The complete evaluation of Rogers Ramanujan and other continued fractions with elliptic functions

For |q| < 1, the Rogers Ramanujan continued fraction (RRCF) is defined as R(q) := q 1/5 1+

We also define (a; q) n := n-1 k=0

(1 -aq k ) (2)

Φ(-q) := ∞ n=1

(1 + q n ) = (-q; q) ∞ (4)

and also hold the following relations of Ramanujan 1 R(q) -1 -R(q) = f (-q 1/5 ) q 1/5 f (-q 5 ) (5) 1 R 5 (q) -11 -R 5 (q) = f 6 (-q) qf 6 (-q 5 ) (

From the Theory of Elliptic Functions we have: Let

It is known that the inverse elliptic nome k = k r , k ′2 r = 1 -k 2 r is the solution of

We also denote

In what it follows we assume that r ∈ R * + . When r is rational then k r is algebraic. k r = 8q 1/2 Φ(-q) 12 1 + 1 + 64qΦ(-q) 24 (9)

We can write the functions f and Φ using elliptic functions. It holds Φ(-q) = 2 5/6 q -1/24 (k r ) 1/12 (k ′ r ) 1/6 (10) f (-q) 8 = 2 8/3 π 4 q -1/3 (k r ) 2/3 (k ′ r ) 8/3 K(k r ) 4 (11) also holds f (-q 2 ) 6 = 2k r k ′ r K(k r ) 3 π 3 q 1/2 (12)

From [19] it is known that R ′ (q) = 1/5q -5/6 f (-q) 4 R(q) 6 R(q) -5 -11 -R(q) 5 (13)

2 Evaluations of Rogers Ramanujan Continued Fraction Theorem 2.1 If q = e -π √ r and

Then

Where M 5 (r) is root of: (5x -1)

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 ( 6) and (11) and the above consequence of the Theory of Elliptic Functions, we get:

Solving with respect to R(q) we get the result.

The relation between k 25r and k r is

We will try to evaluate k 25r . For this we set

Then setting directly to (18) the following parametrization of w (see also [8] pg.280):

we get

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

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 by (20) and (21).

The result is: We can set a number L and from this calculate the two inverse elliptic nome’s, or equivalently, find easy solutions of (18). But we don’t know the r. One can see (from the definition of k r ) that the r can be evaluated from equation

or

However there is no way to evaluate the r in a closed form, such as roots of polynomials, or else. Some numerical evaluations as we will see, indicate as that even k r are algebraic numbers, the r are not rational. where the r is given by

Now we can see (The results are known in the Theory of Elliptic Functions) how we can found evaluations of R(q) when r is given and k r is known:

from the relation between M and L we get

Hence from ( 22)

where y = M/L. Hence (k = k r ):

and

But

where the equation for finding

We give the complete polynomial equation of p arising from (35):

It is evident that the Rogers Ramanujan Continued Fraction is root of polynomial equation with coeficients depending by k r . From (33) we have

where p is root of (37). Using Mathematica we get the following simplification formula for

and

r )w 5 + w 6 = 0 (39b) Once we know k r we can calculate w from the above equation and hence the k 25r . Hence the problem reduces to solve 6-th degree equations. The first is (17) the second is (39b) and for (RRCF) (39a).

and M 5 (r) satisfies (5x -1) 5 (1 -x) = 256(k r ) 2 (k ′ r ) 2 x. After elementary algebraic calculations we get the result. ii) From (35) we get:

and also

Hence solving the system we obtain the 6-th degree equation (39b).

The solution of (39b) with respect to k r when we know w is

Where

Proof.

Combining ( 11) and ( 13) and Theorem 2.1 we get the proof.

Evaluations.

4

Sumarizing our results we can say that: 1) Theorem 2.1 is quite usefull for evaluating R(q) when we know k r and k 25r . But this it whas known allrady to Ramanujan by using the function X(-q) = (-q; q 2 ) ∞ , (see [8]).

2. Theorem 2.2 is more kind of a Lemma rather a Theorem and it might help for further research.

3. Theorem 2.3 is a proof that the Rogers Ramanujan continued fraction is a root of a polynomial equation with coeficients the k r where r positive real. The polynomial equation (39a), is a version of icosahedral equation for the (RRCF) (see [20]). 4) Theorem 2.4 is a consequence of a Ramanujan integral first proved by Andrews (see [5]) and it is usefull for evaluations of R ′ (e -π √ r ), r ∈ Q. The above theorems can used to derive also modular equations of R(q), from the modular equations of k r . More precisely we can guess an equality with the help of a methematical pacage (for example in Mathematica there exist the command ‘recognize’) and then proceed to proof using Theorems which we presented in this article. We follow this prosedure with other continued fractions (the Rogers Ramanujan is litle dificult), such as the cubic or Ramanujan-Gollnitz-Gordon. These two last continued fractions are more easy to handle. The elliptic function theory and the sigular moduli k r will exctrac and give us severa

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