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Bisection Method Calculator – Bracketing Roots

Quick answer

Halve the bracket each step: x²−2 on [1,2] walks 1.5 → 1.25 → 1.375 →…→ √2, gaining one binary digit per iteration.

  • Guaranteed.

Figures are estimates from this calculator’s standard formula — adjust the inputs in the tool below for your exact number.

How the Bisection Method Calculator Works

Keep the half where f changes sign: if f(m)·f(b) < 0, root ∈ [m, b]

Bisection exploits the intermediate value theorem: if a continuous function changes sign over [a, b], a root lies inside, and testing the midpoint tells which half keeps it. For f(x) = x² − 2 on [1, 2] (f(1) = −1, f(2) = 2): midpoint 1.5 gives f = 0.25 > 0, so the root sits in [1, 1.5]; next midpoint 1.25 has f = −0.4375, moving the bracket to [1.25, 1.5]; then 1.375, and so on toward √2 ≈ 1.41421. Each step halves the bracket — 10 iterations shrink uncertainty below 1/1024 ≈ 0.00098, and the required count is exactly ⌈log₂((b−a)/tolerance)⌉. The calculator tabulates every midpoint, its sign, and the surviving interval, and rejects brackets whose endpoints do not straddle zero (f(a)·f(b) > 0). Slow but unstoppable: unlike Newton, bisection cannot diverge once a valid bracket is given.

  1. 1

    Enter f(x) and bracket [a, b]

    Endpoints must straddle the root: f(a)·f(b) < 0

  2. 2

    Test the midpoint

    Sign of f((a+b)/2) picks the surviving half

  3. 3

    Repeat

    Each iteration halves the interval width

  4. 4

    Stop at tolerance

    Error ≤ (b−a)/2ⁿ⁺¹ after n steps

Use Cases

Robust Root Finding

Guaranteed convergence when Newton might diverge

Teaching IVT

Watch sign changes bracket √2 to 5 decimals

Engineering Specs

Locate thresholds inside certified bounds

Tips

  • 1

    One binary digit of accuracy per iteration — 10 steps ≈ 3 decimal digits

  • 2

    Invalid bracket (no sign change) = no guarantee

  • 3

    Slow but certain: convergence never fails on continuous f

Common Mistakes

  • Passing a bracket without a sign change and expecting convergence

  • Keeping the wrong half by comparing f(m) with 0 sloppily

  • Forgetting the error bound (b−a)/2ⁿ⁺¹ when choosing iterations

  • Mixing bisection guarantees with Newton’s speed claims

FAQs

How many iterations for 3-decimal accuracy on [1,2]?

Width must fall below 0.001: 2⁻ⁿ < 0.001 gives n ≥ 10 iterations.

Why must f(a) and f(b) differ in sign?

The intermediate value theorem only guarantees a root inside a sign-changing bracket.

What are the first midpoints for √2?

1.5 → 1.25 → 1.375 → 1.4375 → 1.40625, bracketing down on 1.41421.

How does bisection compare with Newton?

Bisection gains one binary digit per step and cannot fail; Newton doubles digits but may diverge from bad starts.

When is the answer guaranteed within tolerance?

After n steps the root lies within (b−a)/2ⁿ⁺¹ of the reported midpoint.