NETMaths
The bookUnit 3 · Numerical Analysis61 / 83

Root finding: bisection, Newton–Raphson, fixed point, order of convergence

Why this is asked: Order of convergence is the whole topic: bisection 1, secant φ ≈ 1.618, Newton 2 (but only 1 at a multiple root), fixed point 1 unless g′(r) = 0.

The comparison table

Method Order Needs Guaranteed?
Bisection 1 (linear, factor ½) sign change ✓ always
Regula falsi 1 sign change
Fixed point xnx_{n}1=g(xn)_{1} = g(x_{n}) 1 if 0 < |g′(r)| < 1; 2 if g′(r) = 0 |g′| < 1 near r local
Secant (1+5)/21.618(1+\sqrt{5})/2 \approx 1.618 two starting points local
**Newtonxn** x_{n}1=xnf/f_{1} = x_{n} - f/f' 2 at a simple root, 1 at a multiple root f′ ≠ 0 local only
Modified Newton x − m f/f′ 2 at a root of known multiplicity m local

Newton's caveats

  • Quadratic convergence is local: a poor start can diverge (f = x1/3)x^{1/3}), cycle (f=x32x+2(f = x^{3} - 2x + 2 from x0=0)x_{0} = 0), or run away.
  • At a root of multiplicity m ≥ 2 the convergence degrades to linear with factor (m−1)/m.
  • Error recursion: en+1e_{n+1} (f/2f)en2\approx (f''/2f') e_{n}^{2}.

Fixed point iteration

Converges locally iff |g′(r)| < 1; the error satisfies en+1e_{n+1} g(r)en\approx g'(r)e_{n}. Banach's contraction principle gives a global statement on an interval where |g′| ≤ k < 1 and g maps the interval into itself.

Bisection count

To reach tolerance ε\varepsilon from an interval of length L:nlog2(L/ε)L: n \ge log_{2}(L/\varepsilon) steps. Bisection never fails but never accelerates.

Key takeaways

  • Newton is quadratic only at simple roots, and only locally.
  • Secant beats bisection in order but loses the guarantee.
  • Fixed point: the derivative at the root is the convergence factor.

See it move

Newton's method: racing — and cyclinginteractive

Slide the starting point: quadratic convergence one moment, an eternal 2-cycle the next. Local convergence, felt.

Unlock interactive visuals

The trap here

“Newton's method converges quadratically to any root” — false

f(x)=x2f(x) = x^{2} at the root 0

The root is double: xnx_{n}1=xn/2_{1} = x_{n}/2, only linear convergence.

More on this →

Next: Interpolation and numerical integration with error terms

Create a free account to keep your place and have this feed your study plan.

Open this in the full syllabus view · Unit 3