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Oh my god, I'm stupid!
I get it now, I thought it was something fancy. The double norm screwed me up, lmao.
If that's the case, it cannot be analytic in \(z\) as stated. It fails the CauchyRiemann equations by construction. It's a perfectly weird \(C^\infty\) construction though. Weird looking.
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Quote:Weird looking.
Hey, I'll show you that is very natural as a construction, and it is something very general relating extensions on iterations and the glueing of iterations. I'm preparing a post explaining all that business from zero to hero.
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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(06/18/2022, 02:55 PM)MphLee Wrote: ... Reaaaally fascinates me!
to rectify myself, I wrote \(\mathbb{C}^0\) for discontinuous infinitely many "slices" of parellel lines, so that f cannot have a complex derivative, but does have a directional derivative.
Your claim is right about that we'd apply the study of group or monoid structures to this but I now have more a strong (belief?)sense that even that can be still inadequate, and we may need more complex algebra system to extend the iterations of a map,
what I mean is that, for instance if we really do have redefined the complex numbers as \(z=\{Re(z),Im(z),0\}\in\mathbb{R}^3\) and then\(f^t: \{Re(z),Im(z),0\}\to\{Re(z),Im(z)cos(\pi t),sin(\pi t)\}\) as a common function (linear transformation by rotation matrices) which map \(\mathbb{R}^3\) to \(\mathbb{R}^3\), then we can have naturally an extended iteration of conjugate function...... so I feel no surprise that there's limitation on the operators like discontinuous iterations, lol I may give up on continuous iterations on conj(z) only in \(\mathbb{C}\)
I consider that any partition that has the same in "count" (2 partitions has this "count" 2) can be transfered to each other if they can be measured, by some underlying function \(g:X_i\to Y_i\) for different partition X and Y, or at least analogous.
For the conj(z) as you partitioned the plane into \(\bigcup_{r\ge0} rS^1\), and you brought up about K_r function, I think its structure can be similar to the iterations of \(f:[0,2\pi]\to[0,2\pi],f(\theta)\equiv\theta \pmod{2\pi}\), or maybe more \(f:\mathbb{R}\to\mathbb{R},f(x)=x\), the iterations then be transferred into a question that whether there's a continuous function \(F:\mathbb{R}\times[0,2\pi]\to[0,2\pi],F(t+s,\theta)=F(t,F(s,\theta)),F(0,\theta)=\theta,F(1,\theta)=2\pi\theta\), closely to a continuous iteration of \(f:\mathbb{R}\to\mathbb{R},f(x)=x\), I don't know if such function exists, but at least it can be shown that there is no solution to \(f(f(x))=x\) on real axis
Regards, Leo
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07/07/2022, 12:55 AM
(This post was last modified: 07/07/2022, 01:01 AM by JmsNxn.)
Your posts are fantastic, Leo. As far as I'm concerned that's a novel proof of \(f(f(x)) = x\) can't be real valued. We have a couple of these problems on these forums. I hope you post more. You have a great brain.
To iterate \(\text{conj(z)} = z^* = \overline{z}\) you need to do Grothendiek level shit. Iterates only exist in exotic metric spaces, that are probably better represented using padics, or something like that. It'll never be analytic/continuous/differentiable. Nothing good is to come from looking at it like that. But as an abstract algebraic idea, it does exist. Assuming the axiom of choice there are uncountably many automorphisms of \(\mathbb{C}\to \mathbb{C}\) as a field. But the only continuous one is conjugation and the identity mapping. Not to rain on your parade, but this question is pretty much settled. I admire your spirit. Love your energy.
Regards, James
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Meanwhile I have two minutes. Conj is a field automorphism of the complexes but that doesn't mean that we have to look for it's fractional iterates inside the group \({\rm Aut}(\mathbb C)\) unless we ask for an \(\mathbb R\)action over \(\mathbb C\) that acts by automorphisms. Sure since conjugation is involution, hence isomorphism, we'd like its fractional iterates to be iso, since composition preserves bijectivity... and also because we'd like fractional iterations to respect the field structure...This however seems too restrictive. We'd get something very rigid and similar to a weakened scalar multiplication...
Let \(\alpha(r,z)\) be an action over the complexes: \(\alpha(0,z)=z\) and \(\alpha(r+s,z)=\alpha(r,\alpha(s,z))\) s.t. \(\alpha(1,z)={\rm conj}(z)\).
First notice that if we ask it to act by automorphisms then we induce a group homomorphisms \(\alpha:\mathbb R \to \rm{ Aut}(\mathbb C) \). The "intrinsic time" of the iteration is given by the group quotient \(\mathbb R / {\rm ker}\alpha\) where \(2\mathbb Z\subseteq {\rm ker}\alpha =\{k\in\mathbb R\, : \,\alpha(r,z)=z \} \), since conjugation being an involution \({\rm conj}^{2n}={\rm id}_{\mathbb C}\). If the kernel is the even numbers then the intrinsic iterates looks like the circle \(S^1 \simeq \mathbb R/2\mathbb Z\).
In other words we are looking for \(S^1\)actions \(\alpha:{}S^1\times \mathbb C\to \mathbb C\) over \(\mathbb C\). I wonder what this entails at the level of James' remarks. If we consider also wild automorphisms... maybe this wold entails some contradiction... maybe idk. I was thinking about failure of injectivity in \(z\) of iterates \(\alpha(r,z)={\rm conj}^r(z)\) when the time belongs to the circle \(r\in S^1\)... while all of them should be injective since are field morphisms....
this seems fascinating but got no time to study it more.
The second critical point is that we would also have \(\alpha(t,0)=0\) and \(\alpha(r,z+w)=\alpha(r,z)+\alpha(r,w)\), and \(\alpha(t,1)=1\) and \(\alpha(r,zw)=\alpha(r,z)\alpha(r,w)\). This seems too rigid to have any room to play around... probably it is too rigid that ruels out every nontrivial solution....
Just a feeling...
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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(06/26/2022, 05:37 AM)JmsNxn Wrote: Mphlee just made me more confused.
Hi James, I apologize, I went to fast assuming too much.
Idk what part was confusing so I'd start from zero and I'll try to bring you up to speed. I've made a long expository post that belongs to my note dumps on iteration and actions.
GO HERE
@Leo, also made this post for you too. The decomposition mechanism is really general, but is just pasting together known stuff. In the post linked you can find an introduction to it.
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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(07/14/2022, 07:18 PM)MphLee Wrote: (06/26/2022, 05:37 AM)JmsNxn Wrote: Mphlee just made me more confused.
Hi James, I apologize, I went to fast assuming too much.
...
@Leo, also made this post for you too. The decomposition mechanism is really general, but is just pasting together known stuff. In the post linked you can find an introduction to it.
Thx Mphlee, I sensed this by your previous posts, you're reaaaaaaaaaaally good at rigorousify random statements XD, excellant!
Regards, Leo
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(08/10/2022, 01:49 PM)Leo.W Wrote: Thx Mphlee, I sensed this by your previous posts, you're reaaaaaaaaaaally good at rigorousify random statements XD, excellant!
Mphlee is a wizard, he's going to do great things!
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(06/18/2022, 10:38 AM)Leo.W Wrote: Idk if it's time to talk about continuous iterations for nonanalytic functions that can still be defined on the whole complex plane, but these ideas did disastrously harm to my brain lol.
I worked out some cases last year, these are my results.
Consider any function g(x):R>R, then we consider a function f(z), which geometrically rotates the original plane in a way that cut the plane into infinitely many slices in the shape of ring, all locates its center at the origin, each ringshaped slice has its own rotation angle, which can be written as:
$$f(z)=ze^{ig(\z\)}$$
Then, we can easily prove and get a continuous iteration(but only to real orders) of f(z) following these steps:
$$\text{For real t, }f^t(z)=ze^{itg(\z\)}$$
Remember abs(e^(it))=1 for real t,
$$\text{proof: For real s and t, }f^{s+t}(z)=ze^{i(s+t)g(\z\)}=ze^{itg(\z\)}e^{isg(\z\)}=\bigg(ze^{itg(\z\)}\bigg)e^{isg\big(\ze^{itg(\z\)}\\big)}=f^s(f^t(z))$$
it's easy to prove, and also satisfies the geometric intuition that the tth iteration of a rotation is a rotation whose angular velocity is t times the original rotation's.
Another idea, came by trying to figure out whether there's continuous iteration of the conjugate function conj(z)
We start by conj(z)=abs(z)*e^(i*arg(z)), consider a set of functions $$f_u(z)=\z\e^{iu\arg(z)}$$ where u is real,
Then $$\text{For real t and u^t, }f_u^t(z)=\z\e^{iu^t\arg(z)}$$
Because $$\text{For real s and t, }f_u^{s+t}(z)=\z\e^{iu^{s+t}\arg(z)}=\ze^{iu^tz}\e^{iu^s\arg(\z\e^{iu^tz})}=f_u^s(f_u^t(z))$$
if u^t and u^s are both real.
But since u=1, t=1/2, u^t is not real, we cannot find a functional square root of conj(z) through this.
Do you have more examples, I basically based all this on the Polar decomposition of complex numbers and don't have so much time to take more investigation into it.
i was thinking
f^[t + s](z) satisfies the semigroup addition homomo as does f^[t i + s i](z)
but I think f^[t + s i](z) does not.
( i^2 = 1 )
this relates to my recent posts/conjectures about semigroup addition homomo and analyticity.
regards
tommy1729
