Hey, Leo!
I get your confusion. I stand by my point though. What you have done is shown there is no taylor expansion at \( 0 \); which I explained does not exist. The iterate can only exist on a petal about zero; not a neighborhood of zero. So your first point is null to my construction.
It helps to realize in the neutral case; that no iteration exists in a disk about zero; therefore, there is not a taylor series at zero. But; in the petals about zero; there are half iterates. But in a neighborhood of \( 0 \) intersecting with the attracting petals; and then excluding \( 0 \); we have a holomorphic function. Where we likely have a removable singularity at zero; but it is not holomorphic at zero.
May I switch to \( \sin(-\xi) \)? Because I can explain your confusion using this function. The same principle applies to \( f(\xi)= - \xi(1-\xi) \)--forgive the change to a simpler case, please. But using an odd function is easier (same principle applies); and \( \sin \) is easier to reference numerically.
The function,
\(
\sin(-\xi) = f(\xi)\\
\)
has an iteration for all \( k \in \mathbb{Z} \),
\(
f^{\circ z}(\xi) = e^{(2k+1)\pi iz} \xi+\sum_{n=2}^\infty b_n \xi^n\\
\)
Where this is only an ASYMPTOTIC SERIES about zero. if we assume that \( \sin^{\circ z}(-\xi) = - \sin^{\circ z}(\xi) \)--which is the standard iteration. Then there are two square root
\(
f^{\circ 1/2}(\xi) = i\xi + \sum_{n=2}^\infty c_n \xi^n\\
-f^{\circ 1/2}(\xi) = -i\xi - \sum_{n=2}^\infty c_n \xi^n\\
\)
Now notice; neither of these map the real line to the real line; like the iteration \( \sin^{\circ t}(\xi) \) for \( t \in \mathbb{R}^+ \); but they are iterates nonetheless. They are not holomorphic at zero; we just get a good asymptotic expansion when approaching zero. In this sense, the summation, and taylor series, is an *abuse of notation*; these things don't converge. Only asymptotically to zero are they meaningful. They are actually holomorphic on a petal. It's a bit more complicated with \( f(\xi) = -\xi(1-\xi) \) but it's the same scenario--as far as I can tell. They're not holomorphic at zero; only on a petal about zero. What the petal looks like? I have no f**n clue; but there exists a petal where this converges.
Unless I'm missing something large; which I don't think I am from your posts; this still has a square root on the attractive petals. But please, keep posting. I'm happy to be proven wrong
. Your posts are becoming exponentially more educational; I appreciate your posts a lot, Leo. Thank you for your contributions.
Regards, James
EDIT: I fixed my rough analysis to say what I really meant.
I get your confusion. I stand by my point though. What you have done is shown there is no taylor expansion at \( 0 \); which I explained does not exist. The iterate can only exist on a petal about zero; not a neighborhood of zero. So your first point is null to my construction.
It helps to realize in the neutral case; that no iteration exists in a disk about zero; therefore, there is not a taylor series at zero. But; in the petals about zero; there are half iterates. But in a neighborhood of \( 0 \) intersecting with the attracting petals; and then excluding \( 0 \); we have a holomorphic function. Where we likely have a removable singularity at zero; but it is not holomorphic at zero.
May I switch to \( \sin(-\xi) \)? Because I can explain your confusion using this function. The same principle applies to \( f(\xi)= - \xi(1-\xi) \)--forgive the change to a simpler case, please. But using an odd function is easier (same principle applies); and \( \sin \) is easier to reference numerically.
The function,
\(
\sin(-\xi) = f(\xi)\\
\)
has an iteration for all \( k \in \mathbb{Z} \),
\(
f^{\circ z}(\xi) = e^{(2k+1)\pi iz} \xi+\sum_{n=2}^\infty b_n \xi^n\\
\)
Where this is only an ASYMPTOTIC SERIES about zero. if we assume that \( \sin^{\circ z}(-\xi) = - \sin^{\circ z}(\xi) \)--which is the standard iteration. Then there are two square root
\(
f^{\circ 1/2}(\xi) = i\xi + \sum_{n=2}^\infty c_n \xi^n\\
-f^{\circ 1/2}(\xi) = -i\xi - \sum_{n=2}^\infty c_n \xi^n\\
\)
Now notice; neither of these map the real line to the real line; like the iteration \( \sin^{\circ t}(\xi) \) for \( t \in \mathbb{R}^+ \); but they are iterates nonetheless. They are not holomorphic at zero; we just get a good asymptotic expansion when approaching zero. In this sense, the summation, and taylor series, is an *abuse of notation*; these things don't converge. Only asymptotically to zero are they meaningful. They are actually holomorphic on a petal. It's a bit more complicated with \( f(\xi) = -\xi(1-\xi) \) but it's the same scenario--as far as I can tell. They're not holomorphic at zero; only on a petal about zero. What the petal looks like? I have no f**n clue; but there exists a petal where this converges.
Unless I'm missing something large; which I don't think I am from your posts; this still has a square root on the attractive petals. But please, keep posting. I'm happy to be proven wrong
. Your posts are becoming exponentially more educational; I appreciate your posts a lot, Leo. Thank you for your contributions.Regards, James
EDIT: I fixed my rough analysis to say what I really meant.