# Thread: Closed Form of a Recursion with a Sqared Term

1. ## 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!

2. Originally Posted by MollyMillions
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 $1,0,0,0,0,\cdots$

3. 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

$\chi$ $\sigma$

4. Originally Posted by chisigma
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?