Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know how to prove the bound on the error after $k$ steps of the Bisection method.

I.e. $$|\tau - x_{k}| \leq \left(\frac{1}{2}\right)^{k-1}|b-a|$$

where $a$ and $b$ are the starting points.

But does this imply something about the order of convergence of the Bisection method? I know that it converges with order at least 1, is that implied in the error bound?


I've had a go at showing it, is what I am doing here correct when I want to demonstrate the order of convergence of the Bisection method?

$$\lim_{k \to \infty}\frac{|\tau - x_k|}{|\tau - x_{k-1}|} = \frac{(\frac{1}{2})^{k-1}|b-a|}{(\frac{1}{2})^{k-2}|b-a|}$$



Show this shows linear convergence with $\frac{1}{2}$ being the rate of convergence. Is this correct?

share|cite|improve this question
up vote 1 down vote accepted

For the bisection you simply have that $\epsilon_{i+1}/\epsilon_i = 1/2$, so, by definition the order of convergence is 1 (linearly).

share|cite|improve this answer

the $\frac12$ you get is called 'asymptotic error constant $\lambda$'. for any method, it's in form $\frac{|p_{n+1}-p|}{(|p_n-p|)^\alpha}=\lambda$. $\lambda$ is called asymptotic error constant, and $\alpha$ is the order of convergence. 1: linearly, 2:quadratically. and usually it converges faster as $\alpha$ gets bigger; and $\lambda$ also effects the speed of convergence but not extend to the order.

source: Numerical Analysis 9th edition, by Richard L. Burden & J.Douglas Fairs.

ISBN-13: 978-0-538-73351-9 (page 79 definition 2.7)

share|cite|improve this answer




if x=p then g(p)=p




this implies that

g'(p) does not equal to zero and hence order is one,

therefore bisection method is linearly convergent.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.