Some "Theorem" on the generalized superfunction
#43
(08/10/2021, 04:49 AM)JmsNxn Wrote: Hey, Leo!

...

EDIT: I fixed my rough analysis to say what I really meant.
Thank you James

As a matter of fact, what I intended is that there lies no holomorphic function g having property g(0)=0 whose 2nd iteration is \( g(g(z))=f(z)=-z(1-z) \), because if g is holomorphic, g(z) must have a Taylor series due to definition of "holomorphic", and by our calculation before, doesn't exist. But something is bizarre, because for some cases, g exists, this can be observed by simply assuming the taylor series of g(z) and then we have g(g(z)). Same phenomena appears for all nonconstructable cases. Another point of view of this, is that g must not be holomorphic at 0, which is your statement, is very correct, and can be proved easily by differentiating the equation g(g(z))=f(z) and plug z=0 in.
But, if you're considering asymptotic expansion, it's exactly showing such expansion does not exist. We can only declare the linear term is correct: \( g(z)\sim\pm{i}z+O(z^{1+\epsilon}) \) for any \( \epsilon>0 \)


Let's take further investigation into this equation (assuming g is differentiable infinitely many times at most part of the complex plane)(but useless... g must be multivalued)
We've already known g(0) can't be 0 if singlevalued.
1. If g(z) is singlevalued, g(0) exists but is not 2, g(z) doesn't exist.
This is very easy to check. Since we already made the assumption g(0)=a, and due to definition g(g(0))=g(a)=0, hence
g(g(a))=g(0)=a, but remember g(g(a))=f(a), so f(a)=a. And since C is not 0 nor 2, a can't exist.
This implies if g(0) exists, the function g(z) is then multivalued.
2. If g(0) exists but is not 2, single-valued g(z) would behave oscillative, and thus won't exist.
Let g(0)=a, according to the equation, g(a)=g(g(0))=f(0)=0, g(f(a))=f(g(a))=f(0)=0, g(f(f(a)))=f(f(g(a)))=0, etc.
this shows that for all positive integers k, each f^k(a) is a zero of g(z).
If a lies within the attracting pedal, then according to our discussion before, \( f^{2k}\(a\)\sim\sqrt{\frac{1}{4k}}+o(\frac{1}{k}) \), so as k gets larger, f^{2k}(a) gets closer to 0 at the rate \( \sqrt{\frac{1}{4k}} \), getting denser and denser, indicating g(z) has oscillative behavior at z=0.
If a lies within the repelling pedal, the same pattern still applies, because we can extend "this shows that for all positive integers k, each f^k(a) is a zero of g(z)" to "for all integers k, each f^k(a) is a zero of g(z)", which can be done easily by manipulating the same way in the equation g(a)=0. And dependent on the definition of "attracting/repelling pedals", a is thus lying within the attracting pedal of \( f^{-1}(z) \), hence we finished the proof.
And an oscillative behavior can cause many problems, including the ill behavior of g(g(z)) around each zero of g(z), but since f has no such behavior, a singlevalued g doesn't exist.
From another horizon, g(g(z))=0 should only contain 2 roots due to f(z)=-z(1-z), but anytime g(z)=a,f(a),f^2(a),etc., another root is spawned, or z=g^-1(a),g^-1(f(a)),g^-1(f(f(a))).
3. g(0)=2 may be a wise choice, but then the question arises: which function satisfies h(h(z))=g(z)?
it follows immediately the same procedure in #2, implying that h must be multivalued, so does g(z)
And g(0)=2 singlevalued can cause the same issue in 2., we see that f(-1)=2, and thus 0=g(2)=g(f(-1))=f(g(-1)), showing g(-1)=0 or 1. If g(-1)=0, -1 is in the attracting pedal of f^-1, so g oscillative. If g(-1)=1, due to symmetry law, g(z)=g(1-z), g(2)=1, a contradiction. So g should be multivalued.
4. Does g(z) has a pole at z=0?
No. If we set g(0)=infty, we can tell informally \( g'(\infty)g'(0)=f'(0)=-1 \) so \( g(z)\sim{\pm{i}z} \) when z gets large, and because g(g(0))=g(infty)=0, this is impossible.
A formal clamation can be done by considering a neighbourhood of z=0, \( g'(g(z))g'(z)=f'(z)=2z-1 \).
5. A singlevalued abel function of f(z) is never constructed
it follows my last post
And we can detect that, as we already solved the series, \( \alpha\{f\}(z)=\frac{1}{2z^2}+\frac{1}{2z}+\frac{11ln(z)}{4}+O(z) \), we will always get \( \alpha(f(z))-\alpha(z)=1+\frac{11(ln(-z)-ln(z))}{4} \) which is not exactly 1. Whenever we treats the abel function as single valued, this \( \alpha(f(z))-\alpha(z)=1+\frac{11(ln(-z)-ln(z))}{4} \) will hold, and thus not helpful to construct g(z).
And resulted from these, we should consider the method to build up an inverse abel function, which is sufficient to give us each branch cut of g(z).
Now I have the idea to make use of theta mapping to solve g(z), as there may lie a function P... and if not, we can still generate the inverse abel function, with two fixed points 0 and 2. But this is then not what we originally wanted: g to be constructed only from z=0.

It's a good start with considering the simple odd functional pieces to generalize another second iterative root. But it arises that how can we get a third iterative root? or forth?
I Do believe the half iterate exists, but I have no idea how to compute, and all old tricks(bell matrix, fixed point series) don't work for this... so I suggest to construct the inverse abel function.
Sorry I have to disappoint u, I got no more useful information... But I do have some clues.
Viewing from my last post the generalization of superfunction always coverges to 0 or diverges, this is because we used \( T(z)=f^{-1}(T(z+1)) \), in this procedure, f(z) has less absolute value than f(f(z)), and so T(z+2) has always less absolute value than T(z), comparing the descendent rate, we can claim that T must converge to 0. This shows that P(z) must include some poles, but then it'll be hard to do the theta-mapping...
Another point, is that we can always claim that when z is approaching infinity, g is asymptotic to \( g(z)\sim{z^{\sqrt{2}}} \). I tried to make this converge but failed, this is so strange.
Almost all details I can tell are in previous posts, maybe you can do some calculation about this g(z) or the superfunction, I guess?

Your functional family approach may be helpful, I'll do it some day, but to be noticed, the approach from only the real line is not ample, we need to consider a sequence of multiplifier \( e^{2\pi{i}s_0},e^{2\pi{i}s_1},e^{2\pi{i}s_2},\cdots \) where \( s_n \) are all in [0,1] and all irrational numbers, approaching 0 as n diverges. In this manner the multiplifier equation[Image: mimetex.cgi?\left.\frac%7B\mathrm%7Bd%7Df%5Et(z)...%7D\right)%5Et] will hold.
But all these work, are implying that even for constructable cases, there lies n nth-iterative roots, I may test it out later.

I'm so sorry I don't have so much time recently.(Coronovirus, flood, entering university, classfying most theorems etc) I may edit this brand new soon.
Firstly I need to write a formal introduction to EDE, as MphLee invited, I put it off so long...(totally forgotten XD)
I have a long long to-do list now. I wanted to write a post about how to transform from one tetration to another... but, sooner, maybe...
Maybe you could set up a new post about \( g(g(z))=f(z)=-z(1-z) \)? Appreciate! Wink

Regards
Leo


Messages In This Thread
RE: Some "Theorem" on the generalized superfunction - by Leo.W - 08/10/2021, 11:09 PM

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