I'm trying to make images of Fatou coordinate for some polynomial maps. If I'm not wrong there is no explicit general formula/method for computing Fatou coordinate near parabolic fixed point.
I am very new to theory, so I am looking for a simple, intuitive answer. Also I would like to switch from theory to computations and making images.
On
Adam,
See Henryk Trapmann's post on Perturbed Fatou Coordinates at the tetration forum
Given $f=x^2+x+0.01$, where the $\epsilon=0.01$ is the perturbation of the parabolic point, the link shows the existence of the Fatou Coordinate defined between the two fixed points, $\alpha(x)$. Here is my computation for the inverse Fatou Coordinate, $\alpha^{-1}(x)$, for $f(x)$. My solution requires that $\alpha^{-1}(i\infty)=0.1i$, and $\alpha^{-1}(-i\infty)=-0.1i$, the two fixed points for $f(x)$. The algorithm is based on the program I wrote to implement Kneser's solution for tetration, also see the tetration forum. For positive real values of Epsilon, the computation requires a Riemann mapping, which is difficult to compute.
$$\alpha^{-1}(x) = f^{ox}(0)$$
There is a branch singularity at $\alpha^{-1}(-11.701)\approx-0.5$. For normal Mandelbrots iterating $x^2+c$, $\epsilon=0.01 \mapsto c=0.26$, and the branch singularity at $-0.5 \mapsto 0$, where $0$ is the Mandelbrot starting point. The $\alpha^{-1}(x)$ function increases to infinity after the branch point, and is analytic to the right of that branch point in the complex plane for all imaginary values.
= 0
+x^ 1* 0.010016699568842293
+x^ 2* -0.000050418367246083195
+x^ 3* 0.000034178173550238976
+x^ 4* -0.00000060255845085496712
+x^ 5* 0.00000014728664516498046
+x^ 6* -0.0000000047635866181065753
+x^ 7* 0.00000000068914925444492225
+x^ 8* -3.2193920179857513 E-11
+x^ 9* 3.4764175712195070 E-12
+x^10* -2.0345413419304956 E-13
+x^11* 1.8714590352026207 E-14
+x^12* -1.2518981757319611 E-15
+x^13* 1.0595108736880268 E-16
+x^14* -7.6520364024012775 E-18
+x^15* 6.2226912074002442 E-19
+x^16* -4.6927195239006338 E-20
+x^17* 3.7527923737871100 E-21
+x^18* -2.9008512593943895 E-22
+x^19* 2.3081020995208996 E-23
+x^20* -1.8107093728645096 E-24
I have also done calculations for complex values of $\epsilon$, where the function is defined in between the two fixed points; I haven't posted them.
The op mentioned c=-0.75, where iterating $y \mapsto y^2-3/4$ corresponds to iterating $f(x)= x^2-x$, where $x=y+1/2$. The two fixed points of $f(x)$ are 0, which has $\lambda=-1$, and is rationally indifferent, and the other fixed point is 2. The claim is that if you start with iterating $x\mapsto x^2+0.26$, which is equivalent to the Taylor series above. Now slowly change the perturbation, staying in the upper half of the complex plane, until you get to $x\mapsto x^2-0.75$, which is equivalent to $x\mapsto x^2-x$, then you arrive at this merged solution for the iterated function. Below, I generate the Taylor series for $f^{ox}(0.3125)$, accurate to about a thirteen decimal digits. The novel algorithm starts with our solution for merged tetration for complex bases, and is modified to allow computing results where $\lambda$ is rationally indifferent. The algorithm is partially described at this link; tetration forum for e^-e. Convergence for the computation algorithm is not proven, but is based on what I understand about perturbed fatou coordinates. Here is the Taylor series of this interesting, but non standard inverse Fatou Coordinate for this function. I don't know if anyone else has ever computed these solutions or not.
f^x = 0.31250000000000
+x^ 1* ( 0.043182808486163 + 1.0774060901730*I)
+x^ 2* (-1.9578359017697 + 0.23834297940039*I)
+x^ 3* (-0.68144363913437 - 2.5067898772210*I)
+x^ 4* ( 2.4811891207280 - 1.3195628373657*I)
+x^ 5* ( 1.9060571037156 + 1.8788044842752*I)
+x^ 6* (-0.91423964176752 + 2.1371762776650*I)
+x^ 7* (-1.8715957209642 + 0.051769912154558*I)
+x^ 8* (-0.68715768653121 - 1.2344260154422*I)
+x^ 9* ( 0.51937894568058 - 0.86283969405588*I)
+x^10* ( 0.68276123212209 - 0.0090904579966679*I)
+x^11* ( 0.24910214163161 + 0.36602794490275*I)
+x^12* (-0.098162261449123 + 0.25779714775530*I)
+x^13* (-0.15990546838427 + 0.043838294866587*I)
+x^14* (-0.076655855358702 - 0.058560468392113*I)
+x^15* (-0.00086748306390932 - 0.054473578651271*I)
+x^16* ( 0.022902729497259 - 0.019288163563319*I)
+x^17* ( 0.015810454394089 + 0.0027550230200182*I)
+x^18* ( 0.0043134980471593 + 0.0072127736838463*I)
+x^19* (-0.0013064327373118 + 0.0041039541577587*I)
+x^20* (-0.0019690404251635 + 0.00089522635627084*I)
And here is a graphical image of iterating $f(x)=x^2-x$, where I show $f^{ox}(0.3125)$ in the complex plane. You can see the indifferent rationally indifferent fixed point of 0, which is the limiting value as $x\to i\infty$, with a pseudo period=2. The fixed point of 2 is on the left, with its pseudo period=$\frac{2\pi i}{\ln(3)}=5.719i$; the solution converges to this standard Schroeder function solution as imaginary(z) goes to $-\infty$. There is a branch at -2, where f(x)=0.5. The grid lines are every two units. The graph range shown is real -4 to +6, and imaginary -6.6 to +2.
On
I have probably found the example of function for which explicit formula of Fatou coordinate is known ( where c is a parabolic parameter ) On the derivative of the Hausdorff dimension of the quadratic Julia sets by Ludwik Jaksztas
$$f(z) = z^2-3/4$$
$$\widehat{z} = h_c(z) = -i(z+\frac{1}{2})$$
then Fatou coordinate Z :
$$Z_c (\widehat{z}) = -\frac{1}{2\widehat{z}} $$
I have made image of small attracting petals P using such coordinate, using property that a right half-plane in the Z coordinates are called small pertals
$$P = \{ z : Z_c(h_c(z))>=0 \}$$
Shape of these "petals" ( triangles ) suggest that above coordinate are prepared coordinate or pre-Fatou coordinate
I have checked my code for creating image and found some errors. Good function for computing real part of Fatou coordinate
$$ux = re(Z_c)$$
is :
double complex GiveU(complex double Z)
{
double Hx, Hy;
double t; // temp
// Z = Zx +Zy*I
double Zx = creal(Z);
double Zy = cimag(Z);
// U = Ux +Uy*I
double Ux; // re(U)
double Uy; // im(U)
// from z to h
Hx= Zy;
Hy= -Zx - 0.5;
// from h to u
t = Hx*Hx+Hy*Hy;
Ux = -Hx*Hx - (Hy*Hy)/(2*t*t);
Uy = (Hx*Hy)/(t*t);
return (Ux+Uy*I);
}
To check it find result for points on real axis from critical point z=0 to fixed point z = -0.5 ( all points shlould be inside attracting petal ) :
i = 0; z.x = 0.000000; u.x = -2.000000
i = 1; z.x = -0.187500; u.x = -5.120000
i = 2; z.x = -0.238998; u.x = -7.339791
i = 3; z.x = -0.269918; u.x = -9.445030
i = 4; z.x = -0.291475; u.x = -11.498873
i = 5; z.x = -0.307719; u.x = -13.523753
i = 6; z.x = -0.320570; u.x = -15.530317
i = 7; z.x = -0.331087; u.x = -17.524447
i = 8; z.x = -0.339912; u.x = -19.509720
i = 9; z.x = -0.347460; u.x = -21.488463
i = 10; z.x = -0.354018; u.x = -23.462264
i = 11; z.x = -0.359786; u.x = -25.432250
i = 12; z.x = -0.364912; u.x = -27.399247
i = 13; z.x = -0.369510; u.x = -29.363871
i = 14; z.x = -0.373664; u.x = -31.326596
i = 15; z.x = -0.377442; u.x = -33.287790
i = 16; z.x = -0.380898; u.x = -35.247743
i = 17; z.x = -0.384076; u.x = -37.206688
i = 19; z.x = -0.389733; u.x = -41.122270
i = 20; z.x = -0.392266; u.x = -43.079188
As you see all values are below zero ( ?) and ux(critial point) is not a zero. On can normalize it using :
double complex GiveNormalizedU(complex double Z, double ux0)
{
// U = Ux +Uy*I
double complex U;
double Ux; // re(U)
double Uy; // im(U)
U = GiveU(Z);
Ux = fabs(creal(U)) - fabs(ux0); // re(U)
Uy = cimag(U);
return (Ux+Uy*I);
}
where ux0 is a ux(critical point ) = -2.0; Then the results are :
i = 0; z.x = 0.000000; u.x = 0.000000 ;
i = 1; z.x = -0.187500; u.x = 3.120000 ;
i = 2; z.x = -0.238998; u.x = 5.339791 ;
i = 3; z.x = -0.269918; u.x = 7.445030 ;
i = 4; z.x = -0.291475; u.x = 9.498873 ;
i = 5; z.x = -0.307719; u.x = 11.523753 ;
i = 6; z.x = -0.320570; u.x = 13.530317 ;
i = 7; z.x = -0.331087; u.x = 15.524447 ;
i = 8; z.x = -0.339912; u.x = 17.509720 ;
i = 9; z.x = -0.347460; u.x = 19.488463 ;
i = 10; z.x = -0.354018; u.x = 21.462264 ;
i = 11; z.x = -0.359786; u.x = 23.432250 ;
i = 12; z.x = -0.364912; u.x = 25.399247 ;
i = 13; z.x = -0.369510; u.x = 27.363871 ;
i = 14; z.x = -0.373664; u.x = 29.326596 ;
i = 15; z.x = -0.377442; u.x = 31.287790 ;
i = 16; z.x = -0.380898; u.x = 33.247743 ;
i = 17; z.x = -0.384076; u.x = 35.206688 ;
i = 18; z.x = -0.387011; u.x = 37.164812 ;
i = 19; z.x = -0.389733; u.x = 39.122270 ;
i = 20; z.x = -0.392266; u.x = 41.079188 ;
Using such function I get different image of attracting petals :
Here is another test :
i = 0; z.x = 0.000000; u.x = 0.000000 ; du = NA
i = 1; z.x = -0.187500; u.x = 3.120000 ; du = 3.120000
i = 2; z.x = -0.238998; u.x = 5.339791 ; du = 2.219791
i = 3; z.x = -0.269918; u.x = 7.445030 ; du = 2.105239
i = 4; z.x = -0.291475; u.x = 9.498873 ; du = 2.053843
i = 5; z.x = -0.307719; u.x = 11.523753 ; du = 2.024880
i = 6; z.x = -0.320570; u.x = 13.530317 ; du = 2.006564
i = 7; z.x = -0.331087; u.x = 15.524447 ; du = 1.994130
i = 8; z.x = -0.339912; u.x = 17.509720 ; du = 1.985273
i = 9; z.x = -0.347460; u.x = 19.488463 ; du = 1.978743
i = 10; z.x = -0.354018; u.x = 21.462264 ; du = 1.973801
i = 11; z.x = -0.359786; u.x = 23.432250 ; du = 1.969987
i = 12; z.x = -0.364912; u.x = 25.399247 ; du = 1.966996
i = 13; z.x = -0.369510; u.x = 27.363871 ; du = 1.964624
i = 14; z.x = -0.373664; u.x = 29.326596 ; du = 1.962725
i = 15; z.x = -0.377442; u.x = 31.287790 ; du = 1.961194
i = 16; z.x = -0.380898; u.x = 33.247743 ; du = 1.959953
i = 17; z.x = -0.384076; u.x = 35.206688 ; du = 1.958945
i = 18; z.x = -0.387011; u.x = 37.164812 ; du = 1.958125
i = 19; z.x = -0.389733; u.x = 39.122270 ; du = 1.957458
i = 20; z.x = -0.392266; u.x = 41.079188 ; du = 1.956918
Here :
so when z is changing from $z_i$ to $f^2(z_i)$ then u is changing $u_i$ to $u_i+2$ aproximately ( close to translation by 2 ):
$$f^2 : z \mapsto f(f(z))$$ $$Z_c : u(z) \mapsto u(z) + 2 $$
Will Jagy has an overview of the formal Fatou coordinate (Jean Ecalle at Orsay) for a parabolic point at mathoverflow; I posted some pari-gp code to implement Ecalle's solution below. Near, but not exactly at a parabolic point, the problem is much more difficult. Give a function $f(z)$, we call the Fatou Coordinate/Abel function $\alpha(z)$.
$$\alpha(f(z)) = \alpha(z)+1$$
Then the iterated function would be $f^{o z} = \alpha^{-1}(z)$, and I have written a program that calculates $\alpha^{-1}$ for tetration, see the tetration forum, and investigated the properties near the parabolic fixed point (which is a branch point for this family of complex functions). I have used the same method to calculate $\alpha^{-1}(z)$ for $x^2+c$ near the parabolic branch point, c=0.25, but have not posted it anywhere. I would also be interested in any other responses.
Some other thoughts. Consider the case where $f(x)=x^2+0.26$, which has two fixed points, $0.5+/-0.1i$. The solution I was looking for treats both of these fixed points symmetrically and is based on extending Kneser's solution for tetration, which involves a Riemann mapping, which helps explain why computing such solutions is difficult. If you only want to calculate $\alpha(z)$ for one of the two fixed points, then the Schroeder function provides a simple well defined solution for $\alpha(z)$.
Finally, nearby c=0.25, there are also much more complicated parabolic points, where $f^{on}(z)$ is a parabolic point. Near such a point do we compute the Fatou coordinate for $f(z)$, or $f^{on}(z)$? Will Jagy's link gives a solution for the Fatou coordinate of $f^{on}(z)$. I also now know to compute the solution for $\alpha^{-1}(z)$ for $f(z)$ using both fixed points; I tried asking a question on math overflow, but I didn't get any relevant responses :(
You could also search parabolic implosion on the web, but I haven't seen any papers showing how to calculate $\alpha(z)$.
EDIT Here is a pari-gp program to implement Jean Ecalle's formal Abel Series, Fatou Coordinate solution for parabolic points with multiplier=1. This is an asymptotic non-converging series, so there is an optimal number of terms to use, so you may have to iterate $f$ or $f^{-1}$ a few times to get optimally accurate results, so that the coeffient is closer to the fixed point of zero.