(08/05/2022, 11:22 PM)bo198214 Wrote: I think its not yet as mature as you wish, but keep going.
In the mean time I was constructing another counter example, this time without linear fractional mappings.
The idea is the following, we know that the Schröder iteration is given by:
\[ f^{\mathfrak{R} t}(x) = \chi^{-1}(c^t \chi(x)) \]
Instead for looking for functions \(f\) I just took \(\chi\) as my start of research, how does \(\chi\) need to look, such that \(f^{\mathfrak{R}t}\) is still analytic at a second fixed point.
For a second fixed point \(z_2\) it must be valid that:
\[ c^t \chi(z_2) = \chi(z_2) \]
which is basically satisfied for \(\chi(z_2)=0\) or \(\chi(z_2)=\pm\infty\).
So I took a function that satisfies all 3 values: the tangent! And constructed the following function
\[ f(x) = \arctan(c\tan(x)) \] with the iteration \( f^{\mathfrak{R}t}(x) = \arctan(c^t\tan(x)) \). One has to be a bit careful with the choosen branch, so we choose:
\[ \arctan_0(c^t\tan(x)) + \pi\left\lfloor \frac{x+\frac{\pi}{2}}{\pi}\right\rfloor \]
This looks like:
So the main question is: Is this function analytic at \(\frac{\pi}{2} + k\pi \)?
Expressing tan with sin:
\[ \tan(x)=\frac{\sin(x)}{\sqrt{1-\sin(x)^2}}, \quad\arctan(x) = \arcsin\left(\frac{x}{\sqrt{1+x^2}}\right) \]
For brevity, write \(s\) for \(\sin(x)\)
\[\arctan(c\tan(x)) = \arcsin\left(\frac{\frac{cs}{\sqrt{1-s^2}}}{\sqrt{1+\frac{c^2s^2}{1-s^2}}}\right) =
\arcsin\left(\frac{cs}{\sqrt{1-s^2+c^2s^2}}\right) = \arcsin\left(\frac{c\sin(x)}{\sqrt{1+\sin(x)^2(-1+c^2)}}\right)\]
So we see: nothing scary happens around \(x\approx \frac{\pi}{2}\), \(\sin(x)\approx 1\) just combination of analytic functions.
Hence \(f\) and \(f^{\mathfrak{R}t}\) is analytic on whole \(\mathbb{R}\) at infintely many fixed points.
BO! You are a beautiful soul! That is fucking unreal! But please hear me, when I say that is not a counter example!
You are using \(\tan\) which is a MEROMORPHIC function. This is something I'm finding far deeper in what I have. I swear, my next argument will be as solid as your last few posts. You've just gotta let me ruminate. We need that these are EUCLIDEAN maps, as Milnor calls them. This means, they are maps from the complex plane to itself. The function \(\tan\) is a map from the complex plane TO THE RIEMANN SPHERE. I know this sounds like a cop out on my part. But this is no different than the LFT results. There's something going on that is very different here. The exponential, and any transcendental entire function (Euclidean mapping--as Milnor calls it) won't suffer these results.
You again, primarily solve on the real line, where poles appear as you add imaginary argument. This forces what won't be a local iteration; it'll be a whole nother beast entirely. I just want to show that.
\[
f : \mathbb{C} \to \mathbb{C}\\
\]
\[
f^{\circ t}(z) : D \times H \to H\\
\]
Then \(H\) can't contain two fixed points of \(f\). And what you have shown to me is that EUCLIDEAN is necessary. Also, recall that \(D\) is a semi group under addition. So this is a semi group translation. And it is not \(\mathbb{R}\), it is a DOMAIN in \(\mathbb{C}\). These restrictions I think are the exact amounts.
I really want to get this write bo, and I promise I'll try my fucking hardest. But you still ain't provided a true true counter example to what I mean. I've been rereading milnor again to make sure I can argue from scratch. lol
I mean, with all due respect, not trying to be contrarian:
\[
\arctan(c\tan(x))\\
\]
Is not Euclidean. You tell me what world that's an entire function.
I know I'm adding these points now, but they were always my assumptions, lol.
So you are absolutely right, we can iterate about multiple fixed points. My point is that this fails for Euclidean cases. Which is again entire functions. Which is a specific branch of complex dynamics which relates to Tetration the most. Meromorphic dynamics is far more whacky. And as I see it, this is meromorphic dynamics.
Again, not trying to be snarky. Just saying how I see it. I love being proven wrong though--because that would make all this shit so much simpler.
I hope I don't seem like I'm moving the goal posts. Like I'm trying not to be proven wrong. You've definitely proven me wrong on my general statement. But that has helped me hone it a lot more. And I'm really fuckign sure there is a way using Euclidean maps \(\mathbb{C} \to \mathbb{C}\) to explain why you can't iterate about two fixed points. The moment you allow for the compact domain \(\widehat{\mathbb{C}}\) to enter the picture; or branching in some manner; everything becomes far more of a free for all, and multiple fixed point iteration exists.
But for \(\mathbb{C} \to \mathbb{C}\), I don't think it's possible. And I am at least 75% there in a much better argument. Just let me ruminate and think and jot down some shit; reread milnor a couple times, lol.
