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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.

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

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  • $\begingroup$ 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
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$c_n$ 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

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

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