TITLE: How to define transfinite derivatives of a function?
QUESTION [5 upvotes]: There are all manners of theories generalizing the notion of derivative. Amongst them is the fractional calculus, a rich theory which gives a sense to the derivation and integration of non-integer (i.e. rational, real, complex) order, that are the real/complex number powers of the differentiation ($D$) and integration ($J$) operators: 
$Df(x)=\frac{d}{dx}f(x)$
$Jf(x)=\int_{0}^{x}f(x)dx$
Along these lines, one may think about transfinite iteration of $D$ and $J$ as well. While $D^{(n)}$ and $J^{(n)}$ are quite well-defined and well-behaved operators for the natural number $n$, I haven't found any convenient definition of $D^{(\alpha)}$ or $J^{(\alpha)}$ for the ordinal $\alpha\geq \omega$ in the literature if there is any. 
Due to the similarity between natural and ordinal numbers, it seems the only difficulty is to formulate a definition of the differentiation operator in the limit steps like $D^{\omega}$ or $D^{\omega+\omega}$. One straightforward (but not necessarily well-defined, natural or fruitful) way to do so is to think about $D^{\omega}$ as a functional limit of $D^{n}$s in a certain function space. Though, I am not sure if it is the most clever approach. Anyway, it is somehow "natural" to expect that any $D^{\omega}$ operator demonstrates certain properties like: $D^{\omega}x^{n}=0$ for every $n\in \omega$.  
Also, the spaces of smooth and analytic functions, $C^{\infty}$ and $C^{\omega}$, don't seem to capture the essence of the very notion of the $\omega$-the derivative of a function, particularly because they don't suggest a clear way of calculating transfinite successor differentiation operators, $D^{\omega+1}$, $D^{\omega+2}$, $D^{\omega+3}$, etc.

Question. Is there any paper in which transfinite derivatives of (real/complex) functions are defined/used? If so, what sort of applications do they have? 


Update. Due to the answer that Andrés mentioned in his comment, it turned out that defining $D^{\omega}$ operator as the limit of $D^{n}$s gives rise to a trivial notion. So maybe a more direct approach is needed here.

REPLY [2 votes]: If there is a such a notion of transfinite derivative, I don't think it will be very robust. Here's the problem I see with indexing repeated differentiation with ordinals. We would almost certainly want the following natural property: $$D^{\beta} \circ D^{\alpha} = D^{\alpha+\beta}$$.
This would give us, for example, that $$D^{\omega} \circ D = D^{1+\omega} = D^{\omega}$$ and $$D \circ D^{\omega} = D^{\omega+1}$$ But now let's try some of the nicest analytic functions we have: $sin(t)$ and $cos(t)$. We should expect $D^{\omega}sin(t) = sin(t)$ and $D^{\omega}cos(t) = cos(t)$, since the ordinal $\omega$ is "0 (mod 4)", if anything. But then $$sin(t) = D^{\omega}sin(t) = (D^{\omega} \circ D)sin(t) = D^{\omega}cos(t) = cos(t)$$ which is a problem.
To make this objection more precise: what is the largest "natural class" of functions $\mathbb{C} \rightarrow \mathbb{C}$ for which we can define transfinite derivatives? Let's assume that by "natural class", we mean it is a collection of functions closed under scaling, addition of functions, and composition of functions. Then there is no natural class $\mathcal{A}$ containing both $e^t$ and all polynomials for which we can make a reasonable definition of transfinite derivative. Why? Because if $e^t, it \in \mathcal{A}$, then $cos(t) = \frac{e^{it}+e^{-it}}{2} \in \mathcal{A}$ and $sin(t) = \frac{e^{it}-e^{-it}}{2i} \in \mathcal{A}$, and we have already argued there is no reasonable way to define the $\omega$-th derivative of $sin(t)$ and $cos(t)$.

Here's a rough sketch of an idea using non-standard analysis (as motivated by Dávid Natingga's comment):
There probably is a reasonable notion of taking "non-standardly" many derivatives, however. One way of thinking about this is to start with a (standard) analytic function $f$ whose Taylor coefficients are form a nice definable sequence (like $a_n = \frac{1}{n!}$) which decays sufficiently quickly. Then for a non-standard natural number N, we can look at the non-stanard polynomial $p_N(x) = \sum_{n=0}^{N} a_n x^n$. If we plug in a standard real number $r$ into $p_N(x)$ and take the standard part, we will just get $\sum_{n=0}^{\infty} a_n r^n = f(r)$, roughly because the terms $a_M r^M$ for M non-standard are infinitessimal.
Now let $M$ be some non-standard natural number (here we imagine N is much bigger than M), we can certainly define $p_{N}^{(M)}(x) = D^{M}\sum_{n=0}^{N} a_n x^n$. Keeping $M$ fixed, as long as $N$ is big enough, the standard part of $p^{(M)}_N(r)$ for $r$ a standard real number does not depend on $N$. So we can define $f{(M)}(r) = p_N^{(M)}(r)$ for any $N$ sufficiently big. This gives us a well-defined notion of of the $M$-th derivative $f^{(M)}$ of a function $f$ with a "nice" Taylor series.
(Caveat: I don't really do non-standard analysis, so there may be some conceptual errors here)