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After searching about why $\int e^{x^2}$ is not an elementary function, I was disappointed that I should understand about Galois theory, but then I started to think about a concept that treats elementary functions, anti-derivative of elementary functions, and operations (+-*/, composition) of those functions, which includes non-elementary functions like $\int e^{x^2}$, $\int \frac{sin(x)}x$, or $\int log(log(x))$.

I do believe that some people in past thought about that, but I simply don't know how to search.

What's the name of the type of functions that can be derived by differentiate, integrate, and composite of elementary functions? Plus, just out of curiosity, is there any function that's infinitely differentiable but not in the type of functions I asked?

JiminP
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You might be interested in a more extreme type of non-elementary function. A function $y$ of $x$ is said to be "transcendentally transcendental" on an interval $(a,b)$ if $P(x,y,y', y'',...,y^{n})$ is not identically zero on $(a,b)$ for every positive integer $n$ and every nonzero polynomial $P$ of $n+2$ variables. In other words, $y$ doesn't satisfy any algebraic differential equation, including non-linear algebraic differential equations. None of the elementary transcendental functions (trigonometric, exponential, logarithmic) are transcendentally transcendental on any open interval, and most of the higher functions in mathematical physics aren't either (elliptic functions, Bessel functions, etc.). However, in 1887 Hölder proved that the gamma function is transcendentally transcendental, which incidentally gives a naturally occurring example of an infinitely differentiable function that is far more non-elementary than what you were asking about. (I'm not sure, but I think Hölder was also the first to formulate the property of being "transcendentally transcendental".) A nice survey paper of this topic is:

Lee Albert Rubel, "A survey of transcendentally transcendental functions", American Mathematically Monthly 96 (1989), 777-788.

Many of the older papers on this topic, including Hölder's original paper and some papers by E. H. Moore (in a 1897 paper, E. H. Moore introduced the name "transcendentally transcendental") and some papers by J. F. Ritt (1923, 1926) are in Math. Annalen, and thus are freely available on the internet. There's also a 1902 paper by Edmond Maillet in Bulletin de la Societe Mathematique de France (Vol. 30, pp. 195-201) that is freely available on the internet.

azimut
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Dave L. Renfro
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  • In case anyone is interested, I had no idea this was an old question (asked 2 months ago) when I submitted my reply. For some reason this question was near the top of the list of questions when I visited StackExchange today (first visit in 2 or 3 days), and I never thought to look at the date. – Dave L. Renfro Aug 01 '11 at 19:41
  • It was probably bumped by the Community user. That means it didn't have an upvoted answer, so don't worry about it: it's good for questions like that to get answers. – Qiaochu Yuan Aug 01 '11 at 20:52
  • Thanks for the good answer and information! – JiminP Aug 02 '11 at 11:03
  • @Dave: +1 Nice answer indeed! I was wondering if gamma function being "transcendentally transcendental" implies factorial also being "transcendentally transcendental", given that factorial is restriction of gamma function to nonnegative integers? – Tim Jan 12 '12 at 17:35
  • @Tim: Since transcendentally transcendental involves satisfying a differential equation of a certain type, this seems like a non-issue (perhaps like asking if there exists a prime number that is a Hausdorff topological space). Also, restrictions of functions typically have nicer properties than the corresponding un-restricted function. For example, $\chi_{\mathbb Q}:{\mathbb R} \rightarrow {\mathbb R}$ is nowhere continuous while $\chi_{\mathbb Q}:{\mathbb Q} \rightarrow {\mathbb R}$ is constant, and $\sin$ is transcendental while its restriction to integer multiples of $\pi$ is algebraic. – Dave L. Renfro Jan 12 '12 at 18:37
  • @Dave: Thanks! (1) What type of differential equations do transcendentally transcendental functions satisfy? (2)Since you said transcendentally transcendental functions are defined not to satisfy algebraic differential equations, I wonder what is the definition of an algebraic differential equation? Is it also an ordinary differential equation? I don't find its definition at wikipedia http://en.wikipedia.org/wiki/Algebraic_differential_equation – Tim Jan 12 '12 at 19:43
  • Tim: Any $n$th order differential equation has the form $f\left(x,y,y',...,y^{(n)}\right); = ; 0$ for some function $f$ of $n+2$ variables. If it is possible to choose $f$ to be a polynomial, then the differential equation is algebraic. For example, $xy^{2} + yy' - \left(y''\right)^{3} ; = ; x^{2}$ is an algebraic differential equation and $y' = e^{y}$ is not an algebraic differential equation. Also, $\sqrt[3]{x^{5}y''} + \sqrt[5]{xy'} - \sqrt{x^{2} + y^{4}} ; = ; x^{2} y^3y'$ is an algebraic differential equation (via the "it is possible" clause). – Dave L. Renfro Jan 12 '12 at 20:14
  • @Dave: Thanks! (1) Does the definition of algebraic differential equations require the coefficients of polynomial $f$ to be rational numbers, as you said in your original post? Can it be reals, or complex numbers? (2) What type of differential equations do transcendentally transcendental functions satisfy? – Tim Jan 13 '12 at 00:22
  • @Tim: In the answer I said the coefficients have to be rational functions, not rational numbers. However, by multiplying through by a common denominator, we can incoporate that aspect as an additional polynomial variable (as I did in a comment just above). For example, $\frac{x}{x+1}\left(y'\right)^{3} - \left(\frac{2}{x-1}\right)y^{2} = 0$ is a polynomial in the variables $y$ and $y'$ with rational function coefficients, whereas $3x(x-1)\left(y'\right)^{3} - 2(x+1)y^{2} = 0$ is a polynomial in the variables $x,$ $y,$ and $y'$ with real number coefficients. I don't know an ODE for gamma. – Dave L. Renfro Jan 13 '12 at 19:38
  • @Tim: I've changed the original slightly to reflect the simpler (in my opinion) view of looking at the situation as a polynomial in $n+2$ variables as I said in a comment above. – Dave L. Renfro Jan 13 '12 at 19:41
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The function class that contains all elementary functions as defined by Liouville and all antiderivatives of the elementary functions are the Liouvillian functions. The Liouvillian functions are the closure under indefinite integration of the set of the elementary functions. This function classes are defined e.g. in section 1 of Davenport, J. H.: What Might "Understand a Function" Mean. In: Kauers, M.; Kerber, M., Miner, R.; Windsteiger, W.: Towards Mechanized Mathematical Assistants. Springer, Berlin/Heidelberg, 2007, page 55-65.

The Liouvillian functions are only a small part of the closed-form functions. Not all named Special functions are Liouvillian functions. E.g. the power series or the generalized hypergeometric functions are infinitely differentiable but not all in the Liouvillian functions.

IV_
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The point of Liouville's theorem is that functions that have elementary integrals must have a special form, no matter how you define elementary function. See How can you prove that a function has no closed form integral?. My answer there contains a list of readable references.

Community
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lhf
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  • Liouville's theorem does depend crucially on the definition of 'elementary function'. 2) The question concerns a much broader class of functions than the elementary functions, where Liouville's theorem does not apply.
    • – Bill Dubuque Jun 02 '11 at 16:28
  • @Bill, ok, I wasn't too precise. I meant that Liouville's theorem depends on the base field and it still says that functions in extensions of the base field that have an integral in the base field are restricted in nature. – lhf Jun 02 '11 at 19:22
  • No, that's not what it says. For a precise statement of Liouville's theorem (and a proof) I recommend reading Max Rosenlicht's Monthly article - see the link in my answer in the thread that you linked to. – Bill Dubuque Jun 02 '11 at 19:47
  • @Bill, I did. My interpretation is wrong then. Wikipedia seems to say something similar: Thus, on an intuitive level, the theorem states that the only elementary antiderivatives are the "simple" functions plus a finite number of logarithms of "simple" functions. – lhf Jun 02 '11 at 20:17