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Numerical root finding – change of sign, bisection, iteration and Newton–Raphson

MathematicsCalculusAges 16–17

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Choose a function f and see a table of values and signs on an interval, zoom in step by step to trap a root, and meet the cases where the change-of-sign method fails (a vertical asymptote, a repeated root, two roots in one step). Step through interval bisection, fixed-point iteration xₙ₊₁ = g(xₙ) with staircase or cobweb diagrams and the condition |g′(α)| < 1, and the Newton–Raphson method with its tangent lines, Heron's method for √a and its failure cases; the iteration table shows how fast each method converges.

Lesson: Solving equations numerically: change of sign, bisection, fixed-point iteration and the Newton–Raphson method

What it shows

If f is continuous on [a, b] and f(a), f(b) have opposite signs, the intermediate value theorem guarantees a root between a and b. Decimal search and bisection shrink that interval; bisection always works but adds only about 0.3 correct digits per step. Rearranging f(x) = 0 as x = g(x) gives the iteration xₙ₊₁ = g(xₙ), which converges near a root α when |g′(α)| < 1. Newton–Raphson uses the tangent: xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ), and near a simple root the number of correct digits roughly doubles each step.

How to use

On Change of sign, choose f(x), set a, b and the Step h, then press Zoom into sign change or tap a green band. On Bisection, drag a and b and press Next step, Run or To the end. On Iteration, pick a rearrangement x = g(x) and set x₀. On Newton–Raphson, drag on the graph to move x₀ and watch the tangents and the table.

Parameters you can change

  • Screen Change of sign, Bisection, Iteration, Newton–Raphson
  • Function f(x) x³ − x − 1, x³ − 3x + 1 (three roots), x³ − 2x + 2, x² − a (Heron's method for √a), cos x − x, eˣ − 3x (two roots), x² − 4.6x + 5.28 (two close roots), (x − 1)² (repeated root), 1/(x − 1) (vertical asymptote), ∛x
  • Number a in f(x) = x² − a 1–50
  • Left end a of the interval -10–10
  • Right end b of the interval -10–10
  • Step h of the table of values (Change of sign screen) 0.0001–1
  • Rearrangement x = g(x) (Iteration screen) x = ∛(x + 1), x = √(1 + 1/x), x = 1/x + 1/x², x = x³ − 1, x = cos x, x = (x² + 1)/3
  • Starting value x₀ (Iteration screen) -5–5
  • Starting value x₀ (Newton–Raphson screen) -10–10

Questions to explore

  1. Why does 1/(x − 1) change sign between 0.7 and 1.2 although the equation 1/(x − 1) = 0 has no root?
  2. Which rearrangements of x³ − x − 1 = 0 converge, and how does the value of g′(α) explain it?
  3. Why does Newton–Raphson for (x − 1)² converge much more slowly than for x³ − x − 1?

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