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Is there any meaning behind infinitely differentiating or integrating a function? Also, sure, this would cause most functions to either collapse to zero, blow up to infinity, oscillate infinitely (e.g., $sin(x)$ and $cos(x)$), or never change, but are there any functions that would converge to some other function that is continuous at least somewhere upon doing this? Moreover, if there is such a function, how would one go about finding a closed-form expression for its infinite derivative/integral?

senormittens7
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    I think if you restrict your class of functions to most reasonable choices you will always end up at an exponential due to the uniqueness of the limit in many fixed-points theorems. – whpowell96 Sep 06 '23 at 01:41
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    Maybe this is helpful: https://math.stackexchange.com/q/4250457/2838 – J126 Sep 06 '23 at 04:00
  • @J126 I don't fully understand the jargon in that post, could you give me a brief summary of what's going on there? Thanks – senormittens7 Sep 06 '23 at 13:56
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    The main point is the two answers given. Let the $n$th derivative be denoted $f_n$. Suppose that ${f_n}$ converges uniformly to some function $f$. Since $f_n'(x) = f_{n + 1}(x)$, you would also have that ${f_n'}$ will also converge uniformly to $f$. The first answer tells you that if you assume $f$ is continuous, then you'll get $f' = f$. The second answer tells you that you need some assumptions like uniform convergence or continuity of $f$, or you get counterexamples to $f' = f$. – J126 Sep 06 '23 at 16:16
  • https://math.stackexchange.com/questions/13815/infinitely-differentiable/3145390#3145390 – IV_ Sep 06 '23 at 17:39
  • Look for the terms "n-th derivative" and "n-th antiderivative". – IV_ Sep 06 '23 at 17:39

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the differentiating part:

Let $f(x)$ be such a function that for some function, differentiating it infinitely many times yields it. The immediate consequence is that differentiating $f(x)$ one more time gives precisely $f(x)$, that is, $\frac{df}{dx}(x)=f(x)$. An elementary differential equation argument shows that $f(x)$ must be of the form $Ce^x$, where C any constant.

the integrating part:

Since doing indefinite integrals leave an annoying constant, it's not well-defined that you integrate a function 'infinitely'. Please provide further information :)

SalutaFungo
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  • We have to be careful to define what differentiating infinitely many times means. If it just means we are taking a limit of infinitely many derivatives in some topology, then the derivative might not be continuous with respect to that topology. For example, differentiating is not continuous with respect to topology of uniform convergence. So it is not immediate that "differentiating one more time gives precisely $f(x)$". – M W Sep 06 '23 at 02:15