Ramanujan's Class Invariant $G_{625}$

Question

How to calculate the Ramanujan Class Invariant $G_{625}$?

Equation is:

$x^6-2\varphi^5x^5+2\varphi x+\varphi^6=0$.

$\varphi$ is the golden ratio.

Answer 1

I wish to give another form for $G_{625}$ that I found using Ramanujan's modular equation of the 25th order. Let $G_{25} = \varphi$ with golden ratio $\varphi$, then,

$$\frac{2\,G_{625}}{(G_{25})^5}+1=\frac{1}{\varphi^3}\left(1+(4\varphi)^{1/5}\Big(\big(3+\frac{5^{1/4}}{\varphi^{3/2}}\big)^{1/5}+\big(3-\frac{5^{1/4}}{\varphi^{3/2}}\big)^{1/5}\Big)\right)^2$$

It seems very elegant!

Along with this expression I'd like to write the inverse:

$$\frac{2 G_{25}^{5}} {G_{625}} – 1=\varphi^{3}\left(1-(\frac{4} {\varphi} )^{1/5}\Big(\big(3+5^{1/4}\varphi^{3/2}\big)^{1/5}+\big(3-5^{1/4} \varphi^{3/2}\big)^{1/5}\Big)\right)^{2}$$

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Method: Squaring

$$5\frac{w(q^{25})}{w(q)} = 1+\sqrt[5]{u}+\sqrt[5]{v}$$

we have

$$5\frac{w^2(q^{25})}{w^2(q^{5})} 5\frac{w^2(q^{5})}{w^2(q)} = \big(1+\sqrt[5]{u}+\sqrt[5]{v} \big)^2$$

Or

$$5\frac{w^2(q^{25})}{w^2(q^{5})} = \frac15\frac{w^2(q)}{w^2(q^5)} \big(1+\sqrt[5]{u}+\sqrt[5]{v} \big)^2$$

But the modular equation of 5th degree has the expression

$$\frac{w^2(q)}{w^2(q^{5})}=1+2\frac{G_{n}}{(G_{25n})^5}=\frac {5} {1+2\frac{G_{25n}} {(G_{n})^5}}\tag{1}$$

Set

$$(uv)^\frac 1 5=\frac {5}{\frac {w^2(q)}{w^2(q^5)}}-1=2\frac{G_{25n}} {(G_{n})^5}=x$$

and $y=(x-1)^2+7,\;\Delta=y^2-4x^3=(x-4)^2(x^2+4),\,$ hence

$$u=\frac x 2(y+\sqrt \Delta),\quad v=\frac x 2(y-\sqrt \Delta)$$

Since $w(q)$ has $q = e^{\pi i\sqrt{-n}}$, then using $(1)$,

$$1+2\frac{G_{625n}}{(G_{25n})^5}=\frac 1 5\Big(1+2\frac{G_{n}} {(G_{25n})^5}\Big)\big(1+\sqrt[5]{u}+\sqrt[5]{v} \big)^2\tag2$$

Since $G_1 = 1,\; G_{1/5}=G_5 = \varphi^{1/4},\; G_{5^2} = \varphi$, then $(2)$ is an iterative method to express in radicals all $G_{5^m}$ in terms of the golden ratio $\varphi$.

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$\color{green}{Added}$ (by TP):

And a similar one found here also by Manco for $G_{125}$. Let $G_{5} = \varphi^{1/4}$, then

$$\frac{2\,G_{125}}{(G_{5})^5}+1=\frac{1}{\sqrt5}\left(1+(4/\varphi)^{1/5}\Big( \big(4-\sqrt{5}+\frac{5^{3/4}}{\phi^{3/2}}\big)^{1/5}+\big(4-\sqrt{5}-\frac{5^{3/4}}{\phi^{3/2}}\big)^{1/5}\Big)\right)^2$$

These two were derived from an entry in Notebook 2, Chapter XIX, Entry 12 (iii). (Also in Berndt's Ramanujan's Notebooks, Vol III, Entry 12(iii), p.269.) Given the Jacobi theta function $\vartheta_3(0,q)=w(q)$. Then, $$5\frac{w(q^{25})}{w(q)} = 1+\sqrt[5]{u}+\sqrt[5]{v}$$ where,
$$uv = \left(5\frac{w^2(q^5)}{w^2(q)}-1\right)^5$$ $$u+v = \left(5\frac{w^2(q^5)}{w^2(q)}-1\right)\left(11-20\frac{w^2(q^5)}{w^2(q)}+25\frac{w^4(q^5)}{w^4(q)} \right)$$

Answer 2

The Ramanujan $G_n$ and $g_n$ functions can be computed in Mathematica using the Dedekind eta function. Let $\tau=\sqrt{-n}$, then, $$G_n=\frac{2^{-1/4}\,\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\,\eta(2\tau)}\quad \text{odd}\; n$$ $$g_n=\frac{2^{-1/4}\,\eta\big(\tfrac{\tau}{2}\big)}{\eta(\tau)}\quad \text{even}\; n$$ There is a modular equation between $u=G_{25n}$ and $v = G_{n}$ given by,

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

or expanded out, $$u^6 - 2 u^5 v^5 + 2 u v + v^6 = 0$$

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Your post seeks $u=G_{625}$ if we are given $v = G_{25} = \phi$ which is the golden ratio. The sextic factors nicely as

$$(u-1)\big(u^5 - 5 \phi^3 (u^4 + u^3 + u^2 + u) - \phi^6\big)=0$$

Thus, the problem is to solve that quintic. Define,

$$w_1=\big(6+4\phi+3\cdot 5^{1/4}\sqrt{\phi} \big)^{1/5}+\big(6+4\phi-3\cdot 5^{1/4}\sqrt{\phi} \big)^{1/5}$$ $$w_2 = \big(12\phi+4\cdot 5^{1/4}\sqrt{\phi^{-1}} \big)^{1/5} +\big(12\phi-4\cdot 5^{1/4}\sqrt{\phi^{-1}} \big)^{1/5} $$

then $u$'s radical expression is, $$u = G_{625}=\phi^3+\phi^2 (w_1+w_2) = 22.180323\dots$$