Consider the logistic family $$f_\alpha(x) = \alpha x (1 - x), \; \alpha > 0.$$ I want to find the periodic points with period 4. So, I have to solve the following equation: $$Q_\alpha(x) : =f_\alpha^4(x) - x = 0 \hspace{15pt} (1)$$ $0$ and $p_\alpha = \frac{\alpha - 1}{\alpha}$ are the fixed points for $f_\alpha$. Hence $0$ and $p_\alpha$ are solutions for (1)and then $$Q_\alpha(x) = x(x - p_\alpha) R(x).$$ It remains to find the solutions of $$R(x) = 0.$$ To do this, is there a way simpler than calculating $R(x)$ and trying to solve $R(x) = 0$?
Thank you!
Your map is conjugate to a map of the form $z \mapsto z^2 + c$, (with $z = -\alpha x+\alpha/2$ and $c = \alpha/2-\alpha^2/4$), and I feel more comfortable with this because then we will get polynomials in $c$ rather than rational fractions in $\alpha$.
There are three $4$-cycles, whose elements are root of the degree $12$ polynomial $P(z,c) = ((((z^2+c)^2+c)^2+c)^2-z+c)/((z^2+c)^2-z+c)$.
If $c$ is finite, the $12$ roots are finite, and as $c \to \infty$, all the roots tend to $\infty$ and they are all a $O(\sqrt {|c|})$
Let $x_1$ be a root. Then you can form its iterates $x_2=x_1^2+c, x_3=x_2^2+c, x_4=x_3^2+c$, and the Galois group of $x_1$ over the the cycle is cyclic of order $4$.
To solve for $x_1$ knowing the cycle (which means assuming you have the value of any polynomial expression in the $x_i$ that's invariant by the $4$-cycle $x_1 \mapsto x_2 \mapsto x_3 \mapsto x_4 \mapsto x_1$), you can follow the Galois group structure as follows :
Assume you know $(x_1+x_2+x_3+x_4), (x_1+x_3)(x_2+x_4), x_1x_3+x_2x_4, x_1x_2x_3x_4$, and $(x_1x_3 - x_2x_4)((x_1+x_3)-(x_2+x_4))$.
Then you know $((x_1+x_3) - (x_2x_4))^2$, $((x_1+x_3)-(x_2+x_4))^2$, and $((x_1+x_3) - (x_2x_4))((x_1+x_3)-(x_2+x_4))$, which means you can find the values of $(x_1+x_3) - (x_2x_4)$ and $(x_1+x_3)-(x_2+x_4)$ by extracting a square root.
Once you have them you can deduce the values of $x_1+x_3$ and $x_1x_3$, thus of $(x_1-x_3)^2$. By taking another square root you get $x_1-x_3$ and finally you get all the $x_i$ from there.
So this is pretty easy, assuming you know the five polynomial expressions listed above.
Now, knowing them is equivalent to knowing the cycle, so is equivalent to breaking the $S_3$ symmetry between the three cycles, so is equivalent to solving a general cubic over $\Bbb C(c)$.
Looking at the first coefficient, $s = x_1+x_2+x_3+x_4$, its minimal polynomial is $(S-s)(S-s')(S-s'')$ where $s',s''$ are the sum of the two other cycles.
The coefficients in this expression should be in $\Bbb C(c)$. Furthermore the only poles of $s,s',s''$ can only happen when $c= \infty$ so they are in $\Bbb C[c]$.
Since the $x_i$ are "of degree $1/2$", the first coefficient must be a constant (in fact it must be the coefficient of $z^{11}$ in $P$, which is $0$), and the other two can only be polynomials of degree at most $1$ in $c$.
There are several methods to do the following (expressing $s$ as a polynomial in $x_1$, then looking for a relation formally modulo $P$, or computing the $x_i$ at enough points, or using a power expansion around $c=0$, etc ),
but you should obtain the relation $s^3 + (4c+3) s + 4 = 0$.
Now looking at this cubic, we can see that its Galois group is $S_3$, and more importantly, that the value of $s$ completely determines the value of $c$ (since the coefficients are of degree at most $1$). This means that $\Bbb C(c,s) = \Bbb C(s)$ is the $S_3$-covering of $\Bbb C(c)$ that corresponds to the knowledge of cycles, and that all the other expressions should be rational fractions in $s$.
As a function of $s$, $c$ has a pole of order $2$ at $s=\infty$ and of order $1$ at $s=0$.
Supposing you have solved for $s$ you should get all the other expressions are rational fractions in $s$. Since they have degree $2,2,4,3$ in the $x_i$, their poles are at worst of degree $2,2,4,3$ at infinity and $1,1,2,1$ at zero with respect to $s$.
This tells you their denominator and a bound on the degree of the numerator.
Then you should get the following relations :
$(x_1+x_3)(x_2+x_4) = -1 \\ x_1x_3+x_2x_4 = -(s^2+s+4)/2s \\ x_1x_2x_3x_4 = (-s^6-2s^5-4s^4-6s^3+5s^2+8s+16) / 16s^2 \\ ((x_1+x_3)-(x_2+x_4))(x_1x_3-x_2x_4) = (s^2+4)(s+1)/2 $
Then
$((x_1+x_3)-(x_2+x_4))^2 = s^2+4 \\ (x_1x_3-x_2x_4)^2 = (s^+4)(s+1)^2/4 $
The first square root to extract is thus $u = \pm \sqrt{s^2+4}$, which gives you
$x_1+x_3 = (s+u)/2 \\ x_2+x_4 = (s-u)/2 \\ x_1x_3 = -(s²+s+4-u(s^2+s))/4s\\ x_2x_4 = -(s²+s+4+u(s^2+s))/4s\\ $
Then
$(x_1-x_3)^2 = (s^3 + (2-u)s^2 + (4-2u)s + 8)/ 2s$
And after taking a square root of this you can obtain everything.