07/05/2022, 12:27 PM
(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 re-defined 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

