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Thread: Limit 1

  1. #1
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    Limit 1

    Sequences $\displaystyle \{x_n\}_{n=1}^\infty$ is defined with
    $\displaystyle \displaystyle
    x_0=\frac{1}{3},\quad x_n=\frac{1}{2}x_{n-1}^2-1,\quad n=1,2,3,\dots$

    Find $\displaystyle \lim\limits_{n\to\infty}x_n$
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  2. #2
    MHF Contributor chisigma's Avatar
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    The possible finite limits of the sequence are the solutions of the equation...

    $\displaystyle x_{n}=x_{n-1} \rightarrow x^{2} -2\cdot x -2 =0$ (1)

    ... that are...

    $\displaystyle x= 1 \pm \sqrt{3}$ (2)

    A quickly inspection reveals that if $\displaystyle x_{0}= \frac{1}{3}$ is...

    $\displaystyle \lim_{n \rightarrow \infty} x_{n} = 1- \sqrt{3} = -.732050807568 \dots $ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Jan 21st 2010 at 12:23 PM.
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  3. #3
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    Yes, that is limit. But how to prove convergence of this, using therem of monoton and bounded sequences? It's not quite easy, it must be considered, I think, subsequences
    $\displaystyle x_{2n}$ and $\displaystyle x_{2n-1}$
    One is increasing, second decreasing, and both oscilate about $\displaystyle 1-\sqrt 3$. This may be a strict proof.
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  4. #4
    MHF Contributor chisigma's Avatar
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    It is not at all difficult to prove that...

    $\displaystyle x_{n+1} > x_{n}$ if $\displaystyle x_{n} > 1 + \sqrt {3}$ or $\displaystyle x_{n}< 1 - \sqrt{3}$

    $\displaystyle x_{n+1} = x_{n}$ if $\displaystyle x_{n} = 1 \pm \sqrt {3}$

    $\displaystyle x_{n+1} < x_{n}$ if $\displaystyle 1-\sqrt{3} < x_{n} < 1 + \sqrt {3}$ (1)

    On the basis of these unequalities we have the following distinct cases...

    a) if $\displaystyle x_{0} > 1 + \sqrt{3}$ the sequence diverges at $\displaystyle + \infty$

    b) if $\displaystyle x_{0} = 1 + \sqrt{3}$ is $\displaystyle x_{n}= 1 + \sqrt{3}$ , $\displaystyle \forall n $

    c) if $\displaystyle x_{0} < 1 + \sqrt{3}$ the sequence converges at $\displaystyle 1 - \sqrt{3}$ in any case with 'oscillatory behavior' unless is $\displaystyle x_{0} = 1-\sqrt{3}$ that produces the constant sequence $\displaystyle x_{n} = 1 - \sqrt{3}$ , $\displaystyle \forall n$ ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    $\displaystyle 1-\sqrt 3=-0.73205...$

    $\displaystyle x_1=-0.94444...$,
    $\displaystyle x_2=-0.55401...$,
    $\displaystyle x_3=-0.84653...$,
    $\displaystyle x_4=-0.64168...$,
    $\displaystyle x_5=-0.79411...$,
    $\displaystyle x_6=-0.68468...$,
    $\displaystyle x_7=-0.76560...$,
    $\displaystyle x_8=-0.70692...$,
    $\displaystyle x_9=-0.75012...$,
    $\displaystyle x_{10}=-0.71865$
    $\displaystyle \dots$

    and you can "see " that $\displaystyle x_{2n-1}>1-\sqrt 3$, and "is" monotonicaly increasing, and $\displaystyle x_{2n}<1-\sqrt 3$, and "is" monotonicaly decreasing.
    So, here is neccessary only strict proof of this monotonity. Maybe this relation
    $\displaystyle
    x_{n+1}-x_n=\frac 12\left[x_n-(1-\sqrt 3)\right]\left[x_n-(1+\sqrt 3)\right ]
    $
    Observe that $\displaystyle -1<x_n< 0$, (by inicial value) for all $\displaystyle n=1,2,3\dots$, so second parenteze is negative, and somehow here must be used induction.
    I belive similar can be done for
    $\displaystyle x_{2n+2}-x_{2n}$ and $\displaystyle x_{2n+1}-x_{2n-1}$

    Update: Hear is another relation
    $\displaystyle x_{n+2}-x_n=\frac {1}{4}\, (x_{n+1}-x_n)[\underbrace{(x_n+1)^2+1}_{> 0}]$, so this could be used maybe (I think) for variant of induction (with assumptation for $\displaystyle n-1$ and $\displaystyle n$, and proff for $\displaystyle n+1$ and $\displaystyle n+2$).
    Last edited by ns1954; Jan 30th 2010 at 01:23 PM.
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