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Math Help - Limit 1

  1. #1
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    Limit 1

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

    Find \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...

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

    ... that are...

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

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

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

    Kind regards

    \chi \sigma
    Last edited by chisigma; January 21st 2010 at 01:23 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
    x_{2n} and x_{2n-1}
    One is increasing, second decreasing, and both oscilate about 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...

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

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

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

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

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

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

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

    Kind regards

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

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

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

    Update: Hear is another relation
    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 n-1 and n, and proff for n+1 and n+2).
    Last edited by ns1954; January 30th 2010 at 02:23 PM.
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