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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 $\frac{\eta(13i)} {\eta(i)}$ that is missing.

Can someone help me solve (in radical form) the following equation, whose solution is the value of Dedekind's modular $\frac{\eta(13i)} {\eta(i)}$ function? $$x^{12}+\frac{10}{13} x^{10}+\frac{46}{13^{2}}x^{8}+\frac{108}{13^{3}} x^{6}+\frac{122}{13^{4}}x^{4}+\frac{38}{13^{5}}x^{2}-\frac{1}{13^{6}}=0$$ where $$x=\frac{\eta(13i)}{\eta(i)}.$$ Thank you.

Today I found the solution in radical form see Additional values of Dedekind's $\eta$ function in radical form Thank you.

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. Thank you.

giuseppe mancò
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  • Do you care only of the result ? Once you know how we obtain the polynomial whose $\eta(Ni)/\eta(i)$ is a root then expressing it as a radical is the easy part. If the root of $f$ is radical then compute the splitting field and Galois group of $f$ to find it. – reuns Apr 28 '19 at 23:26

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Using Mathematica I found a solution. Let $\ p = 13,\ $ $s = \sqrt{p},\ $ $x = \eta(p\ i)/\eta(i),\ $ $x_0 = p\,x^2,\ $ $f(x) := 2x^3 + 10x^2 + (21+s)x + (3-s).\ $ Then $\ f(x_0) = 0.\ $ Solve $\ f(x)=0\ $ using $$ c_1 = 91+18s,\ c_2 = 19+5s,\ c_3 = 13+3s,\ r_2 = 3\sqrt{78c_2},\ r_3 = \sqrt[3]{c_1+r_2}. $$ Then $\ x_0 = (-10-c_3/r_3+2r_3)/6 \approx 0.024367851307181284. $

Somos
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  • I translated what you wrote in radicals and I found an approximation of $10 ^ {-41} $(always compared with PARI/GP). Thank you. – giuseppe mancò May 02 '19 at 08:05