Bisection method inequality

**arbolis** · June 3rd 2008, 12:04 PM

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 $c_j=\frac{a_j+b_j}{2}$ then $|c_n-c_{n+1}|=\frac{b_0-a_0}{2^{n+2}}$ .
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 $c_n-c_{n+1}=\frac{c_n}{2}$ 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...

**CaptainBlack** · June 3rd 2008, 10:51 PM

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 $c_j=\frac{a_j+b_j}{2}$ then $|c_n-c_{n+1}|=\frac{b_0-a_0}{2^{n+2}}$ .
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 $c_n-c_{n+1}=\frac{c_n}{2}$ 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 $c_{n+1}$ is the mid point of the interval on the previous step and $c_{n}$ is one end of that interval so $|c_n-c_{n+1}|$ 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:

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

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

RonL

**arbolis** · June 4th 2008, 06:53 AM

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.

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