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Math Help - Recursive sequence (Fibonacci Sequence) & Limits

  1. #1
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    Recursive sequence (Fibonacci Sequence) & Limits

    Hey, I would like some help with this question:

    A recursive sequence is a sequence where the nth term, t_{n}, is defined in terms of preceding terms,  t_{n-1}, t_{n-2}, etc

    one of the most famous recursive sequences is the Fibonacci sequence, created by Leonardo Pisano (1170-1250). The terms of this sequence are define as follows  f_{1} = 1, f_{2} = 1, f_{n} = f_{n-1} + f_{n-2} where n \geq 3

    It asks me to graph some things, did those, but then it asks me to write an expression, using a limit, to represent the value of the ratios of consecutive terms of the Fibonacci sequence. How would I go about doing this?

    Thanks in advance.
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  2. #2
    MHF Contributor
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    Hi

    f_{n} = f_{n-1} + f_{n-2}

    \frac{f_{n}}{f_{n-1}} = 1 + \frac{f_{n-2}}{f_{n-1}}

    \frac{f_{n}}{f_{n-1}} = 1 + \frac{1}{\frac{f_{n-1}}{f_{n-2}}} = 1 + \frac{1}{1 + \frac{1}{\frac{f_{n-2}}{f_{n-3}}}}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Setting \rho_{n} = \frac{f_{n}}{f_{n-1}} the sequence of the \rho_{n} is generated recursively by the relation...

    \rho_{n}= 1 + \frac{1}{\rho_{n-1}} (1)

    ... that is equivalent to the difference equation...

    \Delta_{n} = \rho_{n} - \rho_{n-1}= 1 - \rho_{n-1} + \frac{1}{\rho_{n-1}} (2)

    The (2) indicates that the limit when n tends to infinity of the \rho_{n} is given by the positive solution of the equation...

    \rho^{2} - \rho - 1 = 0 (3)

    ... so that is...

    \lim_{n \rightarrow \infty} \rho_{n} = \frac{1 + \sqrt {5}}{2} = \varphi = 1.61803398874989\dots (4)

    The constant \varphi is known as 'golden ratio'...

    Kind regards

    \chi \sigma
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