A rational function

Define $f_{\alpha}$ by
$$f_{\alpha}(z) = \frac{z^5+\alpha}{z^3}.$$
The figure below shows the basins of attractions of $f_{\alpha}$ for $\alpha=0.01$. The figure also shows an orbit that appears to be attracted to an orbit of period 4. You can play with that function a little further using the last demo on this page.

temp.png1254x1259 1.32 MB

• Show that, for every $\alpha$, the point at $\infty$ is always a super-attractive fixed point.
• Find a high precision estimate to a numerical value of $\alpha$ such that $f_{\alpha}$ has four super-attractive orbits.

@mark Do you want four super-attractive orbits or a super attractive orbit of period four?

@RedCrayon There happen to be four attractive orbits - two of which have period 4. As it turns out, for the correct choice of the parameter, they happen to all be super-attractive.

So, coincidentally - both!

I'll begin with the first part.

Let $f_\alpha[z]=\frac{z^5+\alpha}{z^3}$, and let $\phi(z)=\phi^{-1}(z)=\frac{1}{z}$. Let $\alpha \in \mathbb{C}$.

Then note, $\phi \circ f_\alpha \circ \phi^{-1} (z)=\frac{z^2}{1+ \alpha z^5}$. We denote this function, which is conjugate to the original function, $g(z)$.

Considering the derivative $g'(z)=\frac{2z-3\alpha z^6}{(1+\alpha z^5)^2)}$, we have $g'(0)=\frac{0}{1}$, and further $g(0)=0$, and so zero is a super attractive fixed point of $h(z)\Rightarrow \infty $ is a super-attractive fixed point for $f(z)$.

We now seek a value of $\alpha$ such that $f_\alpha(z)$ has four super attractive orbits.

Utilizing mathematica, we write

f[a_][z_] = z^2/(1 + a z^5)
F[a_][z_] = Nest[f[a], z, 4]
eqs = {F[a][z] == z, F[a]'[z] == 0}
as = a /. NSolve[eqs, {a, z}]
as = Union[as, SameTest -> (Abs[#1 - #2] < 0.0000001 &)]
as

This gives us a list of candidate $\alpha$ values as a list, namely ${-0.209273, -0.130092 + 0.0227936 I, -0.130092 -
0.0227936 I, -0.115265 + 0.108163 I, -0.115265 -
0.108163 I, -0.110725 - 0.0885781 I, -0.110725 +
0.0885781 I, -0.0262079 - 0.0777137 I, -0.0262079 +
0.0777137 I, -0.00939762 + 0.150968 I, -0.00939762 -
0.150968 I, -0.0090919 + 0.00613058 I, -0.0090919 -
0.00613058 I, -0.00525272 - 0.00793709 I, -0.00525272 +
0.00793709 I, -0.0038916 + 9.54729*10^-9 I, -0.00389148 -
1.18249*10^-7 I, -0.00389147 - 2.59888*10^-10 I, -0.00389145 +
1.49564*10^-7 I, -0.00389132 -
1.06469*10^-8 I, -0.00321798, 0.000302474,
0.000838845 - 0.0424907 I, 0.000838845 + 0.0424907 I, 0.0049318,
0.00978064 + 2.3801*10^-7 I, 0.00978155 + 1.59318*10^-9 I,
0.00978165 + 6.84955*10^-7 I, 0.0097823 - 1.77586*10^-7 I,
0.0479572 - 0.0569088 I, 0.0479572 + 0.0569088 I,
0.0486073 - 0.00326855 I, 0.0518073 - 5.39964*10^-6 I,
0.0518099 - 1.86444*10^-6 I, 0.0518134 - 6.74436*10^-7 I,
0.051815 + 5.80434*10^-7 I, 0.0518151 - 4.75716*10^-6 I,
0.0518323 - 0.0000344687 I, 0.0518362 - 0.0000323141 I,
0.05184 - 0.0000279354 I, 0.051841 + 0.0000264454 I,
0.051842 + 0.0000278376 I, 0.0518455 - 0.0000244157 I,
0.0518472 + 0.0000290001 I, 0.0518507 - 0.0000329779 I,
0.0518533 - 0.0000368313 I, 0.0518684 - 7.00661*10^-6 I,
0.0518691 - 0.000011351 I, 0.0518725 - 5.441*10^-6 I,
0.0548728 + 0.00301302 I, 0.0551607 - 0.0031988 I}$.

Using our rational julia set tool, we set out to test some of these $\alpha$ candidates for the desired behavior.

Let us consider $\alpha= -0.110725 - 0.0885781 i$. Plugging this value into our rational julia set tool, we generate the following graphic, which shows some interesting periodic behavior:

4Per4SA.png700x700 143 KB

We again utilize mathematica to find values of $z$ which exhibit the periodic behavior.

F4[z_] = FP[FP[FP[FP[z]]]]
Solve[F4[z] == z, z]
JuliaSetPlot[FP[z], z]

This generates a relatively large list of values of $z$, and the following image, with which we use the get coordinates function to get approximations of the values of $z$ we are looking for. Armed with this, we select particular candidates from our list of $z$ values.

We settle upon the following four points as representatives for each of the four period four orbits:
Red,$(0.5302005551614561 + 0.5072329008030877 i)$,
Green: $(-0.727085 - 0.098716 i)$,
Pink $(0.646248 - 0.347507 i)$,
Yellow $(-0.130797 - 0.722004 i)$.

For each of these points, it can be seen that they return to their initial value after 4 iterations of our function. Additionally, $f'_\alpha(z)=0$ at all four of these values, and so they are super attractive orbits of period 4.

It is worth noting that this particular value of $\alpha$ also has a period $2$ orbit, denoted in blue.