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Math Help - Contractive Sequence

  1. #1
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    Contractive Sequence

    Let x1 be greater than zero. x(n+1)=1/(2+xn) for all n greater than or equal to 2. Show that (xn) is a contractive sequence and find its limit.

    Stuck on the proof. Pretty sure the limit would just be the roots of (x^2)+2x-1=0. Any help appreciated.
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  2. #2
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    I just need a little help. I have that:
    abs(x(n+1)-x(n))=abs(1/(2+xn)-1(2+x(n-1))).

    I just don't know how to simplify this in terms of abs(xn-x(n-1)) and find what C in the contractive sequence is.

    Could anyone just steer me in the right direction?
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  3. #3
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    We need to show,
    |x_{n+1}-x_n|\leq k|x_n-x_{n-1}|
    For some k\in (0,1)
    Now,
    Let a=x_{n-1} then we have,
    x_{n+1}=\frac{1}{2+x_n}=\frac{1}{2+\frac{1}{2+a}}=  \frac{a+2}{2a+5}
    x_n=\frac{1}{a+2}

    Thus,
    \left| \frac{a+2}{2a+5}-\frac{1}{a+2} \right| \leq k \left| \frac{1}{a+2}-a\right|
    If und only if,
    \left| \frac{a^2+2a-1}{(2a+5)(a+2)}\right| \leq k \left|\frac{-a^2-2a+1}{2a+5} \right|
    If und only if,
    0<\frac{1}{a+2}\leq k<1
    Thus, by choosing this konstant to be that value we can form a contractive sequence.
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