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Thread: limit

  1. #1
    Senior Member Dinkydoe's Avatar
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    limit

    Hoi guys,

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

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

    Now, this seems quite obvious maybe...as $\displaystyle p_{k+1}>p_k$ if $\displaystyle p_0>0.968$
    But how can we see $\displaystyle p_k\to 1$?
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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 $\displaystyle p_{k+1}\geq p_k^9+9p_k^8(1-p_k)$

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

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

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

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

    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 $\displaystyle p_0 \in (0.968, 1] $

    I wonder how we can see $\displaystyle p_k\to 1$...as $\displaystyle p_k$ 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 $\displaystyle p_k$ can not exceed the value 1??
    Last edited by Dinkydoe; Jun 29th 2011 at 07:26 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 $\displaystyle p_{k+1}\geq p_k^9+9p_k^8(1-p_k)$

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

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

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

    The (1) can be written as...

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

    The $\displaystyle f(x)$ is represented here...



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

    Kind regards

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

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