Iteration and numerical methods · Year 13

In Year 13 you learn: numerical methods.

Many equations, such as x³ + x − 5 = 0, cannot be solved exactly with algebra. Numerical methods find the solution approximately, as accurately as you need: first trap the root between two values, then close in on it with an iteration formula or the Newton–Raphson method.

Key facts

  • Change of sign: if f(a) and f(b) have opposite signs (and f is continuous), there is a root between a and b.
  • Iteration: rearrange f(x) = 0 into x = g(x), then use xₙ₊₁ = g(xₙ) again and again.
  • An iteration converges near a root α when |g′(α)| < 1.
  • Newton–Raphson: xₙ₊₁ = xₙ − (f(xₙ))/(f′(xₙ)).
  • To show a root is 1.52 to 2 decimal places, show that f changes sign between 1.515 and 1.525.
  • Keep full calculator accuracy (use the ANS key) and round only at the end.

Worked examples

Example 1

Use Newton–Raphson on f(x) = x³ + x − 5 from x₀ = 1.5.

Answer: x₁ = 1.5161, x₂ = 1.5160 (root ≈ 1.5160)

Other years