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As proposed in this answer, I wonder if the answer to following question is known.

Let $E = E_0$ be the set of elementary functions. For each $i > 0$, inductively define $E_i$ to be the closure of the set of functions whose derivative lies in $E_{i-1}$ with respect to multiplication, inversion, and composition. Does there exist an integer $n$ such that $E_n = E_{n+1}$?

This seems like such a natural generalization of Liouville's theorem, it has to have been asked before. After a couple of quick internet searches, I can't seem to find anything.

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I feel like you need to throw some kind of closure operation in there -- something like "Define $E_i$ to be the set of all functions that can be formed by a composition of functions whose derivatives lie in $E_{i-1}$." But maybe that is unnecessary. – mweiss Jun 17 '14 at 0:01
Indeed, it is necessary. Thanks. – RghtHndSd Jun 17 '14 at 0:07
Similar question has been asked before… but alas no answers. – Conifold Jun 17 '14 at 1:03
And how would this relate to notions of computability (e.g Turing Machine), since an elementary functin can be a symbol along with the other (finite) symbols of "addition", "nth power", "division" etc.. This may indeed be related.. And provide sth analogous to an "algebraic closure", but if taken in a computable sense, this would be non-decidable.. – Nikos M. Jun 17 '14 at 20:36
@NikosM.: I don't understand how "notions of computability" can be related to my question. What does this mean? Perhaps you are asking "do the E_i's consist of computable functions"? Also regarding the last sentence, precise what is it that "would be non-decidable"? – RghtHndSd Jun 17 '14 at 21:04

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