Periodic points of $F(x)=\lambda x + \beta x^3$ where $\lambda >1$ and $\beta <0$

Question

So, when the coefficients are somewhat arbitary, I'm quite confused. Obviously there are 3 fixed points. There are other quite obvious ones, by letting $\lambda x+ \beta x^3 = -x$.

However when we look at its graph versus $y=x$, it seems for some choice of $\lambda$ and $\beta$, we can find a point that is greater than the positive fixed point, go between $0$ and the positive fixed point, and go back to this point by graphical analysis. The period is 2 (if possible, I just look at the graph and made an educated guess). Seems higher period is possible too if there are more steps between $0$ and positive fixed point. But I don't know if it's possible and how to restrict $\lambda$ and $\beta$.

The most straight forward idea is to compute $F^n$. However $F$ is already cubic and the power immediately goes out of control.

Answer 1

By the period-3 theorem, if you can find a period-3 point, then there are fixed points cycles of any period. So you only need to explore the roots of $(F^3(x)−x)/(F(x)−x)$ of $F(x)=ax-x^3$.

In the CAS Magma (online calculator) this can be computed as

Q<a>:=PolynomialRing(Rationals());
P<x>:=PolynomialRing(Q);
F:=func<x|a*x-x^3>;
G:=Quotrem(F(F(F(x)))-x,F(x)-x);"Polynomial G for period 3 points", G;

To find the parameters for which the number of real roots changes, compute the discriminant relative to the variable $x$. As these roots come in multiples, of 3 for the period-3 points, the difficulty of the polynomial can be reduced by computing its factorization.

D:=Discriminant(G); 

Dfac := Factorization(D); "Factors of the x-discriminant of G",Dfac;
for fac in Dfac do 
     rts :=Roots(fac[1],ComplexField(60));
    "Real roots of ",fac[1]," are ",[ Real(r[1]) : r in rts | Abs(Imaginary(r[1])) lt 10^-50 ]; 
end for;

The result of this small script is

Polynomial G for period 3 points x^24 + (-8*a - 1)*x^22 + (28*a^2 + 7*a + 1)*x^20 + (-56*a^3 - 21*a^2 - 9*a - 1)*x^18 + (70*a^4 +
    35*a^3 + 33*a^2 + 8*a + 1)*x^16 + (-56*a^5 - 35*a^4 - 65*a^3 - 25*a^2 - 7*a - 1)*x^14 + (28*a^6 + 21*a^5 + 75*a^4 + 40*a^3 + 
    21*a^2 + 6*a + 1)*x^12 + (-8*a^7 - 7*a^6 - 51*a^5 - 35*a^4 - 34*a^3 - 15*a^2 - 5*a - 1)*x^10 + (a^8 + a^7 + 19*a^6 + 16*a^5 +
    31*a^4 + 19*a^3 + 10*a^2 + 4*a + 1)*x^8 + (-3*a^7 - 3*a^6 - 15*a^5 - 12*a^4 - 10*a^3 - 6*a^2 - 4*a - 1)*x^6 + (3*a^6 + 3*a^5 
    + 5*a^4 + 4*a^3 + 5*a^2 + 3*a + 1)*x^4 + (-a^5 - a^4 - 2*a^3 - 2*a^2 - 2*a - 1)*x^2 + a^2 + a + 1

Factors of the x-discriminant of G [
    <a^2 - 7/2*a + 13/4, 4>,
    <a^2 + a + 1, 5>,
    <a^8 + 4*a^7 - 35/4*a^6 - 103/2*a^5 - 113/4*a^4 + 94*a^3 + 715/4*a^2 + 845/2*a + 2197/4, 6>
]

Real roots of  a^2 - 7/2*a + 13/4
 are  []
Real roots of  a^2 + a + 1
 are  []
Real roots of  a^8 + 4*a^7 - 35/4*a^6 - 103/2*a^5 - 113/4*a^4 + 94*a^3 + 715/4*a^2 + 845/2*a + 2197/4
 are  [ -2.92487335244582053342287937237620388106853002896463015195, 
2.98177187596001825613614486451934187127600281476921585297, 
-2.69964153155259029627024496000852060358898268448327991427, 
2.45044096448787517281550248691820215755941414877568026892 ]

so that the relevant roots for the given task are at $2.4504409644878751728..$ and $2.9817718759600182561..$. Plotting $G$ around these values for $a=2.4, 2.5, 2.95, 3.05$ gives the graphs which demonstrate that each of the 2 transitions introduces 4 period-3 cycles