Entry 97

Level 5 In Entry 96, we asserted that \(a=q^{11/60}H(q)\) and \(b= q^{-1/60}\,G(q)\) are radicals for appropriate \(\tau\). Proof: Given the j-function \(j(q)\) and let \(R(q)=r=a/b\)

$$\begin{align}a &= q^{11/60}H(q) = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{11/20}}{j(q)^{11/60}\,(r^{10}+11r^5-1)}\\ b &= q^{-1/60}G(q) =  \frac{j(q)^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{1/20}}\end{align}$$ Getting their ratio \(a/b\), then it simplifies to

$$r = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{3/5}}{j(q)^{1/5}\,(r^{10}+11r^5-1)}$$ Raising to the fifth power and moving \(j(q)\) to the LHS 

$$j(q)  = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{3}}{r^5(r^{10}+11r^5-1)^5}$$ and we get the icosahedral equation for the j-function in terms of the Rogers-Ramanujan continued fraction \(R(q)=r\). Thus if \(j(q)\) and \(R(q)\) are radicals, then so are \((a, b)\) while the integer \(60\) of \(b= q^{-1/60}G(q)\) reflects the order \(60\) of the icosahedral group.