Can we give an explicit${^*}$ example of a real algebraic number that provably cannot be represented as an expression built from integers and elementary${^{**}}$ functions only?

${^*}$ explicit means we can write down a polynomial equation with integer coefficients having the algebraic number as a root, and an interval with rational bounds that isolates that root.

${^{**}}$ an expression built from integers and elementary function only means any valid expression in the set of elementary expressions $\mathcal{E}$ (as defined in that question at MO). Briefly, it is any finite combination of the following:

  • the imaginary unit $i$,
  • the exponent $x\mapsto e^x$,
  • the principal branch of the natural logarithm $x\mapsto\ln x$, provided $x\ne0$, and
  • the multiplication function $(x,y)\mapsto x\cdot y$.

Note that it allows to express constants $\pi$, $e$, integers, rationals, sums, powers, radicals, and also trigonometric and hyperbolic functions and their inverses, e.g. $$\pi=i\cdot i\cdot i\cdot \ln(i\cdot i).$$

Update: I reposted this question at MO.

  • $\begingroup$ If the field of numbers that can be expressed in terms of integers and elementary functions is algebraically closed, then the answer is no $\endgroup$ Oct 14 '13 at 18:05
  • $\begingroup$ Well, that is to say, for some subset of elementary functions and their compositions. I.e. things like $\cos ((p/q) \arctan (a/b))$ like what you wrote. If any such set is algebraically closed then the answer is no. $\endgroup$ Oct 14 '13 at 18:16
  • $\begingroup$ My understanding is that the background to Hilbert's 13th Problem was a result that the general sixth degree polynomial's roots cannot be expressed in terms of functions of one argument. I'll try to find a reference, but I take it your "elementary functions" are of one argument. $\endgroup$
    – hardmath
    Oct 14 '13 at 18:37
  • 3
    $\begingroup$ @hardmath The solution to Hilbert's $13^{th}$ Problem found by Kolmogorov and Arnold states that 2-argument functions are sufficient to solve algebraic equations of $7^{th}$ degree. For example, addition, multiplication and raising an expression to a power are all 2-argument functions. $\endgroup$ Oct 14 '13 at 18:51
  • 3
    $\begingroup$ It wouldn't hurt to be more explicit about what you mean by 'elementary functions' - in particular, whether you just mean the exp-based functions (cos, sin, arctan, log, etc.) or whether you mean to explicitly allow e.g. things like hypergeometrics. Also, the Kolmogorov/Arnold result is not, AFAIK, speaking specifically in terms of elementary functions and so may not be (entirely) germane here. $\endgroup$ Oct 14 '13 at 19:48

This may not be what your are looking for but, after some tinkering, I found your example in fact can be expressed in radicals. Let,

$$x = 2\cos \frac{2\arctan k}{5}$$

then $x$ is a root of,

$$x^5-5x^3+5x+2\left(\frac{k^2-1}{k^2+1}\right) = 0$$

This is the DeMoivre quintic in disguise,


and is solvable in radicals. Your $\alpha$ then has the radical expression,

$$\alpha = 2\cos \frac{2\arctan 2}{5} =\left(\frac{-3-4i}{5}\right)^{1/5}+\left(\frac{-3+4i}{5}\right)^{1/5} = 1.807059\dots$$

  • $\begingroup$ Thanks! I fixed my question. I hope the new example should work. Welcome to try to prove me wrong this time as well :) $\endgroup$ Oct 14 '13 at 19:51
  • 2
    $\begingroup$ @VladimirReshetnikov: Actually, it is as well. :) All $2\cos\frac{2\arctan k}{n}$ can be expressed in radicals. The new one is just $$2\cos \frac{2\arctan 2}{7} =\left(\frac{-3-4i}{5}\right)^{1/7}+\left(\frac{-3+4i}{5}\right)^{1/7} = 1.900768\dots$$ $\endgroup$ Oct 14 '13 at 19:58
  • $\begingroup$ Oops... I need to think deeper to find a real example. $\endgroup$ Oct 14 '13 at 20:18

If I'm not mistaken, this is a profound problem and little is known about it. Some years back there was an article in the American Mathematical Monthly by Timothy Chow, What is a Closed-Form Number? (pdf here). I believe what Chow calls EL numbers are the same as the numbers you have identified.

This area is closely connected with Schanuel's conjecture. Chow proves one conditional result that is relevant here:

  • If Schanuel's conjecture is true, then the algebraic numbers $\alpha$ belonging to the class EL are precisely those whose equations are solvable over $\mathbb{Q}$ (meaning the Galois group of the splitting field of the minimal polynomial for $\alpha$ is solvable).

See corollary 1 on page 444 of Chow's paper. I'm not sure a single explicit example that answers your question is known, although I'd be absolutely delighted to be shown otherwise.

  • $\begingroup$ That PDF is interesting but I only have a vague understanding of the terminology. Are the terms "transcendence degree", "fields over $\mathbb{Q}$", and "irreducible polynomial" topics that would be covered in undergrad algebra? $\endgroup$
    – user18862
    Oct 16 '13 at 23:14
  • 3
    $\begingroup$ Definitely the terms "fields over $\mathbb{Q}$" and "irreducible polynomial" would be covered in a standard undergraduate algebra course that covers the elements of Galois theory. "Transcendence degree" might be a different story, but the concept itself is not unduly difficult: it is the maximal number of algebraically independent elements in a field extension, very analogous to how the dimension of a vector space is the maximal number of linearly independent elements. All of this would certainly be covered in a respectable graduate algebra book like Lang's. $\endgroup$
    – user43208
    Oct 16 '13 at 23:25
  • $\begingroup$ user43208 is right. The elementary numbers (see Wikipedia) can be generated by applying elementary functions to the rational numbers. Chow's numbers EL are the explicit elementary numbers. $\endgroup$
    – IV_
    Aug 21 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.