On a General Sextic Equation Solved by the Rogers Ramanujan Continued Fraction

using the j-invariant and the Rogers-Ramanujan continued fraction.

For |q| < 1, the Rogers Ramanujan continued fraction (RRCF) (see [2], [3], [4]) is defined as R(q) := q 1/5 1+

From the Theory of Elliptic functions the j-invariant (see [5], [8]) is

where

is the Dedekind’s eta function and

, τ = √ -r , r positive real.

We have also in the q-notation f (-q) := ∞ n=1

(1 -q n ).

(

In what follows we use the following known result (see Wolfram pages for ‘Rogers-Ramanujan Continued Fraction ’ and [17]): If R = R(e -2π √ r ), then:

From ( [3], [4]) we have 1 R 5 (q) -11 -R 5 (q) = f 6 (-q) qf 6 (-q 5 ) (

The general hypergeometric function is defined as The standard definition of the elliptic integral of the first kind (see [7], [8], [15]) is:

In the notation of Mathematica we have

The elliptic singular modulus k = k r is defined to be the solution of the equation:

In Mathematica’s notation

The complementary modulus is given by k ′ r = 1 -k 2 r . (For evaluations of k r see [5], [15], [16]). Also we call w r := √ k r k 25r noting that if one knows w = w r then (see [2]), knows k r and k 25r .

2 Theorems Proposition 1. (see [2]) If q = e -π √ r and r real positive then we define

then

Theorem 1.

Let a, b, C 1 be constants. One can solve the equation

finding r > 0 such that

Then (15) have solution

Proof.

For to solve the equation ( 15) find r such that

Consider also the transformation of the constants

and

, with inverse

.

where

Relation ( 19) is equivalent to equation ( 6), in view of (7). Hence from Proposition 1

and the proof is complete.

The j-invariant is connected with the singular modulus from the equation

We can solve (21) and express k r in radicals to an algebraic function of j r . The 5th degree modular equation which connects k 25r and k r is (see [3]):

We will evaluate the root of (1) first with parametrization and second with Rogers-Ramanujan continued fraction and the Elliptic-K function.

For this it have been showed (see [19]) that if

setting the following parametrization of w:

we get

(25) where M = 18 + L 64 + 3L From the above relations we get also

Hence we can consider the above equations as follows: Taking an arbitrary number L we construct an w. Now for this w we evaluate the two numbers k 25r and k r . Thus when we know the w, the k r and k 25r are given from (24),(25),(26).

The result is: We can set a number L and from this calculate the two inverse elliptic nome’s. But we don’t know the r. One can see (from the definition of k r ) that the r can evaluated from equation

Hence we define

However is very difficult to evaluate the r in a closed form, such as roots of polynomials or else when a number x is given. Some numerical evaluations indicate us that even if x are algebraic numbers, (not trivial as with k (-1) 2 -1/2 = 1 or the cases x = k r , r = 1, 2, 3, . . .) the r are not rational and may even not algebraics.

We know that (see [2]):

where x L = k r is the singular modulus which corresponds to some L.

The procedure is to select a number L and from (24),(25) evaluate w L , x L and

The solution X = X(L) of ( 1) is (29) and for this L holds

(33)

Hence we get the next:

One can find parametric solutions of (1) if for a given L construct the w L , x L and the complementary w ′ L (these values are given from ( 23),( 24),( 25),(31)). Also

The solution X = X L is given from

Note.

i) The above solution (37) works for parametric solutions (setting a L), as also for solutions which we know r, k r and k 25r . (For a related method on solving the quintic see Wolfram pages ‘Quintic Equation’) ii) In [16] it have been shown that when one knows for some r 0 the k r0 and k r0/25 then can evaluate any k 25 n r0 in radicals closed form for all n positive integers. But in general the values k r and k 25r can given from tables or with a simple PC (see [4],[5],[13], [15], [17]).

From the analysis in [2], the solution X of (1) can reduced with inverse functions as follows: Consider the function

The equation U (x) = t have known solution with respect to x, which we will call x = U (-1) (t). Hence

or

The function k (-1) (x) is that of (28).

The equation ( 1) have solution

where

Notes.

1. Observe here that we don’t need the value of w and the class invariant j.

2. From [10] we have

Hence the solution can expressed also in theta functions. That is if α = k r , r = 1, 2, 3, . . . then the solution of (1) reduced to that of evaluation of Rogers-Ramanujan continued fraction R(q) with q = e -π √ r . In view of [10] we have

The equation

hence a solution is

where

For this r the X is a solution.

Continuing one can set to

any value X = X 0 and C 1 = 1 then evaluate

equation ( 42) holds always and we get that

The result is the following parametrized evaluation of the Rogers-Ramanujan continued fraction Theorem 4.

and

Corollary.

where A, B, C, D rationals Theorem 5. If for a certain r > 0 we know the value of R(e -π √

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