Let's get some notation clear right now.
We will call \(\mathcal{P}\) an attracting petal of \(f(z)\) at \(f(0) = 0\) where \(|f'(0)| = 1\). We will call \(\cup \mathcal{P}\) the collection of all attracting petals. We will say \(f\) is an entire function (mostly for convenience). From which, we have for all \(\mathcal{P} \subset \mathbb{C}\) that are petals:
\[
\lim_{n\to\infty} f^{\circ n} \Big{|}_{\mathcal{P}} \to 0\\
\]
Whereby, we are guaranteed now, that the union of such domains:
\[
\lim_{n\to\infty} f^{\circ n} \Big{|}_{\cup\mathcal{P}} \to 0\\
\]
Identify that \(f : \cup \mathcal{P} \to \cup \mathcal{P}\). Recall a "domain" is an open and connected set in \(\mathbb{C}\). \(\mathcal{P}\) is a domain, the set \(\cup \mathcal{P}\) is not a domain, it is a collection of domains... There's a bunch of branch cuts/julia set shit which separates everything.
As I'm too lazy to write the latex code every time:
\[
\frac{d^s}{dw^s}h(w) = H(s) = \frac{d^s}{dw^s}h(w)\Big{|}_{w=0}\\
\]
This is just a fancy mellin transform.
By the above renditions, we know that:
\[
\vartheta(w,z) = \sum_{n=0}^\infty f^{\circ n+1}(z) \frac{w^n}{n!}\\
\]
Is holomorphic for:
\[
\vartheta(w,z) : \mathbb{C} \times \cup\mathcal{P} \to \mathbb{C}\\
\]
Now, what we're trying to show; is that \(\frac{d^s}{dw^s} \vartheta(w,z)\) always exists.
This is highly non trivial. We know when \(f(z) = e^z-1\) and \(\cup \mathcal{P} = \mathbb{C}/[0,\infty)\); this absolutely happens. When we have a negative:
\[
f_-(z) = f(-z) = e^{-z}-1
\]
This instantly becomes a non obvious problem. The domain \(\cup \mathcal{P}\), now has a cycle permutation, for everytime we apply that negative. If we take:
\[
f_2(z) = e^{iz} - 1\\
\]
We now have a permutation of order 4. But no matter what these things shuffle around. They still produce holomorphic \(\vartheta\). And they produce integrals, though they may be on exotic domains.
To take \(f_-\) for example. The manner we take this integral is rather strange. We have our function:
\[
\vartheta(w,z): \mathbb{C} \times \cup\mathcal{P} \to \mathbb{C}\\
\]
And then we have the standard abel function on the attracting cup petal \(\cup\mathcal{P}\):
\[
\alpha(f_-(z)) = \alpha(z) + 1\\
\]
We can restrict this right away to a connected petal \(K \subset \cup \mathcal{P}\). The question, is whether the differintegral can do this.
To start this debauchery:
\[
b(w) = \sum_{k=0}^\infty f_-^{\circ 2(k+1)} \frac{w^k}{k!}\\
\]
Which:
\[
\frac{d^s}{dw^s} b(w) = \left(f_- \circ f_-\right)^{\circ s}\\
\]
Well:
\[
\frac{\vartheta(iw,z) + \vartheta(-iw,z)}{2} = g(w,z)
\]
Where:
\[
\sum_{k=0}^\infty f_-^{\circ 2(k+1)} \frac{(-1)^k w^{2k}}{2k!} = g(w,z)\\
\]
The function \(\frac{d^s}{dw^s} g(w,z) = (f_- \circ f_-)^{\circ s}(z)\frac{\Gamma(\frac{s}{2})\Gamma(1-\frac{s}{2})}{\Gamma(1-s)}\). Using old knowledge:
\[
\frac{d^s}{dw^s}\varphi(\lambda w) = \lambda^s \frac{d^s}{dw^s} \varphi(w)\\
\]
So we can pull out:
\[
\frac{i^s \frac{d^s}{dw^s}\vartheta(w,z) + (-i)^s \frac{d^s}{dw^s} \vartheta(w,z)}{2} = (f_- \circ f_-)^{\circ s}(z)\frac{\Gamma(\frac{s}{2})\Gamma(1-\frac{s}{2})}{\Gamma(1-s)}
\]
This means something I've worked towards for a long time. You do see this, even with the geometric case. But to pull this out in a boundary case has always escaped me. I hope I'm not fucking anything up too egregiously. I love me some hypergeometric reasoninig.
The value:
\[
f_-^{\circ s}(z) = \frac{d^s}{dw^s}\vartheta(w,z)\\
\]
Is given by the formula:
\[
f_-^{\circ 2s}(z)\cos\left(\pi s\right) = \left(f_-\circ f_-\right)^{\circ s}(z)\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}\\
\]
(I've gotten this formula for the past year and I always thought it was too good to be true. So It might be too good to be true, but there's something like this at least appearing locally)
I hate to be the take it with a grain of salt kind of guy; but take it with a grain of salt. I might have screwed up some things.
BUT FUCKING SERIOUSLY! this my Euler equation!
\[
f_-^{\circ 2s}(z)\cos\left(\pi s\right) = \left(f_-\circ f_-\right)^{\circ s}(z)\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}\\
\]
We will call \(\mathcal{P}\) an attracting petal of \(f(z)\) at \(f(0) = 0\) where \(|f'(0)| = 1\). We will call \(\cup \mathcal{P}\) the collection of all attracting petals. We will say \(f\) is an entire function (mostly for convenience). From which, we have for all \(\mathcal{P} \subset \mathbb{C}\) that are petals:
\[
\lim_{n\to\infty} f^{\circ n} \Big{|}_{\mathcal{P}} \to 0\\
\]
Whereby, we are guaranteed now, that the union of such domains:
\[
\lim_{n\to\infty} f^{\circ n} \Big{|}_{\cup\mathcal{P}} \to 0\\
\]
Identify that \(f : \cup \mathcal{P} \to \cup \mathcal{P}\). Recall a "domain" is an open and connected set in \(\mathbb{C}\). \(\mathcal{P}\) is a domain, the set \(\cup \mathcal{P}\) is not a domain, it is a collection of domains... There's a bunch of branch cuts/julia set shit which separates everything.
As I'm too lazy to write the latex code every time:
\[
\frac{d^s}{dw^s}h(w) = H(s) = \frac{d^s}{dw^s}h(w)\Big{|}_{w=0}\\
\]
This is just a fancy mellin transform.
By the above renditions, we know that:
\[
\vartheta(w,z) = \sum_{n=0}^\infty f^{\circ n+1}(z) \frac{w^n}{n!}\\
\]
Is holomorphic for:
\[
\vartheta(w,z) : \mathbb{C} \times \cup\mathcal{P} \to \mathbb{C}\\
\]
Now, what we're trying to show; is that \(\frac{d^s}{dw^s} \vartheta(w,z)\) always exists.
This is highly non trivial. We know when \(f(z) = e^z-1\) and \(\cup \mathcal{P} = \mathbb{C}/[0,\infty)\); this absolutely happens. When we have a negative:
\[
f_-(z) = f(-z) = e^{-z}-1
\]
This instantly becomes a non obvious problem. The domain \(\cup \mathcal{P}\), now has a cycle permutation, for everytime we apply that negative. If we take:
\[
f_2(z) = e^{iz} - 1\\
\]
We now have a permutation of order 4. But no matter what these things shuffle around. They still produce holomorphic \(\vartheta\). And they produce integrals, though they may be on exotic domains.
To take \(f_-\) for example. The manner we take this integral is rather strange. We have our function:
\[
\vartheta(w,z): \mathbb{C} \times \cup\mathcal{P} \to \mathbb{C}\\
\]
And then we have the standard abel function on the attracting cup petal \(\cup\mathcal{P}\):
\[
\alpha(f_-(z)) = \alpha(z) + 1\\
\]
We can restrict this right away to a connected petal \(K \subset \cup \mathcal{P}\). The question, is whether the differintegral can do this.
To start this debauchery:
\[
b(w) = \sum_{k=0}^\infty f_-^{\circ 2(k+1)} \frac{w^k}{k!}\\
\]
Which:
\[
\frac{d^s}{dw^s} b(w) = \left(f_- \circ f_-\right)^{\circ s}\\
\]
Well:
\[
\frac{\vartheta(iw,z) + \vartheta(-iw,z)}{2} = g(w,z)
\]
Where:
\[
\sum_{k=0}^\infty f_-^{\circ 2(k+1)} \frac{(-1)^k w^{2k}}{2k!} = g(w,z)\\
\]
The function \(\frac{d^s}{dw^s} g(w,z) = (f_- \circ f_-)^{\circ s}(z)\frac{\Gamma(\frac{s}{2})\Gamma(1-\frac{s}{2})}{\Gamma(1-s)}\). Using old knowledge:
\[
\frac{d^s}{dw^s}\varphi(\lambda w) = \lambda^s \frac{d^s}{dw^s} \varphi(w)\\
\]
So we can pull out:
\[
\frac{i^s \frac{d^s}{dw^s}\vartheta(w,z) + (-i)^s \frac{d^s}{dw^s} \vartheta(w,z)}{2} = (f_- \circ f_-)^{\circ s}(z)\frac{\Gamma(\frac{s}{2})\Gamma(1-\frac{s}{2})}{\Gamma(1-s)}
\]
This means something I've worked towards for a long time. You do see this, even with the geometric case. But to pull this out in a boundary case has always escaped me. I hope I'm not fucking anything up too egregiously. I love me some hypergeometric reasoninig.
The value:
\[
f_-^{\circ s}(z) = \frac{d^s}{dw^s}\vartheta(w,z)\\
\]
Is given by the formula:
\[
f_-^{\circ 2s}(z)\cos\left(\pi s\right) = \left(f_-\circ f_-\right)^{\circ s}(z)\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}\\
\]
(I've gotten this formula for the past year and I always thought it was too good to be true. So It might be too good to be true, but there's something like this at least appearing locally)
I hate to be the take it with a grain of salt kind of guy; but take it with a grain of salt. I might have screwed up some things.
BUT FUCKING SERIOUSLY! this my Euler equation!
\[
f_-^{\circ 2s}(z)\cos\left(\pi s\right) = \left(f_-\circ f_-\right)^{\circ s}(z)\frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}\\
\]