Half-iterate exp(z)-1: hypothese on growth of coefficients

[Leo.W]

Hey, Leo.

I'm disagreeing with you--and until I see concrete evidence I will continue to do so. This is not the function in question at all. You are talking about a different function entirely.

You have developed a Taylor series about \(0\). This is a question about the divergent series that Gottfried has developed. You have made a function \(g\) that is holomorphic near zero (which I'm not sure about, as I'm pretty sure this contradicts Baker--Which Gottfried mentions). There exists no half iterate that is holomorphic near zero. So you've made a mistake, or Baker has made a mistake. We cannot expand an iteration in a neighborhood of a parabolic fixed point, unless that function is an LFT. Bo and myself even saw this recently in the Karlin Mcgregor paper, where it is written in the beginning.

By your logic, we've now solved \(f(z) = e^{z}-1\) and found a function \(f^{\circ t}(z)\) which is holomorphic about \(z = 0\). That just doesn't happen. Because you are brushing the paths of the Julia set and the attracting/repelling petal. The abel function cannot be expanded in a neighborhood of zero. And through common transformations, we can turn your iteration into an abel function, and now we have an abel function which is holomorphic for \(z \neq 0\) and \(|z| < \delta\). That doesn't happen. Abel functions are not holomorphic like this at parabolic points.

I can't speak to the accuracy of this graph, because I do not understand how you are constructing it. Additionally, a graph doesn't mean holomorphy. Just because it looks holomorphic, doesn't make it holomorphic. What is far more likely, is that you have expanded an asymptotic series, that converges fairly well, and iterated functional relationships to express how close of an asymptotic it is.

This is equivalent to saying that you've found a holomorphic iteration \(\exp^{\circ t}_{\eta}(z)\) such that this expression is holomorphic in \(z\) about \(e\). It's well known this isn't possible.

I apologize, but until I see some form of hard evidence I'll continue to disagree with you. Baker himself has a paper about no Taylor expansion for a half iterate of \(e^z -1\) about \(z=0\). Straight from the horses mouth. See the top answer here https://mathoverflow.net/questions/4347/...ar-and-exp

So unless you want to contradict baker, I'm sorry.

No need to feel sorry, James bro. It's nice to exchange ideas with brilliant guys~

I'd continue to contradict baker. So please show me evidences that showing there ain't a single function that's holomorphic about \(z=0\).
It's insufficient if you say that the Taylor series diverges or there can't be one single Taylor expansion at z=0. It exists and always exists, as we can compute it. A function has a divergent series don't mean the function doesn't exist, for example the series \(f(z)=\sum_{n=0}^{\infty}{n!z^n}\), it is simply the expansion of the function \(\frac{1}{z}e^{-1/z}\mathrm{Ei}(\frac{1}{z})\) at z=0, choosing the right branch cut and asymptotic expansion at z=0 will also give you \(1+z+2!z^2+3!z^3+\cdots\). And beyond that we have ways to analytic continue the series, at least you can't say it's not holomorphic because its series diverge, since there are tons of analytic continuations, especially the hypergeometric functions \(_p\mathrm{F}_q\).
But okay, maybe the definition of the term "holomorphic" diverges?

Yes, you're right, the photo doesnt speak a single truth, I only see a photo as a hint or something that showing there's probabilities, about the existance. From my main branch cut of half iterate \(g\) of \(e^z-1\), p1 shows \(g(g(z))-e^z+1\), because it's multivalued only one branch cut won't cover the whole plane, but still covers around z=0. p2 shows its derivative, and thus hinting it's analytic, real-to-real. 
and *if, just if, you brush off the multivaluedness, you're also banishing the existence of the Julia solution, abel solution and so on, because they're all multivalued, that would be nonsense for us to do researches into flows and iterations.

Why can't we expand parabolic iterations?
I think I have also a similar question to this, for example, it's known that \(f(z)=z+\frac{1}{z}\) has parabolic fixed point at infinity, then why \(f^2(z)=z+\frac{2}{z}+O(z^{-3})\) is parabolic and works and still holomorphic at infinity(what I mean is by conjugation it's still holomorphic at parabolic points)?
And why this 2nd iteration's asymptotic expansion, can't satisfy that \(f^2(z)=f^2(\frac{1}{z})\)?
It's just because the information is localized, and yet the series can't show that \(f^2(z)=f^2(\frac{1}{z})\)
I think there's something wrong, The paper will claim in a singlevalued style, and also it claims

However, there are two half-iterates (or associated Fatou coordinates \(\alpha(e^z - 1) = \alpha(z) + 1\)) that are holomorphic with very large domains. One is holomorphic on the complex numbers without the ray \(\left[ 0,\infty \right)\) along the positive real axis, the other is holomorphic on the complex numbers without the ray \(\left(- \infty,0\right]\) along the negative real axis. And both have the formal power series of the half-iterate \(f(z)\) as asymptotic series at 0.

Only taking singlevaluedness will cause so many limitations when you want a single iteration. Expanding the asymptotic series and Taylor series is equivalent if they're differentiable and has no poles or essential singularities and not at infinity, just because Taylor series is only a special type of wider series. I don't see any reason why it can't be holomorphic around z=0.

I'll just say, there's no holomorphic iterates, if singlevalued, and there won't even be any holomorphic inverse functions, because all inverse functions are multivalued or not-holomorphic except for linear functions, how will you invert z^-1 with holomorphic functions? The singlevaluedness is a limitation, so is the holomorphity. You can't get an iteration always holomorphic, but at least at some points, or in some domains they do, that's enough.
I can show you how to get a main branch cut of it if you're interested. And please show me some evidence that there's not a single half iterate(multivalued or partially-holomorphic). I'll be very appreciative.

@bo: Thank you bo, seems helpful. I think the asymp be proved very soon, looking forward~

[tommy1729]

ok here is my idea.

oh well or someone elses ofcourse there is alot of overlap.

anyway.

Lets say a taylor series with radius 0 but being analytic elsewhere is 

" analytic at some infinitesimals h "

As example f(x) = 2^^1 + 2^^2 x + 2^^3 x^2 + 2^^4 x^3 + ... 
is probably nowhere analytic.

But f(x)  = 2! + 3! x + 4! x^2 + 5! x^3 + ... is.

Now let that infinitesimal be a positive real infinitesimal.

that implies or assumes the taylor at 0 is an expansion at the positive axis.

now let tg(x) be the same as g(x) but the coefficients in absolute value.

( yes fake function ideas again )

so now we want to express tg(h) for positive real infinitesimal h...and have it as an upper bound.

To do so consider tg2(x) where tg2 is the same as g2 but the coefficients in absolute value.

g2(x) is equal to g(x) but expanded at 1 , so the taylor at 1 or g(x+1).

now

tg(h) coefficients a_n are given by the bounds 

tg(h) = tg2(x+1) with h = x , expanded at 0 by using binomium summing from tg2(x).

then 

g(x) coefficients are bounded by this tg(h) in absolute value.

that is the idea.

i know alot of assumptions and stuff.
but some logic.


regards

tommy1729

[bo198214]

I'd continue to contradict baker. So please show me evidences that showing there ain't a single function that's holomorphic about \(z=0\).

You can have a look into the paper of Baker.
It has something to do with that the parabolic fixed point is a point of non-normality (Julia set) and that the Julia set is a perfect set - means it has no isolated points.
I attach the file here for archiving purpose.

The general theory there is anyways that there are these Leau-Fatou-flower petals around a fixed point of multiplier 1, where you can define Abel-Function (or also called Fatou-Coordinates), hence you have fractional iterates there. And these are *asymptotically* analytic in the fixed point, i.e. the powerseries coefficients converge to those of the formal solution at the fixed point. But the normal thing is that these Abel-Functions/Fatou-Coordinates are different from each other (and hence the iterates). Only in exceptional cases (e.g. LFTs) they agree.

[bo198214]

Interestingly the iterative logarithm of \(e^x-1\) follows a similar pattern.
The iterative logarithm is an indicator whether all iterates at the fixed point are analytic, the defining equation is \(j(f(x))=f'(x)j(x)\).
It also satisfies \(j(x)=\frac{\partial f^{\circ t}(x)}{\partial t}\big|_{t=0}\).
It is called iterative logarithm because if you consider it a functional mapping f to j, i.e. \(j=\text{logit}[f]\) then you have \(\text{logit}[f^{\circ t}] = t\;\text{logit}[f]\).

Anyways this is the pattern:
   

with the advantage that it doesn't depend on t, which we took as t=1/2.

[Gottfried]

Interestingly the iterative logarithm of \(e^x-1\) follows a similar pattern.
(...)
with the advantage that it doesn't depend on t, which we took as t=1/2.

Wow. This makes me happy to see :-) !
Now the question comes up, why, for fractional multiplier \( h \) the exponential of the logit() keeps this form while for integer \( h \) the sinusoidal effect disappears and the known -entire- exponentail series comes to light... Hmmm. well, in my other picture we see, that the index \( \kappa \) "from where divergence begins", goes to infinity as the fractional part of \( h \) goes to zero... Perhaps we're going to find "the key"...

[Leo.W]

I'd continue to contradict baker. So please show me evidences that showing there ain't a single function that's holomorphic about \(z=0\).

You can have a look into the paper of Baker.
It has something to do with that the parabolic fixed point is a point of non-normality (Julia set) and that the Julia set is a perfect set - means it has no isolated points.
I attach the file here for archiving purpose.

The general theory there is anyways that there are these Leau-Fatou-flower petals around a fixed point of multiplier 1, where you can define Abel-Function (or also called Fatou-Coordinates), hence you have fractional iterates there. And these are *asymptotically* analytic in the fixed point, i.e. the powerseries coefficients converge to those of the formal solution at the fixed point. But the normal thing is that these Abel-Functions/Fatou-Coordinates are different from each other (and hence the iterates). Only in exceptional cases (e.g. LFTs) they agree.

Thank you, bo. I read the paper.

In fact, I don't think my idea contradicts Baker's.

First, when you take singlevalued functions into consideration, like I said, is a limitation for iterations, thus there aint any iterations for any functions except for linear functions and their conjugacies, and because only singlevalued holomorphic invertible functions are linear, only linear functions have iterations. If you banish holomorphity, then you can include rational linear functions.
I also knew some other theorems, one claims: if a singlevalued function f has a n-periodic fixed point a, alternatively \(f(a)\ne a,f^2(a)\ne a,\cdots,f^{n-1}(a)\ne a, f^n(a)=a\), then there lies no n-th iteration root of f. This theorem is enough to prove that \(e^z-1\) has no singlevalued half-iterate.
Just don't ingore the multivaluedness. Because these statements are not puissant when it comes to the real world where almost all functions are multivalued.

Second, as in the paper he claims the statement with the assumption \(f(z),f^2(z),\cdots,f^n(z)\) are all singlevalued, and then applied the theorem to \(f(z)=e^z-1\). But what if applied to \(f^{-1}(z)=\log(z+1)\)? In the inverting case, it doesnt satisfy the assumption anymore, and we only ask for \((f^{-1})^{-\frac{1}{2}}\). This is the way to generate a half-iterate of f.

I understand there's fatal contradictions about parabolic iterations especially involves \(\forall z\in\mathbb{U}_0, f^n(z)\to0\rightarrow (f^n)'(0)\to0\) but \((f^n)'(0)=1\)
This is a classic contradict, I think there're plenties of ways to explain this.
1) This is because the iterations converge so slowly. In other cases when multiplier isn't 1, all points in a neighborhood of a fixed point converges to the fixed point after t iterations, they converge at a exponential speed \(O(s^t)\). But multiplier=1 implies they converge at a very slow speed about \(O(t^{1-\delta})\) for some real \(delta>0\).
2) \((f^n)'(0)=1\) doesnt imply \(\lim_{n\to\infty}{(f^n)'(0)}=1\), all points in that neighborhood, by our converging speed would still culminate at the fixed point, this is pretty like another contradict:
All functions \(f_n(x)=\left\{\begin{matrix}0& x<0\\nx& x\in[0,\frac{1}{n}]\\1& x>\frac{1}{n}\end{matrix}\right.\) are continuous, but as n goes to infinity, it isnt continuous anymore.

I'd follow on. btw I wanna mention up about the iterative logarithm or Julia function, about its multivaluedness.
Now consider an iteration \(f^t\) generated by the schroder function, with multiplier s that \(|s|\ne1,0\), ignore the multivaluedness of schroder function or its inverse, the Julia function must be multivalued since \(f^t(z)=\sigma^{-1}(s^t\sigma(z))\) implies \(f^t(z)=f^{t+2k\pi i\log(s)^{-1}}(z)\), k any integer.
by petals, there should lies halfiterate of a function whose multiplier is -1, me and James discussed earlier. Then in this case, the iteration logarithm gets at a fixed point \(J(f(L))=J(L)=-J(L)=J(L)f'(L)\), showing \(J(L)=0\), then the logarithm fails in a neighborhood of L because \(f^t(z)\sim f^{t+2k}(z)\), the Julia function then should have 2 branch cuts at L.
Also, you can't define julia's function only by \(j(f)=jf'\), to distinct a julia function, you'd need to define an initial value, and claim it's not multiplied by any of the form \(\theta(\alpha(z))\) where theta is 1-periodic and alpha is the abel funct.

But the normal thing is that these Abel-Functions/Fatou-Coordinates are different from each other (and hence the iterates). Only in exceptional cases (e.g. LFTs) they agree.

Please lemme know what is this and why~ 

[JmsNxn]

Interestingly the iterative logarithm of \(e^x-1\) follows a similar pattern.
The iterative logarithm is an indicator whether all iterates at the fixed point are analytic, the defining equation is \(j(f(x))=f'(x)j(x)\).
It also satisfies \(j(x)=\frac{\partial f^{\circ t}(x)}{\partial t}\big|_{t=0}\).
It is called iterative logarithm because if you consider it a functional mapping f to j, i.e. \(j=\text{logit}[f]\) then you have \(\text{logit}[f^{\circ t}] = t\;\text{logit}[f]\).

Anyways this is the pattern:


with the advantage that it doesn't depend on t, which we took as t=1/2.

The half iterate of \(\log(z+1)\) has a similar asymptotic expansion at zero.

There are two half iterates at \(0\), they are holomorphic for \(g^-(z)\) for \(\Re(z) < 0\) and \(g^+(z)\) is holomorphic for \(\Re(z) > 0\) (which is developed using the attracting petal of \(\log(z+1)\)). Both functions produce the same asymptotic series at \(0\). And we are given \(g^+(g^+(z)) = g^{-}(g^{-}(z))= e^z-1\) while being two different solutions. But if you expand the asymptotic series about \(z \approx 0\) they produce the same asymptotic series. Per, again, Baker.

-----

@Leo

Also, I have no idea what you are talking about Leo. If a function isn't holomorphic at \(0\), there is no taylor series. If a function is holomorphic at \(0\), there is a taylor series. You are trying to say that it's holomorphic at 0 is in direct contradiction to baker. I literally just showed you the paper. Not to be harsh leo, but your solution is an asymptotic expansion. Unless you want to rewrite a hundred years of complex dynamics, you can't be right.

[Leo.W]

Interestingly the iterative logarithm of \(e^x-1\) follows a similar pattern.
The iterative logarithm is an indicator whether all iterates at the fixed point are analytic, the defining equation is \(j(f(x))=f'(x)j(x)\).
It also satisfies \(j(x)=\frac{\partial f^{\circ t}(x)}{\partial t}\big|_{t=0}\).
It is called iterative logarithm because if you consider it a functional mapping f to j, i.e. \(j=\text{logit}[f]\) then you have \(\text{logit}[f^{\circ t}] = t\;\text{logit}[f]\).

Anyways this is the pattern:


with the advantage that it doesn't depend on t, which we took as t=1/2.

The half iterate of \(\log(z+1)\) has a similar asymptotic expansion at zero.

There are two half iterates at \(0\), they are holomorphic for \(g^-(z)\) for \(\Re(z) < 0\) and \(g^+(z)\) is holomorphic for \(\Re(z) > 0\) (which is developed using the attracting petal of \(\log(z+1)\)). Both functions produce the same asymptotic series at \(0\). And we are given \(g^+(g^+(z)) = g^{-}(g^{-}(z))= e^z-1\) while being two different solutions. But if you expand the asymptotic series about \(z \approx 0\) they produce the same asymptotic series. Per, again, Baker.

-----

@Leo

Also, I have no idea what you are talking about Leo. If a function isn't holomorphic at \(0\), there is no taylor series. If a function is holomorphic at \(0\), there is a taylor series. You are trying to say that it's holomorphic at 0 is in direct contradiction to baker. I literally just showed you the paper.

So the halfiterate of e^z-1 is holomorphic at 0? Maybe my sentences are a lil bit tangled up, I apologize.
I'm not saying holomorphic at 0 is in direct contradiction to baker. I read the paper, it just says about holomorphity in the whole plane, not at a single point. Also it says the series diverge, but diverge is not the same as not holomorphic. Also the halfiterate can be analytic continued through 0, so to the whole plane.
Does "holomorphic" mean the same to both of us? I meant the analytic continued version has the same series(asymp or even Taylor) at 0.
I think the 2 different solutions should be different branch cuts of the same halfiterate. Plus there're more halfiterates.
I'm very curious now about how 

There are two half iterates at \(0\), they are holomorphic for \(g^-(z)\) for \(\Re(z) < 0\) and \(g^+(z)\) is holomorphic for \(\Re(z) > 0\) (which is developed using the attracting petal of \(\log(z+1)\)).

is derived? maybe you could give me a hint

Ps. what if you don't consider the petals of \(\log(1+z)\)? because \(e^z-1\) maps \(Re(z)<-M\) for very large M into the neighborhood of 0, that is why I defined the branch cuts, so that taking both petals of \(\log(1+z)\) and a "petal" strip from \(Re(z)<0\wedge |Im(z)|<1\) produces a holomorphic halfiterate at 0

[JmsNxn]

Leo, If a function is holomorphic in a neighborhood of \(0\), then it is expandable in a taylor series at \(0\).

If a functions IS NOT holomorphic in a neighborhood of \(0\), then it IS NOT expandable in a taylor series at \(0\).

What Baker is saying, is that there is a divergent series at \(0\)--THEREFORE, it is not holomorphic in a neighborhood of \(0\).

Which is correct, there are two solutions \(g^+\) and \(g^-\). One is analytically continuable to \(\mathbb{C}/(-\infty,0]\) and the other for \(\mathbb{C}/[0,\infty)\). Both have the same asymptotic expansion at \(z=0\). WHICH IS A DIVERGENT SERIES, THEREFORE THE FUNCTION ISN'T HOLOMORPHIC AT ZERO.

As to your question on the construction of these functions, I would point to milnor's book Dynamics in One Complex Variable, which describes Ecalle's construction of Abel functions. Which further, displays the limits of where we can define Abel functions.

I don't have the ability to read off to you every piece of literature I've read, you'd just have to read the literature. We cannot have an iteration \(f^{\circ t}(z)\) which is holomorphic in a \(\epsilon\)-ball about a parabolic fixed point \(f(p) = p,\,f'(p) = e^{2\pi i \theta}\) (*). It just cannot happen. And it is a very deep result to describe, that I would only butcher.

(*) Unless it is an LFT-- by which it wouldn't be a Euclidean mapping

[Leo.W]

Leo, If a function is holomorphic in a neighborhood of \(0\), then it is expandable in a taylor series at \(0\).

If a functions IS NOT holomorphic in a neighborhood of \(0\), then it IS NOT expandable in a taylor series at \(0\).

What Baker is saying, is that there is a divergent series at \(0\)--THEREFORE, it is not holomorphic in a neighborhood of \(0\).

(*) Unless it is an LFT-- by which it wouldn't be a Euclidean mapping

But James, holomorphic refers to differentiable, not the convergent Taylor series, it can have divergent series, as mostly hypergeometric functions do?
Here's another example, consider \(f(z)=\sum_{n\ge0}{_2\mathrm{F}_0(n,a;;c)z^n}\), this function diverges because all its coefficients refers to a divergent series, \(_2\mathrm{F}_0(n,a;;c)=\sum_{k\ge0}{\frac{(a)_k(n)_k}{k!}c^n}\), but still can be continued by \(f(z)=-cze^{\frac{z-1}{c}}\big(\frac{z-1}{c}\big)^{a-1}\Gamma(1-a,\frac{z-1}{c})\), choosing proper a, the function is holomorphic, has a divergent series in some sense but indeed a convergent series(Borel sum).

Ps. what if you don't consider the petals of log⁡(1+z)? because e^z−1 maps Re(z)<−M for very large M into the neighborhood of 0, that is why I defined the branch cuts, so that taking both petals of log⁡(1+z) and a "petal" strip from Re(z)<0∧|Im(z)|<1 produces a holomorphic halfiterate at 0

I wonder if you'd try this?