This article discusses the theoretical background for generating Ramanujan-type formulas for and constructs series for and . We also study the elliptic alpha function, whose values are useful for such evaluations.
In Mathematica, these are and .
We also have
The elliptic singular moduli is defined to be the solution of the equation
In Mathematica, is computed using .
We need the following relation satisfied by the elliptic alpha function (see ):
Our method requires finding derivatives of powers of the elliptic integrals and that can always be expressed in terms of , , and . This article uses Mathematica to carry out these evaluations.
where is a root of the polynomial equation
In the next section, we review and extend the method for constructing a series for based on . These Ramanujan-type formulas for , are presented here for the first time. The only formulas that were previously known are of orders 1, 2, and 3 (, ). There are few general formulas of order 2 and only one for order 3, due to B. Gourevitch (see references , , , , , :
In the last section we prove a formula for the evaluation of in terms of .
The General Method and the Construction of Formulas for and
We have (see ):
This is the Mathematica definition.
Define , , such that
It turns out that
Here are the Mathematica definitions for for .
Consider the following equation for the function :
Set ; then and , for suitable values of , is a function of and , so is an algebraic number when . The and can be evaluated from (13). Higher values of and give more accurate and faster formulas for and .
The general formula produced by our method for is
This computes the polynomial in the variable in the sum (13).
To find the , the function Arules replaces by and by and sets all the Taylor expansion coefficients with respect to to 0.
Choose M large enough to get a solution for all the for . (Here and .)
Now that we have the A[i], this computes the sum on the left-hand side of (13).
This computes the right-hand side of (13).
We verify this numerically.
Example 2. Here is another example for that we verify numerically.
The coefficients of and the parameters for the formula are obtained using the same method as for . (The same can be done as well for , of course.) Higher values of and give more accurate and faster formulas for and .
For we get
This calculates the .
Example 3. For ,
Example 4. For , we have and ; then
We verify this numerically.
Example 5. For , we have and ; then
Evaluating the Elliptic Alpha Function
It is clear from the results in the previous section that getting rapidly convergent series for and its even powers requires values of the alpha function for large , say (see , , ). In this section we address this problem.
From the duplication formula
equation (20) becomes
gives the following proposition.
Entry 4 of , p. 436 is
where and .
and hence the evaluation
But for the evaluation of the Rogers-Ramanujan continued fraction, from  we have
If and is a positive real, then
In some cases, the next formula from  is very useful:
Here the function is , where , , and are as defined in  and is the iterate of .
If we know and , we can evaluate from (31) and then we can evaluate .
The following conjecture is most compactly expressed in terms of the quantity
Numerical results calculated with Mathematica indicate that whenever , then .
For a given and , , or , if the smallest nested root of is , then we can evaluate the Rogers-Ramanujan continued fraction with integer parameters.
1. When ,
with , where , , are positive integers.
2. When ,
a) If , then
and where is the positive integer solution of . Hence and is a positive integer. The parameter is a positive rational and can be found directly from the numerical value of .
b) If , then
where we set . Then a starting point for the evaluation of the integers , is
the square of an integer.
3. When , then we can evaluate .
The degree of is 8 and the minimal polynomial of is of degree 4 or 8 and symmetric. Hence the minimal polynomial can be reduced to at most a fourth-degree polynomial and so it is solvable. With the help of step 2, we can evaluate .
where , , , , , are integers, and
Here are some values of that can found with the Mathematica built-in function Recognize or by solving Pell’s equation and applying the conjecture.
Example 6. If , from (54) we have , hence
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About the Author
Nikos D. Bagis is a mathematician with a PhD in Mathematical Informatics from Aristotle University of Thessaloniki.
N. D. Bagis
Stenimahou 5 Edessa Pellas