Iteration and numerical methods

Some equations can’t be solved exactly, so we get closer and closer to the answer with a rule (an iteration). A change of sign shows a root lies between two values; Newton–Raphson uses tangents to close in fast.

Key facts

  • If f(a) and f(b) have different signs (and f is continuous), there’s a root between a and b.
  • Iteration: xₙ₊₁ = g(xₙ). Keep feeding the answer back in.
  • Newton–Raphson: xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ).

Worked examples

Example 1

Show x³ − 3x − 1 = 0 has a root between 1 and 2.

  1. f(1) = 1 − 3 − 1 = −3
  2. f(2) = 8 − 6 − 1 = 1
  3. The sign changes, so a root lies between 1 and 2.

Answer: Sign change from −3 to 1

Example 2

Use xₙ₊₁ = ∛(3xₙ + 1) with x₀ = 2 to find the root to 2 d.p.

  1. x₁ = ∛7 = 1.9129
  2. x₂ = 1.8888, x₃ = 1.8821, x₄ = 1.8801, x₅ = 1.8796
  3. Settles at about 1.88.

Answer: x ≈ 1.88