@James
I think this totally does not work. One can only talk about petals for multiplier 1 - and Milnor only does that - or show me where he does differently. For other unit roots it does not work, i.e. you can not have continuous iterates or petals there.
Just consider the following, let the multiplier be a q-th root of unity, then by Milnor we know that \({\rm valit}[f^{\circ q}] = qm \) for some integer m>0.
Say for this case just m=1. We have q attractive petals of \(f^{\circ q}\), so if we start with \(z_0\) in one attracting petal then the sequence of iterates of \(z_0\) \(f^{\circ 0}(z_0)\dots f^{\circ q-1}(z_0)\) orbits the fixed point 1 time and \(f^{\circ q}(z_0)\) arrives in the same petal as we started.
If we would now have continuous iterates \(f^{\circ t}(z_0)\) for t=0...q it would be a curve \(\gamma(t)\) connecting all the previously mentioned integer iterates. Particularly it would cross all the attractive and repelling petals of \(f^{\circ q}\).
For the next round t=q...2q the image of the curve would just be \(\gamma(q..2q)=f^{\circ q}(\gamma(0..q))\) and so on for further rounds.
So with higher t it would bulge out at the repelling petals and sucked in at the attractive petals.
So this curve gets more wild the greater t is and the closer parts of the curve come to the fixed point.
In no way could \(\lim_{t\to\infty}f^{\circ t}(z_0)=0\) - but wouldnt that be required if you want some regular iteration at 0?
In the following picture I show the iteration of \(f(z)=-z+z^2\) on \(z_0=0.25\) where I connect the first two iterates with a *hyptothetical* curve (but any curve would suffer from the effects I described above) and continue this curve by applying f. One can clearly see that the petals are those of \(f^{\circ 2}(z)=z - 2z^3 + z^4\), i.e. 2 attractive petals on the real axis and two repelling petals along the imaginary axis.
@Daniel
The irrational indifferent case is not so much my concern in the moment because it can - from the formal powerseries standpoint - be treated like the hyperbolic case.
I think this totally does not work. One can only talk about petals for multiplier 1 - and Milnor only does that - or show me where he does differently. For other unit roots it does not work, i.e. you can not have continuous iterates or petals there.
Just consider the following, let the multiplier be a q-th root of unity, then by Milnor we know that \({\rm valit}[f^{\circ q}] = qm \) for some integer m>0.
Say for this case just m=1. We have q attractive petals of \(f^{\circ q}\), so if we start with \(z_0\) in one attracting petal then the sequence of iterates of \(z_0\) \(f^{\circ 0}(z_0)\dots f^{\circ q-1}(z_0)\) orbits the fixed point 1 time and \(f^{\circ q}(z_0)\) arrives in the same petal as we started.
If we would now have continuous iterates \(f^{\circ t}(z_0)\) for t=0...q it would be a curve \(\gamma(t)\) connecting all the previously mentioned integer iterates. Particularly it would cross all the attractive and repelling petals of \(f^{\circ q}\).
For the next round t=q...2q the image of the curve would just be \(\gamma(q..2q)=f^{\circ q}(\gamma(0..q))\) and so on for further rounds.
So with higher t it would bulge out at the repelling petals and sucked in at the attractive petals.
So this curve gets more wild the greater t is and the closer parts of the curve come to the fixed point.
In no way could \(\lim_{t\to\infty}f^{\circ t}(z_0)=0\) - but wouldnt that be required if you want some regular iteration at 0?
In the following picture I show the iteration of \(f(z)=-z+z^2\) on \(z_0=0.25\) where I connect the first two iterates with a *hyptothetical* curve (but any curve would suffer from the effects I described above) and continue this curve by applying f. One can clearly see that the petals are those of \(f^{\circ 2}(z)=z - 2z^3 + z^4\), i.e. 2 attractive petals on the real axis and two repelling petals along the imaginary axis.
@Daniel
The irrational indifferent case is not so much my concern in the moment because it can - from the formal powerseries standpoint - be treated like the hyperbolic case.