# Limit of sequence $a_{0} \in (0,1)$, $a_{n+1} = \mu a_{n}(1-a_{n})$, $\mu \in (1,3)$

I'm considering the sequence $$a_{0} \in (0,1)$$, $$a_{n+1} = \mu a_{n}(1-a_{n})$$, where $$\mu \in (1,3)$$ is a constant. I found that the fixed points of the function $$f(x) = \mu x (1-x)$$ are $$0$$ and $$1-\frac{1}{\mu}$$ and by drawing a lot of examples, I've concluded that the limit is generally $$1-\frac{1}{\mu}$$. However, I haven't been able to prove this.

I know that IF a limit exists, then by letting $$n \to \infty$$ in $$a_{n+1} = \mu a_{n}(1-a_{n})$$, it must be $$0$$ or $$1-\frac{1}{\mu}$$. I've tried to prove that in some cases the sequence is increasing and in others, decreasing, but the situation gets much too messy to be useful.

Is there a nice way to show that, for whichever choice of $$a_{0}$$, this sequence converges?

• – lhf
May 5, 2019 at 19:32
• $f(x)=\mu x(1-x) \Rightarrow f'(x)=\mu (1-2x)$ and $\left|f'\left(1-\frac{1}{\mu}\right)\right|=|2-\mu|<1$. This makes $x=1-\frac{1}{\mu}$ an attractive fixed point. Finding attraction interval is not so easy though. May 5, 2019 at 19:41
• @lhf is there a proof of the claim: With r between 1 and 2, the population will quickly approach the value r − 1/r, independent of the initial population. With r between 2 and 3, the population will also eventually approach the same value r − 1/r, but first will fluctuate around that value for some time. The rate of convergence is linear, except for r = 3, when it is dramatically slow, less than linear (see Bifurcation memory). May 5, 2019 at 21:19
• You can also try to look at the problem from the conjugate functions perspective. For example $\varphi(x)=-\mu x+\frac{\mu}{2}$ and $g(x)=x^2+\frac{\mu}{2}\left(1-\frac{\mu}{2}\right)$. It is easy to see that $f(x)=\varphi^{-1}\circ g \circ\varphi$ and $a_n=f^{\circ n}(a_0)=(\varphi^{-1}\circ g^{\circ n} \circ\varphi)(a_0)$. May 5, 2019 at 21:23
• If I were to rephrase the question as: find the set of non-wandering points of this dynamical system, would it be easier to look at $g$ instead of $f$? Where, of course, $f(x)=\mu x (1-x)$. May 5, 2019 at 21:33

Let $$x_n = a_n -(1-\frac{1}{\mu})$$. Then the recurence for $$x_n$$ is $$x_{n+1} = -\mu x_n^2 + x_n(2-\mu)$$ $$\frac{|x_{n+1}|}{|x_n|} = |2 - \mu - \mu x_n| \le |2-\mu| + \mu|x_n|$$ For $$\mu\in(1,3)$$, $$|2-\mu|<1$$. So you only need to prove that at some point $$|x_n| < \frac{1-|2-\mu|}{\mu}$$, so that $$|2-\mu| + \mu|x_n| < 1$$, and from that point on $$|x_n|$$ will be monotonically decreasing (approximately geometrically), and will tend to $$0$$, that is $$a_n$$ will tend to $$1-\frac{1}{\mu}$$.
• $|x_{n}|<\frac{1-|2-\mu|}{\mu}$ is a pretty strong condition for $\mu$ close to $3$ or close to $1$. Do you know of any references on the logistic map for $\mu \in (1,3)$? May 5, 2019 at 22:29
• I know no references. But it doesn't surprise me that $\mu\approx 1$ and $\mu \approx 3$ are problematic. First is $0\approx 1-\frac{1}{\mu}$ so the two stable points get close together; the other is a point of bifurcation. May 6, 2019 at 10:08