1. ## Bisection method inequality

Today I had a very important exam about many topics and as I didn't study enough, I felt like playing chess against an opponent who was playing without a King and with 17 queens in his favor (I'm more than sure I'll get a 2/10).
I had a painful headache and couldn't even do this exercise :
Show that in the bisection method, if $\displaystyle c_j=\frac{a_j+b_j}{2}$ then $\displaystyle |c_n-c_{n+1}|=\frac{b_0-a_0}{2^{n+2}}$.
I've tried many times to solve it, but I just can't. I prefer not to show my work, but I can say I put $\displaystyle c_n-c_{n+1}=\frac{c_n}{2}$ but now that I've turned around the answer during more than 1 hour, I'm not even sure of it. I doubt about all what I know, as did Descartes one day.
I'd be happy and grateful if you could help me to solve this problem...

2. Originally Posted by arbolis
Today I had a very important exam about many topics and as I didn't study enough, I felt like playing chess against an opponent who was playing without a King and with 17 queens in his favor (I'm more than sure I'll get a 2/10).
I had a painful headache and couldn't even do this exercise :
Show that in the bisection method, if $\displaystyle c_j=\frac{a_j+b_j}{2}$ then $\displaystyle |c_n-c_{n+1}|=\frac{b_0-a_0}{2^{n+2}}$.
I've tried many times to solve it, but I just can't. I prefer not to show my work, but I can say I put $\displaystyle c_n-c_{n+1}=\frac{c_n}{2}$ but now that I've turned around the answer during more than 1 hour, I'm not even sure of it. I doubt about all what I know, as did Descartes one day.
I'd be happy and grateful if you could help me to solve this problem...
As at each itteration $\displaystyle c_{n+1}$ is the mid point of the interval on the previous step and $\displaystyle c_{n}$ is one end of that interval so $\displaystyle |c_n-c_{n+1}|$ is the length of the interval after n+2 subdivisions of the original interval. But the length of the interval halves at each subdivision so:

$\displaystyle |c_n-c_{n+1}|=\frac{b_0-a_0}{2^{n+2}}$

The only difficulty here is convincing yourself that we have n+2 for the number of subdivisions and hence the exponent.

RonL

3. Thanks a lot. In fact I thought the problem in terms of lengths. I didn't pay attention to
and is one end of that interval
. You helped me a lot.
Now I will have to pass a Numerical Analysis final exam this July (if I succeed in an exam the 20th of June), which has a written part and a computer (Fortran 90) one.
So I'll have to study this exercise and many more interesting stuff.