I actually don't disagree with you, Tommy!
We have to apply the fractional derivative/integral to an AUXILIARY function; not the function itself. So if I write:
\[
\vartheta(w,\xi) = \sum_{n=0}^\infty f^{\circ n}(\xi) \frac{w^n}{n!}\\
\]
Then the fractional derivative:
\[
\frac{d^{s}}{dw^s} \Big{|}_{w=0} \vartheta(w,\xi) = f^{\circ s}(\xi)\\
\]
This converges to the standard regular iteration (schroder iteration for geometric fixed points; and abel iteration for neutral fixed points). The trouble with this method, is that it's very non-trivial to show that this integral transform converges.
ALSO! It's very important to note; that we must use the exponential differintegral (Or the Riemann-Liouville differintegral); because this is the sole differintegral that satisfies:
\[
\frac{d^s}{dw^s} e^{\lambda w} = \lambda^s e^{\lambda w}\\
\]
Which is the differintegral:
\[
\frac{d^s}{dw^s} f(w) = \frac{1}{\Gamma(-s)} \int_0^\infty f(w-y)y^{-s-1}\,dy\\
\]
Here the integral is taken along SOME PATH \(\gamma(0) = 0\) and \(\gamma(\infty) = \infty\). This definition is not path independent.
Which in our language is: \(f(\xi) = \lambda \xi\) then:
\[
\vartheta(w,\xi) = e^{\lambda w} \xi = \sum_{n=0}^\infty f^{\circ n}(\xi) \frac{w^n}{n!}\\
\]
And:
\[
\frac{d^{s}}{dw^s} \Big{|}_{w=0} \vartheta(w,\xi) = \lambda^s \xi = f^{\circ s}(\xi)\\
\]
This is pretty fucking trivial here; but if we allow for more advanced \(f\) (and not dilations); the same result happens. I've proven it in as general a manner as I could. There are still edge cases I'm not certain on; but I am certain they "converge in some manner". For example take:
\[
\vartheta(w,\xi) = \sum_{n=0}^\infty \sin^{\circ n +1}(\xi) \frac{w^n}{n!}\\
\]
Then:
\[
\frac{d^s}{dw^s}\Big{|}_{w=0} \vartheta(w,\xi) = \sin^{\circ 1+s}(\xi)\\
\]
But this is only true for \(\xi \in \mathbb{R}\). It's nowhere holomorphic at \(\xi = 0\)--where we get an asymptotic series. And outside of this area; I'm still not sure what happens--but it looks like it does converge.
If \(0<|f'(\xi_0)|<1\) then this result is "generally true"--it gets really tricky. For example; let:
\[
f(\xi) = -\frac{\xi}{2} + \xi^2\\
\]
Then we have to take:
\[
\vartheta(w,\xi) = \sum_{n=0}^\infty f^{\circ 2n}(\xi) \frac{w^n}{n!}\\
\]
Then:
\[
\frac{d^s}{dw^s}\Big{|}_{w=0} \vartheta(w,\xi) = \left(f^{\circ 2}\right)^{\circ s}(\xi)\\
\]
This is not the iteration, but it's close enough you can recover the iteration. We have to take into account the period of \(-1\). The function \(e^{\pi i s}\) has period \(2\).
\[
\frac{d^{s/2+ \pi i s}}{dw^{s/2 + \pi i s}}\Big{|}_{w=0} \vartheta(w,\xi) = f^{\circ s}(\xi)\\
\]
I may have made a typo here( hope to god not); but this creates a fractional iteration that is exactly the Schroder iteration. It's actually pretty easy to identify too. A lot of my notes are scattered on the subject, but I do have a few papers on my arxiv dealing with specific instances. I never found a "global theory" that worked; so I abandoned a lot of the work. It ended up being a case by case kind of theory; and that's ugly as fuck; so I moved away from it.
There's a lot, and I fucking mean a lot, of similarities between the exponential differintegral and local iteration (or regular iteration).
Regards, James
We have to apply the fractional derivative/integral to an AUXILIARY function; not the function itself. So if I write:
\[
\vartheta(w,\xi) = \sum_{n=0}^\infty f^{\circ n}(\xi) \frac{w^n}{n!}\\
\]
Then the fractional derivative:
\[
\frac{d^{s}}{dw^s} \Big{|}_{w=0} \vartheta(w,\xi) = f^{\circ s}(\xi)\\
\]
This converges to the standard regular iteration (schroder iteration for geometric fixed points; and abel iteration for neutral fixed points). The trouble with this method, is that it's very non-trivial to show that this integral transform converges.
ALSO! It's very important to note; that we must use the exponential differintegral (Or the Riemann-Liouville differintegral); because this is the sole differintegral that satisfies:
\[
\frac{d^s}{dw^s} e^{\lambda w} = \lambda^s e^{\lambda w}\\
\]
Which is the differintegral:
\[
\frac{d^s}{dw^s} f(w) = \frac{1}{\Gamma(-s)} \int_0^\infty f(w-y)y^{-s-1}\,dy\\
\]
Here the integral is taken along SOME PATH \(\gamma(0) = 0\) and \(\gamma(\infty) = \infty\). This definition is not path independent.
Which in our language is: \(f(\xi) = \lambda \xi\) then:
\[
\vartheta(w,\xi) = e^{\lambda w} \xi = \sum_{n=0}^\infty f^{\circ n}(\xi) \frac{w^n}{n!}\\
\]
And:
\[
\frac{d^{s}}{dw^s} \Big{|}_{w=0} \vartheta(w,\xi) = \lambda^s \xi = f^{\circ s}(\xi)\\
\]
This is pretty fucking trivial here; but if we allow for more advanced \(f\) (and not dilations); the same result happens. I've proven it in as general a manner as I could. There are still edge cases I'm not certain on; but I am certain they "converge in some manner". For example take:
\[
\vartheta(w,\xi) = \sum_{n=0}^\infty \sin^{\circ n +1}(\xi) \frac{w^n}{n!}\\
\]
Then:
\[
\frac{d^s}{dw^s}\Big{|}_{w=0} \vartheta(w,\xi) = \sin^{\circ 1+s}(\xi)\\
\]
But this is only true for \(\xi \in \mathbb{R}\). It's nowhere holomorphic at \(\xi = 0\)--where we get an asymptotic series. And outside of this area; I'm still not sure what happens--but it looks like it does converge.
If \(0<|f'(\xi_0)|<1\) then this result is "generally true"--it gets really tricky. For example; let:
\[
f(\xi) = -\frac{\xi}{2} + \xi^2\\
\]
Then we have to take:
\[
\vartheta(w,\xi) = \sum_{n=0}^\infty f^{\circ 2n}(\xi) \frac{w^n}{n!}\\
\]
Then:
\[
\frac{d^s}{dw^s}\Big{|}_{w=0} \vartheta(w,\xi) = \left(f^{\circ 2}\right)^{\circ s}(\xi)\\
\]
This is not the iteration, but it's close enough you can recover the iteration. We have to take into account the period of \(-1\). The function \(e^{\pi i s}\) has period \(2\).
\[
\frac{d^{s/2+ \pi i s}}{dw^{s/2 + \pi i s}}\Big{|}_{w=0} \vartheta(w,\xi) = f^{\circ s}(\xi)\\
\]
I may have made a typo here( hope to god not); but this creates a fractional iteration that is exactly the Schroder iteration. It's actually pretty easy to identify too. A lot of my notes are scattered on the subject, but I do have a few papers on my arxiv dealing with specific instances. I never found a "global theory" that worked; so I abandoned a lot of the work. It ended up being a case by case kind of theory; and that's ugly as fuck; so I moved away from it.
There's a lot, and I fucking mean a lot, of similarities between the exponential differintegral and local iteration (or regular iteration).
Regards, James