I'm going to continue this train of thought now, while I have it.
This essentially reduces "under the mellin transform" transformations, into the standard "regular iteration" but only for specific values. Namely, when:
\[
\Im \theta >0\,\,\Re \theta = k \pi\,\,\text{for}\,k \in \mathbb{Z}\\
\]
Because, \(f_\theta\) in this instance, satisfies \(0 < e^{i2\theta} < 1\)--is real positive.
But there is nothing stopping us from extending this for all \(\Im \theta > 0\). We just have to think a bit more clever. Let's take:
\[
\theta \in 2\pi \mathbb{Q}(i)\\
\]
Where, still, \(\Im(\theta) > 0\). So that \(\theta = 2 \pi a + 2 \pi b i\), where \(a,b \in \mathbb{Q}\) and \(b > 0\).
And let's assume that \( q \theta = \in 2\pi \mathbb{Z}(i)\), and \(q \in \mathbb{N}\) is the minimal element to do so. The set \(2 \pi \mathbb{Q}(i)\) is dense in \(\Im(\theta) > 0\). This is nothing new.
The function \(f_\theta^{\circ q}(z) = h(z)\). Then... \(h'(0) = e^{i\theta q}\), satisfies, \(0 < e^{i \theta q} < 1\) is real positive.
This means that \(h^{\circ s}\) has the same period as \(e^{i\theta q s}\), which is precisely \(\frac{2 \pi}{\theta q}\). We are taking a \(q\)'th of this, which says that:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2 \theta}}(z)\\
\]
BUT!!!!!!! Here is the huge fucking but. When we take a square root, like I did in my last post; we only have \(1\) or \(-1\). In this case, we have precisely \(q\) roots. Since we have chosen that \(q\) is the minimal value for \(q \theta \in 2 \pi \mathbb{Z}(i)\)--there are \(q\) other values which satisfy this equation. (The above equation is fine, but it can be tricky to answer...) So the correct answer, and the more blanket version is...
There exists some \(0 \le t < q\) such that:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s t}{q^2 \theta}}(z)\\
\]
Now each one is not a solution, but sum are (Fucking Gaussian sum shit); but we don't care. We've found the Regular iteration solution.
So:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2 \theta}}(z)\\
\]
Where: \(h(z) = f_\theta^{\circ q}(z)\)
Where \(q \in \mathbb{N}\) is the minimal integer, so that \(q\theta \in 2\pi \mathbb{Z}(i)\)...
Again........ the Gamma's cancel out, but in the Mellin transform we use Gauss's Gamma multiplication formula....
We can also see a cool looking Gauss like sum kinda appearing.....
We're going to have a lot of trouble taking \(\Im\theta \to 0\). But honestly, not that much trouble. So long as we don't send \(\theta \to 2 \pi k\) we should be fine. So the singularity action is precisely at \(f_0(z) = e^z -1\), which we know how to iterate. At \(f_\pi(z) = e^{-z} -1\) we're not gonna see such a singularity. It should behave better-- and near zero we can accurately say that:
\[
f_\theta^{\circ s}(z) = e^{i\theta s}z + O(z^2)\\
\]
As we limit \(\Im \theta \to 0\), this becomes an asymptotic series; so it is not holomorphic. But it still looks like this expression.
THE BIG WHODUNNIT ABOUT THIS IS THAT WE CAN WRITE THIS UNDER A MELLIN TRANSFORM.
\[
\int_{1/2 - i\infty}^{1/2 + i\infty} h^{\circ s/q}(z) \frac{\Gamma(s/q)\Gamma(1-s/q)}{\Gamma(1-s)}x^{-s}\,dx = \int_{1/2 - i \infty}^{1/2+i\infty} f_\theta^{\circ s}(z)\frac{\Gamma(s/q)\Gamma(1-s/q)}{\Gamma(1-s)}x^{-s}\,dx\\
\]
Which is basically the statement that:
\[
\sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^k}{(qk)!} = \sum_{k=0}^\infty f_\theta^{\circ qk}(z) \frac{(-x)^k}{(qk)!}\\
\]
But now we've written it with the Fourier transform, but not just the Fourier transform; a Distributional understanding of the Fourier Transform...
But this only happens in a distributional sense. The right hand integral does not converge \(f_\theta^{\circ s}\) does not decay properly. But there's only a single other function which satisfies this in a distributional sense (once we've extended the Fourier transform linearly).... \(f_\theta^{\circ s} = h^{\circ \frac{s}{q} + \frac{2 \pi t}{q^2 \theta}}\) for \(t\) coprime to \(q\)--but the regular iteration is given as \(t =1\). The other iterations will fail a unification, monodromy theorem, but will act as mock solutions. As they will not give \(f'_\theta(0) = e^{i\theta}\).
So....... the Gamma function stuff I wrote above, it actually cancels out.
I apologize. I saw some Gammas and got excited and ahead of myself lmao.
This essentially reduces "under the mellin transform" transformations, into the standard "regular iteration" but only for specific values. Namely, when:
\[
\Im \theta >0\,\,\Re \theta = k \pi\,\,\text{for}\,k \in \mathbb{Z}\\
\]
Because, \(f_\theta\) in this instance, satisfies \(0 < e^{i2\theta} < 1\)--is real positive.
But there is nothing stopping us from extending this for all \(\Im \theta > 0\). We just have to think a bit more clever. Let's take:
\[
\theta \in 2\pi \mathbb{Q}(i)\\
\]
Where, still, \(\Im(\theta) > 0\). So that \(\theta = 2 \pi a + 2 \pi b i\), where \(a,b \in \mathbb{Q}\) and \(b > 0\).
And let's assume that \( q \theta = \in 2\pi \mathbb{Z}(i)\), and \(q \in \mathbb{N}\) is the minimal element to do so. The set \(2 \pi \mathbb{Q}(i)\) is dense in \(\Im(\theta) > 0\). This is nothing new.
The function \(f_\theta^{\circ q}(z) = h(z)\). Then... \(h'(0) = e^{i\theta q}\), satisfies, \(0 < e^{i \theta q} < 1\) is real positive.
This means that \(h^{\circ s}\) has the same period as \(e^{i\theta q s}\), which is precisely \(\frac{2 \pi}{\theta q}\). We are taking a \(q\)'th of this, which says that:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2 \theta}}(z)\\
\]
BUT!!!!!!! Here is the huge fucking but. When we take a square root, like I did in my last post; we only have \(1\) or \(-1\). In this case, we have precisely \(q\) roots. Since we have chosen that \(q\) is the minimal value for \(q \theta \in 2 \pi \mathbb{Z}(i)\)--there are \(q\) other values which satisfy this equation. (The above equation is fine, but it can be tricky to answer...) So the correct answer, and the more blanket version is...
There exists some \(0 \le t < q\) such that:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s t}{q^2 \theta}}(z)\\
\]
Now each one is not a solution, but sum are (Fucking Gaussian sum shit); but we don't care. We've found the Regular iteration solution.
So:
\[
f_\theta^{\circ s}(z) = h^{\circ \frac{s}{q} + \frac{2 \pi s}{q^2 \theta}}(z)\\
\]
Where: \(h(z) = f_\theta^{\circ q}(z)\)
Where \(q \in \mathbb{N}\) is the minimal integer, so that \(q\theta \in 2\pi \mathbb{Z}(i)\)...
Again........ the Gamma's cancel out, but in the Mellin transform we use Gauss's Gamma multiplication formula....
We can also see a cool looking Gauss like sum kinda appearing.....
We're going to have a lot of trouble taking \(\Im\theta \to 0\). But honestly, not that much trouble. So long as we don't send \(\theta \to 2 \pi k\) we should be fine. So the singularity action is precisely at \(f_0(z) = e^z -1\), which we know how to iterate. At \(f_\pi(z) = e^{-z} -1\) we're not gonna see such a singularity. It should behave better-- and near zero we can accurately say that:
\[
f_\theta^{\circ s}(z) = e^{i\theta s}z + O(z^2)\\
\]
As we limit \(\Im \theta \to 0\), this becomes an asymptotic series; so it is not holomorphic. But it still looks like this expression.
THE BIG WHODUNNIT ABOUT THIS IS THAT WE CAN WRITE THIS UNDER A MELLIN TRANSFORM.
\[
\int_{1/2 - i\infty}^{1/2 + i\infty} h^{\circ s/q}(z) \frac{\Gamma(s/q)\Gamma(1-s/q)}{\Gamma(1-s)}x^{-s}\,dx = \int_{1/2 - i \infty}^{1/2+i\infty} f_\theta^{\circ s}(z)\frac{\Gamma(s/q)\Gamma(1-s/q)}{\Gamma(1-s)}x^{-s}\,dx\\
\]
Which is basically the statement that:
\[
\sum_{k=0}^\infty h^{\circ k}(z) \frac{(-x)^k}{(qk)!} = \sum_{k=0}^\infty f_\theta^{\circ qk}(z) \frac{(-x)^k}{(qk)!}\\
\]
But now we've written it with the Fourier transform, but not just the Fourier transform; a Distributional understanding of the Fourier Transform...
But this only happens in a distributional sense. The right hand integral does not converge \(f_\theta^{\circ s}\) does not decay properly. But there's only a single other function which satisfies this in a distributional sense (once we've extended the Fourier transform linearly).... \(f_\theta^{\circ s} = h^{\circ \frac{s}{q} + \frac{2 \pi t}{q^2 \theta}}\) for \(t\) coprime to \(q\)--but the regular iteration is given as \(t =1\). The other iterations will fail a unification, monodromy theorem, but will act as mock solutions. As they will not give \(f'_\theta(0) = e^{i\theta}\).
So....... the Gamma function stuff I wrote above, it actually cancels out.
I apologize. I saw some Gammas and got excited and ahead of myself lmao.