Some "Theorem" on the generalized superfunction

[Leo.W]

Hey, Leo

Welcome to the forum! It's always nice to get fresh blood.

This is interesting; I'm quite the fan of the notation \( \zeta(f|h) h = f\zeta(f|h) \).

I'm wondering, do you have any general idea on how to define the character,

\(
\zeta(f|h)
\)

For general instances? This is something I've been stuck on; developing a general theory to handle arbitrary \( f,h \); and I keep hitting dead-ends. I know MphLEE is a fan of black boxing it. I'm curious to hear what he'll have to say about this. As to conjugating fixed points, this is pretty standard, and many authors have done that before. The trouble is doing it for exotic scenarios, and not just when it's convenient. I do believe that generally finding well behaved solutions to,

\(
f\phi = \phi g\\
\)

Is an open problem. And really, the famous examples are the Schroder/Abel/Bottchner equations. Doing this, for say \( \phi = \zeta(tan(z) | \exp(\exp(z))) \) would be something else though, lol.

Anyway, welcome to the forum! I hope we can be of service, and we can all learn together.

Regards, James

Thank you, James! I'm really happy to join you all, and I'm a big fan of your respectable elaborate previous work, so impressive!


The case \( \phi = \zeta(tan(z) | \exp(\exp(z))) \) seems really horrible
However, you can utilize the "Cancel Law" I mentioned in Section II:
Letting h(z)=tan(z), f(z)=exp(exp(z)), and simply add that g(z)=z+1,
then the phi{h|f} function can be represented as phi{h|g}(phi{g|f}(z)), where phi{h|g}=phi{tan(z)|z+1} is the superfunction of tan(z), and phi{g|f}=phi{z+1|exp(exp(z))} is the abel function of exp(exp(z)).
The Cancel Law can produce infinitely many ways of solving the same equation by choosing different g(z).

Another Law I forgot to mention is that, for function f, and some iteration f^t where t is nonzero,
we can always generate one's superfunction family(abel,schroder,bottcher and so on) to another's.
To see this, suppose Z satisfies Z f=g Z, then notice that
Z (f^t)=(Z f) f^(t-1)=g Z f^(t-1)=g (Z f) f^(t-2)=g (g Z) f^(t-2)=g^2 Z f^(t-2)=....=(g^t) Z, so Z also satisfies the identity: Z (f^t)=(g^t) Z, unless t is 0.
Then we arrive at Z=Z{g|f}~=Z{g^t|f^t}, where A~=B represents a relation that, A and B satisfies the same equation, and there's some probability that A=B (so it's also possible they're not equal). This is the ''Invariant Law''
Then we use the Cancel Law:
Z{g|f^t}(z)=Z{g|g^t}(Z{g^t|f^t}(z))~=Z{g|g^t}(Z{g|f}(z))
where Z{g|g^t} is solvable, using the cancel law one more time, Z{g|g^t} can be transformed into the case g(z)=s*z
then it's easy to check Z{g|g^t}=c*z^t where c in a nonzero constant.
I prefer to call this the "Combination Law".

I've been work on the exotic cases for so long, let's call the function f having fixed point L, and f'(L)=s.
For the classic exotic case s=1, the Julia equation is always solvable in coeffients(if you assume a series of it), thus we have an approach to abel function. Then assume the asymptotic expansion of the superfunction at infinity, solve the coefficients term by term, and the superfunction is solved. And we should be careful with the direction of complex infinity, since we may generate different branch cuts of the superfunction. This method is already well-known.

When I was attempting to solve the "general exotic case"/parabolic case, where s=exp(2*pi*I*q) and q is a real rational number(and not an integer), then this case has no solution able to be generated from L, which is also pretty well-known.

However, let's just look on the 2-periodic case f(z)=-z(1-z), which is the logistic recurrence when s=-1:

Firstly I noticed that though the fixed point 0 is unsolvable, but the function has another, f(2)=2 and f'(2)=3, hence we can generate the superfunction of f at L=2.

Then I calculated that, all quadratic functions F(z)=a z^2+b z+c is always conjugate to the simplest case f(z)=z^2+v where v=a*c-b(b-2)/4, and the "conjugator" is a linear function.
So I asked if all general exotic cases in f(z)=z^2+v  have another fixed point which is solvable, and it turned out this is true leaving out v=0.25.
(The way to check is simply solve f(z)=z, denoting solution as z1 and z2, then use Vieta's theorem, or Newton's identities about polynomial roots, we see that f'(z1)+f'(z2)=2, so if f'(z1) is on the unit complex circle, then f'(z2) is on the outside.)
Now I really wonder if there exists a function whose derivative at its fixed points are all unsolvable.
The case: f(z)=z+(z-c)^n is obviously parabolic, with an nth order multi-fixed point c.

Secondly I tried to calculate the asymtotic expansion of the superfunction T, generated from L=0,f(z)=-z(1-z)
Applying the "Combination Law" I mentioned above, we see that, (since f(f(z))=z-2*z^3+z^4 which is classic parabolic case, we have the leading asymptotic term of the superfunction of f(f(z)), which is z^(-1/2)/2) the superfunction of f should have asymptotic expansion with leading term (2 z)^(-1/2).
So I assumed that T(z)~(2 z)^(-1/2)~f((2 z)^(-1/2)), and by slight rearrangement, I used the initial guess
T_0(z)=cos(pi z/2)^2/sqrt(2 z+6)+sin(pi z/2)^2*f(1/sqrt(2 z+4))
and applied T(z)=f^-1(T(z+1)) to make it converge.
Though the function is rather converging to 0 than converging to a smooth function,
I plot the T function after 50 iterations, T_50(z)=f^-51(f(T_0(z+50))), and the function did satisfy the relation:
T(z+1)=f(T(z)) up to about 7 decimal places.
So I guess the superfunction can be generated but in a rather hard method?

Lastly here's the plot of the T_50 function
The image is exported by the Wolfram Mathematica function AbsArgPlot.
(I'm sorry for my poor English and if I did say some words that is offensive or impolite, I'm really sorry for that)
(ps I wonder how you type a math formula in your reply? could u plz help me in it?)

Regards
Leo