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WARNING: Religious topic discussed!!!
Is real tetration dependent on complex tetration? Could real tetration exist without complex tetration?
Daniel
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(08/16/2022, 09:55 AM)Daniel Wrote: WARNING: Religious topic discussed!!!
Is real tetration dependent on complex tetration? Could real tetration exist without complex tetration?
Very difficult. There exists Smooth real tetration; but it isn't analyticand there are uncountably many of these. The trouble is, we can also make uncountably many real analytic tetrations by just adding a well behaved enough \(\theta\) mapping. By which; these wouldn't succumb to the complex uniqueness conditions though.
So No, real tetration is not dependent on complex tetration. It's just that complex tetration allows us the only confirmed constructionand allows us the uniqueness conditions we so love.
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(08/17/2022, 01:56 AM)JmsNxn Wrote: (08/16/2022, 09:55 AM)Daniel Wrote: WARNING: Religious topic discussed!!!
Is real tetration dependent on complex tetration? Could real tetration exist without complex tetration?
Very difficult. There exists Smooth real tetration; but it isn't analyticand there are uncountably many of these. The trouble is, we can also make uncountably many real analytic tetrations by just adding a well behaved enough \(\theta\) mapping. By which; these wouldn't succumb to the complex uniqueness conditions though.
So No, real tetration is not dependent on complex tetration. It's just that complex tetration allows us the only confirmed constructionand allows us the uniqueness conditions we so love.
I want to confirm something.
1. Is \( R^\infty \) not enough to uniquely determine an real tetration?
2. Are there two (mainstream) real tetration sexp(base, height) defined in two different ways, one of which satisfies \( e^{e} \leq b \leq e^{\frac{1}{e}} \) and the other satisfies \( b > 1 \), and the natural extension of the latter is Kneser?
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Atheist answer :
Everything is related by an appropriate oneperiodic function theta(s).
sexp_1(s) = sexp_2(s + theta(s))
There is no superior or magic.
Spiritual answer :
It depends on what properties you want and if they exist and are proven for a given case.
All is one.
Everything is connected.
Common answer :
tet what ?
My answer :
very vague and general question.
Thats ok , but for math we often prefer precise questions, despite the love of generalizations.
In particular if we want formal answers.
But one more thing :
The semigroup isom is a uniqueness criterion !
So the question is, is the semigroup isom solution analytic or not ?
In fact for dynamics in general , I am unaware of examples of nowhere analytic semigroup iso solutions.
I posted about this many times.
If our conditions are not uniqueness conditions then usually they can both be analytic or nonanalytic.
For more info, read the whole forum
regards
tommy1729
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03/31/2023, 07:05 PM
(This post was last modified: 03/31/2023, 07:08 PM by tommy1729.)
(03/31/2023, 02:01 PM)Ember Edison Wrote: (08/17/2022, 01:56 AM)JmsNxn Wrote: (08/16/2022, 09:55 AM)Daniel Wrote: WARNING: Religious topic discussed!!!
Is real tetration dependent on complex tetration? Could real tetration exist without complex tetration?
Very difficult. There exists Smooth real tetration; but it isn't analyticand there are uncountably many of these. The trouble is, we can also make uncountably many real analytic tetrations by just adding a well behaved enough \(\theta\) mapping. By which; these wouldn't succumb to the complex uniqueness conditions though.
So No, real tetration is not dependent on complex tetration. It's just that complex tetration allows us the only confirmed constructionand allows us the uniqueness conditions we so love.
I want to confirm something.
1. Is \( R^\infty \) not enough to uniquely determine an real tetration?
2. Are there two (mainstream) real tetration sexp(base, height) defined in two different ways, one of which satisfies \( e^{e} \leq b \leq e^{\frac{1}{e}} \) and the other satisfies \( b > 1 \), and the natural extension of the latter is Kneser?
1. No
We can pick a one periodic function theta(s) such that
sexp_1(s + theta(s)) = sexp_2(s)
where one of them is analytic and the other is only C^{oo}.
2. Depends what you consider mainstraim.
For the bases you mention we have 2 fixpoints , so we have 2 expansions with koenigs function at each fixpoint.
I recently added a way to unify the two fixpoints so that we have tetration between them , but not analytic at them and not beyond them (*).
And then we have kneser which uses the smallest (nonreal) fixpoint(s) of ln_b(z) = z.
( * https://math.eretrandre.org/tetrationfor...p?tid=1652 )
and alot of methods that do not consider fixpoints ...
( my fav : the gaussian method and Peter Walkers method )
And ofcourse many open questions.
regards
tommy1729
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03/31/2023, 11:43 PM
(This post was last modified: 03/31/2023, 11:45 PM by Ember Edison.)
(03/31/2023, 07:05 PM)tommy1729 Wrote: 1. No
We can pick a one periodic function theta(s) such that
sexp_1(s + theta(s)) = sexp_2(s)
where one of them is analytic and the other is only C^{oo}.
2. Depends what you consider mainstraim.
For the bases you mention we have 2 fixpoints , so we have 2 expansions with koenigs function at each fixpoint.
I recently added a way to unify the two fixpoints so that we have tetration between them , but not analytic at them and not beyond them (*).
And then we have kneser which uses the smallest (nonreal) fixpoint(s) of ln_b(z) = z.
Thank you very much.
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(03/31/2023, 11:43 PM)Ember Edison Wrote: (03/31/2023, 07:05 PM)tommy1729 Wrote: 1. No
We can pick a one periodic function theta(s) such that
sexp_1(s + theta(s)) = sexp_2(s)
where one of them is analytic and the other is only C^{oo}.
2. Depends what you consider mainstraim.
For the bases you mention we have 2 fixpoints , so we have 2 expansions with koenigs function at each fixpoint.
I recently added a way to unify the two fixpoints so that we have tetration between them , but not analytic at them and not beyond them (*).
And then we have kneser which uses the smallest (nonreal) fixpoint(s) of ln_b(z) = z.
Thank you very much.
your welcome
regards
tommy1729
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04/02/2023, 03:22 AM
(This post was last modified: 04/02/2023, 03:23 AM by JmsNxn.)
(03/31/2023, 02:01 PM)Ember Edison Wrote: (08/17/2022, 01:56 AM)JmsNxn Wrote: (08/16/2022, 09:55 AM)Daniel Wrote: WARNING: Religious topic discussed!!!
Is real tetration dependent on complex tetration? Could real tetration exist without complex tetration?
Very difficult. There exists Smooth real tetration; but it isn't analyticand there are uncountably many of these. The trouble is, we can also make uncountably many real analytic tetrations by just adding a well behaved enough \(\theta\) mapping. By which; these wouldn't succumb to the complex uniqueness conditions though.
So No, real tetration is not dependent on complex tetration. It's just that complex tetration allows us the only confirmed constructionand allows us the uniqueness conditions we so love.
I want to confirm something.
1. Is \( R^\infty \) not enough to uniquely determine an real tetration?
2. Are there two (mainstream) real tetration sexp(base, height) defined in two different ways, one of which satisfies \( e^{e} \leq b \leq e^{\frac{1}{e}} \) and the other satisfies \( b > 1 \), and the natural extension of the latter is Kneser?
I'd just like to add to Tommy's correct answer.
Say I take a function which is infinitely differentiable on \(\mathbb{R}\)... let's call it \(f\).
Then \(f\) is unique because it has unique derivatives everywhere on \(\mathbb{R}\).
Now let's take a nowhere holomorphic function \(g : \mathbb{R} \to \mathbb{R}\); such that \(g :\mathbb{C}_{\Im(z) > 0} \to \mathbb{C}_{\Im(z)> 0}\) is holomorphic. Let's assume that \(g(\mathbb{R}) = 0\). Well then; if we try to move away from the real line with \(f\), then \(f+g\) is also a candidate (this is the same thing as the theta discussion). So we lose uniqueness.
This is pretty much the trouble with the beta method outside of the ShellThron regionand why for \(b = e\) we're actually NOWHERE holomorphic, despite finding infinitely differentiable solutions.
This is one of those moments where you have to remember that COMPLEX DIFFERENTIABILITY > DIFFERENTIABILITY. Once we lose complex differentiability; a lot of tetration can be pretty bananas. Especially; if say; I try to approximate a complex differentiable \(F(s)\) using \(f(x)\); near \(x \in \mathbb{R}\); \(f(x) \approx F(s)\), but maybe it's actually \(f(x) + g(s) \approx F(s)\)... where now we have an equivalence class of \(g\) which may or may not actually make a holomorphic tetration...
Shit's such a fucking head ache. Tommy's right though, just thought I'd add my two cents.
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04/03/2023, 01:58 AM
(This post was last modified: 04/03/2023, 02:05 AM by marcokrt.
Edit Reason: Fixing typos
)
A naive reply to Daniel's original question: also integer tetration depends in a certain way on complex tetration.
In order to try to explain my point, let us just take a look at Corollary 3.3./Equation (26) of my paper https://arxiv.org/pdf/2208.02622.pdf .
Basically (by assuming the standard decimal numeral system), it shows that if the tetration base is congruent to \( 5\pmod{10} \), then a peculiar property of integer tetration (i.e., the constancy of its congruence speed, which in the present case is guaranteed by the constraint that the height of the power tower is at least equal to three) uniquely describes a subset of those bases ending with the digit \( 5 \) returned by \( 90 \) degrees rotations on the complex plane (of course, we can achieve the same goal by using goniometric functions as showed by Equation (21)).
In the mentioned paper, in order to provide the inverse map of all the tetration bases with any given congruence speed, I invoked decadic integers and we know that \( \mathbb{Z}_{10} \) has the countability of the continuum, while this is not true for \( \mathbb{Z} \) and for \( \mathbb{Q} \).
Thus, I think that, if we wish to clearly see the whole picture and understand its true meaning (i.e., congruence speed \( \rightarrow \) phase displacement/"sfasamento" involving the rightmost unfrozen digits comparison \( \rightarrow \) chaos theory related stuff), including intrinsic properties characterizing hyper4 itself (not only holomorphic functions constructed by us following tetration rules that we have previously defined, I mean), we need to look at \( \mathbb{C} \) before turning again our eyes on the "real" axis to the ground.
Let \(G(n)\) be a generic reverseconcatenated sequence. If \(G(1) \notin \{2, 3, 7\}\), then \(^{G(n)}G(n) \pmod {10^d}≡^{G({n+1})}G({n+1}) \pmod {10^d}\), \(\forall n \in \mathbb{N}\{0\}\)
("La strana coda della serie n^n^...^n", p. 60).
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(04/03/2023, 01:58 AM)marcokrt Wrote: A naive reply to Daniel's original question: also integer tetration depends in a certain way on complex tetration.
In order to try to explain my point, let us just take a look at Corollary 3.3./Equation (26) of my paper https://arxiv.org/pdf/2208.02622.pdf .
Basically (by assuming the standard decimal numeral system), it shows that if the tetration base is congruent to \( 5\pmod{10} \), then a peculiar property of integer tetration (i.e., the constancy of its congruence speed, which in the present case is guaranteed by the constraint that the height of the power tower is at least equal to three) uniquely describes a subset of those bases ending with the digit \( 5 \) returned by \( 90 \) degrees rotations on the complex plane (of course, we can achieve the same goal by using goniometric functions as showed by Equation (21)).
In the mentioned paper, in order to provide the inverse map of all the tetration bases with any given congruence speed, I invoked decadic integers and we know that \( \mathbb{Z}_{10} \) has the countability of the continuum, while this is not true for \( \mathbb{Z} \) and for \( \mathbb{Q} \).
Thus, I think that, if we wish to clearly see the whole picture and understand its true meaning (i.e., congruence speed \( \rightarrow \) phase displacement/"sfasamento" involving the rightmost unfrozen digits comparison \( \rightarrow \) chaos theory related stuff), including intrinsic properties characterizing hyper4 itself (not only holomorphic functions constructed by us following tetration rules that we have previously defined, I mean), we need to look at \( \mathbb{C} \) before turning again our eyes on the "real" axis to the ground.
Talking about mod tetration, what do you think about this ?
https://math.eretrandre.org/tetrationfor...p?tid=1735
regards
tommy1729
