Bisection Iteration & Error Bound Calculator
Enter the starting interval and tolerance to see how many bisection midpoint evaluations are required. The calculator also shows whether a planned budget is enough and what decimal-place guarantee that budget supports.
Minimum midpoint evaluations10
- Initial interval width
- 1
- Error bound after planned evaluations
- 0.0009765625
- Resulting bracket width
- 0.0009765625
- Additional evaluations required
- 0
- Maximum allowable starting width
- 1.024
- Evaluations for requested decimal places
- 11
Assumes a continuous function with at least one root bracketed by the two endpoints.
This planner does not evaluate the function or check that endpoint function values have opposite signs.
How to use this calculator
- Enter the two endpoints of the starting interval.
- Enter the target absolute error for the midpoint approximation.
- Enter the number of midpoint evaluations you plan to run.
- Enter the decimal places you want guaranteed by absolute error.
How the bisection bound is calculated
The bisection method starts with an interval [a, b] that brackets a root, evaluates the midpoint, and keeps the half interval that still brackets the root. After each completed midpoint evaluation, the bracket width is cut in half.
This calculator uses the standard planning bound for the nth evaluated midpoint:
E_n = |p_n - p| <= W / 2^n
- W is the initial width,
|b - a|. - n is the number of midpoint evaluations.
- p_n is the nth midpoint and p is the true root.
To find the smallest number of evaluations for a target absolute error epsilon, solve W / 2^n <= epsilon. That gives n = ceil(log2(W / epsilon)), with a minimum of one because at least one midpoint has to be produced. If the target error is already as large as the starting width, the calculator still reports one evaluation.
The planned-budget rows use the same bound in the forward direction. For m planned evaluations, both the error bound and the remaining bracket width are W / 2^m. The maximum starting width that can meet the target within that budget is epsilon * 2^m. If your actual width is larger than that value, the budget is short by the reported number of additional evaluations.
Decimal places
For d correct decimal places, this calculator uses the usual absolute-error test epsilon_d = 0.5 * 10^-d. For example, three decimal places require an absolute error no larger than 0.0005. The tool then applies the same ceil(log2(W / epsilon_d)) rule.
What changes the answer
The initial interval width and the tolerance move the result most. Doubling the interval width usually adds one evaluation. Cutting the tolerance in half also usually adds one evaluation. Tight decimal-place requests grow quickly because each decimal place asks for a ten-times-smaller absolute error, which is about 3.32 more bisection evaluations.
What this calculator leaves out
This is an iteration planner, not a root finder. It does not evaluate your function, choose the correct half interval, check continuity, or verify that the endpoint function values have opposite signs. The guarantee applies only when the bisection hypotheses are true: a continuous function on the interval and at least one root bracketed by the endpoints.
Worked example
Suppose the starting interval is [1, 2], so W = 1. With a target absolute error of 0.001, the required evaluations are ceil(log2(1 / 0.001)) = ceil(log2(1000)) = 10.
If the planned budget is 10 midpoint evaluations, the error bound is 1 / 2^10 = 0.0009765625, and the resulting bracket width is the same. No additional evaluations are required. The largest starting width that could meet the 0.001 target in 10 evaluations is 0.001 * 2^10 = 1.024. For three guaranteed decimal places, the tolerance is 0.0005, so the decimal-place requirement is ceil(log2(1 / 0.0005)) = 11 evaluations.
Common questions
How many iterations does the bisection method need?
Use n = ceil(log2(W / epsilon)), where W is the initial interval width and epsilon is the target absolute error. This calculator reports at least one midpoint evaluation because the method must produce a midpoint before it has an approximation.
What is the bisection method error bound?
With the indexing used here, after n midpoint evaluations the midpoint error is bounded by W / 2^n, where W is the initial interval width. Some textbooks index the first interval before the first midpoint, which can shift the formula by one.
Why does each bisection add about one binary digit of accuracy?
Each successful bisection halves the interval that can contain the root. Halving is one bit of information, so the absolute error bound falls by a factor of two per midpoint evaluation.
How many bisections guarantee a number of decimal places?
For d decimal places, require the absolute error to be no more than 0.5 * 10^-d. The calculator converts that decimal-place tolerance into the same bisection formula, ceil(log2(W / epsilon_d)).
Does the interval width equal the root error?
The final bracket width is a bound on where the root can lie inside the bracket. The midpoint error is no larger than the bound used by this calculator, but the actual error can be smaller depending on where the root sits.
What if the endpoints do not bracket a root?
Then the bisection guarantee does not apply. You need a continuous function with endpoint values of opposite sign, or another reason to know a root is bracketed, before these iteration counts are meaningful.