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Thread: Closed Form of a Recursion with a Sqared Term

  1. #1
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    Closed Form of a Recursion with a Sqared Term

    Hello, I have this recursion: $\displaystyle x_n = rx_{n-1}(1- x_{n-1})$. 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: $\displaystyle x_n = rx_{n-1}(1- x_{n-1})$. 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 $\displaystyle r=x_0=1$ then you have the sequence $\displaystyle 1,0,0,0,0,\cdots$
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    MHF Contributor chisigma's Avatar
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    The sequence is the solution of the difference equation...

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

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

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

    $\displaystyle x_{+} = 0$ (2)

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



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

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

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

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

    Kind regards

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

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

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

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

    $\displaystyle x_{+} = 0$ (2)

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



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

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

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

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

    Kind regards

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