Newton-Raphson Method

Overview

• Objective: Numerical methods for solving equations
• Equation: x^3 + 2x - 2 = 0
• Root: Known to exist between 0 and 1

Previous Methods Covered

• Bisection Method
• New Interpolation

Newton-Raphson Method

• Characteristics: Powerful and quick, but may not always converge

• Formula (without proof):

  

  • Next approximation: x_(n+1) = x_n - F(x_n)/F'(x_n)

Application

1. Define the function f(x):
   • f(x) = x^3 + 2x - 2
   • Ensure equation equals 0 (already satisfied)
2. Find the derivative f'(x):
   • f'(x) = 3x^2 + 2

Example

Initial Approximation

• Start with x_1 = 1
• Calculate x_2:
  • x_2 = x_1 - f(x_1)/f'(x_1)
  • f(1) = 1^3 + 2*1 - 2 = 1
  • f'(1) = 3*1^2 + 2 = 5
  • x_2 = 1 - 1/5 = 0.8

Iterate for More Accuracy

• Calculate x_3:
  • x_3 = x_2 - f(x_2)/f'(x_2)
  • f(0.8) = 0.8^3 + 2*0.8 - 2 ≈ 0.592
  • f'(0.8) = 3*0.8^2 + 2 ≈ 3.92
  • x_3 = 0.8 - 0.592/3.92 ≈ 0.7714

Further Iteration

• Use calculator for quick repetition:
  • Input initial value x_1 (e.g., 1)
  • Use ANS (answer) functionality to iterate quickly
  • Example results:
    • x_4 ≈ 0.7709
    • x_5 ≈ 0.7709
  • Converges to x = 0.77 (2 decimal places)

Alternative Initial Value

• Try x_1 = 0.5:
  • Results:
    • x_2 ≈ 0.818
    • x_3 ≈ 0.7722
    • x_4 ≈ 0.7709
    • x_5 ≈ 0.7709
    • Also converges to x = 0.77 (2 decimal places)

Conclusion

• Newton-Raphson method is efficient but sometimes doesn't converge
• Always verify results
• Practice with different initial values to see variance