08/05/2021, 04:36 PM
(This post was last modified: 08/05/2021, 06:20 PM by Leo.W.
Edit Reason: typo
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(08/04/2021, 03:58 AM)Jms Nxn Wrote: Taking a similar approach as Leo; this is largely Leo's discussion; but we're going to switch to \( f(z) = z - 1/z \); the reason for this switch is to reduce the group structure.Indeed. I brought up this example because it shows what originally the symmetry law talks about, is that, the functions
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I enjoy your posts, Leo; I hope this makes sense.
\( f(z),\frac{1}{f(z)} \) by the original symmetry \( f(z)=f(z^{-1})\to{f^{-1}(z)=f^{-1}(z)^{-1}\to{f(z)=f(z)^{-1}}} \)
and their iterative family
\( f^t(z),(z\to\frac{1}{f(z)})^t(z)(t\ne0) \)
are all sharing the same Riemann surface in a sense that whenever they're considered as multivalued, or sharing each other's iterative family in some certain domain, in this case, \( f(z),\frac{1}{f(z)} \) shared their second iterative root, and if you check, they even share their inverse.
Or to say, that these 2 functions, are inseperable when considering Riemann surface or multivalued function.
However, please let me have some different opinions from that "If we take
We can check out this simply in the construction of the real-natural-tetration, where the 2 Abel functions are taken from the 0.318+1.337i fixed point pair, using the limit definition, we can define the Abel functions only in halfplane, but slightly change the limit definition by using ln(z)+2pi*k*i with k integer, we can then continue the definition and generate every branch cuts of Abel function. I also tried this strategy out in the theta-mapping of the quadratic function \( f(z)=z^2+i \), where the original inverse function \( f^{-1}(z)=\sqrt{z-i} \) CAN NOT be used to construct the Abel function, so Can not construct the theta-mapping. But changing the branch cuts of inverse of f, it works.
I lately worked out the proof, that any polynomials having no multiple fixed points
(\( f(z)=z+(z-L)^kg(z) \), g is polynomial, k>1 integer then L is called a multiple fixed point),
always have a fixed point L which is constructable(\( \frac{ln(f'(L))}{\pi*i}\notin\mathbb{Q}\setminus{0} \)),
or is available to generalize the polynomial function's iterations from that point.
I wonder if you have any suggestion on this very important aforementioned question: (i'm appreciative)
Do any analytic functions, or differentiable Riemann surfaces, have at least one fixed point L which is constructable?
And do you already solved the noncontructable cases? I'm like, stuck in the very initial case \( f(z)=-z(1-z) \) at \( f(0)=0,f'(0)=-1 \), I wonder if anyone had already successfully construct a second iterative root?
Though if you assume the second iterative root has a Taylor series at z=0, quickly sooner you'll notice it's not helpful, contradicting the fundamental laws of iterated functions.
Regards
Leo

