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I don't know if this is exactly research level, as I am only starting college. But I feel like this is the best place to ask the question. We all know of 1st, 2nd, 3rd, nth derivatives. Is there a way of extending it to the transfinite ordinals? More specifically, can anyone show me a link to a webpage or paper that discuss this concept? Even if this question gets closed, I would still like a link to one or a few papers about this concept. I have searched google, but I can't find anything.

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migrated from mathoverflow.net Dec 6 '13 at 8:45

This question came from our site for professional mathematicians.

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    $\begingroup$ I can't see that you'll get anything reasonable by trying to do this. Consider the example of the function $f(x) = e^{ax}$. Its $n^{th}$ derivative is $a^n e^{ax}$. At least when $a$ is positive we might allow $n$ to be an arbitrary complex number here (en.wikipedia.org/wiki/Fractional_calculus). But the limit as $n \to \infty$ isn't defined for $a \neq 1$. $\endgroup$ – Qiaochu Yuan Dec 5 '13 at 23:05
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    $\begingroup$ One way to make sense of this would be to consider functions given by 'taylor series' indexed by ordinals greater than the first countable one. But what, for example, does the function $f(x) = x^{\omega+2}$ mean in the context of actual functions $\mathbb{R} \to \mathbb{R}$? If one is considering formal sums, perhaps it's ok; but as Qiaochu points out, the usual stuff of calculus falls over. $\endgroup$ – David Roberts Dec 5 '13 at 23:06
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    $\begingroup$ It's possible that this makes sense when doing calculus over the surreal numbers. $\endgroup$ – Sam Hopkins Dec 5 '13 at 23:27
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    $\begingroup$ Well, I believe transfinite ordinals were originally introduced for the purpose of indexing transfinite derivatives. Cantor-Bendixson derivatives, that is. $\endgroup$ – bof Dec 6 '13 at 1:51
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    $\begingroup$ Cantor-Bendixon derivatives have nothing to do with derivatives... $\endgroup$ – Mariano Suárez-Álvarez Dec 6 '13 at 2:27
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In Ralph P. Boas's A primer of real functions, page 118, this is discussed in the following way: The derivative of infinite order of $f$ is defined on an interval $I$ iff the sequence $(f^{(n)})$ converges uniformly on $I$ (it is enough to require uniform convergence on compact subsets of $I$). Call $L$ the limit of this sequence, so $L$ is continuous and therefore Riemann integrable over bounded intervals. Fix $a\in I$, and for $x\in I$ consider $\lim_n\int_a^x f^{(n)}$. Note that, on the one hand, this is $$\lim_n (f^{(n-1)}(x)-f^{(n-1)}(a))= L(x)-L(a),$$ and on the other it is $\int_a^x L$, that is $L(x)=ce^x$ for some constant $c$, which ends up trivializing the notion. The reference given there is Lee Lorch, Derivatives of infinite order, Pacific Journal of Mathematics, 3, (1953), 773-778, MR0060553 (15,689c).

We used uniform convergence on compact subsets twice. First, to ensure that $L$ is integrable, though we in fact get more, as $L$ ends up being $C^\infty$. And second, to ensure that the limit of the integrals is the integral of the limit. We can avoid referring to integration by instead using the following well-known fact:

If $(f_n)_n$ is a sequence of differentiable functions on an interval $I$, $a\in I$, $(f_n(a))_n$ converges, and $(f_n')_n$ converges uniformly, then $(f_n)$ converges uniformly to a differentiable function $f$ such that $f_n'\to f'$.

From this it follows that if $(f^{(n)})_n$ converges uniformly, then the limit $L$ is $C^\infty$ and satisfies $L=L'$.

Lorch's paper provides additional references discussing various notions of derivatives of infinite order (for $C^\infty$ and specially for analytic functions), and discusses a few variants as well. The paper and some of the references it lists (particularly the Boas-Chandrasekharan papers), discuss some of the issues with the different suggestions, and the kind of results one can expect in each case.

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  • $\begingroup$ And this result holds true even if the interval I is the whole real line? Because, that is the case I am most interested in. $\endgroup$ – user107952 Dec 6 '13 at 0:22
  • $\begingroup$ And also, does this result hold if the convergence is pointwise? That is the case I am really interested in. $\endgroup$ – user107952 Dec 6 '13 at 0:26
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    $\begingroup$ The interval $I$ can be the whole line, in fact we only need uniform convergence on compact sets, which is less restrictive that uniform convergence on $\mathbb R$. But we do need some assumption (such as uniform convergence) beyond pointwise convergence to ensure the limit of the integrals is the integral of the limit. $\endgroup$ – Andrés E. Caicedo Dec 6 '13 at 0:34
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    $\begingroup$ So you are saying if we just have pointwise convergence, there could still be a non-trivial theory of transfinite derivatives? Meaning, there could be a function f such that the w_0 derivative of f is different from the w_0 + 1 derivative of f? $\endgroup$ – user107952 Dec 6 '13 at 0:44
  • $\begingroup$ I think you may want to read both Lorch's paper and some of the listed references there (particularly the Boas-Chandrasekharan papers), as they discuss some of the issues with the different suggestions, and the kind of results one can expect in each case. That said, I do not know how malleable one can truly make the notion if only pointwise convergence is assumed. $\endgroup$ – Andrés E. Caicedo Dec 6 '13 at 0:49
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As Qiaochu and David pointed out, the limit $d^n f(x)/dx^n$ doesn't make much sense. However, you may want to consider formal power series $a(f):=\Sigma_{n=0}^\infty a_nt^n$ over indeterminate variable $t$, with coefficients $a_n=\frac{1}{n!}\frac{d^n f(x)}{dx^n}$.

Such formal series, with the formal product rule $(ab)_n=\Sigma_{k=0}^n a_kb_{n-k}$ would encapsulate all the derivatives. You can view it this infinite formal power series as a "transfinite limit" for the finite derivatives.

No doubt you recognized Taylor series, but the point here is not to produce series that converge to the function given a specific value of $t$, the point is to produce an object that would encapsulate, in some way, the limit of $n^{th}$ derivative for $n\to\infty$. And the appropriate object for that is not a function that is the "infinite derivative", the appropriate object is an expression that encapsulates infinitely many derivatives.

In fact, something like that appears in often enough in Algebraic Geometry: if the object itself (such as $n^{th}$ derivative in this case) does not generalize directly into the desired domain (in this case $n=\infty$) then replace the object with a functor from a more general category (in this case the category is the formal power series, the functor is the $n^{th}$ term of the series times $n!$, and the intuition that guides this categorification is Taylor series), and then proceed working within the more general category.

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