped changing to 12 decimal places.

10. x \approx 1.52137971

    Quadratic convergence: the number of correct digits roughly doubles each step.