## Hypergeometric formulas for Ramanujan’s continued fractions 2

(continued from yesterday’s post)

III. Icosahedral group

Given the Rogers-Ramanujan identities (see also here),

\begin{aligned} G(q) &= \sum_{n=0}^\infty \frac{q^{n^2}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\H(q) &= \sum_{n=0}^\infty \frac{q^{n^2+n}}{(q;q)_n} = \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}\end{aligned}

I observed that,

\begin{aligned}&q^{-1/60}G(q) = j^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j}\big) = (j-1728)^{1/60}\,_2F_1\big(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{1728-j}\big)\\[2.5mm]&q^{11/60}H(q) = j^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j}\big) = (j-1728)^{-11/60}\,_2F_1\big(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{1728-j}\big)\end{aligned}

where, as in the previous post, $j=j(\tau)$ is the j-function, $q = e^{2\pi i \tau} = \exp(2\pi i \tau)$, $\tau = \sqrt{-N}$, and $N>1$.  Since it is known that,

\begin{aligned}&r(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \cfrac{q^3}{1 + \ddots}}}} = \frac{q^{11/60}H(q)}{q^{-1/60}G(q)} = \frac{q^{11/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})}}{q^{-1/60}\prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(q^{5n-4})}}\end{aligned}

this implies that,

\begin{aligned}r(q) &=\frac{j^{-11/60}\,_2F_1\big(\frac{31}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{j}\big) }{j^{1/60}\,_2F_1\big(\frac{19}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{j}\big)}\\[3mm]&=\frac{(j-1728)^{-11/60}\,_2F_1\big(\frac{41}{60},\frac{11}{60},\frac{6}{5},\frac{1728}{1728-j}\big) }{(j-1728)^{1/60}\,_2F_1\big(\frac{29}{60},\frac{-1}{60},\frac{4}{5},\frac{1728}{1728-j}\big)}\end{aligned}

Example. Let $\tau = \sqrt{-4}$, hence $j = j(\sqrt{-4}) = 66^3$. Then,

$1/r(q) -r(q) = \left(\frac{1+\sqrt{5}}{2}\right)^4+\left(\frac{1+\sqrt{5}}{2}\right)5^{3/4} = 12.2643\dots$

Furthermore, since Ramanujan established that,

$G(q^{11})H(q)-q^2G(q)H(q^{11}) = 1$

if we define the two functions,

\begin{aligned}U(\tau) &= \big(j(\tau)\big)^{1/60}\,_2F_1\big(\tfrac{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j(\tau)}\big)\\V(\tau) &= \big(j(\tau)\big)^{-11/60}\,_2F_1\big(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j(\tau)}\big) \end{aligned}

then the counterpart hypergeometric identity is also beautifully simple and given by,

$U(11\tau)V(\tau)-U(\tau)V(11\tau)=1$

In the next post, we will use one of the hypergeometric formulas to solve the general quintic.

Advertisements

### 2 responses to this post.

1. Posted by Nikos Bagis on May 4, 2013 at 3:15 am

This is extremely remarkable formulation of the Rogers-Ramanujan Continued fraction.
How you get it?
I personally searching such formulas for years. Can you find $R(q)^{-5}-11-R(q)^5$ in hypergeometric functions? This may lead to the solution of a sextic equation
$b^2/(20a)+bx+ax^2=c x^{5/3}$. I want to ask you, can help me find the reduced form of the above equation $x^6+ax^2+bx+c=0$?

Reply

2. You can derive that from Raimundas Vidunas’ 2008 paper, “Transformations of Algebraic Gauss hypergeometric functions” (p.17 and 20), though he does not connect it to the Rogers-Ramanujan cfrac.

The reduced form of the sextic you mentioned does not have a concise form. You have to go through TWO Tschirnhausen transformations, a quadratic and quartic one. I described an easy approach in “A New Way to Derive the Bring-Jerrard Quintic”, the 18th file in http://sites.google.com/site/tpiezas/ramanujan. (While I discuss there the quintic, I trust you can extrapolate it for the sextic.)

Reply