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Thread: Alternating Sequence

  1. #1
    Member Haven's Avatar
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    Alternating Sequence

    Fix $\displaystyle \alpha>1$, take $\displaystyle x_1 > \sqrt{\alpha}$ and define

    $\displaystyle x_{n+1} = \frac{\alpha+x_n}{1+x_n} = x_n + \frac{\alpha-x_n^2}{1+x_n}$

    I know this sequence should alternate about $\displaystyle \sqrt{\alpha}$. I.E., the odd terms are greater than $\displaystyle \sqrt{\alpha}$ and the even terms are less than $\displaystyle \sqrt{\alpha}$. However, I cannot seem to show this.

    I've tried to use the fact that $\displaystyle \alpha>1$ and $\displaystyle \sqrt{\alpha} < x_n$ but I can't seem to get anything to work.

    I would appreciate a hint very much.
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  2. #2
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    You may consider a sequence
    $\displaystyle
    y_n=x_n-\sqrt{\alpha}
    $

    and

    $\displaystyle
    y_n \rightarrow 0.
    $

    Now you may examine

    $\displaystyle
    y_{n+1}-y_n
    $

    how it behaves around zero.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Exclamation

    Quote Originally Posted by Haven View Post
    Fix $\displaystyle \alpha>1$, take $\displaystyle x_1 > \sqrt{\alpha}$ and define

    $\displaystyle x_{n+1} = \frac{\alpha+x_n}{1+x_n} = x_n + \frac{\alpha-x_n^2}{1+x_n}$

    I know this sequence should alternate about $\displaystyle \sqrt{\alpha}$. I.E., the odd terms are greater than $\displaystyle \sqrt{\alpha}$ and the even terms are less than $\displaystyle \sqrt{\alpha}$. However, I cannot seem to show this.

    I've tried to use the fact that $\displaystyle \alpha>1$ and $\displaystyle \sqrt{\alpha} < x_n$ but I can't seem to get anything to work.

    I would appreciate a hint very much.
    Problems of this type have been solved a lot of times, recently in...

    http://www.mathhelpforum.com/math-he...it-170087.html

    The recursive relation can be written as...

    $\displaystyle \displaystyle \Delta_{n}= x_{n+1}- x_{n} = \frac{\alpha-x_{n}^{2}}{1+x_{n}} = f(x_{n})$ (1)

    Supposing $\displaystyle \alpha>0$ there is an 'attractive fixed point' at $\displaystyle x_{0}=\sqrt{\alpha}$ and the condition for monotonic convergence at $\displaystyle x_{0}$ is that...

    $\displaystyle |f(x)| \le |x_{0}-x|$ (2)

    It is easy to verify that the condition (2) is satisfied for $\displaystyle 0<\alpha<1$ so that in this case the sequence will be increasing if $\displaystyle x_{1}< \sqrt{\alpha}$ and decreasing if $\displaystyle x_{1}> \sqrt{\alpha}$. For $\displaystyle \alpha>1$ the sequence remains convergent but is 'oscillating'...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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