Newton's Method Examples

Numerical Analysis · Axiom Academy

EXAMPLE Newton's Method: Finding √3 Step-by-step walkthrough with multiple choice questions Use Newton's method to find √3 (the positive root of f(x) = x² - 3). Start with x₀ = 2 and perform 3 iterations. For f(x) = x² - 3, what is the derivative f'(x)? Correct! The derivative of x² - 3 is f'(x) = 2x . Step 2: First Iteration (n = 0) With x₀ = 2, calculate x₁ using the simplified formula x₁ = (x₀ + 3/x₀)/2 x₁ = (2 + 3/2)/2 = (2 + 1.5)/2 = 3.5/2 = 1.75 Error: |1.75 - √3| ≈ |1.75 - 1.732| ≈ 0.018 Step 3: Second Iteration (n = 1) With x₁ = 1.75, calculate x₂ = (x₁ + 3/x₁)/2 x₂ = (1.75 + 3/1.75)/2 = (1.75 + 1.7143)/2 ≈ 1.732143 Error: |1.732143 - √3| ≈ 0.000092 — we've gained several digits of accuracy! Step 4: Third Iteration (n = 2) With x₂ ≈ 1.732143, calculate x₃. How many correct decimal places do we have now? Correct! After just 3 iterations: We have about 8 correct decimal places — that's quadratic convergence! Quadratic convergence: Correct digits roughly doubled each iteration (0 → 1 → 3 → 8) Fast convergence: Only 3 iterations for 8-digit accuracy Simple formula: For √a, Newton's method simplifies to xₙ₊₁ = (xₙ + a/xₙ)/2 This is called the Babylonian method — one of the oldest algorithms, known for over 4,000 years!

This is the written version of the interactive lesson above. See the full Numerical Analysis course.