Theorem in fractional calculus needed for hyperoperators

[MphLee]

For the first time I think I reached a general understanding of your method. Ignoring the fine details, and by focusing on the skeleton of your method I think I have a schematic image of how your method works.
Sure... to know WHY it works I should know all those fine details (and understand the proof you gave in your paper) but the fact that I can follow your idea is enough exciting at the moment, even with a tiny knowledge of the analysis.


Let me try to use a high-order funtion notation (it's easier to me) and tell me if my understanding is wrong.

Given a function \( f \), lets call \( F_{\beta} \) its \( \beta \)-based superfunction over the naturals

\( F_{\beta}(n):=f^{\circ n}(\beta) \)

your method gives an extension \( F'_{\beta} \) of \( F_{\beta} \) to the complex numbers.

\( F_{\beta}\subset F'_{\beta} \)
---------------------------------
Let me define

1-\( \mathcal{S}_{\beta}\{f\}(n) := f^{\circ n}(\beta) \)

2-\( \mathcal{T}\{f\}(x) := \sum_{n=0}^\infty f(n) x^n/n! \)

3-\( \mathcal{F}\{f\}(z) :=\frac{d^{z}}{dx^z}|_{x=0}f(x) \)

4- \( f \cdot g(x):=\cdot \{f,g\}(x):=f(x)g(x) \); \( \frac{f} { g}(x):=/ \{f,g\}(x):=\frac{f(x)}{g(x)} \)
------------------------------------

You say that if \( \mathcal{S}_{\beta}\{f\} \) has some nice properties then your method give an extension of it to the complex:

if \( \mathcal{S}_{\beta}\{f\} \) has some properties then

\( \mathcal{S}_{\beta}\{f\}
\subset
\mathcal{F}\{
\mathcal{T}\{
\mathcal{S}_{\beta}\{f\}
\}
\}
\) and \(
f\circ
\mathcal{F}\{
\mathcal{T}\{
\mathcal{S}_{\beta}\{f\}
\}
\}
=
\mathcal{F}\{
\mathcal{T}\{
\mathcal{S}_{\beta}\{f\}
\}
\}
\circ S \)


to be more precise we have that
\(
\mathcal{F}\{
\mathcal{T}\{
\mathcal{S}_{\beta}\{f\}
\}
\}(n)=\mathcal{S}_{\beta}\{f\}(n)
\)

Now of course, the problem is that when \( a_n \) is something like tetration, or pentation, or whatever, this doesn't converge and we are stuck in the mud.

[...]
So without further ado, here is the theorem we need.

Assume \( a_n \) is a sequence of complex numbers such that \( f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!} \) is entire. Then, there always exists \( b_n \) such that, \( g(x) = \sum_{n=0}^\infty b_n \frac{x^n}{n!} \) is entire and Weyl differintegrable on all of \( \mathbb{C} \) and

\( h(x) = \sum_{n=0}^\infty a_n b_n \frac{x^n}{n!} \)

is such that \( \frac{d^z}{dx^z}{|}_{x=0} h(x) \) exists for all z.


If this theorem is shown, then... define \( G(z) = \frac{\frac{d^z}{dx^z}|_{x=0} h(x)}{\frac{d^z}{dx^z}|_{x=0} g(x)} \)

and

\( F(G(z)) = G(z+1) \) and we are done.



Any one have any advice on how I can show this theorem? this is quite a stump.

So you say that from every \( f \) we can find an unique function that we call \( f^{\psi} \) such that

i) - \( \mathcal{T}\{f^{\psi}\} \) " is entire and Weyl differintegrable on all of \( \mathbb{C} \)"

ii) - \( \mathcal{F}\{ \mathcal{T} \{f \cdot f^{\psi}\} \} \) is defined for all the complex numbers


.................................
from the existence and the uniqueness (it has to be bijective?) of such map \( {-}^{\psi}:f\mapsto f^{\psi} \) and if it satiefies i) and ii) then

\( \frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} }
{\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}}

(n)=\mathcal{S}_{\beta}\{f\}(n)
\)

and

\(
f(\frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} }
{\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}}(z)
)=
\frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} }
{\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}}(z+1)
\)


-------------------------------
I hope that this is fine...then you have to work on the map \( {-}^{\psi}:f\mapsto f^{\psi} \)

Questions

A- I'm sure that I can't understand how you got there (with my knowledge) but where is the link betwen the fractional iteration and the fractional calculus and how you discovered this link?

B- The way you use \( \mathcal{T}\{f\}(x) := \sum_{n=0}^\infty f(n) x^n/n! \) ... it reminds me the taylor series (when is centred at a=0 but with the derivative index replaced by the iteration...) I'm sure it is not a case... where is the link?

C- the results depends alot on the choice of \( \beta \) ?

D-What is the relation betwen \( \frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} }
{\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}} \) and \(

\mathcal{F}\{
\mathcal{T}\{
\mathcal{S}_{\beta}\{f\}
\}


\}
\)?
When the two expression coincides (if they do)?

E -About the hyperoperations...how you can get them using this method?
If I have to think about it..the only way to reach the fractional hyperoperations i see is the fractional iteration of the operator \( \mathcal{J}_{\beta}^{\circ z} \)

where

\( \mathcal{J}_{\beta}\{f\}:=
\frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} }
{\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}} \)


...where is the trick to avoid that?


ps: i did a mistake in the tex code... ill fix it soon (is important)
pps: fixed