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I've been thinking about this: say you have an infinitely differentiable function. Then you can form a sequence $f(x), f'(x), f''(x), \cdots, f^{(n)}(x), \cdots$ and attempt to take its limit. For some functions this is easy: for $e^x$ the limit will be itself, $\sin x$ and $\cos x$ won't have a limit, every polynomial will eventually be $0$, etc.

I was wondering, does this have any use? Is there any interesting interpretation of the infinite derivative, and does it existing or not tell us anything about the function?

Javier
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    I find it curious how many answers this question is drawing that talk about "infinitely differentiable" and have nothing to do with the question you ask. –  Apr 15 '13 at 23:28
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    @Hurkyl There are two questions he asks: one, "does this [infinite differentiability] have any use;" two, "is there any interesting interpretation of the infinite derivative." Most answers focused on the first question, which is related to properties of $C^\infty$. The second question has also been answered, in that it is a meaningless notion. – Emily Apr 15 '13 at 23:32
  • My question says "Is there any interesting interpretation of the infinite derivative", not "infinite differentiability". I know what the latter is and what it means, I was asking about the first. – Javier Apr 15 '13 at 23:35
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    Any "infinite derivative" of a function as you've defined it will almost certainly be a fixed point of the map $f \mapsto f'$. There aren't very many of those... – Micah Apr 15 '13 at 23:38
  • @JavierBadia You should reword your question. Saying "does this have any use" is ambiguous: do you refer to the notion of taking the limit of a sequence, or classifying a function as being infinitely differentiable? I see now that you mean the former. – Emily Apr 15 '13 at 23:45
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    @Arkamis: I think it's pretty clear; I spend half a paragraph talking about the limit of the sequence of functions before asking "does this have any use". I don't see a better way to put it. I mention the concept of infinite differentiability exactly once, because it's necessary for the question to make any sense. – Javier Apr 15 '13 at 23:51
  • While I agree that the OP seems clear, it's hard to argue with the cold empirical evidence that people have gotten confused. While my comment was meant to ward off further confusion, it might help further to replace "does this" with "does this limit" or something similar. OTOH we haven't seen any more answers -- the existing discussion may be sufficient, or maybe it could have just been an artifact of people responding quickly to a new question. –  Apr 16 '13 at 00:01
  • See here for a more general argument that does not require analyticity of $f$. – Andrés E. Caicedo Apr 26 '14 at 22:12

1 Answers1

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If we consider just analytic functions....

Suppose $f(x)$ is analytic at $x=0$. Then we can write it as a Taylor series

$$ f(x) = \sum_{n=0}^{\infty} f^{(n)}(0) \frac{x^n}{n!} $$

Furthermore, we can express its derivatives

$$ f^{(k)}(x) = \sum_{n=0}^{\infty} f^{(n+k)}(0) \frac{x^n}{n!} $$

If we insist that the convergence of the sequence $f^{(k)}$ be sufficiently nice (I'm quite rusty, but I think we want the limit to converge uniformly), then the limit of the sequence of Taylor series (if it exists) can be computed by taking the limit of the coefficients, when :

$$ f^{(\infty)}(x) = \sum_{n=0}^{\infty} \left( \lim_{k\to \infty} f^{(n+k)}(0) \right) \frac{x^n}{n!} $$

Of particular note is that the limit is completely independent of $n$; if we define

$$a = \lim_{k \to \infty} f^{(k)}(0) $$

then, when it exists, we have

$$ f^{(\infty)}(x) = a e^x$$

Hopefully you'll find this partial analysis useful.

This answer has a reference that asserts the same conclusion with just the hypothesis that $f$ is infinitely differentiable and the sequence of derivatives converges uniformly.

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