Polynomials with degree $5$ solvable in elementary functions?

Question

Quadratic, cubic and quartic polynomials are solvable in radicals, so there is no question here.

What about the polynomials of degree $5$ (quintic)? Do we know all such polynomials (classes of polynomials) with all (or some) of the roots expressible as elementary functions of the coefficients?

If we do, I would be glad for a reference. If we don't (or it hasn't been proven that we do) then it might be possible to find a new class of solvable quintics?

Elementary functions are understood in the usual sense: algebraic functions, exponents, logarithms, trigonometric and inverse trigonometric functions and some finite combination of the above.

Which means that I'm not interested in the solution using Bring radicals, Hypergeometric functions etc.

A related question

Information on polynomials of higher degree would be appreciated as well.

The other (more practical) question - is there a complete list of known quintics solvable in elementary functions? (Not necessarily in radicals).

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Edit

The importaint point: If at least one root of the quintic can be expressed as elementary function, then all its roots can be expressed as elementary functions, since the general quartic is solved in radicals.

I think this greatly simplifies the problem.

For example, the following quintic has a root expressed as elementary function:

$$x^5-5ax^4-10x^3+10ax^2+5x-a=0 \tag{1}$$

$$x_1=\tan \left( \frac{1}{5} \arctan a \right)$$

Then we can divide $(1)$ by $(x-x_1)$ and obtain a quartic equation with all the roots expressible in radicals. (The example is not ideal though, since this quintic can most likely be solved in radicals).

Answer 1

These two are relatively well-known.

I. For the DeMoivre quintic: $$x^5+5ax^3+5a^2x+b = 0\tag1$$

$$x = \left(\frac{-b+\sqrt{D}}{2}\right)^{1/5}-a\left(\frac{-b+\sqrt{D}}{2}\right)^{-1/5},\quad D=b^2+4a^5\\ \color{blue}{\text{or}}\\ \\ x_k = 2\sqrt{-a}\;\sin\left(\tfrac{1}{5}\,\arcsin\big(\tfrac{-b}{2\sqrt{-a^5}}\big)-\tfrac{2\pi\,k}{5}\right)\\ \color{blue}{\text{or}}\\ \\ x_k = 2\sqrt{-a}\;\cos\left(\tfrac{1}{5}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^5}}\big))-\tfrac{2\pi\,k}{5}\right)$$

for all five roots $x_k$ with $k =0,1,2,3,4$. Note that all quintics can be reduced, in radicals (using a quadratic Tschirnhausen transformation), to the form,

$$x^5+5ax^3+5bx+c=0\tag2$$

so the general quintic is tantalizingly close to, but not quite, solvable.

P.S. This is directly analogous to the soln of the depressed cubic which all cubics can be reduced to,

$$x^3+3ax+b = 0\tag3$$

where,

$$x_k = 2\sqrt{-a}\;\cos\left(\tfrac{1}{3}\,\arccos\big(\tfrac{-b}{2\sqrt{-a^3}}\big))-\tfrac{2\pi\,k}{3}\right)$$

II. For the tangent quintic: $$x^5 + 5a x^4 + 10 b x^3 + 10a b x^2 + 5b^2 x + a b^2 = 0\tag4$$

$$x = \sqrt{b}\;\frac{1+R^{1/5}}{1-R^{1/5}},\quad R=\frac{a+\sqrt{b}}{a-\sqrt{b}}\\ \color{blue}{\text{or}}\\ \\ x_k =\sqrt{-b}\,\tan\left(\tfrac{1}{5}\,\arctan\big(\tfrac{-a}{\sqrt{-b}}\big)-\tfrac{2\pi\,k}{5}\right)$$

There does not seem to be a commonly known quintic with roots that can be expressible in terms of elementary functions (trigonometric, logarithmic, hyperbolic, etc) that cannot be expressed in terms of radicals as well.

Answer 2

The explicit elementary numbers are generated from the rational numbers by applying finite numbers of $\exp$, $\ln$ and/or radicals.
Clearly, the elementary numbers contain all algebraic numbers and the explicit elementary numbers contain all explicit algebraic numbers (the numbers representable by radicals).

Chow proves in [Chow 1999] his
Corollary 1:
"If Schanuel's conjecture is true, then the algebraic numbers in" the explicit elementary numbers "are precisely the roots of polynomial equations with integer coefficients that are solvable in radicals."

That means, the quintics that are not solvable in radicals cannot be solved by elementary numbers (means by applying elementary functions).

There is an extension:
Consider that the roots of algebraic equations of one unknown, the algebraic numbers, are values of algebraic constant functions.
"Corollary 6.24. 1) If the monodromy group of an algebraic equation over the field of rational functions is not solvable, then the solution of this equation does not belong to the class of functions representable by single-valued S-functions and quadratures." [Khovanskii 2004]
"Theorem. If the monodromy group of an algebraic function is solvable one can represent it via radicals. But if it is unsolvable one cannot represent it by a formula which involves meromorphic functions and elementary functions and uses integration, composition and meromorphic operations." [Khovanskii]
"Theorem 20. If the monodromy group of an algebraic function is unsolvable one can not represent it by a formula which involves meromorphic functions and elementary functions and uses integration, composition and meromorphic operations." [Khovanskii 2019]
I couldn't find if this theorem was published already earlier by Khovanskii or Vladimir Arnold. $\ $

for the second question:
The algebraic functions considered in [Ritt 1922] give i.a. the algebraic equations that have solutions that are radicals.

$a,b,c,d\in\overline{\mathbb{Q}}$

For the beginning, we get i. a. the following quintics from Ritt's paper.

$$a(z+b)^5+c=0$$

$$16a(bz+c)^5-20a(bz+c)^3+5abz+5a+d=0$$ $\ $

[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448

[Khovanskii 2004] Khovanskii, A.: On solvability and unsolvability of equations in explicit form. Russian Math. Surveys 59 (2004) (4) 661–736

[Khovanskii 2014] Khovanskii, A.: Topological Galois Theory. Solvability and Unsolvability of Equations in Finite Terms. Springer 2014

[Khovanskii] Khovanskii, A.: Topological Galois Theory

[Khovanskii 2019] Khovanskii, A.: Topological approach to 13th Hilbert problem

[Ritt 1922] Ritt, J. F.: On algebraic functions which can be expressed in terms of radicals. Trans. Amer. Math. Soc. 24 (1922) (1) 21-30