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I encountered an interesting function which is called "Eulerian" by the Wolfram's MathWorld:

$$\phi(q)=\prod_{k=1}^{\infty} (1-q^{k})$$

It is interesting because it is claimed that roots of any polynomial can be expressed in this function and elementary functions. Is this true and how the roots of arbitrary polynomial can be expressed?

P.S. In Mathematica this function is inplemented as QPochhammer[q]

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    $\begingroup$ Could you please link to where this statement is made? $\endgroup$
    – David E Speyer
    Commented Feb 22, 2012 at 0:11

1 Answer 1

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This Euler function is essentially the same as Dedekind's eta function (Wikipedia, Mathworld). The usual use of the $\eta$ function is to express various modular forms. In particular, you should be able to rewrite Hermite's solution of the quintic by modular functions in terms of the $\eta$ function.

I don't know whether you can express roots of higher degree polynomials in terms of $\eta$, but I would guess not. Here are my two hazy arguments:

  • Hilbert conjectured that the roots of a general sextic could not be expressed using functions of one variable. I am told that this conjecture appears in Über die Gleichung neunten Grades, Mathematische Annalen Volume 97, Number 1, 243-250; I have not read this article. Abhyankar proves a version of Hilbert's conjecture in this paper, which I discussed in my answer here. Unfortunately, to my limited understanding of Abhyankar's result, it is about algebraic functions of one variable, so it is not clear to me that it answers your question.

  • According to Wikipedia, Hermite's solution of the quintic by modular forms was finally generalized to equations of arbitrary degree by Umemura, using Siegel modular forms. Siegel modular forms are analytic functions of many variables. This suggests to me that a generalization using modular forms in a single variable was found unworkable.

As you can tell by the hesitant style of this answer, I find the results on solving equations by transcendental functions rather hard to follow; they all seem to be written in very equation heavy nineteenth century style. Since these questions come up fairly often on MO, it would be great if someone could recommend a good survey which translates them into the modern language.

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    $\begingroup$ So basically, the answer should be 'yes' for degree 5 or less and 'no' for higher degree? $\endgroup$
    – J.C. Ottem
    Commented Feb 22, 2012 at 0:40
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    $\begingroup$ Should be. But there are a lot of caveats in the above. I'm hoping someone will give a clear reference to clean this all up. $\endgroup$
    – David E Speyer
    Commented Feb 22, 2012 at 1:35
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    $\begingroup$ Looks right. The $\eta$ function gives access to classical modular functions, which give field extensions with Galois group contained in a quotient of some $\text{GL}_2({\bf Z}/N{\bf Z})$. That's enough to deal with the generic quintic, but not sextics and beyond (though the sextics only barely fail: $A_6$ is of $\text{GL}_2$ type, but over the field of $9$ elements!). $\endgroup$
    – Noam D. Elkies
    Commented Feb 22, 2012 at 2:16
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    $\begingroup$ @D.Speyer: because the kernel is not a congruence subgroup. $\endgroup$
    – Noam D. Elkies
    Commented Feb 22, 2012 at 19:59
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    $\begingroup$ In his Traité p. 378, Jordan proves the following theorem : The solution of the general equation of degree > 5 cannot be reduced to that of equations arising from circular or elliptic functions. As far as I can see the proof boils down to show that the alternating group $\mathcal{A}_n$ with $n \geq 6$ is not isomorphic to $\mathrm{PSL}_2(\mathbf{Z}/p\mathbf{Z})$ for any prime $p$. After that, Jordan also remarks that any equation can be solved by bisecting periods of hyperelliptic functions, as Noam said above. $\endgroup$
    – François Brunault
    Commented Feb 23, 2012 at 10:31

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