TITLE: !Dynamical Systems! Show the existence of a non-fixed periodic point of $q_\mu$ of period 3 for $\mu>4$ and $q_\mu=\mu x(1-x)$.
QUESTION [5 upvotes]: 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 $q_\mu=\mu x(1-x)$.

REPLY [4 votes]: 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 $\frac{\mu^2-3\mu-1}{\mu^2(\mu-3)}$.