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Regarding the Post Additional values of Dedekind's $\eta$ function in radical form I wrote the equation that has as root the value $\frac{\eta(19i)}{\eta(i)}$ that is missing.

Can someone help me solve (in radical form) the following equation, whose solution is the value of Dedekind’s modular $\frac{\eta(19i)}{\eta(i)}$ function?

$x^{40}-\frac{4}{19}x^{36}+\frac{138}{ 19^{3}}x^{32}-\frac{154}{19^{4}}x^{28}+\frac{41035}{19^{7}}x^{24}-\frac{359820}{19^{9}}x^{20}+\frac{1743935}{19^{11}}x^{16}-\frac{4798430}{19^{13}}x^{12}+\frac{16921266}{19^{15}}x^{8}+\frac{422140}{19^{17}}x^{4}-\frac{1}{19^{19}}=0$

where

$$x=\frac{\eta(19i)}{\eta(i)}.$$

This equation comes from the work of L. Kiepert and specializes for the value reported in the title of the application. My intent is to find the solution in closed form.

1 Answers1

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After substitution $x^4\to x,$ we get a 10-degree polynomial that factors over the rationals extended with $\sqrt{19}$ and a root of a solvable quintic $$\!\!\!\!\!\!\!\!\!\!\small 322687697779 \,z^5-67934252164 \,z^4+5974826887 \,z^3-229104318 \,z^2+3132036 \,z+216=0,\tag{$\small\spadesuit$}$$ so your 40-degree root can be represented in radicals. Let $$\small\begin{align} \alpha &=30 \sqrt{198616747730+65513019062 \sqrt{5}},\\ \beta &=30 \sqrt{198616747730-65513019062 \sqrt{5}},\\ \eta &=\sqrt[5]{11410567+2790935 \sqrt{5}+\alpha\;}+\sqrt[5]{11410567+2790935 \sqrt{5}-\alpha\;}\\ &+\sqrt[5]{11410567-2790935 \sqrt{5}+\beta\;}-\sqrt[5]{2790935 \sqrt{5}-11410567+\beta\;},\,\text{and}\\ \gamma& =8-\left(\frac{2}{19}\right)^{4/5} \eta, \end{align}$$ then $$\small\begin{align}&\!\!\!\frac{\eta(19i)}{\eta(i)}\\ &=\frac{\sqrt[4]{100680000 +7361892000{\tiny\text{ }}\gamma+76992000 \sqrt{19}\,\gamma -1888138300{\tiny\text{ }}\gamma ^2+145028140{\tiny\text{ }}\gamma ^3-4533799{\tiny\text{ }}\gamma ^4}}{20\cdot 2^{3/4} \cdot19^{3/8}\cdot \sqrt[4]{1203\,}}. \end{align}$$ I found the quintic $\small(\spadesuit)$ using my computer program that uses a combination of brute force and heuristics to find an extension of rationals in which a given polynomial can be factored.


Mathematica expression for the root:

With[{α = 30 Sqrt[198616747730 + 65513019062 √5], 
      β = 30 Sqrt[198616747730 - 65513019062 √5]},
  With[{η = (11410567 + 2790935 √5 + α)^(1/5) 
          + (11410567 + 2790935 √5 - α)^(1/5) 
          + (11410567 - 2790935 √5 + β)^(1/5) 
          - (2790935 √5 - 11410567 + β)^(1/5)}, 
    With[{γ = 8 - (2/19)^(4/5) η}, 
       (100680000 + 7361892000 γ + 76992000 √19 γ 
         - 1888138300 γ^2 + 145028140 γ^3 - 4533799 γ^4)^(1/4)
            /(20 2^(3/4) 19^(3/8) 1203^(1/4))]]]
  • My comment should have stated no solution in radicals in general, but obviously there are exceptions. Nice work. – Somos Sep 11 '19 at 22:47
  • For completeness, I would like to know the link between z and x. – giuseppe mancò Nov 04 '19 at 14:16
  • $z$ is just a fresh variable for the quintic polynomial equation $({\small\spadesuit})$, independent of $x$. – Vladimir Reshetnikov Nov 04 '19 at 17:18
  • May I ask what some of the brute force and heuristic methods are for the root finding procedure ? I started working on a root finding algorithm that attempts to find roots by polynomial algebra methods and by adding a parameter and searching for the differential resolvent. In the most basic case I will use the differential resolvent to find parameters of an ansatz involving hypergeometric functions by equating the differential resolvent of the two. – userrandrand Dec 21 '22 at 08:46
  • $\sqrt{19}$ was just a hunch based on that $19$ appears in the original problem statement. To factor it further, I use a mix of brute force search, and a few algorithms based on answers to these questions: https://mathematica.stackexchange.com/q/105933/7288, https://mathematica.stackexchange.com/q/133550/7288, https://mathematica.stackexchange.com/q/34011/7288. If you are interested, I can share my Mathematica notebooks that contain functions implementing those, but they are a big mess, and I forgot how they worked. – Vladimir Reshetnikov Dec 27 '22 at 06:54