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Math Help - Convergence problem

  1. #1
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    Convergence problem

    Hello

    Let f be a function [0 ,1] ->[0,1] and

    |f(x)-f(z)| <= |x-z| x,z in [0,1].

    I need to proove that the sequence

    x_n+1 = 1/2 * [x_n + f(x_n)], n in N

    converges to a fixed point of f.

    Thanks

    /Mark
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  2. #2
    MHF Contributor
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    Is it sufficient to prove that |x_{n+1} - x_{n}| \le |x_{n} - x_{n-1}|? What do you think? The successive absolute differences are non-increasing.

    Or will we also have to prove that it's bounded? (Which seems pretty obvious.)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by TKHunny View Post
    Is it sufficient to prove that |x_{n+1} - x_{n}| \le |x_{n} - x_{n-1}|? What do you think? The successive absolute differences are non-increasing.

    Or will we also have to prove that it's bounded? (Which seems pretty obvious.)
    Consider the sequence of partial sums of the harmonic series:

    <br />
S_n=\sum_{r=1}^n \frac{1}{r}<br />

    This sequences satisfies:

    |S_{n+1} - S_{n}| < |S_{n} - S_{n-1}|,

    but does not converge.

    RonL
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  4. #4
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    I believe TKHunny was referring to something called a "contractive sequence" which is |x_{n+1}-x_n|\leq k|x_n-x_{n-1}| where k\in (0,1). (Note TKHunny did not fullfil this condtion). The theorem says that contractive sequences converges. See Here.

    This is Mine 68th Post!!!
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  5. #5
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    I wasn't quite seeing it. I was hoping I could get one of you'all to talk me into it.

    It was rather weak to rely on "non-increasing". Since the sequence would quikly leave the Domain if it had to stick with equality, it would be stronger to state "decreasing", but that remains only as convincing as the Harmonic example.

    Well, Mark84, what are your thoughts? What is it about that slope that ensures convergence?

    \frac{|f(x) - f(z)|}{|x - z|} < 1
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