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(06/06/2022, 08:55 AM)Catullus Wrote: (06/06/2022, 08:52 AM)Daniel Wrote: (06/06/2022, 08:46 AM)Catullus Wrote: (06/06/2022, 08:23 AM)Daniel Wrote: JmsNxn
Yes, I agree with you, except I believe there are at least \( \aleph_1 \) solutions, the countable infinity of period 1 fixed points, the uncountable infinity of nperiod fixed points like the period 2 fixed point of \( 0^0=1, 0^1=0 \). Then there are the countable infinite rationally neutral fixed points on the Shell Thron boundary. Why would n period points, for n ∈ \mathbb{N}? How would you use the limit formula on them? Please restate your question. Thanks. How would you use periodic points to do continuous iterations of exponentials?
The limit formula uses fixed points usually, if not always. Consider a period 3 fixed point of \( f^n(z) \). But to \( f^{3n}(z) \) the period 3 fixed points just become 3 fixed points.
Daniel
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06/06/2022, 09:12 AM
(This post was last modified: 06/18/2022, 08:32 AM by Catullus.)
(06/06/2022, 09:02 AM)Daniel Wrote: (06/06/2022, 08:55 AM)Catullus Wrote: (06/06/2022, 08:52 AM)Daniel Wrote: (06/06/2022, 08:46 AM)Catullus Wrote: (06/06/2022, 08:23 AM)Daniel Wrote: JmsNxn
Yes, I agree with you, except I believe there are at least \( \aleph_1 \) solutions, the countable infinity of period 1 fixed points, the uncountable infinity of nperiod fixed points like the period 2 fixed point of \( 0^0=1, 0^1=0 \). Then there are the countable infinite rationally neutral fixed points on the Shell Thron boundary. Why would n period points, for n ∈ \mathbb{N}? How would you use the limit formula on them? Please restate your question. Thanks. How would you use periodic points to do continuous iterations of exponentials?
The limit formula uses fixed points usually, if not always. Consider a period 3 fixed point of \( f^n(z) \). But to \( f^{3n}(z) \) the period 3 fixed points just become 3 fixed points. Happy 100th post!
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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Daniel
Wow, so it is. Thanks
Daniel
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06/06/2022, 09:30 AM
(This post was last modified: 06/18/2022, 08:33 AM by Catullus.)
(06/06/2022, 09:28 AM)Daniel Wrote: Daniel
Wow, so it is. Thanks You're welcome.
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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06/07/2022, 11:54 PM
(This post was last modified: 06/07/2022, 11:55 PM by JmsNxn.)
(06/06/2022, 08:58 AM)Catullus Wrote: (06/06/2022, 07:48 AM)JmsNxn Wrote: This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.
There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.
This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea. I said tetration has to be unique.
You are saying using different fixed points to iterate off of would produce different tetrations. Then maybe it would also happen with higher hyperoperations. If tetrations of a converge, then a^^^x as x approaches infinity approaches a fixed point. And a^^^x as x approaches negative infinity approaches a different fixed point. Maybe you could do polynomial interpolation (Having points to interpolate on the left and right of the yaxis.) on the integer pentations, and higher hyperoperations with two fixed points approachable using real iteration.
For bases not in the ShellThron region, maybe you could analytically continue the hyperoperations there.
Oh it definitely would happen for all the hyperoperations. I'll do you one better. It's different for every function \(f\).
If \(f\) has a fixed point \(A\) and a fixed point \(B\) and assume there is a path connecting them \(\gamma\). Then \(f^{\circ s}\) can't be analytic on \(\gamma\). It has to pass the Julia set (of either \(f\) or \(f^{1}\)), and it'll diverge.
So if \(f^{\circ s}(z)=\exp^{\circ s}(z)\) is holomorphic about \(z \approx L\), it cannot be holomorphic in the neighborhood of any other fixed point/periodic point. This throws a huge wrench in the gearsespecially because the orbits of \(\exp\) are dense in \(\mathbb{C}\), and therefore arbitrarily close to a periodic point. The only solution to this, is to admit singularities/branchcuts and the such into the discussion.
So for example, when you write \(\exp^{\circ s}(z)\) using Kneser's construction we have the function \(\text{slog}(z)\) which is holomorphic for all \(z\) not a periodic point (incl fixed points), with singularities of some kind at these points. And Kneser's tetration \(\text{tet}_K(z)\) which is holomorphic for \(z \not \in (\infty,2]\). Then \(\exp^{\circ s}(z) = \text{tet}_K(s+\text{slog}(z))\), which has to account for the branching/singularities in both variables. And produces a whole bunch of nonsense.
One thing though, is that it is only analytic at the fixed point pair \(L,L^*\), and not analytic at any other fixed point/periodic point; but this is only true if \(\Im(s) > 0\) or \(\Im(s) < 0\). So again, you can't trace a path as I wrote above interweaving the two domains. In fact \(\exp^{\circ s}(L)\) will heavily depend on the value of \(s\), as \(\text{slog}(L)\) looks like \(\infty\). I don't know too much about the dynamics, of this tbh, I might have mixed something up. But this is what I recall.
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06/08/2022, 03:39 AM
(This post was last modified: 06/08/2022, 03:58 AM by Catullus.)
(06/07/2022, 11:54 PM)JmsNxn Wrote: (06/06/2022, 08:58 AM)Catullus Wrote: (06/06/2022, 07:48 AM)JmsNxn Wrote: This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.
There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.
This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea. I said tetration has to be unique.
You are saying using different fixed points to iterate off of would produce different tetrations. Then maybe it would also happen with higher hyperoperations. If tetrations of a converge, then a^^^x as x approaches infinity approaches a fixed point. And a^^^x as x approaches negative infinity approaches a different fixed point. Maybe you could do polynomial interpolation (Having points to interpolate on the left and right of the yaxis.) on the integer pentations, and higher hyperoperations with two fixed points approachable using real iteration.
For bases not in the ShellThron region, maybe you could analytically continue the hyperoperations there.
Oh it definitely would happen for all the hyperoperations. I'll do you one better. It's different for every function \(f\).
If \(f\) has a fixed point \(A\) and a fixed point \(B\) and assume there is a path connecting them \(\gamma\). Then \(f^{\circ s}\) can't be analytic on \(\gamma\). It has to pass the Julia set (of either \(f\) or \(f^{1}\)), and it'll diverge.
So if \(f^{\circ s}(z)=\exp^{\circ s}(z)\) is holomorphic about \(z \approx L\), it cannot be holomorphic in the neighborhood of any other fixed point/periodic point. This throws a huge wrench in the gearsespecially because the orbits of \(\exp\) are dense in \(\mathbb{C}\), and therefore arbitrarily close to a periodic point. The only solution to this, is to admit singularities/branchcuts and the such into the discussion.
So for example, when you write \(\exp^{\circ s}(z)\) using Kneser's construction we have the function \(\text{slog}(z)\) which is holomorphic for all \(z\) not a periodic point (incl fixed points), with singularities of some kind at these points. And Kneser's tetration \(\text{tet}_K(z)\) which is holomorphic for \(z \not \in (\infty,2]\). Then \(\exp^{\circ s}(z) = \text{tet}_K(s+\text{slog}(z))\), which has to account for the branching/singularities in both variables. And produces a whole bunch of nonsense.
One thing though, is that it is only analytic at the fixed point pair \(L,L^*\), and not analytic at any other fixed point/periodic point; but this is only true if \(\Im(s) > 0\) or \(\Im(s) < 0\). So again, you can't trace a path as I wrote above interweaving the two domains. In fact \(\exp^{\circ s}(L)\) will heavily depend on the value of \(s\), as \(\text{slog}(L)\) looks like \(\infty\). I don't know too much about the dynamics, of this tbh, I might have mixed something up. But this is what I recall. I wonder how the analytical continuation of the Schröder tetraion behaves.
Please remember to stay hydrated.
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06/08/2022, 03:50 AM
(This post was last modified: 06/08/2022, 04:01 AM by JmsNxn.)
(06/08/2022, 03:39 AM)Catullus Wrote: I wonder how the analytical continuation of the schröder tetraion behaves.
It's fairly well studied. It looks essentially like \(f^{\circ s}(z)\) for any transcendental function \(f\) about a repelling fixedpoint \(L\). For any transcendental function, \(f: \mathbb{C} \to \mathbb{C}\) with a repelling fixed point \(L\) has a transcendental inverse Schroder function \(\Psi^{1}:\mathbb{C} \to \mathbb{C}\). Then:
\[
f(\Psi^{1}(z)) = \Psi^{1}(\lambda z)\\
\]
Where, \(\lambda= f'(L)\). Then, so long as \(1\) is in the image of \(f\), there always exists a function:
\[
f^{\circ s}(1) = \Psi^{1}(\lambda^{ss_0})\\
\]
It will be entire in \(s\) (transcendental), and will satisfy \(f^{\circ 0}(1) = 1\).
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06/09/2022, 11:19 PM
(This post was last modified: 06/10/2022, 09:43 AM by Catullus.)
(06/07/2022, 11:54 PM)JmsNxn Wrote: (06/06/2022, 08:58 AM)Catullus Wrote: (06/06/2022, 07:48 AM)JmsNxn Wrote: This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.
There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.
This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea. I said tetration has to be unique.
You are saying using different fixed points to iterate off of would produce different tetrations. Then maybe it would also happen with higher hyperoperations. If tetrations of a converge, then a^^^x as x approaches infinity approaches a fixed point. And a^^^x as x approaches negative infinity approaches a different fixed point. Maybe you could do polynomial interpolation (Having points to interpolate on the left and right of the yaxis.) on the integer pentations, and higher hyperoperations with two fixed points approachable using real iteration.
For bases not in the ShellThron region, maybe you could analytically continue the hyperoperations there.
Oh it definitely would happen for all the hyperoperations. I'll do you one better. It's different for every function \(f\).
If \(f\) has a fixed point \(A\) and a fixed point \(B\) and assume there is a path connecting them \(\gamma\). Then \(f^{\circ s}\) can't be analytic on \(\gamma\). It has to pass the Julia set (of either \(f\) or \(f^{1}\)), and it'll diverge.
So if \(f^{\circ s}(z)=\exp^{\circ s}(z)\) is holomorphic about \(z \approx L\), it cannot be holomorphic in the neighborhood of any other fixed point/periodic point. This throws a huge wrench in the gearsespecially because the orbits of \(\exp\) are dense in \(\mathbb{C}\), and therefore arbitrarily close to a periodic point. The only solution to this, is to admit singularities/branchcuts and the such into the discussion.
So for example, when you write \(\exp^{\circ s}(z)\) using Kneser's construction we have the function \(\text{slog}(z)\) which is holomorphic for all \(z\) not a periodic point (incl fixed points), with singularities of some kind at these points. And Kneser's tetration \(\text{tet}_K(z)\) which is holomorphic for \(z \not \in (\infty,2]\). Then \(\exp^{\circ s}(z) = \text{tet}_K(s+\text{slog}(z))\), which has to account for the branching/singularities in both variables. And produces a whole bunch of nonsense.
One thing though, is that it is only analytic at the fixed point pair \(L,L^*\), and not analytic at any other fixed point/periodic point; but this is only true if \(\Im(s) > 0\) or \(\Im(s) < 0\). So again, you can't trace a path as I wrote above interweaving the two domains. In fact \(\exp^{\circ s}(L)\) will heavily depend on the value of \(s\), as \(\text{slog}(L)\) looks like \(\infty\). I don't know too much about the dynamics, of this tbh, I might have mixed something up. But this is what I recall. Then, when the tetration converges the Schröder tetration would be different depending on which fixed point you use.
If only you could use every fixed and n period point at the same time, to do Schröder iteration on. Or at least the two primary fixed points.
Please remember to stay hydrated.
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(06/09/2022, 11:19 PM)Catullus Wrote: (06/07/2022, 11:54 PM)JmsNxn Wrote: (06/06/2022, 08:58 AM)Catullus Wrote: (06/06/2022, 07:48 AM)JmsNxn Wrote: This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.
There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.
This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea. I said tetration has to be unique.
You are saying using different fixed points to iterate off of would produce different tetrations. Then maybe it would also happen with higher hyperoperations. If tetrations of a converge, then a^^^x as x approaches infinity approaches a fixed point. And a^^^x as x approaches negative infinity approaches a different fixed point. Maybe you could do polynomial interpolation (Having points to interpolate on the left and right of the yaxis.) on the integer pentations, and higher hyperoperations with two fixed points approachable using real iteration.
For bases not in the ShellThron region, maybe you could analytically continue the hyperoperations there.
Oh it definitely would happen for all the hyperoperations. I'll do you one better. It's different for every function \(f\).
If \(f\) has a fixed point \(A\) and a fixed point \(B\) and assume there is a path connecting them \(\gamma\). Then \(f^{\circ s}\) can't be analytic on \(\gamma\). It has to pass the Julia set (of either \(f\) or \(f^{1}\)), and it'll diverge.
So if \(f^{\circ s}(z)=\exp^{\circ s}(z)\) is holomorphic about \(z \approx L\), it cannot be holomorphic in the neighborhood of any other fixed point/periodic point. This throws a huge wrench in the gearsespecially because the orbits of \(\exp\) are dense in \(\mathbb{C}\), and therefore arbitrarily close to a periodic point. The only solution to this, is to admit singularities/branchcuts and the such into the discussion.
So for example, when you write \(\exp^{\circ s}(z)\) using Kneser's construction we have the function \(\text{slog}(z)\) which is holomorphic for all \(z\) not a periodic point (incl fixed points), with singularities of some kind at these points. And Kneser's tetration \(\text{tet}_K(z)\) which is holomorphic for \(z \not \in (\infty,2]\). Then \(\exp^{\circ s}(z) = \text{tet}_K(s+\text{slog}(z))\), which has to account for the branching/singularities in both variables. And produces a whole bunch of nonsense.
One thing though, is that it is only analytic at the fixed point pair \(L,L^*\), and not analytic at any other fixed point/periodic point; but this is only true if \(\Im(s) > 0\) or \(\Im(s) < 0\). So again, you can't trace a path as I wrote above interweaving the two domains. In fact \(\exp^{\circ s}(L)\) will heavily depend on the value of \(s\), as \(\text{slog}(L)\) looks like \(\infty\). I don't know too much about the dynamics, of this tbh, I might have mixed something up. But this is what I recall. Then, when the tetration converges the Schröder tetration would be different depending on which fixed point you use.
If only you could use every fixed and n period point at the same time, to do Schröder iteration on. Or at least the two primary fixed points.
That would be quite the function!
I'm doubtful it'd be possible though, it'd be really cool if it was. The trouble is every tetration induces an iteration, and as I remarked that iterations \(f^{\circ s}(z)\) can't be holomorphic in the neighborhood of two fixed points.
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06/10/2022, 11:08 PM
(This post was last modified: 06/11/2022, 11:42 PM by Catullus.)
(06/10/2022, 08:41 PM)JmsNxn Wrote: I'm doubtful it'd be possible though, it'd be really cool if it was. The trouble is every tetration induces an iteration, and as I remarked that iterations \(f^{\circ s}(z)\) can't be holomorphic in the neighborhood of two fixed points. What if the different Schröder iterations flowed together nicely?
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