08/02/2022, 10:18 PM
(This post was last modified: 08/02/2022, 10:20 PM by bo198214.
Edit Reason: typo
)
And just as we are at the topic, I also have an example for a function having analytic iteration at a *parabolic* fixed point: \(f(z)=\frac{z}{1+z}\)
I was always wondering when that case would occur.
This function has a parabolic fixed point at 0 and one can write an explicit formula for the regular iteration:
\(f^{\circ t}(z) = \frac{z}{1+tz}\) which is analytic at 0 for all t!
So this makes it a bit clearer in my head:
It seems that a regular iteration at a parabolic fixed point is analytic is as rare as an iteration is analytic at 2 hyperbolic fixed points.
It would even say that they are related:
if and only if the regular iteration at a parabolic fixed point is analytic then the perturbed function (with two (or more?) exploded hyperbolic fixed points is analytic at all fixed points).
And maybe this scenario happens only for fractional linear functions, who knows.
But I really withdraw my conjecture that most parabolic fixed points of \(b^z\) have iterations analytic at the fixed point.
I was always wondering when that case would occur.
This function has a parabolic fixed point at 0 and one can write an explicit formula for the regular iteration:
\(f^{\circ t}(z) = \frac{z}{1+tz}\) which is analytic at 0 for all t!
So this makes it a bit clearer in my head:
It seems that a regular iteration at a parabolic fixed point is analytic is as rare as an iteration is analytic at 2 hyperbolic fixed points.
It would even say that they are related:
if and only if the regular iteration at a parabolic fixed point is analytic then the perturbed function (with two (or more?) exploded hyperbolic fixed points is analytic at all fixed points).
And maybe this scenario happens only for fractional linear functions, who knows.
But I really withdraw my conjecture that most parabolic fixed points of \(b^z\) have iterations analytic at the fixed point.
