3

For the bisection method, we have the following error bound:

$$ |x_n - x| \leq \frac{1}{2^{n+1}} |a_0 - b_0|. $$

This estimate is commonly interpreted in textbooks as indicating that the bisection method is first-order convergent with a convergence rate of $\frac{1}{2}$.

Is it possible to rigorously prove the errors satisfy the definition of the rate and order of convergence given by:

$$ \overline{\lim\limits}_{n \rightarrow \infty}\frac{|x_{n+1} - x|}{|x_n - x|} = \frac{1}{2}? $$

Or is this just a vague (formal) statement? I understand heuristically, one would expect an estimate of that type if the sequence of approximations generated by the bisection method are nice in some sense.

If the root is a dyadic number such estimates are not required, since the root is achieved after a finite number of steps. The interesting case is when the root is a non-dyadic number and the method never achieves the root.

I am looking for a rigorous proof for the order and the rate of convergence of the bisection method(or a reference with rigorous proof)

Or rigorous counter example, where the $\overline{\lim}_{n \rightarrow \infty}\frac{|x_{n+1} - x|}{|x_n - x|} \neq 1/2$.

If not 1/2 can one prove at least the limsup is finite?

A reference where existence or non existence of this limsup is discussed would also be useful for me.

Xander Henderson
  • 32,453
Veronica
  • 552

1 Answers1

3

No, you can not apply this Q-convergence rate formula, in many cases the sequence of quotients has no limit. It is only the interval length, which is an upper error bound, that converges that way to zero.

Specifically, if $x_n$ is the midpoint of the current interval $[a_n,b_n]$, then you know that for any root $x_*$ in this interval you have $$ |x_*-x_n|\le \frac12|b_n-a_n|=\frac14|b_{n-1}-a_{n-1}|=\frac1{2^{k+1}}|b_{n-k}-a_{n-k}|=\frac1{2^{n+1}}|b_0-a_0|. $$

Lutz Lehmann
  • 131,652
  • That was a typo. I corrected the index. The question still remains. Can we now rigorously prove that the sequence of approximations has a rate of ( \frac{1}{2} ) and an order of 1? If so, how? I don't see the proof. Thanks in advance – Veronica Jul 20 '24 at 09:47
  • 1
    No, this fraction usually has no limit. It is only the interval length, which is an upper error bound, that converges that way to zero. – Lutz Lehmann Jul 20 '24 at 09:55
  • Is there an explicit example to show that this limit does not exist in general – Veronica Jul 20 '24 at 10:00
  • 3
    Take any non-dyadic number as root. For instance $f(x)=x-\frac37$ with start interval $[a_0,b_0]=[0,1]$. The sequence of ratios will cycle through a periodic pattern around $\frac12$. – Lutz Lehmann Jul 20 '24 at 10:13
  • Lehman If the limit does not exist, can we find the lim sup? – Veronica Jul 21 '24 at 12:57
  • 1
    Sure you can. However it will not give correct information on the convergence behavior, at least not in the case of such counter-examples. – Lutz Lehmann Jul 21 '24 at 13:13
  • @ Lutz Lehmann How? We have an upper bound on the error but not a lower bound (atleast I don't see it) Could you please elaborate why the ratio is uniformly bounded. I have elaborated the question now. U can have a look at it. – Veronica Jul 22 '24 at 19:23
  • 1
    You can, in rare cases, reach the exact root in a finite number of steps. Thus a lower bound can be problematic. One has the design choice to treat divergence as limit at infinity. – Lutz Lehmann Jul 22 '24 at 22:06