Alright; so it appears I made a fumble. The Gamma functions should almost surely cancel out in some manner; so they appear in the process, but probably won't make an appearance in the final formula. To justify this, let's look at:
\[
f_\theta(z) = e^{e^{i\theta}z} -1\\
\]
But we're going to restrict \(\Im\theta > 0\); and then limit \(\Im\theta \to 0\), to hit the neutral case.
When \(\Im\theta > 0\), the function \(f_\theta(z) = e^{i\theta} z + O(z^2)\) and \(|e^{i\theta}| < 1\); so this is a geometrically attracting fixed point at \(z=0\). This means we can write:
\[
\Psi_\theta(f_\theta(z)) = e^{i\theta} \Psi_\theta(z)\\
\]
For a unique Schroder function \(\Psi_\theta\). Where then the regular iteration is:
\[
f_\theta^{\circ s}(z) = \Psi_\theta^{-1} \left( e^{i\theta s} \Psi_\theta(z)\right)\\
\]
Which is holomorphic for \(|z| < \delta\) and \(|e^{i\theta s}| < 1\)--which is a half plane in \(s\) if we fix \(\theta\).
We are going to focus on \(f_\pi(z) = e^{-z} -1\), but we're going to instead write \(g_n(z) = e^{-z + z/n} -1\). The function \(g_n(z) = (\frac{1}{n} -1) z + O(z^2)\), so that it has an attracting fixed point with multiplier \(\lambda_n = 1/n - 1\). Where the limit \(g_n \to f_\pi\) as \(n\to \infty\). This can also be written as \(f_{\theta_n}(z)\) when \(e^{i\theta_n} = 1/n - 1\)--which is \(\theta_n = -i\log(1/n - 1)\).
The function \(g_n(g_n(z)) = h(z)\), and \(h(z)\) looks like:
\[
h(z) = (1/n-1)^2 z + O(z^2)\\
\]
And thereby; if we write:
\[
\vartheta[h](x,z) = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^k}{k!}\\
\]
Then:
\[
\frac{d^{s}}{dx^s}\Big{|}_{x=0} \vartheta[h](x,z) = h^{\circ s}(z)\\
\]
But this is not an iteration of \(g_n\)! Capitulated by the statement \(\sqrt{h} \neq g_n\)....
So how do we make the square root of \(h\) equal \(g_n\)? How do we take "the negative square root".
This is an old idea of mine, I had never fully developed. But we return now to the Schroder expansion:
\[
h(z) = \Psi_{\theta_n}^{-1}\left( (1/n-1)^2 \Psi_{\theta_n}(z)\right)\\
\]
Whereupon, the formula we have given for \(h^{\circ s}(z)\) is aptly written:
\[
h^{\circ s}(z) = \Psi_{\theta_n}^{-1}\left( |1/n-1|^{2s} \Psi_{\theta_n}(z)\right)\\
\]
If we wanted \(h^{\circ 1/2} (z) = g_n(z)\), we would need to stick a negative somewhere in there. We do this by massaging our exponents.
\[
(1/n-1)^{2s} = e^{2\pi i s} |1/n-1|^{2s} = |1/n-1|^{2s +\frac{2 \pi i}{\log|1/n-1|}}\\
\]
Whereby; we can now deduce that:
\[
h^{\circ \frac{1}{2} + \frac{\pi i}{\log|1/n-1|}}(z) = g_n(z)\\
\]
Where now we can iterate \(g_n(z)\) (not quite under the mellin transform, but close) by writing:
\[
g_n^{\circ s}(z) = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}(z)\\
\]
(THIS CHANGE OF VARIABLES I FORGOT WILL CAUSE THE GAMMA FUNCTIONS TO JUST DISAPPEAR).
I'm going to walk through a simple case here, and explain how we are doing the same thing. This result dates very far back; but probably was never spoke of in this context.
When we take the function:
\[
e^{x}\\
\]
Then the Exponential Differintegral, is precisely itself:
\[
\frac{d^s}{dx^s} e^x = e^x\\
\]
When we change to \(e^{-x}\); then something a little tricky happens...
First of all; The exponential differintegral is defined as follows:
\[
\frac{d^{-s}}{dx^{-s}} f(x) = \frac{e^{-i\tau s}}{\Gamma(s)} \int_0^\infty f(x - e^{i\tau}y)y^{s-1}\,dy\\
\]
Where if a function is differintegrable, there is some open sector \(-\pi \le a < \tau < b \le \pi\), where this thing converges (And it's invariant under our choice of \(\tau\), constant in this sector--thanks to Cauchy). Well, for \(f(x) = e^{-x}\), this happens at \( \tau = \pi\). (It happens as we move in the opposite direction). So that:
\[
\frac{d^{-s}}{dx^{-s}} e^{-x} = \frac{e^{-i\pi s}}{\Gamma(s)} \int_0^\infty e^{-(x-e^{i\pi}y)}y^{s-1}\,dy = e^{-\pi i s} e^{-x}\\
\]
Now we do all of this solely with the principle branch; choosing different solutions to the differintegral outside of the principle branch equates to multiplying by \(e^{2 \pi i k s}\) for some \(k \in \mathbb{Z}\), so bear with me...
This looks like a non result; but it perfectly aligns the equation:
\[
\frac{d^s}{dx^s} e^{\mu x} = \mu^s e^{\mu x}\\
\]
And does so in the complex plane; and does so while accounting for different paths of convergence for each \(\mu\). Where the Exponential Differintegral is not just a mellin transform, it's precisely a generalized mellin transform that behaves as an Extension of a Linear Operator in a Hilbert space...We are extending a Hilbert Space a la Von Neumann. This is just an example of this.
So returning to the problem at hand, we can write:
\[
g_n^{\circ s}(z) = \frac{1}{\Gamma(s)} \int_0^\infty \vartheta[h](x,z)x^{-\frac{s}{2} -\frac{\pi i s}{\log|1/n-1|}-1}\,dx\\
\]
This will not converge on a maximal domain; but if we instead write:
\[
g_n^{\circ s}(z) = \frac{d^{\frac{s}{2} +\frac{\pi i s}{\log|1/n-1|}}}{dx^{\frac{s}{2} +\frac{\pi i s}{\log|1/n-1|}}}\Big{|}_{x=0} \vartheta[h](x,z)
\]
We can achieve the exact same maximal domain as:
\[
g_n^{\circ s}(z) = \Psi_{\theta_n}^{-1} \left( (1/n - 1)^{s} \Psi_{\theta_n}(z)\right)\\
\]
This will produce a Mittag Leffler expansion I'm too lazy to write out; but it'll look a lot like what I wrote above. Essentially, the \(\frac{d^s}{dx^s}\) is holomorphic for \(\Re(s) > 0\), but with this modification we have to map that domain to a new one.
This kind of ties the knot to what I was saying before. Let's try to find the iterate of \(g_n\) using the vartheta function of \(g_n\).
\[
\vartheta[g_n](x,z) = \sum_{k=0}^\infty g_n^{\circ k}(z) \frac{(-x)^k}{k!}\\
\]
Then! Since \(g_n^{\circ 2k}(z) = h^{\circ k}(z)\); we have that:
\[
\frac{\vartheta[g_n](ix,z) + \vartheta[g_n](-ix,z)}{2} = \frac{\vartheta[\sqrt{h}](ix,z) + \vartheta[\sqrt{h}](-ix,z)}{2}
\]
What we should end up with is going to look a lot wackier though; because, in basic principle:
\[
\frac{d^s}{dx^s} \vartheta[g_n](x,z) = g^{\circ s}(z) = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}(z)
\]
This may appear to be a contradiction, but:
\[
h^{\circ s + \frac{2\pi i}{\log|1/n-1|}}(z) = h^{\circ s}\\
\]
So we are utilizing a duplication formula, and the halving of a period, which when duplicated becomes the same function. And we can argue canonically that \(g_n^{\circ 2k} = h^{\circ k}\), with absolutely zero repercussions on our iterated format.
So This actually causes the Gamma function stuff to disappear
; as it should appear on both sides of the equation, and what we our left with is some kind of translation in the original argument.
When we let \(\Im\theta \to 0\), things are definitely going to get complicated. As \(\theta_n \to \pi\), something more clever is needed to do these arguments. But something like it should appear.
To explain where I went wrong is a little tricky, but I'll go as follows:
\[
\frac{1}{2\pi i} \int_{1/2 - i \infty}^{1/2 + i \infty} h^{\circ s}(z) \frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}x^{-s}\,ds = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^{k}}{2k!}\\
\]
But,
\[
\frac{\vartheta[g_n](i\sqrt{x},z) + \vartheta[g_n](-i\sqrt{x},z)}{2} = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^{k}}{2k!}\\
\]
From this point, we should just end up with the formula:
\[
g_n^{\circ s} = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}\\
\]
........ The gamma's should cancel out....
Regards, James
\[
f_\theta(z) = e^{e^{i\theta}z} -1\\
\]
But we're going to restrict \(\Im\theta > 0\); and then limit \(\Im\theta \to 0\), to hit the neutral case.
When \(\Im\theta > 0\), the function \(f_\theta(z) = e^{i\theta} z + O(z^2)\) and \(|e^{i\theta}| < 1\); so this is a geometrically attracting fixed point at \(z=0\). This means we can write:
\[
\Psi_\theta(f_\theta(z)) = e^{i\theta} \Psi_\theta(z)\\
\]
For a unique Schroder function \(\Psi_\theta\). Where then the regular iteration is:
\[
f_\theta^{\circ s}(z) = \Psi_\theta^{-1} \left( e^{i\theta s} \Psi_\theta(z)\right)\\
\]
Which is holomorphic for \(|z| < \delta\) and \(|e^{i\theta s}| < 1\)--which is a half plane in \(s\) if we fix \(\theta\).
We are going to focus on \(f_\pi(z) = e^{-z} -1\), but we're going to instead write \(g_n(z) = e^{-z + z/n} -1\). The function \(g_n(z) = (\frac{1}{n} -1) z + O(z^2)\), so that it has an attracting fixed point with multiplier \(\lambda_n = 1/n - 1\). Where the limit \(g_n \to f_\pi\) as \(n\to \infty\). This can also be written as \(f_{\theta_n}(z)\) when \(e^{i\theta_n} = 1/n - 1\)--which is \(\theta_n = -i\log(1/n - 1)\).
The function \(g_n(g_n(z)) = h(z)\), and \(h(z)\) looks like:
\[
h(z) = (1/n-1)^2 z + O(z^2)\\
\]
And thereby; if we write:
\[
\vartheta[h](x,z) = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^k}{k!}\\
\]
Then:
\[
\frac{d^{s}}{dx^s}\Big{|}_{x=0} \vartheta[h](x,z) = h^{\circ s}(z)\\
\]
But this is not an iteration of \(g_n\)! Capitulated by the statement \(\sqrt{h} \neq g_n\)....
So how do we make the square root of \(h\) equal \(g_n\)? How do we take "the negative square root".
This is an old idea of mine, I had never fully developed. But we return now to the Schroder expansion:
\[
h(z) = \Psi_{\theta_n}^{-1}\left( (1/n-1)^2 \Psi_{\theta_n}(z)\right)\\
\]
Whereupon, the formula we have given for \(h^{\circ s}(z)\) is aptly written:
\[
h^{\circ s}(z) = \Psi_{\theta_n}^{-1}\left( |1/n-1|^{2s} \Psi_{\theta_n}(z)\right)\\
\]
If we wanted \(h^{\circ 1/2} (z) = g_n(z)\), we would need to stick a negative somewhere in there. We do this by massaging our exponents.
\[
(1/n-1)^{2s} = e^{2\pi i s} |1/n-1|^{2s} = |1/n-1|^{2s +\frac{2 \pi i}{\log|1/n-1|}}\\
\]
Whereby; we can now deduce that:
\[
h^{\circ \frac{1}{2} + \frac{\pi i}{\log|1/n-1|}}(z) = g_n(z)\\
\]
Where now we can iterate \(g_n(z)\) (not quite under the mellin transform, but close) by writing:
\[
g_n^{\circ s}(z) = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}(z)\\
\]
(THIS CHANGE OF VARIABLES I FORGOT WILL CAUSE THE GAMMA FUNCTIONS TO JUST DISAPPEAR).
I'm going to walk through a simple case here, and explain how we are doing the same thing. This result dates very far back; but probably was never spoke of in this context.
When we take the function:
\[
e^{x}\\
\]
Then the Exponential Differintegral, is precisely itself:
\[
\frac{d^s}{dx^s} e^x = e^x\\
\]
When we change to \(e^{-x}\); then something a little tricky happens...
First of all; The exponential differintegral is defined as follows:
\[
\frac{d^{-s}}{dx^{-s}} f(x) = \frac{e^{-i\tau s}}{\Gamma(s)} \int_0^\infty f(x - e^{i\tau}y)y^{s-1}\,dy\\
\]
Where if a function is differintegrable, there is some open sector \(-\pi \le a < \tau < b \le \pi\), where this thing converges (And it's invariant under our choice of \(\tau\), constant in this sector--thanks to Cauchy). Well, for \(f(x) = e^{-x}\), this happens at \( \tau = \pi\). (It happens as we move in the opposite direction). So that:
\[
\frac{d^{-s}}{dx^{-s}} e^{-x} = \frac{e^{-i\pi s}}{\Gamma(s)} \int_0^\infty e^{-(x-e^{i\pi}y)}y^{s-1}\,dy = e^{-\pi i s} e^{-x}\\
\]
Now we do all of this solely with the principle branch; choosing different solutions to the differintegral outside of the principle branch equates to multiplying by \(e^{2 \pi i k s}\) for some \(k \in \mathbb{Z}\), so bear with me...
This looks like a non result; but it perfectly aligns the equation:
\[
\frac{d^s}{dx^s} e^{\mu x} = \mu^s e^{\mu x}\\
\]
And does so in the complex plane; and does so while accounting for different paths of convergence for each \(\mu\). Where the Exponential Differintegral is not just a mellin transform, it's precisely a generalized mellin transform that behaves as an Extension of a Linear Operator in a Hilbert space...We are extending a Hilbert Space a la Von Neumann. This is just an example of this.
So returning to the problem at hand, we can write:
\[
g_n^{\circ s}(z) = \frac{1}{\Gamma(s)} \int_0^\infty \vartheta[h](x,z)x^{-\frac{s}{2} -\frac{\pi i s}{\log|1/n-1|}-1}\,dx\\
\]
This will not converge on a maximal domain; but if we instead write:
\[
g_n^{\circ s}(z) = \frac{d^{\frac{s}{2} +\frac{\pi i s}{\log|1/n-1|}}}{dx^{\frac{s}{2} +\frac{\pi i s}{\log|1/n-1|}}}\Big{|}_{x=0} \vartheta[h](x,z)
\]
We can achieve the exact same maximal domain as:
\[
g_n^{\circ s}(z) = \Psi_{\theta_n}^{-1} \left( (1/n - 1)^{s} \Psi_{\theta_n}(z)\right)\\
\]
This will produce a Mittag Leffler expansion I'm too lazy to write out; but it'll look a lot like what I wrote above. Essentially, the \(\frac{d^s}{dx^s}\) is holomorphic for \(\Re(s) > 0\), but with this modification we have to map that domain to a new one.
This kind of ties the knot to what I was saying before. Let's try to find the iterate of \(g_n\) using the vartheta function of \(g_n\).
\[
\vartheta[g_n](x,z) = \sum_{k=0}^\infty g_n^{\circ k}(z) \frac{(-x)^k}{k!}\\
\]
Then! Since \(g_n^{\circ 2k}(z) = h^{\circ k}(z)\); we have that:
\[
\frac{\vartheta[g_n](ix,z) + \vartheta[g_n](-ix,z)}{2} = \frac{\vartheta[\sqrt{h}](ix,z) + \vartheta[\sqrt{h}](-ix,z)}{2}
\]
What we should end up with is going to look a lot wackier though; because, in basic principle:
\[
\frac{d^s}{dx^s} \vartheta[g_n](x,z) = g^{\circ s}(z) = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}(z)
\]
This may appear to be a contradiction, but:
\[
h^{\circ s + \frac{2\pi i}{\log|1/n-1|}}(z) = h^{\circ s}\\
\]
So we are utilizing a duplication formula, and the halving of a period, which when duplicated becomes the same function. And we can argue canonically that \(g_n^{\circ 2k} = h^{\circ k}\), with absolutely zero repercussions on our iterated format.
So This actually causes the Gamma function stuff to disappear
; as it should appear on both sides of the equation, and what we our left with is some kind of translation in the original argument.When we let \(\Im\theta \to 0\), things are definitely going to get complicated. As \(\theta_n \to \pi\), something more clever is needed to do these arguments. But something like it should appear.
To explain where I went wrong is a little tricky, but I'll go as follows:
\[
\frac{1}{2\pi i} \int_{1/2 - i \infty}^{1/2 + i \infty} h^{\circ s}(z) \frac{\Gamma(s)\Gamma(1-s)}{\Gamma(1-2s)}x^{-s}\,ds = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^{k}}{2k!}\\
\]
But,
\[
\frac{\vartheta[g_n](i\sqrt{x},z) + \vartheta[g_n](-i\sqrt{x},z)}{2} = \sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^{k}}{2k!}\\
\]
From this point, we should just end up with the formula:
\[
g_n^{\circ s} = h^{\circ \frac{s}{2} + \frac{\pi i s}{\log|1/n-1|}}\\
\]
........ The gamma's should cancel out....
Regards, James