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Math Help - Bisection method inequality

  1. #1
    MHF Contributor arbolis's Avatar
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    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 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...
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by arbolis View Post
    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
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  3. #3
    MHF Contributor arbolis's Avatar
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    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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