Bisection Method Calculator – Bracketing Roots
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
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
Enter f(x) and bracket [a, b]
Endpoints must straddle the root: f(a)·f(b) < 0
- 2
Test the midpoint
Sign of f((a+b)/2) picks the surviving half
- 3
Repeat
Each iteration halves the interval width
- 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]?
Why must f(a) and f(b) differ in sign?
What are the first midpoints for √2?
How does bisection compare with Newton?
When is the answer guaranteed within tolerance?
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