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Math Help - Closed Form of a Recursion with a Sqared Term

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    Closed Form of a Recursion with a Sqared Term

    Hello, I have this recursion: x_n =  rx_{n-1}(1- x_{n-1}). I don't really know how to solve this kind of recursion... I tried to do it with generating functions but I couldn't effectively deal with the squared term. Could somebody give me a tip on how to get started on finding a closed form (if you want you can start with r=1. and I'll go from there)? Or at least point me in the direction of some useful website? Thanks!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by MollyMillions View Post
    Hello, I have this recursion: x_n =  rx_{n-1}(1- x_{n-1}). I don't really know how to solve this kind of recursion... I tried to do it with generating functions but I couldn't effectively deal with the squared term. Could somebody give me a tip on how to get started on finding a closed form (if you want you can start with r=1. and I'll go from there)? Or at least point me in the direction of some useful website? Thanks!
    I was under the impression that most quadratic recurrence relations do not have closed form solutions....I could be wrong though.

    Note that if r=x_0=1 then you have the sequence 1,0,0,0,0,\cdots
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    MHF Contributor chisigma's Avatar
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    The sequence is the solution of the difference equation...

    \Delta_{n} = x_{n+1} - x_{n} = (r-1)\cdot x_{n} - r\cdot x_{n}^{2} = f(x_{n}) (1)

    ... with some sort of 'initial value' x_{0}. We don't have information regarding r and, only for demonstrative purpose, we will suppose that 0< r < 1. The 'fixed point' of the sequence are the solution of the equation f(x)=0 that are...

    x_{-} = \frac{r-1}{r}

    x_{+} = 0 (2)

    With the hypotheses we have made x_{-} is a repulsive fixed point [i.e. a point from which the x_{n} tend to diverge...] and x_{+} an attractive fixed point [i.e. a point to which the x_{n} tend to converge...]. The figure is the case r=.2 so that is x_{-}=-4...



    From the figure we can derive the asintotic behavior of the x_{n} as function of x_{0}. In particular...

    a) if x_{0} < -4 or x_{0}>5 the sequence will diverge to -\infty...

    b) if x_{0} = -4 or x_{0} = 5 the sequence will converge to -4

    c) if -4 < x_{0} < 5 the sequence will converge to 0 without oscillations...

    Kind regards

    \chi \sigma
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    Quote Originally Posted by chisigma View Post
    The sequence is the solution of the difference equation...

    \Delta_{n} = x_{n+1} - x_{n} = (r-1)\cdot x_{n} - r\cdot x_{n}^{2} = f(x_{n}) (1)

    ... with some sort of 'initial value' x_{0}. We don't have information regarding r and, only for demonstrative purpose, we will suppose that 0< r < 1. The 'fixed point' of the sequence are the solution of the equation f(x)=0 that are...

    x_{-} = \frac{r-1}{r}

    x_{+} = 0 (2)

    With the hypotheses we have made x_{-} is a repulsive fixed point [i.e. a point from which the x_{n} tend to diverge...] and x_{+} an attractive fixed point [i.e. a point to which the x_{n} tend to converge...]. The figure is the case r=.2 so that is x_{-}=-4...



    From the figure we can derive the asintotic behavior of the x_{n} as function of x_{0}. In particular...

    a) if x_{0} < -4 or x_{0}>5 the sequence will diverge to -\infty...

    b) if x_{0} = -4 or x_{0} = 5 the sequence will converge to -4

    c) if -4 < x_{0} < 5 the sequence will converge to 0 without oscillations...

    Kind regards

    \chi \sigma
    Is Binet's formula a solution?
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