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Math Help - Prove result for limits?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove result for limits?

    If a sequence of numbers \{x_n\} is determined by the equality x_n = \frac{x_{n - 1} + x_{n - 2}}{2} and the values x_0 and x_1, prove that:

    \lim_{n\rightarrow \infty} x_n = \frac{x_0 + 2x_1}{3}
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  2. #2
    MHF Contributor chisigma's Avatar
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    Since x_{n} is the mean between x_{n-2} and x_{n-1} is...

    x_{n} = x_{0} + (x_{1}-x_{0}) (1-\frac{1}{2} + \frac{1}{4} - ... + \frac{(-1)^{n}}{2^{n}} ) (1)

    ... so that is...

    \lim_{n \rightarrow \infty} x_{n} = x_{0} + \frac {x_{1}-x_{0}}{1+\frac{1}{2}} = x_{0} -\frac{2}{3}\cdot x_{0} + \frac{2}{3}\cdot x_{1} = \frac{1}{3}\cdot x_{0} + \frac{2}{3}\cdot x_{1} (2)

    Kind regards

    \chi \sigma
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  3. #3
    Super Member fardeen_gen's Avatar
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    Could you explain the second line please? I couldn't understand it.
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  4. #4
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    Rewrite

    Rewrite x_n=x_0+(x_1-x_0)a_n, for a_0=0 and a_1=1. By substituting into the recursive formula, you get a_n=\frac{a_{n-1}+a_{n-2}}2. Since we are taking iterative midpoints, the gap between successive values of a_n are decreasing by half, so [tex]|a_n-a_{n-1}|=\left(\frac12\right)^{n-1}[/MATh] . It should also be apparent that a_{n}>a_{n-1} when n is odd, and a_{n}<a_{n-1} when n is even, again, by virtue of iterating the midpoint formula, making the value of the gap alternate between positive and negative. Therefore, a_n=\sum_{i=1}^{n-1}\left(-\frac12\right)^i for n\geq1, the limit of which as n gets large is \frac23.
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  5. #5
    Senior Member pankaj's Avatar
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    There is no need for such heavy explanation.
    2x_{2}=x_{1}+x_{0}
    2x_{3}=x_{2}+x_{1}
    2x_{4}=x_{3}+x_{2}
    2x_{5}=x_{4}+x_{3}
    2x_{6}=x_{5}+x_{4}
    2x_{7}=x_{6}+x_{5}
    2x_{8}=x_{7}+x_{6}
    2x_{9}=x_{8}+x_{7}
    -----------------------------------
    2x_{n-2}=x_{n-3}+x_{n-4}
    2x_{n-1}=x_{n-2}+x_{n-3}
    2x_{n}=x_{n-1}+x_{n-2}

    Adding
    x_{n-1}+2x_{n}=2x_{1}+x_{0}

    \lim_{n\to \infty}(x_{n-1}+2x_{n})= 2x_{1}+x_{0}

    If \lim_{n\to \infty}x_{n}=x,then \lim_{n\to\infty}x_{n-1}=x

    3x=2x_{1}+x_{0}

     <br />
x=\frac{2x_{1}+x_{0}}{3}<br />
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