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Thread: Prove the limit of this sequence exists

  1. #1
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    Prove the limit of this sequence exists

    Hi, I'm having a little bit trouble applying the Monotone Sequence Theorem to this question:

    Determine if the sequence defined by the recursive relation:

    $\displaystyle x_{k+1} = \sqrt{2+3x_k}$
    $\displaystyle x_1=1$

    has a limit. Present your reasoning by using the Monotone Sequence Theorem.

    Thanks a lot!
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  2. #2
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    Use induction to show that $\displaystyle x_{n}\le x_{n+1}\le 4$.
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  3. #3
    MHF Contributor chisigma's Avatar
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    The difference equation that defines the sequence can be written as...

    $\displaystyle \displaystyle \Delta_{n} = x_{n+1}-x_{n} = \sqrt{2+3\ x_{n}}-x_{n} = f(x_{n})$ (1)

    The function $\displaystyle f(*)$ is represented here...



    There is only one 'attractive fixed point' in $\displaystyle x_{0}=3.5615528128 \dots$ and, because for any $\displaystyle x\ge -\frac{3}{2}$ is $\displaystyle |f(x)| < |x_{0}-x|$ , any $\displaystyle x_{1}\ge -\frac{3}{2}$ will produce a sequence that converges monotonically at $\displaystyle x=x_{0}$...

    Kind regards

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



    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Oct 14th 2010 at 11:12 PM. Reason: The post has been modified...
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