Let $$K(k):=\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}=\frac{\pi}{2}{ _2F_1\bigg(\frac{1}{2},\frac{1}{2},1;k^2 \bigg)}$$

where $0<k<1$.

Let $K, K′, L$ and $L′$ denote the complete elliptic integrals of the first kind associated with the moduli $k, k′, l$ and $l′$, respectively. the complementary modulus is defined as $k':=\sqrt{1-k^2}$.

Then a modular equation of degree $n$ is a relation between the moduli $k$ and $l $


If $\alpha=k^2$ and $\beta=l^2$, then we say $\beta$ is of degree $n$ over $\alpha$. The multiplier $m$ is defined by $m=\frac{K}{L}$.

Ramanujan’s class invariant $G_n$ is defined by


First I calculate $G_5=(\frac{1+\sqrt{5}}{2})^{1/4}$. To calculate $G_5$ we need to use a modular equation of degree 5

$$m=1+2^{4/3}\bigg(\frac{\beta^5(1-\beta)^5}{\alpha(1-\alpha)}\bigg)^{1/24}$$ $$\frac{5}{m}=1+2^{4/3}\bigg(\frac{\alpha^5(1-\alpha)^5}{\beta(1-\alpha)}\bigg)^{1/24} $$

apply the restriction $\beta=1-\alpha$ and $G_5=\lbrace4\alpha(1-\alpha)\rbrace^{-1/24}$. We find that $$G_5=\bigg(\frac{1+\sqrt{5}}{2}\bigg)^{1/4}$$

I tried to calculate $G_{125}$ by using Schläfli's modular equation of degree $5$.

$$\bigg(\frac{u}{v}\bigg)^3+\bigg(\frac{v}{u}\bigg)^3=2 \bigg(u^2v^2-\frac{1}{u^2v^2}\bigg)$$

where $u=G_{125}$ and $v=G_{5}$. this problem reduce to find the real solution of following sextic equation

$$G^6-2\varphi^5G^5+2\varphi G+\varphi^6=0$$ where $\varphi=\phi^{1/4}=\bigg(\frac{1+\sqrt{5}}{2}\bigg)^{1/4}$ and $G=G_{125}$

How to solve this equation in radicals? Or, does there exists easy way to calculate $G_{125}$ by using modular equations?

P.S. I use Mathematica, but there is no solution in terms of radicals.

  • $\begingroup$ I made some edits. Pls follow convention as $\phi=\frac{1+\sqrt5}2$ is usually reserved for the golden ratio, not fractional powers of it. $\endgroup$ – Tito Piezas III Jan 1 '17 at 16:24

First, the Ramanujan $G_n$ and $g_n$ functions can be computed efficiently in Mathematica using the Dedekind eta function. Let $\tau=\sqrt{-n}$, then, $$G_n=\frac{2^{-1/4}\,\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\,\eta(2\tau)}\quad \text{odd}\; n$$ $$g_n=\frac{2^{-1/4}\,\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}\quad \text{even}\; n$$ Regarding your question of how to solve for $G_{125}$ in radicals, we can reduce the problem slightly by solving a quintic instead of a sextic. Define, $$x = \frac{G_5}{G_{125}} + \frac{G_{125}}{G_{5}}$$ and we find it is the real root of, $$x^5-10x^3-10x^2+5x-2=0$$ Since we know $G_5$, all we have to do is solve this. So given the golden ratio $\phi = \frac{1+\sqrt5}2$, then, $$G_{125}= \frac{x+\sqrt{x^2-4}}{2}\,\phi^{1/4}=3.633572\dots$$ $$x =\frac1{5^{1/4}} \big(\phi\,f_1+\phi^{-1}f_2\big)$$ and where, $$f_1=\sqrt[5]{5^{3/4}+\phi^{-5/2}}+\sqrt[5]{5^{3/4}-\phi^{-5/2}}$$ $$f_2=\sqrt[5]{5^{3/4}+\phi^{5/2}}+\sqrt[5]{5^{3/4}-\phi^{5/2}}$$


There is also an iterative method derived by G. Manco in this post using $G_{n/25},\, G_n,\, G_{25n}$,

$$1+2\frac{G_{25n}}{(G_{n})^5}=\frac 1 5\Big(1+2\frac{G_{n/25}} {(G_{n})^5}\Big)\big(1+\sqrt[5]{u_1}+\sqrt[5]{u_2} \big)^2\tag1$$

where the $u_1$ are the two roots of, $$u^2 -\alpha(\alpha^2 - 2\alpha + 8)u + \alpha^5 = 0,\quad \text{with}\;\alpha=2\frac{G_{n}} {(G_{n/25})^5}$$

Since $G_1 = 1,\; G_{1/5}=G_5 = \phi^{1/4},\; G_{25} = \phi$, then $(1)$ is an iterative method to express in radicals all $G_{5^m}$ in terms of the golden ratio $\phi$.

P.S. The similar $G_{n/9},\, G_n,\, G_{9n}$ appears in this post.

  • $\begingroup$ I know how to use Mathematica, it works well for $G(5)$, $G(25)$,$G(625)$ but there is no answer (in RADICALS) for the $G(125)$ see here $\endgroup$ – vito-ვიტო Feb 2 '18 at 20:37
  • $\begingroup$ @vito-ვიტო: Because $G(125)$ involves a quintic root, and Mathematica has difficulty with that case. $\endgroup$ – Tito Piezas III Feb 3 '18 at 4:29

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