# !Dynamical Systems! Show the existence of a non-fixed periodic point of $q_\mu$ of period 3 for $\mu>4$ and $q_\mu=\mu x(1-x)$.

I'm taking a course in dynamical systems and this question popped up in the course-litterature. In the litterature they show a similar example where we do the same thing exept for us instead of showing the existence of a non-fixed periodic point of period 3, we show the existence of a non-fixed periodic point of period 2. What they do in this example is the following: $$q_\mu([1/\mu,1/2])\supset[1-1/\mu,1]$$ $$q_\mu([1-1/\mu,1])\supset[0,1-1/\mu]\supset[1/\mu,1/2]$$ And so at this point we can clearly see that $$q_\mu^2([1/\mu,1/2])\supset[1/\mu,1/2]$$ and then we can see by the intermediate value theorem $$q_\mu^2$$ has a fixed point $$p_2\in[1/\mu,1/2]$$. Thus $$p_2$$ and $$q_\mu(p_2)$$ are non-fixed periodic points of period 2.

What i'm not getting here is how am I supposed to come up with an interval to start with? And if i'm not supposed to come up with an interval to start with, to show the existence of a non-fixed periodic point. Then how am I supposed to show the existence of such a point?

Repost of the question:

Show the existence of a non-fixed periodic point of $$q_\mu$$ of period 3 for $$\mu>4$$ and $$q_\mu=\mu x(1-x)$$.

By letting $$p_{\mu}(x)=\mu x(1-x)$$ we have that $$q_{\mu}(x) = \frac{p_{\mu}(p_{\mu}(p_{\mu}(x))))-x}{x}$$ is a seventh-degree polynomial fulfilling $$q_{\mu}(0)=\mu^3-1$$, $$q_{\mu}(1/\mu) = \mu-2$$ and $$q_\mu(1/(2\mu))=\frac{1}{128} \left(-129-16 \mu +8 \mu ^2+64 \mu ^3-16 \mu ^4\right)<0,$$ hence there is a root of $$q_\mu$$ between $$\frac{1}{2\mu}$$ and $$\frac{1}{\mu}$$. It is not difficult to show that this root cannot be a root of $$\frac{p_\mu(x)-x}{x}$$ or $$\frac{p_\mu(p_\mu(x))-x}{x}$$, hence it is a periodic point of period $$3$$.
With a little refinement, Newton's method locates the first positive 3-periodic point around $$\frac{\mu^2-3\mu-1}{\mu^2(\mu-3)}$$.
For convenience, fix $$\mu>4$$ and drop the subscript $$\mu$$.
Define $$\Lambda=\{x\in[0,1]:f^n(x)\in[0,1]\ \forall n\}$$. If $$x\notin\Lambda$$, then $$f^n(x)\to-\infty$$. The dynamics of $$f$$ are concentrated on $$\Lambda$$. It can be shown that $$\Lambda$$ is a Cantor set. The system $$f\colon\Lambda\to\Lambda$$ is equivalent to $$\sigma\colon\Sigma\to\Sigma$$, where $$\Sigma$$ is the set of sequences of $$0$$ 'sand $$1$$'s and $$\sigma$$ is the shift operator: $$\sigma(a_1a_2a_3\dots)=a_2a_3\dots$$ Finally, it is clear that $$100100100100\dots$$ is a periodic point of $$\sigma$$ of period $$3$$. By equivalence, $$f$$ has a periodic point of period $$3$$ in $$\Lambda$$.