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Math Help - limit

  1. #1
    Senior Member Dinkydoe's Avatar
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    limit

    Hoi guys,

    suppose we have p_{k+1}\geq p_k^9+9p_k^8(1-p_k)

    I'd like to show that p_k\to 1 as k\to\infty if p_0>0.968...

    Now, this seems quite obvious maybe...as p_{k+1}>p_k if p_0>0.968
    But how can we see p_k\to 1?
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  2. #2
    Grand Panjandrum
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    Re: limit

    Quote Originally Posted by Dinkydoe View Post
    Hoi guys,

    suppose we have p_{k+1}\geq p_k^9+9p_k^8(1-p_k)

    I'd like to show that p_k\to 1 as k\to\infty if p_0>0.968...

    Now, this seems quite obvious maybe...as p_{k+1}>p_k if p_0>0.968
    But how can we see p_k\to 1?
    Suppose:

    p_{k+1}=100p_k^9+1000

    It clearly satisfies the inequality but p_k \to \infty as k \to \infty

    So what am I missing?

    (I may be completely misunderstanding the question as I am having to read raw LaTeX as this machine is not rendering any MHF LaTeX correctly)

    CB
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Re: limit

    sorry, I need to add p_0 \in (0.968, 1]

    I wonder how we can see p_k\to 1...as p_k clearly is growing with such a starting value. But, why can it not converge
    to some specific value a in (0.968,1)

    edit: furthermore, is it even obvious that  p_k can not exceed the value 1??
    Last edited by Dinkydoe; June 29th 2011 at 08:26 AM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: limit

    Quote Originally Posted by Dinkydoe View Post
    Hoi guys,

    suppose we have p_{k+1}\geq p_k^9+9p_k^8(1-p_k)

    I'd like to show that p_k\to 1 as k\to\infty if p_0>0.968...

    Now, this seems quite obvious maybe...as p_{k+1}>p_k if p_0>0.968
    But how can we see p_k\to 1?
    Let's suppose to have the sigh = instead of sigh \ge so that the 'recursive relation' is...

    p_{k+1}= 9\ p_{k}^{8} - 8\ p_{k}^{9} (1)

    The (1) can be written as...

    \Delta_{k}= p_{k+1}-p_{k}= -p_{k} + 9\ p_{k}^{8} - 8\ p_{k}^{9}= f(p_{k}) (2)

    The f(x) is represented here...



    There are three zeroes in x_{0}=0, x_{1}= .967689763902... and x_{2}=1. x_{0} and x_{2} are both 'attractive fixed points' and x_{1} is a 'repulsive fixed point'. Any p_{0}< x_{1} will produce a sequence converging at x_{0} and any x_{1}< p_{0} \le x_{2} will produce a sequence converging at x_{2}. The case p_{0} > x_{2} is left to You...

    Kind regards

    \chi \sigma
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  5. #5
    Senior Member Dinkydoe's Avatar
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    Re: limit

    Thanks, I have another question. Dunno If I should make a new thread for this...this was actually part of an article I was reading. (and didn't quite understand all the details).

    Suppose we have p_{k+1}=f(p_k)=p_k^9+9p_k^8(1-p_k) then we have p_k\to 1 as k\to\infty if p_0\in (0.968,1]

    Then the following observation is made: \sum_k(1-p_k)<\infty . Now this very unclear to me...Can
    we see from the recurrence relation p_k=f(p_k) that p_k goes to 1 fast enough?
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