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Math Help - Quadratic recurrence equation

  1. #1
    Junior Member Greg98's Avatar
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    Quadratic recurrence equation

    Hello,
    the problem is following quadratic recurrence equation:
    u_0=1
    u_1=2
    u_n=u_{n-2}^2u_{n-1}, \ n \geq 2

    I know how to solve most of the linear homogeneous and non-homogeneous equations, but I haven't encountered quadratic equations before. I tried googling and using forums search function, but I couldn't found any information, how to solve these. So, any help is appreciated. Thanks!

    WolframAlpha gives following solution, but that doesn't help me either.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting v_{n} = \log_{2} u_{n} the 'recursive relation' becomes...

    \displaystyle v_{n}= v_{n-1} + 2\ v_{n-2}\\ , \\ v_{0}=0 \\, v_{1}= 1 (1)

    ... which is linear...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chisigma's Avatar
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    Very interesting is the case of the 'recursive relation'...

    \displaystyle u_{n}= u_{n-1}\ u_{n-2}\\,\\ u_{0}=1\\,\\ u_{1}=2 (1)

    Setting v_{n}= \log_{2} u_{n} we obtain the solution...

    \displaystyle u_{n}= 2^{f_{n}} (2)

    ... where the f_{n} are the terms of the Fibonacci's sequence...

    Kind regards

    \chi \sigma
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