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Many elementary functions, like $e^{-x^2}$ and $\frac{\sin(x)}{x}$ have antiderivatives that are are nonelementary; is this property generic? That is, does the set of all elementary functions whose antiderivatives are nonelementary form residual (or second category) subset (that is, the complement of a meager or first category subset) of the elementary functions (with some version of the compact-open topology on them, presumably: the elementary functions don't all share a common domain)?

On a related (but still very basic) note, what would be a catch-all term for the collection of all elementary and nonementary functions taken together as one set? "Functions of a real (or complex) variable"?

$\underline{\text{Edit 1}}$: I guess Liouville's Theorem is a partial answer, at least. It appears to yield that the elementary functions whose antiderivatives are elementary are emphatically a meager set, but if I could just get someone who is much more experienced at this game to confirm that for me in simple terms that will be easy for me to understand, I would be most appreciative.

$\underline{\text{Edit 2}}$: In response to an answer from Robert Israel below, I changed the question from asking if the elementary functions whose antiderivatives are nonelementary form an open, dense subset to asking if they form a residual set.

$\underline{\text{Aside}}$: Why isn't Liouville's Theorem part of a standard graduate curriculum for those who want to go on to teach calculus? In what standard (graduate or undergraduate) course would one typically encounter Liouville's Theorem? What book would be a reference for differential algebras and Liouville's Theorem?

Jeffrey Rolland
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  • One might note that the elementary functions are a vector space and the ones with elementary anti-derivatives are a strict subspace thereof. (And, if you're careful, a subspace with infinite codimension). In some (standard, but rather uninteresting) sense, this means that most functions don't have elementary antiderivative. – Milo Brandt Jan 15 '17 at 02:52
  • Thanks, @MiloBrandt ! ("Strict subspace" meaning "proper subspace", of course) – Jeffrey Rolland Jan 15 '17 at 03:05
  • @MiloBrandt Thanks so much again for the reply! I really appreciate it! Unfortunately, the answer to this question http://math.stackexchange.com/questions/496436/construct-dense-subspace-of-codimension-n-for-all-n appears to imply that a subspace can be proper and of infinite codimension and dense (Take the union $W$ of all the spaces $X_n$ in the answer). So, being a proper subspace (even of infinite codimension) does not appear to necessarily imply meager. – Jeffrey Rolland Jan 16 '17 at 01:19
  • Residual in what topology? Unless you have somehow a topology where the elementary functions are complete? Otherwise I would guess the space of elementary functions is of first category in itself... – GEdgar Jan 16 '17 at 02:13
  • A topology on the differential algebra of functions of a real variable? https://en.wikipedia.org/wiki/Liouville%27s_theorem_%28differential_algebra%29#Definitions I'm not sure myself, which topology to use was a part of my question, I'm a little new at this game myself – Jeffrey Rolland Jan 16 '17 at 04:37

1 Answers1

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Not open, no. Polynomials are dense in the compact-open topology on continuous functions on an interval (or a disk in the complex plane), and they always have elementary antiderivatives.

Robert Israel
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  • (Off-topic: Is this the Robert Israel [http://mathforum.org/kb/thread.jspa?threadID=2251729&tstart=0 among oher things]? Wow, thanks for the reply!) – Jeffrey Rolland Jan 15 '17 at 03:21
  • Of course, Stone-Weierstrass https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem applies; thanks so much – Jeffrey Rolland Jan 15 '17 at 03:25
  • Yes, I am that Robert Israel. – Robert Israel Jan 15 '17 at 07:26
  • (I apologize that this question is basic.) So, putting aside the notion of elementary/nonelementary and just looking at W, the set of polynomials, in V, $C^1(I)$/$C^n(I)$/$C^{\infty}(I)$, on the one hand, by Milo Brandt's comment, W is a proper subspace subspace of V (with infinite codimension, in fact), while on the other hand, by Robert Israel's comment, V is dense in W.

    (cont)

     – Jeffrey Rolland Jan 15 '17 at 20:05
  • Intuitively, these two notions don't make sense together: if V were an inner product space, for instance, one could just take v in V with <v|p> = 0 for all p in W, take $k \in \mathbb{R}$ with $k>0$, and take B(v,k/2); that should contain no points of W - but W is dense in V! This idea should work for any inner product one could put on V which induces the compact-open topology on V - but, of course, it can't work for any of them (meaning no inner product on V can induce the compact-open topology on V?) since W is dense in V by Stone-Weierstrass. (cont) – Jeffrey Rolland Jan 15 '17 at 20:05
  • Thanks in any assistance you can provide. – Jeffrey Rolland Jan 15 '17 at 20:11
  • (The v in V with <v|p> = 0 for all p in W in the above should have $v \ne 0$, of course :/ ) – Jeffrey Rolland Jan 15 '17 at 20:17
  • If $W$ is dense in an inner product space $V$, there is no nonzero $v \in V$ with $\langle v | p \rangle = 0$ for all $p \in W$. Only closed subspaces can have nontrivial orthogonal complements. – Robert Israel Jan 15 '17 at 21:05
  • Thanks; I never took a course in Functional Analysis and I don't recall going over closed subspaces of Hilbert/infinite-dimensional inner product spaces in Real Analysis - this http://www.math.colostate.edu/~pauld/M645/BasicHS.pdf helped with the confusion.

    So, elementary functions with nonelementary antiderivatives are definitely not an open, dense subset of all elementary functions; thanks, I really appreciate the help

     – Jeffrey Rolland Jan 16 '17 at 00:02