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Apropos "fix"point: are the fractional iterations from there "fix" as well?
#1
This is a tiny question which bothered me from time to time, but I took it always as somehow given.

Now by the current discussions this comes up again.  Practically this has a relevance, when I thought to ask my brother, who has a 3-D-printer, for a 3 D model of fractional iteration in the complex numbers, where the height \( h \in \mathbb R \) gives the 3rd dimension, and the integer heights are visualized simply like dots at the floors of the stages in a house. The fractional iterates from one point to another follow then some "noodle" from a point in first floor to the according point in second floor. 
(Our chosen interpolation-method define the form of the "noodles".) 

The noodles from a fixpoint in the first floor to the same point in the second floor - how is it shaped? Straight, vertical? I remembered I've never consciously seen a discussion of that, I gave it a chance, that it might be simply taken as an axiom. But perhaps, with the obvious informations around from Milnor or Devaney: they might have derived this, or at least might have explicitely mentioned this. (Note, that the "noodles" connecting periodic points have/must have an "exploding", chaotically divergent, form - although I've as well not seen this discussed in the material I have/had made available for me.)            

So in this sense:
  • are the fractional iterates of a fixpoint "fix" as well/ are they all identically the same coordinate?     
And    
  • is this taken so-to-say axiomatically, or is it a consequence of something?

Gottfried
Gottfried Helms, Kassel
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#2
I'd say that there is nothing algebraically that forces \(f^t(p)=p\) for every \(t\in\mathbb R\) when \(p\) is a fixed point of \(f^1\).

If I understand what you are asking, you are asking if the following conjecture is a theorem

Conjecture: let \(f:X\to X\) be a function and \(f(p)=p\) a fixed point. Then, given a way to non-integer iterating \(f\) using a superfucntion, i.e. a function \(\varphi(t,x)\) s.t. \(\varphi(0,x)=x\) and \(\varphi(t+1,x)=f(\varphi(t,x))\), we have that the orbit of the fixedpoint \(p\) is constant
$$\varphi(t,p)=p$$

I'd say there is nothing forcing this, at least algebraically.

First of all remember that we can build the superfunction piecewise, a la Andrew Robbins, obtaining k-differentiability on the endpoints.

We can define \(\Phi(n):=p\), the fixed point for every integer \(n\in\mathbb N\), then define \(\Phi(t)\) for \(t\in [0,1]\) to be an arbitrary loop in \(\gamma:[0,1]\to X\) that connects \(p\) to itself. You can then evaluate $$\Phi(x):=f^{{\rm floor}(x)}(\gamma (\{x\})) $$.

Note that the constant solution \(\forall x.\,K(x)=p\) is a solution of the superfunction equation.

Maybe only adding constraints can force the orbit, the noodle, that connects the fixed point to itself at different iteration height, to be constant/straight line.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#3
MphLee -

ah, thanks.
Your formulation (the conjecture) sounds good - maybe I should even incorporate it into my question later.
Bythe time of thinking about it I got the vague idea, that at least for attracting fixpoints it should be impossible for any iteration - be it as fractionally small as we want - that it could "swim against the attraction field" - so to say... Surely a very vague imagination and possibly a circular thinking (or a knotted "noodlolar" one Smile )

I'll take some time to think about this further...

Gottfried
Gottfried Helms, Kassel
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#4
Really nice Gottfried... swimming against the attraction field... interesting.
We need an expert opinion on this. I'm sure is some trivial proposition one can find in any book on dynamics... but I'm too ignorant. I'm sure James can settle this in 1 second.

To think about it... if the orbit crosses the basin of attraction of the fixed point... maybe it is not necessary that from that moment each successive point of the orbit has a distance from the fixed point that is monotone decreasing.

Conjecture 2: let \(f:X\to X\), \(X\) a metric space, \(p\in X\) attracting and \(I\subseteq X\) its basin of attraction. Let \(\chi:[0,\infty)\to X\) be a \(x_0\)-initialized superfucntion that is continuous, i.e. \(\chi(0)=x_0\) and $$\chi(t+1)=f(\chi(t))$$
if exists a \(r\in [0,\infty)\) s.t. \(\chi({r})\in I\) is in the basin of attraction of \(p\) then
$$D(t):=d(\chi(t+r),p)$$
is monotone decreasing. I.e. for \(r<t\) we have \(\chi(t)\) gets closer and closer to \(p\) and never oscillates back.

I'd say that this is false and that an orbit that crosses the basin is doomed to reach the fixed point but it can reach it in oscillating ways...I conjecture...
if this is true there will be moments where the orbit is swimming against the attraction field... and we have no problems...

At this point it can be interesting to understand how the map \(f:X\to X\) maps \(p\) based loops \(\gamma:{}S^1\to X\) to loops \(f\circ \gamma:{}S^1\to X\) since each \(p\)-initialized superfunction \(\chi:[0,\infty)\to X\) is completely determined by it's first loop \(\gamma:[0,1)\to \to X\) and all the other loops are determined by iteration \(\chi|_{[n,n+1)}=f^n\circ \gamma:[n,n+1) \to X\).

Question. For example... if \(f\) is continuous... then is \(\gamma\) homotopic to \(f\circ \gamma\)?
It would be interestin to study the map that each superfucntion induces on the fundamental group of \(X\): the map \({\boldsymbol\chi}:\mathbb N \to \pi_1(X) \)
$${\boldsymbol\chi}(n)=[f^n\circ \chi|_{[0,1)}]$$

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#5
So, this is actually a defining property of the standard Schroder iteration. But it's a little difficult to fully flesh out why.  Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them.

If you iterate locally about a fixed point, and your solution satisfies \(f^t(p) = p\), then the iteration is expressible via Shroder iteration (provided that \(|f'(p)| \neq 0,1\)).  For convenience, assume that \(|f'(p)| < 1\).

You can actually prove this pretty fast. Assume that \(f^t(x)\) is a super function in \(t\) about a fixed point \(p\), and \(x\) is in the neighborhood of \(p\). Assume that \(f^t(p) = p\).  Well then:

$$
\Psi(f^{t+1}(x)) = \lambda \Psi(f^t(x))\\
$$

So that:

$$
\theta(t) = \frac{\Psi(f^t(x))}{\lambda^t}\\
$$

And we know that there must be some 1-periodic function \(\theta(t)\), such that:

$$
f^t(x) = \Psi^{-1}\left(\lambda^{t}\theta(t) \Psi(x)\right)\\
$$

In fact, any periodic function will work fine here, and will have a constant noodle, will be a super function, but will not be Schroder iteration.

Now let's add one more constraint, let's say that \(f^{t}(f^{s}(x)) = f^{t+s}(x)\). Well then, we have that \(\theta\) must be constant. By which, we are guaranteed that it's a Schroder iteration...


So to clarify. Any fractional iteration with a constant noodle is a Schroder iteration. But there are plenty of superfunctions of \(f\) which have a constant noodle, but in turn, they aren't fractional iterations then (don't satisfy the semi-group law).

Now to return to Gottfried's question a bit more robustly. It depends on how we qualify fractional iterations, as to whether \(f^t(p)\) is constant or not. But I'll take the classic approach, and say that \(f^t(x) : \mathbb{R}^+ \to \mathbb{R}^+\) and that it satisfies the semigroup property.

Every element \(q \in \mathbb{Q}^+\) has some value \(m\) such that \(f^{mq}(p) = p\).  And therefore:

$$
f^{mq}(x) = \Psi^{-1}(\lambda^{mq} \Psi(x))\\
$$

This is where it gets fun; take an \(|x-p| < \delta\) so that \(x\) is in the basin about \(p\). Well then:

$$
\lim_{m\to\infty} \frac{f^{mq}(x) - p}{\lambda^{mq}} \to \Psi(x)\\
$$

Or rather:

$$
\lim_{t \to \infty} \frac{f^t(x) - p}{\lambda^t} \to \Psi(x)\\
$$

Hmmmmmm.... Well this just means that \(f^t\) is the Schroder iteration... Any non Schroder iteration, this limit will not converge and will continue to oscillate (there'll be a 1-periodic \(\theta\) somewhere in there).

This shows two things, any iteration about a fixed point, which has the semi-group property is Schroder's iteration, and by result has a constant noodle.


BUTTT

Superfunctions, don't have to have a constant noodle--but they don't have the semi-group property then.


Now you may be wondering how Kneser works then... Kneser as a fractional iteration is not holomorphic about any fixed point or periodic point, it has a singularity there. So Kneser sneaks past the above proof by not being a local iteration.
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#6
Man, this is one of the most beautiful pieces of math I discovered recently.
This makes many things really clear. The flow of the logic... is always unpredictable in your posts.. like you user black magics tricks... there is always something unexpected but this time everything is beautiful, like a painting xD

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Reply
#7
(07/11/2022, 08:45 PM)JmsNxn Wrote: So, this is actually a defining property of the standard Schroder iteration. But it's a little difficult to fully flesh out why.  Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them.
(...)
Yes, my take of this has been: by the analysis of the power series including the conjugacy we know, that the powerseries for the Schroder-function has no constant term, and the original function \(f(z) \) at \( z=t \) (where \(t\) is the fixpoint), gives \( \sigma(t-t) = 0\) and is thus zero, so that any fractional iteration has \( \lambda^h \cdot \sigma(t-t) = \lambda^h \cdot 0 = 0 \) and by this \( \sigma°^{-1}(0) +t = t \)
is constant for every \(h \).
So the use of the Schroder-mechanism produces the constant \(t \) at each (integer or fractional or even complex) iteration height. (Interesting that you say it either the other way: if we want have the constant \(t\) at the fractional iteration then that implies that we'll need the Schroeder-mechanism, this looks good, but I'll have to chew on it.)

What is/was the unclear point for me is: is this an arbitrary decision to choose a model which has this constancy or are there more/deeper reasons, that the option for this model is required to keep things consistent.

You introduce then the property of being a semigroup, expressed by \( f°^a(f°^b(z)) = f°^{a+b}(z) \) - that would be something I would say, is demanded for any implementation of fractional iteration, so this would give a convincing argument that we do not have the freedom to choose the model, (it's not like with the axiom of the parallels in geometry, where we simply decide for another (legitimate) geometry if we decide for another version of the axiom).

(07/11/2022, 08:45 PM)JmsNxn Wrote: (...)
Superfunctions, don't have to have a constant noodle--but they don't have the semi-group property then.
I'd like this argument concerning the semigroup-property being valid, which you gave in the following, only at the moment I couldn't get it completely yet, I'll come back to this if I'm clearer with this.

Thanks -
Gottfried
Gottfried Helms, Kassel
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#8
(07/11/2022, 08:45 PM)JmsNxn Wrote: So, this is actually a defining property of the standard Schroder iteration. But it's a little difficult to fully flesh out why.  Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them.

If you iterate locally about a fixed point, and your solution satisfies \(f^t(p) = p\), then the iteration is expressible via Shroder iteration (provided that \(|f'(p)| \neq 0,1\)).  For convenience, assume that \(|f'(p)| < 1\).

You can actually prove this pretty fast. Assume that \(f^t(x)\) is a super function in \(t\) about a fixed point \(p\), and \(x\) is in the neighborhood of \(p\). Assume that \(f^t(p) = p\).  Well then:

$$
\Psi(f^{t+1}(x)) = \lambda \Psi(f^t(x))\\
$$

So that:

$$
\theta(t) = \frac{\Psi(f^t(x))}{\lambda^t}\\
$$

And we know that there must be some 1-periodic function \(\theta(t)\), such that:

$$
f^t(x) = \Psi^{-1}\left(\lambda^{t}\theta(t) \Psi(x)\right)\\
$$

In fact, any periodic function will work fine here, and will have a constant noodle, will be a super function, but will not be Schroder iteration.

Now let's add one more constraint, let's say that \(f^{t}(f^{s}(x)) = f^{t+s}(x)\). Well then, we have that \(\theta\) must be constant. By which, we are guaranteed that it's a Schroder iteration...


So to clarify. Any fractional iteration with a constant noodle is a Schroder iteration. But there are plenty of superfunctions of \(f\) which have a constant noodle, but in turn, they aren't fractional iterations then (don't satisfy the semi-group law).

This is not acceptable for me.
Not formal , detailed and general enough.

It is actually quite simple 

even without fixpoints.

\(f^{t}(f^{s}(x)) = f^{t+s}(x)\). 

let theta(v) be a one periodic function.

so do we get 

\(f^{t+theta(t)}(f^{s+theta(s)}(x)) = f^{t+s+theta(s+t)}(x)\). 

well that would require 

theta(v+1) = theta(v)


and

theta(t + s) = theta(t) + theta(s)

so theta must be constant.

keywords :  cauchy functional equation , axiom of choice , linear function , non-constructive , non-periodic.


So we get a (local ?) uniqueness criterion.

This has many consequences.

(for instance probably if you use iterations of another function to get to iterations of yours , it is required that the other function has the  semi-group property.)

So we have uniqueness up to convergeance speed ofcourse.

This should be on page 1 of any dynamics book !


regards

tommy1729
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#9
(07/13/2022, 12:27 PM)tommy1729 Wrote:
(07/11/2022, 08:45 PM)JmsNxn Wrote: So, this is actually a defining property of the standard Schroder iteration. But it's a little difficult to fully flesh out why.  Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them.

If you iterate locally about a fixed point, and your solution satisfies \(f^t(p) = p\), then the iteration is expressible via Shroder iteration (provided that \(|f'(p)| \neq 0,1\)).  For convenience, assume that \(|f'(p)| < 1\).

You can actually prove this pretty fast. Assume that \(f^t(x)\) is a super function in \(t\) about a fixed point \(p\), and \(x\) is in the neighborhood of \(p\). Assume that \(f^t(p) = p\).  Well then:

$$
\Psi(f^{t+1}(x)) = \lambda \Psi(f^t(x))\\
$$

So that:

$$
\theta(t) = \frac{\Psi(f^t(x))}{\lambda^t}\\
$$

And we know that there must be some 1-periodic function \(\theta(t)\), such that:

$$
f^t(x) = \Psi^{-1}\left(\lambda^{t}\theta(t) \Psi(x)\right)\\
$$

In fact, any periodic function will work fine here, and will have a constant noodle, will be a super function, but will not be Schroder iteration.

Now let's add one more constraint, let's say that \(f^{t}(f^{s}(x)) = f^{t+s}(x)\). Well then, we have that \(\theta\) must be constant. By which, we are guaranteed that it's a Schroder iteration...


So to clarify. Any fractional iteration with a constant noodle is a Schroder iteration. But there are plenty of superfunctions of \(f\) which have a constant noodle, but in turn, they aren't fractional iterations then (don't satisfy the semi-group law).

This is not acceptable for me.
Not formal , detailed and general enough.

It is actually quite simple 

even without fixpoints.

\(f^{t}(f^{s}(x)) = f^{t+s}(x)\). 

let theta(v) be a one periodic function.

so do we get 

\(f^{t+theta(t)}(f^{s+theta(s)}(x)) = f^{t+s+theta(s+t)}(x)\). 

well that would require 

theta(v+1) = theta(v)


and

theta(t + s) = theta(t) + theta(s)

so theta must be constant.

keywords :  cauchy functional equation , axiom of choice , linear function , non-constructive , non-periodic.


So we get a (local ?) uniqueness criterion.

This has many consequences.

(for instance probably if you use iterations of another function to get to iterations of yours , it is required that the other function has the  semi-group property.)

So we have uniqueness up to convergeance speed ofcourse.

This should be on page 1 of any dynamics book !


regards

tommy1729

So we want f^[s+t](z) = f^[s](f^[t](z)) = f^[t](f^[s](z))

2sinh^[s+t](x) has this property for real x. Or 2sinh^[s+t](z) has this property for complex z around the real axis.  

That is if we use the koenings function around the fixpoint 0 of sinh for the construction.

It is easy to show that ln ( 2sinh^[s+t](exp(x)) ) has the same property.

and by induction ln^[n] ( 2sinh^[s+t](exp^[n](x)) ) for any positive integer n also has the property.

notice ln^[n] ( 2sinh^[s+t](exp^[n](x)) ) is also analytic on the real line.

Letting n go to +oo gives the solution on the real line :

exp^[s+t](x) = lim_n  ln^[n] ( 2sinh^[s+t](exp^[n](x)) )

and this limit has the same property !!!


see also :

Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1.
Say is an Abel function of e^x-1, then



is an Abel function of . This should also work for beta being the Abel function of .

This approach is actually equivalent to the "change of base" approach we considered here on the forum, also Walker [2] used a similar method. (But I am in the moment to lazy to detail how exactly they imply each other.) It is still open whether it is analytic, but it is proven to be infinitely differentiable in [2].

[1] Lévy, P. (1927). Sur l'itération de la fonction exponentielle. C. R., 184, 500–502.
[2] Walker, P. L. (1991). Infinitely differentiable generalized logarithmic and exponential functions. Math. Comput., 57(196), 723–733.


let us verify that it is indeed an iteration of exp, i.e that indeed satisfies:
and .

we get
.

and
because towards infinity gets arbitrarily close to .

Basically thats the iteration equivalent of the Abel function Lévy proposes:

where is the Abel function of (or in Lévy's case ).

The superfunction is then (the inverse of ):



which is the same as Tommy's superfunction.


So my 2sinh solution has this uniqueness criterion making it quite special.

Other methods i proposed that are similar * real entire functions with unique real fixpoint at 0 with real derivative larger than 1 , fast asymptotic to exp(x) and some minor details * also carry this property ....

SO they are unique !!

This also explains that if using others than 2sinh(x) , it must give the same function. THEREFORE the nonreal fixpoints of 2sinh are not really an issue.

Also the 2sinh(x) is easy to handle since its maclauren series has all nonnegative derivates. 


***

even more interesting is that the 2sinh method can be extended to bases > exp(1/2).

And probably by analytic continuation to all bases larger than eta ( e^(1/e) ).

but for bases =< exp(1/2) we get issues with multiple fixpoints or derivatives at the fixpoints.

THAT is also the reason why I proposed alternative similar methods.

Since those alternatives agree on bases larger than exp(1/2) by the uniqueness criterion , they must be the 2sinh method extended to lower bases !!

I hope that is clear to everyone.


Regards

tommy1729

ps wiki will not accept the 2sinh method in the tetration section , even when it does mention non-C^oo solutions.
meh !
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#10
Typos and errors fixed.

I hope everyone forgives me, but I need to put all of this on solid, formal grounds. I need to prove every step.
Now I'm doing it for James' post... but I've done only for the first half of it.

Quote:So, this is actually a defining property of the standard Schroeder iteration. But it's a little difficult to fully flesh out why.  Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them.
If you iterate locally about a fixed point, and your solution satisfies \(f^t(p) = p\), then the iteration is expressible via Shroeder iteration (provided that \(|f'(p)| \neq 0,1\)).  For convenience, assume that \(|f'(p)| < 1\).
You can actually prove this pretty fast. Assume that \(f^t(x)\) is a super function in \(t\) about a fixed point \(p\), and \(x\) is in the neighborhood of \(p\). Assume that \(f^t(p) = p\).  Well then:
$$
\Psi(f^{t+1}(x)) = \lambda \Psi(f^t(x))\\
$$
So that:

$$
\theta(t) = \frac{\Psi(f^t(x))}{\lambda^t}\\
$$

And we know that there must be some 1-periodic function \(\theta(t)\), such that:
$$
f^t(x) = \Psi^{-1}\left(\lambda^{t}\theta(t) \Psi(x)\right)\\
$$
In fact, any periodic function will work fine here, and will have a constant noodle, will be a super function, but will not be Schroeder iteration.

Now let's add one more constraint, let's say that \(f^{t}(f^{s}(x)) = f^{t+s}(x)\). Well then, we have that \(\theta\) must be constant. By which, we are guaranteed that it's a Schroeder iteration...

So to clarify. Any fractional iteration with a constant noodle is a Schroeder iteration. But there are plenty of superfunctions of \(f\) which have a constant noodle, but in turn, they aren't fractional iterations then (don't satisfy the semi-group law).

I'm trying to scan the deep meaning behind your moves... what is happening in the background. I'd like a feedback so I can go on to the next step and then to Tommy's observations.

When you ask for superfunction about a fixed point \(p\) you, in fact, are actually taking a complex object. You are not just taking A superfunction but an entire family of superfunctions \({\varphi}_{x\in U}:\mathbb R^+\to X\), superfunctions of \(f:X\to X\), parametrized by a small \(f\)-stable neighborhood of the fixed point \(p\in U\subseteq X\). In other words you asking for a \(\varphi:\mathbb R\times U\to U\), i.e. a family \(\{\varphi_x\}_{x\in U}:\mathbb R^+\to X\) of superfunctions of \(f\) restricted to the neighborhood of the fixed point \(p\) (that should be the local part about it).

Notice that, as recent posts by me, James and Tommy are making clear... being a superfunction is something weaker than being an iteration, like being an interpolation is weaker than being a superfunction. In other words we just ask for the family \(\varphi_x\) to satisfy the mild conditions $$\varphi_x(t+1)=f(\varphi_x(t))\quad \quad \varphi_x(0)=x$$

No other coherence conditions, like the semigroup property... a property that is very strong. Note that what I write as \(\varphi_x(t)\) we could write as \(\varphi(t,x)\) and James denotes it as \(\varphi_x(t)=f^t(x)\). I use \(\varphi_x\) notation to emphasize that it is a FAMILY OF POSSIBLE SOLUTIONS of the equation \(\varphi\circ  S=f\circ \varphi\). We can think of this as an assignment \(\varphi_{-}:U\to [S,f]\) where \([S,f]\) is the space of all the superfunctions of \(f\)...

Now... we want to know if \(\varphi_p(t)\) is a constant path or... if it loops around \(p\in X\) infinitely many times. The answer James gives is... it is constant if the family \(\varphi_x\) satisfies the stronger condition of being an iteration: i.e. $$\varphi_{\varphi_x(t)}(s)=\varphi(s,\varphi(t,x))=\varphi(s+t,x)=\varphi_x(s+t)$$

The reasoning, subject to multipliers conditions to ensure convergence of Schroeder functions, seems to begin in the following way.
Given a family of superfunctions of \(f\) locally about a fixed point \(p\in U\), i.e. given a family of superfunctions \(\varphi_{x\in U}\) and given a function \(\Psi\in [f,{\rm mul}_\lambda]\), a solution of the Schroeder eq. of \(f\) we define another family
$$A_x=\Psi\circ \varphi_x$$
this time this family lies in the space \([S,{\rm mul}_\lambda]\) that is, each \(A_x\) is a superfunction of multiplication \({\rm mul}_\lambda\) by \(\lambda\).
$$A_x(t+1)=\lambda A_x(t)$$
this is what James denotes by \(A_x(t)=\Psi(f^t(x))\).

Remember that we have a family of functions in the space  \([S,{\rm mul}_\lambda]\) and remember, in this space there is a special solution: exponentiation \(\exp_\lambda \in [S,{\rm mul}_\lambda]\). What James does now is measuring the difference between the \(A_x\) and \(\exp_\lambda\). They differ by a periodic function.
$$\theta_x(t)\lambda^t=A_x(t)$$
If the periodic function is constant \(\theta_x(t)=K\) then \(K\lambda^t=\Psi(\varphi_x(t))\) and \(\varphi_x(t)=\Psi^{-1}(K\lambda^t)\) is purely obtained by the Schroeder function. James calls it the Schroeder iteration.



At this level of detail each \(\theta_x\) can be a different periodic function. What I managed to prove is that... if we ask the family \(\phi_x\) tho satisfy the additional condition \(\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)\), weaker than being an iteration but implied by it we obtain that if \(x\sim_f y\) then \(\theta_x=\theta_y\)

Proposition. let \(\varphi_{-}:U\to [S;f|_U]\) be a family of superfunctions of \(f|_U\) for \(U\subseteq X\) an \(f\) stable subset, let \(\Psi \in [f, \lambda ]\) and define the family of 1-periodic function $$\theta_x(t)=\frac{\Psi(\varphi_x(t))}{\lambda^t}$$ we prove that
  • if for each \(x\in U\) we have \(\varphi_x(0)=x\) (a section); This condition is not needed
  • for each \(t\) \(\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)\) (weakly iterative)
then for \(x,y\in U\) are \(f\) connected, i.e. \(f^n(x)=f^m(y)\) for some \(m,n\in\mathbb N\), $$\theta_x(t)=\theta_y(t)$$

Proof. assume \(\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)\). We prove that \(\theta_x=\theta_{f^n(x)}\). Compute \(\theta_{f^n(x)}(t)=\frac{\Psi(\varphi_{f^n(x)}(t))}{\lambda^t}=\frac{\Psi(\varphi_{x}(t+n))}{\lambda^t}=\theta_x(t+n)\) by assumption. \(\theta_{f^n(x)}(t)=\theta_{x}(t)\) by periodicity. If \(f^n(x)=f^m(y)\) then \(\theta_x(t)=\theta_y(t)\) by transitivity of the identity. \(\square\)

At this point I can't wait to translate the rest of the argument...
Maybe \(\phi_x\) satisfies semigroup, is an iteration, iff \(\theta_x\) is constant. If this was the case, as Tommy seemed to suggest elsewhere... this is something like a criterion to select among the space of superfunctions the special ones... something like an uniqueness criterion.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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