Parametric Evaluations of the Rogers Ramanujan Continued Fraction

In this article with the help of the inverse function of the singular moduli we evaluate the Rogers Ranmanujan continued fraction and his first derivative.


Introductory Definitions and Formulas
For |q| < 1, the Rogers Ramanujan continued fraction (RRCF) (see [6]) is defined as We also define (a; q) n := (1 − q n ) = (q; q) ∞ Ramanujan give the following relations which are very useful: 1 R 5 (q) − 11 − R 5 (q) = f 6 (−q) qf 6 (−q 5 ) is the Elliptic integral of the first kind. It is known that the inverse elliptic nome k = k r , k ′2 r = 1 − k 2 r is the solution of the equation where r ∈ R * + . When r is rational then the k r are algebraic numbers. We can also write the function f using elliptic functions. It holds (see [10]): also holds Consider now for every 0 < x < 1 the equation Hence for example With the help of k (−1) function we evaluate the Rogers Ramanujan continued fraction.

Propositions
The relation between k 25r and k r is (see [6] pg. 280): For to solve equation (12) we give the following The solution of the equation when we know the w is given by where w = L(18 + L) 6(64 + 3L) < 1 and If happens x = k r and y = k 25r , then r = k (−1) (x) and The relation (14) can be found using Mathematica. See also [6]. Then

Proof.
Suppose that N = n 2 µ, where n is positive integer and µ is positive real then it holds that Where The following formula for M 5 (r) is known Thus if we use (5) and (8) and the above consequence of the Theory of Elliptic Functions, we get: See also [4], [5].

The Main Theorem
From Proposition 2 and relation w 2 = k 25r k r we get Combining (13) and (21), we get: Also we have The above equalities follow from ([6] pg. 280 Entry 13-xii) and the definition of w. Note that m is the multiplier. Hence for given 0 < w < 1 we find L ∈ R and we get the following parametric evaluation for the Rogers Ramanujan continued fraction Thus for a given w we find L and M from (15) and (16). Setting the values of M , L, w in (14) we get the values of x and y (see Proposition 1). Hence from (24) if we find k (−1) (x) = r we know R(e −π √ r ). The clearer result is: When w is a given real number, we can find x from equation (14). Then for the Rogers Ramanujan continued fraction holds Note. In the case of x = k r , then k (−1) (x) = r and we have the clasical evaluation with k 25r (see [12]).
We see how the function k (−1) (x) plays the same role in other continued fractions. Here we consider also the Ramanujan's Cubic fraction (see [4]), which is completely solvable using k r . Define the function: Set for a given 0 < w 3 < 1 Then as in Main Theorem, for the Cubic continued fraction V (q), holds (see [4]): Observe here that again we only have to know k (−1) (x).
If happens x = k r , for a certain r, then and if we set then holds which is solvable always in radicals quartic equation. When we know w 3 we can find k r = x from x = G(w 3 ) and hence t.
The inverse also holds: If we know t = V (q) we can find T and hence k r = x.
The w 3 can be find by the degree 3 modular equation which is always solvable in radicals: and Setting now values into (36) we get values for k (−1) (.). The function V i (.) is an algebraic function.

Some Evaluations of the Rogers Ramanujan Continued Fraction
Note that if x = k r , r ∈ Q then we have the classical evaluations with k r and k 25r .
2) Assume that x = 1 √ 2 , hence k (−1) 1 √ 2 = 1. From (16) which for this x can be solved in radicals, with respect to w, we find Hence from Setting these values to (25) we get the value of a r and then R(q) in radicals. The result is 3)Set w = 1/64 and a = 1359863889, b = 36855, then We can evaluate all Where ρ 3 can be evaluated in radicals but for simplicity we give the polynomial form.

Then respectively we get the values
. . .
and R e −π √ r0/3 n −5 − 11 − R e −π √ r0/3 n 5 = = 16x 6 n − 26x 4 n − w n x 3 n + 10x 2 n + w n x n x 4 n − 6x 3 n w n − 20x n w 3 n + 15w 2 n x 2 n − 6x n w n + 15w 4 n + w 2 Where x n = V i (b 0 (n)) = known. The w n are given from (13) (in this case we don't find a way to evaluate w n in radicals, but as a solution of (13)). 6) Set now and if we manage to write k r in the form g(a r ) for a certain a r i.e. V i (V (e −π √ r )) = k r = g(a r ), then R e −π √ r −5 − 11 − R e −π √ r 5 = A(a r ) = A g (−1) (k r )