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 00:01
  • Indeed, it is necessary. Thanks. – RghtHndSd Jun 17 '14 at 00:07
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    Similar question has been asked before http://math.stackexchange.com/questions/686445/is-each-elementary-finite-functions-function-elementary-finite-functions but alas no answers. – Conifold Jun 17 '14 at 01:03
  • 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
  • Possibly means that your extensions are 'transcendental' at each stage, so even if $E_n=E_{n+1}$ we won't have a way to check if a given function is not in $E_n$. Construction of elementary functions can be presented so that functions added at each stage starting from rational ones are of special form, exponents and logs of those already in (in complex notation). That's why the Liouville theorem can provide explicit test for non-integrability. – Conifold Jun 18 '14 at 18:49
  • I would like to note that there are closed $E$, e. g. the polynomial ring is closed under integration and differentiation. – chaosflaws Jun 19 '14 at 14:17
  • But not under inversion, and hence division. Once you get rational functions integration takes you beyond them. – Conifold Jun 19 '14 at 19:03
  • Of course; I just wanted to state that it is indeed important what you define as elementary functions. – chaosflaws Jun 19 '14 at 22:02
  • @chaosflaws: Elementary functions have a [standard definition](http://en.wikipedia.org/wiki/Elementary_function). This is what the OP refers to. – RghtHndSd Jun 20 '14 at 01:53
  • I think a more natural generalization would be: does there exist a set of functions containing $E$ which is closed under composition, arithmetical operations, and anti-derivation? – Jack M Jun 25 '14 at 18:35
  • Is the answer known to the similar question for the [ring of periods](http://en.wikipedia.org/wiki/Ring_of_periods)? – Mikhail Katz Aug 07 '14 at 15:28
  • This has to be one of the absolute best questions asked on this site. I've never even considered such a thing but it's so natural. – Cameron Williams Feb 06 '16 at 01:28
  • "Inversion" means: the inverse function, or the reciprocal? It is not postulated that $E_i$ is closed under addition? Well, $E_0$ has exp and log, so addition follows from multiplication. Or, alternatively, multiplication follows from addition, which would be more natural to work with I guess. And reciprocal follows from subtraction. – GEdgar Feb 06 '16 at 15:52
  • Short answer : No. – mick Oct 11 '16 at 20:20
  • Does it make sense to consider the problem without including composition and inversion, but instead allowing algebraic closure operation? The motivation is that there does not seem to be a Liouville theory for function compositions. But reference refuting my claim is highly appreciated. – John Jiang Jan 01 '17 at 16:37

1 Answers1


Liouville's theorem deals with an elementary differential extension, composition is considered there. But your problem contains the additional operation inversion.

Therefore your problem is not a generalization of Liouville's theorem but a different task.

$E_{i+1}\setminus E_{i}$ contains the non-elementary antiderivatives of the functions from $E_{i}$ and the non-elementary inverses of the functions from $E_{i}$.

With a generalization of the theorem of Ritt of Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 which I hope to prove, one could show there are elementary functions in each of your $E_{i}$ that have a non-elementary inverse.

Your $E_{i}$ are therefore no differential fields and you cannot apply Liouville's theorem. Therefore your problem cannot be solved by the Liouville theory treated in the literature.

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