Bisection Method Calculator
Find a root of f(x)=0 by bisection on a bracketing interval with convergence reporting.
About Bisection Method Calculator
The bisection method finds a root by repeatedly halving a bracketing interval [a, b] where f(a) and f(b) have opposite signs. By the Intermediate Value Theorem, a root must exist in the interval. Each iteration halves the interval, guaranteeing convergence at a linear rate of about 1 bit per iteration.
Formulas
Bisection step
mid = (a + b) / 2; if f(a)·f(mid) < 0 then b = mid, else a = mid
- a
- Left bound
- b
- Right bound
- mid
- Midpoint
Error bound
error ≤ (b - a) / 2^n after n iterations
- n
- Iteration count
Worked example
With the default values below, Bisection Method Calculator returns Root of 1.521379706799. Change any input above and every figure here is recalculated in your browser.
Inputs used
- f(x)
- x^3 - x - 2
- Left bound (a)
- 1
- Right bound (b)
- 2
- Tolerance
- 0
- Max iterationsaccepts 1–200
- 100
Results
- Root
- 1.521379706799
- Iterations
- 33
- |f(root)|
- 3.2780 × 10⁻¹¹
- Status
- Converged
Step by step
- 1Values used
- 2Bisection step
- 3Error bound
- 4Root= 1.521379706799
- 5Iterations= 33
- 6|f(root)|= 0.0000
- 7Status= Converged
Assumptions and limits
- The result is derived only from the 5 inputs listed above. Anything not among them is not part of the calculation.
- In particular, this calculator does not take inflation, tax, fees or charges or compounding frequency into account.
- Figures are rounded for display to 12 decimal places, so totals built from several rounded parts can differ slightly from the unrounded arithmetic.
How these calculators are built and tested is described on the Methodology page. If a figure looks wrong, please report it.
Frequently Asked Questions
Why do f(a) and f(b) need opposite signs?
The IVT guarantees a root only when the function crosses zero between a and b. If both have the same sign, the root may not exist in that interval.
How does bisection compare to Newton's method?
Bisection always converges but is slow (linear). Newton's method converges quadratically but can fail if the derivative is zero or the guess is bad.