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:



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.

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    $\begingroup$ See here. $\endgroup$ – Andrés E. Caicedo Jun 26 '18 at 11:52
  • $\begingroup$ @AndrésE.Caicedo (+1) Oh interesting! Let me check your answer over there! Thank you, Andrés! :-) $\endgroup$ – Morteza Azad Jun 26 '18 at 11:56
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    $\begingroup$ Your question may make more sense in a field of Conway's surreal numbers. Have a look at the preprint Surreal numbers, derivations and transseries by Alessandro Berarducci, Vincenzo Mantova, arxiv.org/abs/1503.00315. $\endgroup$ – Dávid Natingga Jun 26 '18 at 13:05
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    $\begingroup$ I don't see how the paper of Berarducci and Mantova answers this question. $\endgroup$ – nombre Jun 26 '18 at 13:32
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    $\begingroup$ @MortezaAzad Sorry, I think this question is quite important and deserves a better answer than I could give at this time. $\endgroup$ – Dávid Natingga Jun 27 '18 at 19:11

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)

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  • $\begingroup$ This argument fails if we use natural addition in the definition of $D^\beta\circ D^\alpha=D^{\beta+\alpha}$ since $1+\omega=\omega+1\neq\omega$, which seems the natural choice to me. I think the main problem you are illuminating here is that noncommutative binary operations don't generalize well to operators indexed over that set if we want the induced operator addition to be commutative, which is fixed by choosing a commutative binary operation to begin with. $\endgroup$ – Alec Rhea Jun 26 '18 at 17:03
  • $\begingroup$ Speaking as a physicist, you can "regulate" $D^\omega \sin t$ by instead considering $D^\omega \sin \lambda t$. While not defined for $\lambda = 1$, for all $|\lambda| <1$ there is no difficulty in defining $D^\omega \sin \lambda t = \lim_{n\rightarrow \infty} D^n \sin\lambda t = 0$. For $|\lambda| > 1$ it is less well-behaved but can be considered as (complex) $\infty$ (there is some difficulty at integral multiples of $\pi/2$ but if more general perturbations are allowed this can be dealt with). I would not call this "robust" but it shows that your example is perhaps a degenerate case. $\endgroup$ – Logan M Jun 26 '18 at 17:08
  • $\begingroup$ @AlecRhea: If you have $D^{\beta} \circ D^{\alpha} = D^{\beta+\alpha}$, then $D^{\omega} f = D^{1+\omega}f=D(D^{\omega}f)$ and so the $\omega$-th derivatives have to be of the form $Ce^x$. I think James' choice is the correct one since $\alpha+\beta$ in ordinal arithmetic means "apply $\beta$-many successors after $\alpha$-many times", so $D^{\alpha+\beta}$ should mean "differentiate $\beta$-many times after $\alpha$-many times". $\endgroup$ – Burak Jun 26 '18 at 18:01
  • $\begingroup$ "(...) since the ordinal $\omega$ is $0(\mathrm{mod}4)$, if anything" - Really? Why? $\endgroup$ – Qfwfq Jun 26 '18 at 18:44
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    $\begingroup$ @Burak Just to be clear, the relation $\omega=1+\omega$ does not hold if we interpret $+$ as the natural sum of ordinals instead of the recursive sum, which is what I am suggesting we should do. I do think that derivative operators indexed in the Grothendieck ring of the ordinals are more likely to satisfy nice duality properties with integration, since we always have successors in both directions. $\endgroup$ – Alec Rhea Jun 26 '18 at 19:54

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