(*continued from yesterday’s post*)

**III. Icosahedral group**

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

I observed that,

where, as in the previous post, is the j-function, , , and . Since it is known that,

this implies that,

Example. Let , hence . Then,

Furthermore, since Ramanujan established that,

if we define the two functions,

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

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

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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 $x^6+ax^2+bx+c=0$?

Posted by tpiezas on May 5, 2013 at 4:25 am

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.)