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1) How can we prove that the logistic sequence

$$x_{n+1}=rx_n(1-x_n),\ x_1=a\in (0,1)$$

converges to $\frac{r-1}{r}$, for $r\in [1,3]$?

2) Also I wonder how can we prove that the sequence $(x_n)_{n\in\mathbb{N}^*}$ has two limit points (the fixed points $\dfrac{r^2+r+\sqrt{(r-3)(r+1)}}{2r^2}$ and $\dfrac{r^2+r-\sqrt{(r-3)(r+1)}}{2r^2}$ of $f\circ f$ where $f(x)=rx(1-x)$) for $r\in (3,1+\sqrt{6})$.

I saw in some articles of Feigenbaum proofs for the fact that these fixed points are attractors, but nothing about convergence. I post this question on math.stackexchange too but without an answer. I saw that it was posed also by other users in the past, still without answer till today. These are the links:

https://math.stackexchange.com/questions/3655733/convergence-of-logistic-map-with-1-mu3

https://math.stackexchange.com/questions/2239522/what-should-mu-be-here-in-order-for-the-logistic-map-to-be-stable

https://math.stackexchange.com/questions/4051605/logistic-map-convergence-r-in-1-3?noredirect=1#comment8368505_4051605

I have a proof for $r\in [0,1]$ that the sequence converges to $0$. Also I made a lot of simulations for $r\in (1,3]$ and I observe that the sequences $(x_{2n})$ and $(x_{2n+1})$ become monotonic from some point on (but I cannot prove it...). This happens even for $r\in (3,1+\sqrt{6})$. It looks obvious when you see the terms of the sequence but seems very hard to prove it.

3) Is there a general method to find all the limit points (i.e. partial limits) for the logistic sequence?

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  • $\begingroup$ Where I can find a proof that it is indeed convergence on those points with $g'(x)<0$ no matter what is our choice of the starting value $a$? So we can take $a$ some distance apart from $x$. $\endgroup$
    – Bogdan
    Commented Mar 7, 2021 at 5:44
  • $\begingroup$ I will see the video. I understand what you say but it's not clear for me... $\endgroup$
    – Bogdan
    Commented Mar 7, 2021 at 5:58
  • $\begingroup$ Practically I want some technical details about the proof of convergence (not the existence of 2,3-cycles). I studied a lot the problem and I know it's interesting story. But I cannot figure out even in these simple cases how to prove the convergence to the same values for any starting point $a\in (0,1)$. $\endgroup$
    – Bogdan
    Commented Mar 7, 2021 at 6:06
  • $\begingroup$ In the earliest of the three stackexchange posts, the user gives the formula for the fixed point, and the reason for its stability for $1<r<3$. A stable fixed point implies convergence to that fixed point, so it seems to me that what you want (for your first question) is already there. $\endgroup$ Commented Mar 7, 2021 at 6:30
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    $\begingroup$ Imply convergence if the initial data is close to the fixed point (or if the sequence come close that point at some $n$). But in general I don't see why stable fixed point imply convergence. Do you know a proof for that? $\endgroup$
    – Bogdan
    Commented Mar 7, 2021 at 6:32

1 Answer 1

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A boundary point of the immediate basin of attraction of an attracting fixed point for a continuous function is either a non-attracting periodic point of period $2$, or a non-attracting fixed point, or a point mapped to a non-attracting fixed point. If $r \in [1,3]$ there are no real points of period $2$, the only other fixed point is $0$, and the only other point mapped to $0$ is $1$. Thus all $a \in (0,1)$ are attracted to the attracting fixed point $(r-1)/r$.

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  • $\begingroup$ Do you have a reference for this type of result? Thanks a lot! $\endgroup$
    – Bogdan
    Commented Mar 7, 2021 at 6:45

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