I know you don't like my local iteration idea, but if:
\[
h(z) = \phi(g(\phi^{-1}(z)))\\
\]
Where \(g\) is holomorphic in a half plane, and \(\phi\) maps the half plane to the unit disk \(\mathbb{D}\). If we ask that \(h(1) = 1\), while \(g(0) = 0\)--we can solve for any petal about a parabolic fixed point. We just have to "map to a halfplane".
Petals are always simply connected, if you localize them enough. To the point they look like:
\[
f : \mathbb{D} \to \mathbb{D}\\
\]
And \(f(1) = 1\)--where there are no more fixedpoints within \(\mathbb{D}\). The fixed point lies on the boundary.
-----------------------------------------------
So if we conjugate arbitrary iterations; such that the petal they are holomorphic on, is mapped to \(\mathbb{C}_{\Re(z) < 0}\) where it satisfies the same asymptotics as Gottfrieds \(g\). Then we're cooking with fire!
\[
h(z) = \phi(g(\phi^{-1}(z)))\\
\]
Where \(g\) is holomorphic in a half plane, and \(\phi\) maps the half plane to the unit disk \(\mathbb{D}\). If we ask that \(h(1) = 1\), while \(g(0) = 0\)--we can solve for any petal about a parabolic fixed point. We just have to "map to a halfplane".
Petals are always simply connected, if you localize them enough. To the point they look like:
\[
f : \mathbb{D} \to \mathbb{D}\\
\]
And \(f(1) = 1\)--where there are no more fixedpoints within \(\mathbb{D}\). The fixed point lies on the boundary.
-----------------------------------------------
So if we conjugate arbitrary iterations; such that the petal they are holomorphic on, is mapped to \(\mathbb{C}_{\Re(z) < 0}\) where it satisfies the same asymptotics as Gottfrieds \(g\). Then we're cooking with fire!