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We know that fifth (or higher) degree polynomial equation has no general solution formula using only the usual algebraic operations (addition, subtraction, multiplication, division) and application of radicals (square roots, cube roots). (Not solvable using radicals).

But can those polynomial equation have general "closed" formula solution if we allow more operations (and functions) in solution, such as: log functions, exponential functions, trigonometric functions, or other type of functions, etc.

Does such "closed" formula solution exist ?

Bill Dubuque
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ben
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    The quintic can be solved with elliptic functions of the coefficients, in addition to the other you mentioned. – egreg Sep 12 '14 at 16:34
  • related: https://math.stackexchange.com/questions/1828551/polynomials-with-degree-5-solvable-in-elementary-functions/4225322#4225322 https://math.stackexchange.com/questions/3538585/solvability-in-radicals-elementary-functions-and-monodromy-galois-groups/4341942#4341942 – IV_ Apr 10 '25 at 20:35

1 Answers1

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I've been waiting until our paper was published to answer this one, and it finally was. We give a formal power series solution to the general polynomial.

Let's start here with the solution to a polynomial in what we call geometric form, meaning a constant of $1$ and a linear coefficient of $-1$.

Theorem The polynomial or power series equation $$0= 1 - \alpha + t_2 \alpha ^2 + t_3 \alpha^3 +t_4 \alpha^4 + \ldots $$ has a formal power series solution: $$ \alpha %={\mathbf S}= {\mathbf S}[t_2,t_3,t_4,\ldots] = \sum_{m_2,m_3,m_4,\ldots \ge 0} \! C[m_2,m_3,m_4,\ldots] \ t_2^{m_2} t_3^{m_3} t_4^{m_4} \cdots \equiv \sum_{{\mathbf m} \ge {\mathbf 0}} C_{\mathbf m} {\mathbf t}^{\mathbf m} $$

where the hyper-Catalan number $C_{\mathbf m}=C[m_2, m_3, m_4, \ldots]$ counts the number of roofed polygons subdivided with non-crossing diagonals into $m_2$ triangles, $m_3$ quadrilaterals, $m_4$ pentagons, etc.

The closed-form for $C_{\mathbf m}$ has been known since Erdelyi and Etherington found it in 1941:

$$C_{\mathbf m} = \dfrac{( 2m_2 + 3m_3 + 4m_4 + \ldots )!}{(1 + m_2 + 2m_3 + 3m_4 + \ldots)! \, m_2! \, m_3! \, m_4! \cdots} $$

That can be written as:

$$C_{\mathbf m}= \dfrac{( E_{\mathbf m}-1)!}{ (V_{\mathbf m} -1) ! \, {\mathbf m}! \ }$$

where $V_{\mathbf m}=2+ m_2 + 2m_3 + 3m_4 + \ldots$ and $E_{\mathbf m} = 1 +2m_2 + 3m_3 + 4m_4 + \ldots $ are the number of vertices and edges of a roofed subdivided polygon of type ${\mathbf m}$ and ${\mathbf m}! \equiv m_2! m_3! m_4! \cdots$.

A straightforward substitution yields the general polynomial formula:

Theorem The polynomial or power series equation $$ \displaystyle 0 = c_0 - c_1 \alpha + c_2 \alpha^2 + c_3 \alpha^3 + c_4 \alpha^4 + \ldots $$ has a formal series solution: \begin{align*} \alpha & = \!\!\! \!\!\! \sum_{\substack{m_2, m_3, \ldots \ge 0 \ \ }} \!\! \!\! \dfrac{( 2m_2 + 3m_3 + 4m_4 + \ldots )!}{(1 + m_2 + 2m_3 + 3m_4 +\ldots)! m_2! m_3! \cdots} \ \dfrac{ c_0^{1 + m_2 + 2m_3 +\ldots} c_2^{m_2 } c_3^{m_3}\cdots } {c_1^{1 + 2m_2 + 3m_3 + 4m_4 + \ldots}} \\ & = \sum_{{\mathbf m} \ge {\mathbf 0}} \ \dfrac { c_0^{V_{\mathbf m} -1}}{(V_{\mathbf m} -1) !} \ \dfrac{( E_{\mathbf m}-1)!} { c_1^{E_{\mathbf m}}} \ \dfrac {{\mathbf c}^{\mathbf m}}{{\mathbf m}! } . \end{align*}

where ${\mathbf c}^{\mathbf m} \equiv c_2^{m_2} c_3^{m_3} c_4^{m_4} \cdots$.

Reference: Wildberger, N. J., & Rubine, D. (2025). A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode. The American Mathematical Monthly, May 2025. https://doi.org/10.1080/00029890.2025.2460966

  • Using transformations and one series can solve a degree 5 equation rather than a quadruple series for the general equation. Also, Is there any way to hypergeometric functions, like the Lauricella functions to evaluate the inner sums? – Тyma Gaidash Apr 10 '25 at 17:30
  • There are related hypergeometric forms; we didn't want to get too far into it in the paper but we reference a couple, including Mayr, 1936. – Dean Rubine Apr 10 '25 at 17:34
  • Re transformation of the quintic, the usual transformation that produces the Bring radical trinomial form may then be solved with this method, yielding a sum over a single variable. The coefficients are the hexagonal Fuss numbers, the generalization of Catalans when we subdivide into all hexagons instead of all triangles. Eisenstein did it; our formula recovers his solution as we discuss in the paper. – Dean Rubine Apr 10 '25 at 17:48