How many bisection steps on [1, 2] guarantee an error below 10⁻³?
- 2⁻ⁿ < 10⁻³ ⇒ n ≥ 10.
The Bisection Method
The bisection method: halving [a, b] on the sign change, stopping criteria and the error bound |pₙ − p| ≤ (b − a)/2ⁿ.
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- 1 Slide 1
- 2 Bisection Technique The first technique, based on the Intermediate Value Theorem, is called the Bisection, or Binary-search, method. Suppose /is a continuous function defined on the interval [a, b], with f(a) and f(b) of opposite sign. The Intermediate Value Theorem implies that a number p exists in (a, b) with f(p) = 0. Although the procedure will work when there is more than one root in the
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Slide 1
Notes
Bisection: pₙ = (aₙ + bₙ)/2; keep the half where f changes sign. Always converges (slowly, linearly).
Tip Error bound: |pₙ − p| ≤ (b − a)/2ⁿ. To reach tolerance ε, need n ≥ log₂((b − a)/ε).
Exam angle
Do one bisection step for f(x) = x³ + 4x² − 10 on [1, 2].
Compare with the model answer
f(1) = −5 < 0, f(2) = 14 > 0. p₁ = 1.5, f(1.5) = 2.375 > 0, so the root is in [1, 1.5].
Bisection Technique The first technique, based on the Intermediate Value Theorem, is called the Bisection, or Binary-search, method. Suppose /is a continuous function defined on the interval [a, b], with f(a) and f(b) of opposite sign. The Intermediate Value Theorem implies that a number p exists in (a, b) with f(p) = 0. Although the procedure will work when there is more than one root in the
![Slide 2: Bisection Technique The first technique, based on the Intermediate Value Theorem, is called the Bisection, or Binary-search, method. Suppose /is a continuous function defined on the interval [a, b], with f(a) and f(b) of opposite sign. The Intermediate Value Theorem implies that a number p exists in (a, b) with f(p) = 0. Although the procedure will work when there is more than one root in the](figures/slide-02.jpg)
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