Some "Theorem" on the generalized superfunction

[Leo.W]

Also; I must've missed the challenge to construct an inverse abel function. I'll be hard pressed if the following method doesn't work. If it doesn't, I'll try another.
...
I've yet to find a function that didn't have exactly n roots about that fixed point though.

Hi James
Thanks for your patient and elaborate detailed explanation

But I'm afraid this case is blowing everyone's mind... because this case is so stubborn to be solved!
It's obvious that if such function has derivative at 0 \( f'(0)=\pm{i} \), then it won't exist.
A quick check, first consider the case \( f'(0)=i \), and thus we can assume f has a Taylor series, we'll denote it as \( f(z)=iz+bz^2+cz^3+O(z^4) \), simply plug it in we get \( f(f(z))=-z+(-1+i)bz^2+2ib^2z^3+O(z^4) \), showing that if we want \( f(f(z))=-z+z^2=-z(1-z) \), we have to let \( (-1+i)b=1,2ib^2=0 \), which is impossible. The case \( f'(0)=-i \) is with the same condition: let \( f(z)=-iz+bz^2+cz^3+O(z^4) \) we have \( f(f(z))=-z-(1+i)bz^2-2ib^2z^3+O(z^4) \) so in order to make it fit \( f(f(z))=-z(1-z) \) we have to solve \( -(1+i)b=1,-2ib^2=0 \) which is self-contradicting.

---

I've done something resembles your dictation before, in which I denoted \( f(z)=-z(1-z),f^2(z)=f(f(z)),\alpha^{-1}\{f^2\}(z)\text{is the superfunction of}f^2(z) \), we are hunting for an explicit calculation formula for \( \alpha^{-1}\left{f\right}(z)=T(z)\text{ for brief and is the superfunction of}f(z) \)

First of all, the function through your generalization, as \( \alpha^{-1}_0\{f^2\}(z)=\frac{1}{2} \sqrt{\frac{1}{z}}\left(\frac{\frac{5}{64}-\frac{11\log(z)}{64}}{z^{3/2}}+\frac{1}{4\sqrt{z}}-\frac{11\log(z)}{32 z}+1+o(\frac{1}{z^2})\right) \), in practical calculation, we can transfer it by shifting to define \( \alpha^{-1}\left{f^2\right}(0) \).
As we use \( \alpha^{-1}\left{f^2\right}(z-1)=f^{-2}(\alpha^{-1}\left{f^2\right}(z)) \), the function has now an upper bound in the positive real axis which is \( \frac{1}{2} \).
We define \( \alpha^{-1}_1\{f^2\}(z)=\lim_{n\to\infty}f^{-2n}(\alpha^{-1}_0\{f^2\}(z+n)) \), in practical calculation, using n=200 000 can promise a precision of 16 degrees except for \( \alpha^{-1}_1\{f^2\}(z) \) is so close to \( \frac{1}{2} \). Then we find for which z, the function's value is exactly 0.5, we'll call it C. \( C\approx{1.666363609223794} \) by mathematica 12.2, so we define the superfunction of f^2(z) as
\( \alpha^{-1}\{f^2\}(z)=\alpha^{-1}_1\{f^2\}(z+C) \), thus the value taken at integers is exactly \( f^{2n}(\frac{1}{2}) \).


Secondly, we can show that by law 3, or more simply calculation, that the super function of f should be asymptotically: \( T(z)\sim\alpha^{-1}\{f^2\}(\frac{z}{2}) \). But to be noticed, function \( \alpha^{-1}\{f^2\}(\frac{z}{2}) \) is always positive whenever z is positive, not T(z). And because T(z) is determined by function f, which has an oscillating multiplifier -1, T(z) should also behave oscillative. A quick check in mind is, Assume T(0)=0.5>0, so then T(1)=f(T(0))=-0.25<0, T(2)=0.3125>0, T(3)=-0.214844<0, and keeps this pattern all along the positive reals. By calculation, we find that denote briefly \( W(z)=\alpha^{-1}\{f^2\}(z) \), W(0)=0.5, W(0.5)=0.358325>0 is not T(1)=-0.25, which also proves they are only asymptotic to each other.

And I got, finally, stuck here. I figured one way to generate T, as stated below:
\( \text{Let a function having period 2 be called }P(z), P(z)=P(z+2)\\\text{and satisfies at least these 2 conditions:}P(0)=1,P(1)=0\\\text{Then we can define}T(z)=P(z)W(\frac{z}{2})+P(z+1)f(W(\frac{z-1}{2})) \)
This function T would fit the condition \( T(k)=f^k(\frac{1}{2}) \) for all integer k, but may not fit \( T(z+1)=f(T(z)) \) for all z. This is as far as I reached.
ps. A generalization of any nonconstructable cases is hense obtainable following the same pattern.

I used both \( T(z)=f(T(z-1))=f^{-1}(T(z+1)) \) to get it converged to fit the equation \( T(z+1)=f(T(z)) \)
But it turned out that most P(z) would lead to the fact that T(z) does not converge to an analytic function except for real integers, instead, the process led T(z)=0 for all non-real-integers z.
This is a graph of T, where I set \( P(z)=(cos(\frac{\pi}{2}z))^2 \)
Blue line- T(z) by P(z) definition
Red line- \( f^5(T(z-5)) \), divergent
Yellow line- \( f^{-5}(T(z+5)) \), converge to 0
Green line- \( f^{-500}(T(z+500)) \), converge to 0, and evidently it's way different from yellow line and others.

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And here's the mathematica codes to generate T and other functions:

Code:

Clear[F, IF, F2, IF2, A0, IAPrec16, IA2, IA]
F[z_] := -z + z^2
IF[z_] := 1/2 (1 - Sqrt[1 + 4 z])
F2[z_] := z - 2 z^3 + z^4
IF2[z_] := 1/2 (1 - Sqrt[3 - 2 Sqrt[1 + 4 z]])
A0[z_] :=
1/2 Sqrt[1/
 z] (1 + 1/(4 Sqrt[z]) + (5/64 - (11 Log[z])/64)/z^(3/2) - (
   11 Log[z])/(32 z) + (
   139/1024 - (231 Log[z])/1024 + (121 Log[z]^2)/1024)/z^(5/2) + (
   77/512 - (121 Log[z])/512 + (363 Log[z]^2)/2048)/z^2 + (
   2293/12288 - (6897 Log[z])/16384 + (1331 Log[z]^2)/4096 - (
    6655 Log[z]^3)/65536)/z^3 + (
   9959/49152 - (891 Log[z])/2048 + (10285 Log[z]^2)/32768 - (
    1331 Log[z]^3)/16384)/z^(7/2) + (
   783787/2359296 - (168355 Log[z])/196608 + (426767 Log[z]^2)/
    524288 - (270193 Log[z]^3)/786432 + (14641 Log[z]^4)/262144)/z^(
   9/2) + (1335835/4718592 - (580723 Log[z])/786432 + (
    765325 Log[z]^2)/1048576 - (1039511 Log[z]^3)/3145728 + (
    512435 Log[z]^4)/8388608)/z^4)
IAPrec16[z_] := N[Nest[IF2, A0[N[z, 50] + 200000], 200000], 16]
IA2[z_] := IAPrec16[z + N[1.666363609223794000000000`50, 50]]
IA[z_] := (Cos[Pi/2 z]^2) IA2[z/2] + (Sin[Pi/2 z]^2) F[
   IA2[z/2 - 1/2]]

And ps, because of the oscillative behavior of T, the inverse of T, the Abel function of T, is multivalued and hard to define. So basically the computation of the second iterative root of f is very hard problem, if generated at the fixed point L=0.

Regards,
Leo

[JmsNxn]

Hey, Leo!

I get your confusion. I stand by my point though. What you have done is shown there is no taylor expansion at \( 0 \); which I explained does not exist. The iterate can only exist on a petal about zero; not a neighborhood of zero. So your first point is null to my construction.

It helps to realize in the neutral case; that no iteration exists in a disk about zero; therefore, there is not a taylor series at zero. But; in the petals about zero; there are half iterates. But in a neighborhood of \( 0 \) intersecting with the attracting petals; and then excluding \( 0 \); we have a holomorphic function. Where we likely have a removable singularity at zero; but it is not holomorphic at zero.

May I switch to \( \sin(-\xi) \)? Because I can explain your confusion using this function. The same principle applies to \( f(\xi)= - \xi(1-\xi) \)--forgive the change to a simpler case, please. But using an odd function is easier (same principle applies); and \( \sin \) is easier to reference numerically.

The function,

\(
\sin(-\xi) = f(\xi)\\
\)

has an iteration for all \( k \in \mathbb{Z} \),

\(
f^{\circ z}(\xi) = e^{(2k+1)\pi iz} \xi+\sum_{n=2}^\infty b_n \xi^n\\
\)

Where this is only an ASYMPTOTIC SERIES about zero. if we assume that \( \sin^{\circ z}(-\xi) = - \sin^{\circ z}(\xi) \)--which is the standard iteration. Then there are two square root

\(
f^{\circ 1/2}(\xi) = i\xi + \sum_{n=2}^\infty c_n \xi^n\\
-f^{\circ 1/2}(\xi) = -i\xi - \sum_{n=2}^\infty c_n \xi^n\\
\)

Now notice; neither of these map the real line to the real line; like the iteration \( \sin^{\circ t}(\xi) \) for \( t \in \mathbb{R}^+ \); but they are iterates nonetheless. They are not holomorphic at zero; we just get a good asymptotic expansion when approaching zero. In this sense, the summation, and taylor series, is an *abuse of notation*; these things don't converge. Only asymptotically to zero are they meaningful. They are actually holomorphic on a petal.  It's a bit more complicated with \( f(\xi) = -\xi(1-\xi) \) but it's the same scenario--as far as I can tell. They're not holomorphic at zero; only on a petal about zero. What the petal looks like? I have no f**n clue; but there exists a petal where this converges.

Unless I'm missing something large; which I don't think I am from your posts; this still has a square root on the attractive petals. But please, keep posting. I'm happy to be proven wrong . Your posts are becoming exponentially more educational; I appreciate your posts a lot, Leo.  Thank you for your contributions.

Regards, James

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

[Leo.W]

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 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!

Regards
Leo

[JmsNxn]

I agree with everything you're saying, Leo. Very well thought out; I never gave it much thought how to actually numerically calculate this function--but I'm sure it exists. It won't have an asymptotic expansion at zero; just how your noting. What's especially interesting, is that it doesn't even have a formal taylor series at zero; like with sine. I've never given it thought how to evaluate this; I only ever really worried about \( \sin \) when talking about the neutral fixed point theory. It's a good function to keep in mind.

I'll give it some thought; and think if I can create a way of calculating this. I'm still busy trying to make my tetration calculator for the beta method; and that is filling up all my "numerical" thought processes. I still think the fractional calculus approach should work; but you'd need to know the shape of the petals to make it work. Which, I don't have the energy to calculate.

I'll give it some thought, though. Good luck at uni; and I hope the floods don't get you too bad. Don't worry about rushing anything--mathematics is a slow beast. I'm the lucky few where coronavirus has kept me comfy at home--so I got nothing but time to work. Don't burn yourself out. I appreciate your contributions .

Regards, James.

---

And for god's sake read John Milnor. His graduate text on complex dynamics is a must read. You're too smart to put that book off. All your questions can be answered by him--and better than I could

[Leo.W]

Having a little tiny break from my tons of homework, I contemplated the very case,
\( f(z)=z+\Gamma(z) \) (Conjugate to more general case \( a\ne0,f_a(z)=z+a\Gamma(z) \))
Since \( \Gamma(z) \) has no zeros, f has no fixed points. But f is not that like the one already constructed: \( z+e^z \) which has a well behaviored fixed point at -infinity.

My first attempt is to map the fixed point at \( \pm{i}\infty \) to 0, using \( T(z)=ln(z)/i,F=T^{-1}fT \). however, F won't have a second derivative at z=0, so I got stuck.

How would someone construct the family of superfunctions of this f? This will be challenging. I'd like to invite everyone to discuss this.

[Catullus]

Why did Leo.W upload his math as images? That does not seem very efficient.

[Leo.W]

Why did Leo.W upload his math as images? That does not seem very efficient.

I had no intention lol, at that time this forum only allows use [ tex ] to type formulae, and I at first wasn't aware of this
And last year I just graduated from high school, I had to take a very important exam called the entrance exam of university at that time I was quite busy to re-type these

[MphLee]

Congratulations Leo!
Now that you are at university you going next level, and you're already quite good!

[Leo.W]

Congratulations Leo!
Now that you are at university you going next level, and you're already quite good!

Thanks MphLee!

[bo198214]

I think Leo has really found some nugget here.
I can verify \(h(h(z)))=-z+z^2\) does not have a formal powerseries solution. Remember writing \((f\circ g)_n = \sum_{m=1}^n f_m {g^m}_n\) where \(g^m\) is the (normal) power of \(g\) and \({g^m}_n\) the nth coefficient of that power and \({g^m}_n = \sum_{k_1+\dots+k_m=n} g_{k_1}\cdots g_{k_m}\).
So in our case h needs to satisfy the equations for at least the first 3 coefficients of h:
\begin{align}
h_1 {h^1}_1 &= -1\\
h_1 {h^1}_2 + h_2 {h^2}_2 &= 1\\
h_1 {h^1}_3 + h_2 {h^2}_3 + h_3 {h^3}_3 &= 0
\end{align}
which is equivalent to:
\begin{align}
h_1^2 &= -1\\
h_1 {h}_2 + h_2 h_1^2 &= 1\\
h_1 {h}_3 + h_2 2h_1 h_2 + h_3 h_1^3 &= 0
\end{align}
Putting the first equation in the 3rd we get \(2h_2^2h_1 = 0\) which means either \(h_1=0\) or \(h_2=0\), which contradicts the first two equations.

As we don't have formal powerseries iteration formulas for \(f_1^q=1\) for some \(q\ge 2\), let's have a look at \(f_1=1\), here we can always find an iterative square root \(h\) with \(h_1=1\) however only in certain cases a root with \(h_1=-1\), which can be seen from the equations indexed by n:
\begin{align}
\sum_{m=1}^n h_m {h^m}_n &= f_n\\
h_1 h_n + h_n h_1^n + \sum{m=2}^{n-1} h_m {h^m}_n &= f_n \\
h_n (h_1+h_1^n) &= f_n - \sum_{m=2}^{n-1} h_m {h^m}_n
\end{align} 

Because with setting \(h_1=1\) the term \(h_1+h^n\) never vanishes, one can always calculate the next \(h_n\).
If however \(h_1=-1\) then e.g. at n=2 must 
\[f_2  = h_1 {h}_2 + h_2 h_1^2 = -h_2 + h_2  = 0\]
This (using \(h_1=1\) and \(h_1=-1\)) would find both roots \(a(z)=-z+z^3\) and \(b(z)=z-z^3\) of \(a\circ a=b\circ b\).
But \(b\) would be the standard solution which generally always exists (\(b_1=1\)) and \(a\) would only be the side solution which in general not always exists (\(a_1=-1\)).

I think - though did not verify - that you can generalize this to q-th roots.
I.e. you can solve the equation \(h^{\circ q}=f\) (for \(f_1=1\)) always with initial setting \(h_1=1\), and occasionally with setting \(h_1=e^{2\pi i k/q}\) for \(2\le k<q\).
This then would mean that one can consider our original \(f(z)=-z+z^2\) a solution of \(h\circ h=f\circ f\) with initial setting \(h_1=-1\) and the square root of \(f\) would be the fourth root of \(f\circ f\) with initial settings \(h_1=i\) and \(h_2=-i\) - which we saw already does not exist, though.

I am still thinking how to express this with the Julia function/logit but didn't find something convincing.
For me there are these analogies of \(b^t=e^{t\log(b)}\) with \(f^{\circ t}={\rm expit}[t\;{\rm logit}[f]]\).
So if one wants to get the other roots of say \(b^{\frac{1}{2}}\) one would do something \(e^{\frac{1}{2}(\log(b)+2\pi i k)}=b^{\frac{1}{2}}e^{\pi i k}=\pm b^{\frac{1}{2}}\). I just don't see yet what would be the equivalent with expit & logit.


Btw. the logit of \(f(z)=-z+z^2\) does not exist - or rather is the 0-function according to my calculations.
Also it quite seems all the solutions of \(f\circ h=h\circ f\) are \(h=f^{\circ n}\), also according to my calculations