The Bisection Method — Mathematics Knowledge System

Overview

The bisection method is a simple, guaranteed-to-converge algorithm for finding a root of a continuous function f on an interval [a, b] where f(a) and f(b) have opposite signs. By the Intermediate Value Theorem, a root must lie somewhere in that bracket. The method repeatedly evaluates f at the midpoint, keeps whichever half still brackets a sign change, and discards the other half — halving the interval's width at every step. It converges linearly (each iteration gains roughly one bit of accuracy) and is slower than Newton's method, but unlike Newton's method it cannot diverge: as long as the initial bracket has a sign change, convergence is certain.

Intuition

Think of searching for a light switch in a long dark hallway where you know it's somewhere between two points, one where the room is dark and one where it's lit. Bisection checks the halfway point: if it's dark there, the switch is in the far half; if it's lit, the switch is in the near half. Either way, you've cut the search region in half with a single check, and repeating this quickly narrows the hallway down to a spot as small as you like — the same guaranteed halving that makes bisection reliable even when you know nothing else about the function beyond its sign at the endpoints.

Formal Definition

Definition

Given f continuous on [a, b] with f(a)·f(b) < 0, the method repeatedly bisects the bracket:

c = \dfrac{a+b}{2}

Midpoint of the current bracket

\text{if } f(a)f(c) < 0,\ \text{set } b = c;\ \text{else set } a = c

Keep the half-interval with the sign change

|c_n - r| \le \dfrac{b_0 - a_0}{2^{\,n}}

Error bound after n iterations

Worked Examples

Check the bracket: f(1) = 1−2 = −1 < 0, f(2) = 4−2 = 2 > 0, so a root lies in [1,2].
$f(1) = -1, \quad f(2) = 2$
Iteration 1: midpoint c = 1.5, f(1.5) = 2.25 − 2 = 0.25 > 0, same sign as f(2), so replace b: new bracket [1, 1.5].
$c_1 = 1.5, \quad f(1.5) = 0.25$
Iteration 2: midpoint c = 1.25, f(1.25) = 1.5625 − 2 = −0.4375 < 0, same sign as f(1), so replace a: new bracket [1.25, 1.5].
$c_2 = 1.25, \quad f(1.25) = -0.4375$

Answer: After 2 iterations the bracket narrows to [1.25, 1.5], containing √2 ≈ 1.41421.

Practice Problems

Difficulty 3/10

For f(x) = x³ − x − 2 on bracket [1, 2], verify the bracket is valid and compute the first midpoint's function value.

Difficulty 5/10

Starting bracket width is 8 units. How many bisection steps are needed before the bracket width first drops below 0.05?

Difficulty 3/10

Which property of the bisection method makes it guaranteed to converge, unlike Newton's method?

It uses derivativesIt always keeps a bracket containing a sign changeIt requires an integer starting pointIt only works for polynomials

Quiz

Summary

Bisection repeatedly halves a bracket [a,b] where f(a) and f(b) have opposite signs, keeping the half with the sign change.
The error bound after n iterations is (b−a)/2ⁿ, giving guaranteed but linear (slow) convergence.
Unlike Newton's method, bisection cannot diverge once a valid sign-changing bracket is established.

References

WebsiteWikipedia — Bisection method