Convergence of Bisection method 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?
Edit
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?
$$\lim_{k \to \infty}\frac{|\tau - x_k|}{|\tau - x_{k-1}|} = \frac{(\frac{1}{2})^{k-1}|b-a|}{(\frac{1}{2})^{k-2}|b-a|}$$
$$=\frac{(\frac{1}{2})^{k-1}}{(\frac{1}{2})^{k-2}}$$
$$=\frac{1}{2}$$
Show this shows linear convergence with $\frac{1}{2}$ being the rate of convergence. Is this correct?

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)
