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Thread: 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 $\displaystyle \{x_n\}$ is determined by the equality $\displaystyle x_n = \frac{x_{n - 1} + x_{n - 2}}{2}$ and the values $\displaystyle x_0$ and $\displaystyle x_1$, prove that:

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

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

    ... so that is...

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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
    Senior Member
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    Rewrite

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

    Adding
    $\displaystyle x_{n-1}+2x_{n}=2x_{1}+x_{0}$

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

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

    $\displaystyle 3x=2x_{1}+x_{0}$

    $\displaystyle
    x=\frac{2x_{1}+x_{0}}{3}
    $
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