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Lets recall Platonic construction in plane geometry. It is impossible to square a circle using only ruler and callipers. But is also known that it is possible to do it with ruler which has a mark on it.

So we know by Gaois theory that we can't write expressions for the solutions to all algebraic equations using the four arithmetic operations and nth roots.

Question: is it possible to write expressions for the solutions to all algebraic equations using the four arithmetic operations and nth roots and something else in finite numbers of factors? For example adding logarithm to the set of operations? ( or other function, or class of functions...)?

If Yes: Do they form any kind of algebraic structure?

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  • $\begingroup$ Without restrictions on the "something else" this question does not make sense, I believe. $\endgroup$
    – Roland Bacher
    Apr 30, 2010 at 11:15
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    $\begingroup$ So feel free to write how to restrict it as You want if You know the answer. For example if You know that adding logarithms do not changes it in any way, write it. I do not ask for solution but for information what is the state of knowledge in this area... $\endgroup$
    – kakaz
    Apr 30, 2010 at 11:40
  • $\begingroup$ Well, the roots of a polynomial of degree n depends on n+1 parameters, so in any case you'd have to add an infinite number of functions to your class in order to solve all algebraic equations. $\endgroup$
    – J.C. Ottem
    Apr 30, 2010 at 12:36
  • $\begingroup$ @J.C Ottern - I do not understand what do You mean but is probably interesting. I presume that in context of Galois theory, when You analyse symmetric polynomials there is relation to solvable groups. Base class here is arithmetic operators plus roots so You get class of rational functions of rational numbers and its roots. It is enough for n=4 but nit enough for n=5. In Galois theory it depends on number of (group of) permutations of roots, parameters are not mentioned as far as I know... But why it depends on number of parameters in Your opinion? $\endgroup$
    – kakaz
    Apr 30, 2010 at 13:16
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    $\begingroup$ I'm sorry, my comment was somewhat vague. I was talking about functions $f(a_0,..,a_n)$ such that $f$ gives a root of the polynomial $P=a_nx^n+\ldots+a_0$. But as Gerald Edgar noted, the set of such functions is generated by a finite subset. $\endgroup$
    – J.C. Ottem
    Apr 30, 2010 at 13:33

4 Answers 4

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The answer is no: you need to keep adding more and more operations. For degree $n=5$ you can use elliptic functions (or the Jacobi $\Theta$ function, or Bring radicals - see below), for $n=6,7$ the Lauricella functions are needed (they are a 2-variable version of hypergeometric functions), and after that you need more and more complicated 'new' functions. You can apparently use the elliptic Siegel functions (aka Siegel modular forms) for the general case, but I've never looked into that. Most of this was all worked out in gruesome detail by analysts for decades, with lots of research on this up until the early 20th century (and then generally forgotten!).

For trinomial equations, you can use hypergeometrics to solve all of them. The derivation of that is fun.

It is hard to find information about this on the web. The 'best' discussion of related ideas is the Wikipedia page on the Bring radical. There is a good timeline at Wolfram's site.

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    $\begingroup$ Not so much hypergeometrics is given on the Wiki site but a few days ago I MR reviewed the paper "On the solution of a fifth-degree equation" by Mikhalkin, E. N. (MR2583995). The hypergeometric formulae (a 6-tuple(!) hypergeometric series is involved) look cumbersome but nice. $\endgroup$
    – Wadim Zudilin
    Apr 30, 2010 at 12:35
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    $\begingroup$ So is there a proof that Siegel modular forms are enough for polynomial equations with rational coefficients? Is that mean that set of algebraic numbers is generated by it? $\endgroup$
    – kakaz
    Apr 30, 2010 at 13:18
  • $\begingroup$ See J.C. Ottem's answer for a reference to Umemura's paper. Apparently, Mellin integrals are enough too. $\endgroup$
    – Jacques Carette
    Apr 30, 2010 at 13:37
  • $\begingroup$ Yes. Interestingly, these integrals are usually evaluated using a residue calculation, and summing over the infinitely many residues, we get a new derivation of Birkeland's series of hypergeometric type (cf. Mellin's original paper) $\endgroup$
    – J.C. Ottem
    Apr 30, 2010 at 13:49
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    $\begingroup$ Try this article: springerlink.com/content/5511407734528566 and Mellin's original paper: gallica.bnf.fr/ark:/12148/… $\endgroup$
    – J.C. Ottem
    Apr 30, 2010 at 15:26
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According to Kolmogorov's 1957 solution of Hilbert's 13th problem, "generalized slide-rules suffice" ... For given $n$, there exists a finite list of functions of one variable so that solutions for polynomial equations of degree $n$ can be written in terms of addition together with these functions. Indeed, any continuous function of $n$ variables can be written this way (no need to mention polynomials).

addition (statement taken from my book Classics on Fractals, p. 335)
There exist fixed continuous functions $\phi_{pq}(x)$ on $I = [0,1]$ so that each continuous function $f$ on $I^n$ can be written in the form $$ f(x_1,\dots,x_n) = \sum_{q=1}^{2n+1}g_q\left(\sum_{p=1}^n \phi_{pq}(x_p)\right) $$ where $g_q$ are properly chosen continuous functions of one variable.

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    $\begingroup$ But was his result constructive or purely existential? If this is a constructive result, I would love to see the details! $\endgroup$
    – Jacques Carette
    Apr 30, 2010 at 13:10
  • $\begingroup$ Do we know what is a class of such function together fro all n? Share they any property? It is very general existence theorem,or do we know any examples for n>5? $\endgroup$
    – kakaz
    Apr 30, 2010 at 13:21
  • $\begingroup$ Thank You a lot! Your answer was very enlightening here, and You point on Kolmogorov-Arnold solution og Hilbert problem which is great achievement. But I was looking for something like Bring Radicals etc. rather here than existence theorems, even if they are great! Thank You very much! $\endgroup$
    – kakaz
    May 1, 2010 at 18:53
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Eisenstein in 1844 found a very simple hypergeometric series solution of the quintic $x^5+x=a$, namely

$x=a-a^5+10\frac{a^9}{2!}-15\cdot 14 \frac{a^{13}}{3!}+20\cdot 19\cdot 18\frac{a^{17}}{4!}-\cdots$

Since it was shown by Bring in 1786 that the general quintic is reducible by radicals to this form, Eisenstein's solution is probably the simplest solution possible for the quintic.

For more on this story, see S.J. Patterson, Eisenstein and the quintic equation, in Historia Mathematica 17 (1990), pp.132--140.

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    $\begingroup$ I was trying to minimize the "something else" in the question. $\endgroup$
    – John Stillwell
    May 1, 2010 at 4:15
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Well, the logarithm can be defined as an integral or locally as an infinite series. So if we are allowed to add functions defined in this matter to our class of functions, then the answer to your problem is trivially true. But of course, we need to add new functions for each degree.

Classically, the quintic equation (degree 5) was solved in this manner using jacobi theta functions, which can be defined using infinite products or Fourier series. In 1923 Birkeland derived hypergeometric series expansions of the roots. Later, in 1983 in an appendix to Mumfords 'Tata lectures on Theta' H. Umemura showed that the general algebraic equation can be solved using Riemann theta functions. This compares to solving the equation $$x^n-a$$ using $$x=\exp\big(\frac1{n} \ln a\big),$$ but with $\ln a$ replaced by an hyperelliptic integral and $\exp$ replaced by a quotient of thetafunctions. For Umemura's article this here.

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  • $\begingroup$ It is interesting o note that in Galois theory we also have infinite functions: nth roots of numbers, but is not enough. So clearly infinite number of functions is needed but not every infinite set is enough. I am asking rather about class of such functions. Your answer is very interesting here because You mention only one function...Riemann function which means probably that it is enough together with arithmetic to define all algebraic numbers... $\endgroup$
    – kakaz
    Apr 30, 2010 at 13:26
  • $\begingroup$ Riemann theta functions are defined using a hyperelliptic curve of the form y^2=P(x) where P is a polynomial of degree 2g+1 or 2g+2 where g is the genus of the curve. In particular, there are one for each genus g. So if we add all these, we are of course done by Umemura's result. – J.C. Ottem 0 secs ago $\endgroup$
    – J.C. Ottem
    Apr 30, 2010 at 13:40
  • $\begingroup$ But do the Riemann theta functions form a ring? Do they have good normal forms? If they do, then the next question is about the inverse problem: what class of zero-finding problems do these solve? [The same question about Mellin-type integrals is interesting too]. Hmmm, there probably is a good MO question in there! $\endgroup$
    – Jacques Carette
    Apr 30, 2010 at 14:32
  • $\begingroup$ Of course you can form the $\mathbb{C}$-algebra generated by the Riemann theta functions, but I don't think it is what you are after. Note also that in Umemuras formulas, we need to take rational functions in the theta functions. In any case, the theta functions solve a variety of differential equations, e.g. the heat equation. $\endgroup$
    – J.C. Ottem
    Apr 30, 2010 at 15:34
  • $\begingroup$ Right, I meant to ask whether the Riemann theta functions naturally form a ring, i.e. if they satisfy some reasonable closure properties under addition and multiplication. I should have remembered (from Gray's wonderful book) that the theta functions solve the Fuchsian class of DEs! It is when you are non-Fuchsian that things get interesting because there are parameters in the monodromy matrix which seem to 'come out of nowhere'. $\endgroup$
    – Jacques Carette
    Apr 30, 2010 at 16:45

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