Bisection Method Examples

Numerical Analysis · Axiom Academy

EXAMPLE Bisection Method: Finding a Root Step-by-step walkthrough of the bisection algorithm Find a root of f(x) = x³ - x - 2 in the interval [1, 2] using bisection. Perform 5 iterations and track the error at each step. First, we need to verify that [1, 2] brackets a root. Calculate f(1) and f(2). f(1) = 1³ - 1 - 2 = -2 (negative) f(2) = 2³ - 2 - 2 = 4 (positive) Since f(1) 0, by the Intermediate Value Theorem, there must be a root in [1, 2]! Calculate the midpoint c = (1 + 2)/2 = 1.5. Now calculate f(1.5). Which half of the interval contains the root? f(1.5) = 1.5³ - 1.5 - 2 = 3.375 - 1.5 - 2 = -0.125 Since f(1.5) 0, the root is in [1.5, 2] New interval: [1.5, 2], width = 0.5 New interval: [1.5, 2]. Calculate midpoint c = 1.75 and evaluate f(1.75). f(1.75) = 1.75³ - 1.75 - 2 = 5.359 - 1.75 - 2 ≈ 1.609 Since f(1.5) 0, the root is in [1.5, 1.75] New interval width = 0.25 (halved again!) Step 4: Complete the Iterations After 5 total iterations, approximately what will the interval width be? Initial width = 1. After n iterations, width = 1/2ⁿ After 5 iterations: 1/2⁵ = 0.03125 Visualization: Interval Shrinking Linear convergence: Error is halved each iteration Guaranteed: Bisection always converges (unlike Newton) Predictable: After n iterations, error ≤ (b-a)/2ⁿ For 10 digits: Need about 34 iterations (since 2³⁴ > 10¹⁰) The root is approximately r ≈ 1.5214

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