Let $c_n={1\over 2}(a_n+b_n)$, $r=\lim_{n\to \infty}c_n$, and $e_n=r-c_n$. Here $[a_n,b_n]$, with $n\geq 0$, denotes the successive intervals that arise in the bisection method when it is applied to a continuous function $f$. Show that $|c_n-c_{n+1}|=2^{-n-2}(b_0-a_0)$.

I'm having a tough time showing this. Any hints or solutions are greatly appreciated.


closed as off-topic by heropup, Daniel Robert-Nicoud, user147263, Daniel W. Farlow, Shailesh Apr 22 '16 at 0:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Daniel Robert-Nicoud, Community, Daniel W. Farlow, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ All the letters might be confusing you. Try drawing a picture to see what this means. Then, to make it rigorous, consider a proof by induction. $\endgroup$ – Reveillark Oct 2 '15 at 1:58

$c_n$ is the mid point of $[a_n,b_n]$.

$c_{n+1}$ is the mid point of either $[a_n, c_n]$ or $[c_n,b_n]$.

Compute $|c_n-c_{n+1}|$ in terms of $a_n,b_n$.

$|c_n-c_{n+1}| = {1 \over 4} (b_n-a_n)$.

You know that $b_{n+1}-a_{n+1} = {1 \over 2} (b_n-a_n)$, hence

$b_n-a_n = {1 \over 2^n} (b_0-a_0)$.


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