Interval bisection Created 2026-10-07 Updated 2026-10-14
Interval bisection solves a target equation by repeatedly halving a closed interval known to contain a solution. It is the bisection method applied to , so the target becomes a zero of a function. This reduction from an equation to root finding is described in Demanet's notes.[ref]
https://ocw.mit.edu/courses/18-330-introduction-to-numerical-analysis-spring-2012/5b325bfa56a599794c7196de926844b0_MIT18_330S12_Chapter4.pdf. Laurent Demanet, MIT 18.330, Chapter 4, §4.1, pp. 1–2, 2012
Here the emphasis is on a continuous strictly increasing function: the target identifies one input, making interval bisection a way to evaluate its inverse function without an explicit inverse formula.Let be a continuous function, with and . The intermediate value theorem guarantees existence; a strictly increasing function also guarantees uniqueness. Check endpoint equality first. Otherwise, for the current root bracket , evaluate its midpoint . If , retain ; if , retain ; if , return that exact solution. These comparisons preserve the target bracket. General bisection method needs continuity and opposite signs of at the endpoints, without requiring monotonicity or a unique zero of a function; see limitations of bisection.
An original dimensionless polynomial example isHere and the derivative , so there is exactly one solution. In fact ; knowing this lets us check the computed absolute error independently of the guarantee. The first four halvings give the following exact rational numbers:
The beauty of interval bisection is that the enclosure itself supplies the certificate: one comparison removes half the possible inputs, regardless of the steepness of the function graph. In exact arithmetic, after completed halvings without exact termination, the width and the bisection error bound for the new midpoint areThe reported midpoint after update four is , rather than the last tested point . Its certified absolute error is at most , and its actual absolute error is . For a requested position tolerance , the bisection stopping criterion requires eleven halvings: ten give , while eleven giveThe actual absolute error is . The new midpoint need not be evaluated to obtain this certificate.
The signed output residual is ; the position approximation error is . They answer different questions: how close is the output to the target, and how close is the input to the solution? Driscoll and Braun explain why a small residual needs information about the derivative to imply small position error.[ref]
https://fncbook.com/rootproblem/. Toby A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, §4.1.1, Conditioning, error, and residual
For this particular polynomial, direct factorization givesThus at the residual is , despite position absolute error . If a differentiable function satisfies throughout the bracket, the mean value theorem gives the rigorous conversionOur example permits and on . Strict increase alone supplies no positive lower derivative bound: is strictly increasing but . A small residual alone is therefore not a universal position certificate; even the target equation has residual at , a full unit from its solution.The guarantees require continuity and correct evaluations, not just narrowing endpoints. For the strictly increasing function when and when , the target lies between and but is never attained: a jump discontinuity defeats the existence argument. With floating-point arithmetic or an approximate evaluation, an incorrect comparison can discard the true solution. A certified interval enclosure of wholly below or above certifies the corresponding decision; an enclosure containing is inconclusive, rather than proof of equality. If rounding makes the midpoint equal an endpoint, report the precision limit; see bisection stopping criterion for the rounded position bound. Both figures are exact mathematical examples, not measured data.

