07/01/2022, 10:23 PM
(07/01/2022, 10:03 PM)JmsNxn Wrote: Well right off the bat I'm going to say that you can't use a fixed point iteration.
If \(f^{\circ s}(z)\) is a holomorphic function on some domain \(\mathcal{A}\), there cannot be any periodic or fixed points in \(\mathcal{A}\) besides one fixed point.
You can go one step further:
If \(f^{\circ s}(z)\) is holomorphic at a fixed point \(L\), it is not holomorphic about any other fixed point/periodic point \(L_2\) of \(f\).
This is because somewhere in this iteration we either hit a Julia set, or we hit different fatou set.
Best case scenario we get something like Kneser's iteration, which is holomorphic everywhere EXCEPT at periodic/fixed points.
What this means for your function is pretty simple.
The iteration \(f^{\circ s}(z)\) at \(0\), has a discontinuity in \(z\) as it tries to approach \(f^{\circ s}(z)\) at \(1\). This is because, even if both are attracting, somewhere \(z\) will hit the Julia set; remember every basin is separated by a julia set. If one is repelling and one is attracting, then the repelling is in the Julia set.
Best case scenario is that they are both repelling, and your holomorphic on deleted disks about the fixed point
For an iteration like this, the best way to do it is to talk about upper half planes and lower half planes with a fixed point at \(\infty\) in both cases, and somehow trying to glue them together on the real line. No fucking clue what that would look like though.
like 2 black holes tearing up the space between them.
well i was poetic ...
