Hey, everyone.
So I've recently started to pay sole attention to:
\[
f(z) = e^z-1\\
\]
And the different petals this object represents (canonically there are 2 petals here). From this identification, I can create as good as possible iterations. Similarly, I managed to produce the same asymptotic series Gottfried wrote of \(f^{\circ 0.5}(z)\) about \(z=0\). My construction, although much slower and less accurate and precise compared to Gottfried, was much simpler. The trouble I want to talk about now is a tad different.
Let's write:
\[
f_\theta(z) = e^{e^{i\theta}z}-1\\
\]
For example, if \(\theta = \pi\):
\[
f_\pi(z) = e^{-z} - 1\\
\]
What is the fractional iteration of this beast!?
Well, if we reduce using basic mathematics this can be found. Let us say that:
\[
f^{\circ 0.5}(f_{\pi}(z)) = f^{\circ 1.5}(-z)\\
\]
Which is absolutely true. Further more, one can find that:
\[
f_\pi(f^{\circ 0.5}(z)) = f^{1.5}(-z)\\
\]
How does this relate to the iteration of \(f_\theta\) for \(\theta \in \mathbb{R}\). How does this relate across neutral fixed points...
Think of this, as us applying \(e^{i\theta}\) while we apply \(f\), and doing so synchronously. We are taking \(f_0 : z \to e^z -1\) and \(\theta: z \to e^{i\theta}z\). Somehow we are taking this in a manner to produce \(f^{\circ 0.5}_\theta\). We are taking the half root of both functions, while doing neither.
To begin this foray, we can say that \(f_\pi(f_\pi(z)) = z + O(z^2)\). Which means, if we iterate \(f_\pi\), we can get a nice function, which has \(1\) as it's multiplier. The trouble here being, any iterate of this, will be an iterate of \(f^{\circ 2}_\pi\) and not an iterate of \(f_\pi\).
So, how do we write an iterate of \(f_\theta(z)\)?
I hope to answer this shortly, but I'm not perfect here. This result is only true for \(|z| < \delta\), while additionally, is not holomorphic at \(z=0\), and additionally has branch cuts stemming from this point.
I am currently making graphs, and checking my code. but I believe I can explain all of this using the mellin transform and the appropriate iterations.
One of the key observations, which is entirely non obvious, is that:
\[
\vartheta_\pi(x,z) = \sum_{n=0}^\infty f_{\pi}^{\circ n}(z)\frac{(-x)^n}{n!}\\
\]
Does not have decay as \(x \to \infty\)--instead it has this decay as \(x \to -\infty\). The affectation of adding a \((-1)\) in the function argument, causes the taylor series to rotate its domain of convergence under the mellin transform. I hope to explain this better.
Write:
\[
\vartheta_\theta(x,z) = \sum_{n=0}^\infty f_{\theta}^{\circ n}(z)\frac{(-x)^n}{n!}\\
\]
This function has decay for \(\arg(x) = \theta\). And in neighborhoods of this. Whereby, results in mellin, describe results in Abel.
I apologize, I hope to make all of this work better. I just do not have the time as of now.
A helpful way to visualize this is with \(f = e^z-1\). To begin:
\[
f^{\circ n}(z) = f^n\\
\]
Are purely positive, and monotone. If I write:
\[
f_\pi^{\circ n} = f^n_\pi\\
\]
Then this monotone nature no longer appears.
When I write \(e^{-1} -1 < 0\), this appears throughout. When I write:
\[
e^1 -1 > 1
\]
This appears throughout. When I adapt this to taylor series we can write:
\[
\sum_{n=0}^\infty f^{\circ n} \frac{(-x)^n}{n!}\\
\]
Where the coefficients are monotone, and therefore the oscillation is only formed by \((-x)^n\). This allows us to take a mellin transform, because everything is nice. When I do this with \(f_\pi\), things are different:
\[
\vartheta_\pi(x) = \sum_{n=0}^\infty f_\pi^{\circ n} \frac{(-x)^n}{n!}\\
\]
Using this function, this does not have a nice decay as we increase \(x \to \infty\). Which is because \(f_\pi^{\circ n}\) looks like \((-1)^n\) itself. It oscilates in a similar manner. Where by, we must take \(x \to - \infty\), and then the integral converges.
FFS I'm tired boys...
To be straight with you guys, we have to think like Ramanujan:
\[
f^{\circ n}(z) = 1^n
\]
whereby:
\[
\vartheta(x) = e^{-x}\\
\]
From which:
\[
f_\pi^{\circ n}(z) = (-1)^n\\
\]
and:
\[
\vartheta_\pi(x) = e^x\\
\]
Where, at this point, all we have to do is change the direction of our integral... And this follows for fucking every direction... We can perform the iteration no matter what, but we have to write the integral correctly. First of all:
\[
\int_0^\infty \vartheta_0(x)\,dx\\
\]
converges, but then to do this when we change the multiplier:
\[
\int_{0}^{-\infty} \vartheta_\pi(x)\,dx\\
\]
We have to move the domain of integration, as we move our petals/initial points, ALL THIS FUCKING
BULLSHIT!!!!!! If I were to take \(\vartheta_{\pi/2}\) then I'd write something like:
\[
\int_0^{-i\infty}\vartheta_{\pi/2}(x)\,dx\\\
\]
And write various transforms attached to this (AGAIN, THIS MEANS WE BELONG TO A HILBERT SPACE). This is again no different than the identity:
\[
\int_0^{-i\infty} e^{-ix}\,dx\\
\]
And the fact this object converges. We typically attach this with a mellin transform, whereby:
\[
e^{-\pi i s /2} \Gamma(s) = \int_0^{-i\infty} e^{-ix}x^{s-1}\,dx\\
\]
We are doing something deeply similar; just without the obvious beauty of the Gamma function.
I really want to hear you guys out, but I'm just lost at how much you guys are forgetting. And ignoring the "Schrodinger" interpretation has left me blind sided. I plan to write a broken, but fairly well interpreted program. And all it uses is basic code.
EDIT:
A good way to conceptualize this, is if \(a_n \in \mathbb{R}^+\), and it has good enough decay, then:
\[
\vartheta(x) = \sum_{n=0}^\infty a_n \frac{(-x)^n}{n!}\\
\]
Can look a lot like \(e^{-x}\). When \(b_n = (-1)^n a_n\), then:
\[
\vartheta_\pi(x) = \sum_{n=0}^\infty b_n \frac{(-x)^n}{n!} \approx e^x\\
\]
BOTH OF THESE FUNCTIONS ARE MELLIN TRANSFORMABLE!!!!!
We just need to take the mellin transform in different manners.........
So I've recently started to pay sole attention to:
\[
f(z) = e^z-1\\
\]
And the different petals this object represents (canonically there are 2 petals here). From this identification, I can create as good as possible iterations. Similarly, I managed to produce the same asymptotic series Gottfried wrote of \(f^{\circ 0.5}(z)\) about \(z=0\). My construction, although much slower and less accurate and precise compared to Gottfried, was much simpler. The trouble I want to talk about now is a tad different.
Let's write:
\[
f_\theta(z) = e^{e^{i\theta}z}-1\\
\]
For example, if \(\theta = \pi\):
\[
f_\pi(z) = e^{-z} - 1\\
\]
What is the fractional iteration of this beast!?
Well, if we reduce using basic mathematics this can be found. Let us say that:
\[
f^{\circ 0.5}(f_{\pi}(z)) = f^{\circ 1.5}(-z)\\
\]
Which is absolutely true. Further more, one can find that:
\[
f_\pi(f^{\circ 0.5}(z)) = f^{1.5}(-z)\\
\]
How does this relate to the iteration of \(f_\theta\) for \(\theta \in \mathbb{R}\). How does this relate across neutral fixed points...
Think of this, as us applying \(e^{i\theta}\) while we apply \(f\), and doing so synchronously. We are taking \(f_0 : z \to e^z -1\) and \(\theta: z \to e^{i\theta}z\). Somehow we are taking this in a manner to produce \(f^{\circ 0.5}_\theta\). We are taking the half root of both functions, while doing neither.
To begin this foray, we can say that \(f_\pi(f_\pi(z)) = z + O(z^2)\). Which means, if we iterate \(f_\pi\), we can get a nice function, which has \(1\) as it's multiplier. The trouble here being, any iterate of this, will be an iterate of \(f^{\circ 2}_\pi\) and not an iterate of \(f_\pi\).
So, how do we write an iterate of \(f_\theta(z)\)?
I hope to answer this shortly, but I'm not perfect here. This result is only true for \(|z| < \delta\), while additionally, is not holomorphic at \(z=0\), and additionally has branch cuts stemming from this point.
I am currently making graphs, and checking my code. but I believe I can explain all of this using the mellin transform and the appropriate iterations.
One of the key observations, which is entirely non obvious, is that:
\[
\vartheta_\pi(x,z) = \sum_{n=0}^\infty f_{\pi}^{\circ n}(z)\frac{(-x)^n}{n!}\\
\]
Does not have decay as \(x \to \infty\)--instead it has this decay as \(x \to -\infty\). The affectation of adding a \((-1)\) in the function argument, causes the taylor series to rotate its domain of convergence under the mellin transform. I hope to explain this better.
Write:
\[
\vartheta_\theta(x,z) = \sum_{n=0}^\infty f_{\theta}^{\circ n}(z)\frac{(-x)^n}{n!}\\
\]
This function has decay for \(\arg(x) = \theta\). And in neighborhoods of this. Whereby, results in mellin, describe results in Abel.
I apologize, I hope to make all of this work better. I just do not have the time as of now.
A helpful way to visualize this is with \(f = e^z-1\). To begin:
\[
f^{\circ n}(z) = f^n\\
\]
Are purely positive, and monotone. If I write:
\[
f_\pi^{\circ n} = f^n_\pi\\
\]
Then this monotone nature no longer appears.
When I write \(e^{-1} -1 < 0\), this appears throughout. When I write:
\[
e^1 -1 > 1
\]
This appears throughout. When I adapt this to taylor series we can write:
\[
\sum_{n=0}^\infty f^{\circ n} \frac{(-x)^n}{n!}\\
\]
Where the coefficients are monotone, and therefore the oscillation is only formed by \((-x)^n\). This allows us to take a mellin transform, because everything is nice. When I do this with \(f_\pi\), things are different:
\[
\vartheta_\pi(x) = \sum_{n=0}^\infty f_\pi^{\circ n} \frac{(-x)^n}{n!}\\
\]
Using this function, this does not have a nice decay as we increase \(x \to \infty\). Which is because \(f_\pi^{\circ n}\) looks like \((-1)^n\) itself. It oscilates in a similar manner. Where by, we must take \(x \to - \infty\), and then the integral converges.
FFS I'm tired boys...
To be straight with you guys, we have to think like Ramanujan:
\[
f^{\circ n}(z) = 1^n
\]
whereby:
\[
\vartheta(x) = e^{-x}\\
\]
From which:
\[
f_\pi^{\circ n}(z) = (-1)^n\\
\]
and:
\[
\vartheta_\pi(x) = e^x\\
\]
Where, at this point, all we have to do is change the direction of our integral... And this follows for fucking every direction... We can perform the iteration no matter what, but we have to write the integral correctly. First of all:
\[
\int_0^\infty \vartheta_0(x)\,dx\\
\]
converges, but then to do this when we change the multiplier:
\[
\int_{0}^{-\infty} \vartheta_\pi(x)\,dx\\
\]
We have to move the domain of integration, as we move our petals/initial points, ALL THIS FUCKING
BULLSHIT!!!!!! If I were to take \(\vartheta_{\pi/2}\) then I'd write something like:
\[
\int_0^{-i\infty}\vartheta_{\pi/2}(x)\,dx\\\
\]
And write various transforms attached to this (AGAIN, THIS MEANS WE BELONG TO A HILBERT SPACE). This is again no different than the identity:
\[
\int_0^{-i\infty} e^{-ix}\,dx\\
\]
And the fact this object converges. We typically attach this with a mellin transform, whereby:
\[
e^{-\pi i s /2} \Gamma(s) = \int_0^{-i\infty} e^{-ix}x^{s-1}\,dx\\
\]
We are doing something deeply similar; just without the obvious beauty of the Gamma function.
I really want to hear you guys out, but I'm just lost at how much you guys are forgetting. And ignoring the "Schrodinger" interpretation has left me blind sided. I plan to write a broken, but fairly well interpreted program. And all it uses is basic code.
EDIT:
A good way to conceptualize this, is if \(a_n \in \mathbb{R}^+\), and it has good enough decay, then:
\[
\vartheta(x) = \sum_{n=0}^\infty a_n \frac{(-x)^n}{n!}\\
\]
Can look a lot like \(e^{-x}\). When \(b_n = (-1)^n a_n\), then:
\[
\vartheta_\pi(x) = \sum_{n=0}^\infty b_n \frac{(-x)^n}{n!} \approx e^x\\
\]
BOTH OF THESE FUNCTIONS ARE MELLIN TRANSFORMABLE!!!!!
We just need to take the mellin transform in different manners.........