1. Introductory Definitions and Formulas
For |
q| < 1, the Rogers-Ramanujan continued fraction (RRCF) (see [
1]) is defined as
(1.1)
We also define
(1.2)
Ramanujan give the following relations which are very useful:
(1.3)
(1.4)
From the theory of elliptic functions (see [
1–
3]),
(1.5)
is the elliptic integral of the first kind. It is known that the inverse elliptic nome
k =
kr,
is the solution of the equation
(1.6)
where
. When
r is rational then the
kr are algebraic numbers.
We can also write the function
f using elliptic functions. It holds (see [
3])
(1.7)
and also holds
(1.8)
From [
4] it is known that
(1.9)
Consider now for every 0 <
x < 1 the equation
(1.10)
which has solution
(1.11)
Hence for example
(1.12)
With the help of
k(−1) function we evaluate the Rogers Ramanujan continued fraction.
2. Propositions
The relation between
k25r and
kr is (see [
1] page 280)
(2.1)
For to solve (
2.1) we give the following.
Proposition 2.1. The solution of the equation
(2.2)
when one knows
w is given by
(2.3)
where
(2.4)
(2.5)
If it happens that
x =
kr and
y =
k25r, then
r =
k(−1)(
x) and
w2 =
k25rkr,
.
Proof. The relation (2.3) can be found using Mathematica. See also [5].
Proposition 2.2. If and
(2.6)
then
(2.7)
where
M5(
r) is root of
.
Proof. Suppose that N = n2μ, where n is positive integer and μ is positive real then it holds that
(2.8)
where
K[
μ] =
K(
kμ)
The following formula for M5(r) is known:
(2.9)
Thus, if we use (
1.4) and (
1.7) and the above consequence of the theory of elliptic functions, we get:
(2.10)
See also [
4,
5].
3. The Main Theorem
From Proposition
2.2 and relation
w2 =
k25rkr we get
(3.1)
Combining (
2.2) and (
3.1), we get
(3.2)
Solving with respect to
arM5(
r)
3, we get
(3.3)
Also we have
(3.4)
The above equalities follow from [
1] page 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
(3.5)
Thus for a given
w we find
L and
M from (
2.4) and (
2.5). Setting the values of
M,
L,
w in (
2.3) we get the values of
x and
y (see Proposition
2.1). Hence from (
3.5) if we find
k(−1)(
x) =
r we know
. The clearer result is as follows.
Main Theorem 3. When w is a given real number, one can find x from (2.3). Then for the Rogers-Ramanujan continued fraction the following holds:
(3.6)
Theorem 3.1. (the first derivative). One has
(3.7)
Proof. Combining (1.7) and (1.9) and Proposition 2.2 we get the proof.
We will see now how the function k(−1)(x) plays the same role in other continued fractions. Here we consider also the Ramanujan′s Cubic fraction (see [5]), which is completely solvable using kr.
Define the function
(3.8)
Set for a given 0 <
w3 < 1
(3.9)
Then as in Main Theorem, for the Cubic continued fraction
V(
q), the following holds (see [
5]):
(3.10)
Observe here that again we only have to know
k(−1)(
x).
If
x =
kr, for a certain
r, then
(3.11)
and if we set
(3.12)
then the follwing holds:
(3.13)
which is solvable always in radicals quartic equation. When we know
w3 we can find
kr =
x from
x =
G(
w3) and hence
t.
The inverse also holds: if we know
t =
V(
q) we can find
T and hence
kr =
x. The
w3 can be found by the degree 3 modular equation which is always solvable in radicals:
(3.14)
Let now
(3.15)
if
(3.16)
then
(3.17)
or
(3.18)
or
(3.19)
Setting now values into (
3.19) we get values for
k(−1)(·). The function
Vi(·) is an algebraic function.
4. Evaluations of the Rogers-Ramanujan Continued Fraction
Note that if x = kr, , then we have the classical evaluations with kr and k25r.
Evaluations (1) We have
(4.1)
(2) Assume that
, hence
. From (
2.5) which for this
x can be solved in radicals, with respect to
w, we find
(4.2)
Hence from
(4.3)
we get
(4.4)
Setting these values to (
3.6) we get the value of
ar and then
R(
q) in radicals. The result is
(4.5)
(3) Set
w = 1/64 and
a = 1359863889,
b = 36855, then
(4.6)
(4) For
(4.7)
we get
(4.8)
Hence
(4.9)
(5) Set
, then from
(4.10)
We can evaluate all
(4.11)
where
(4.12)
hence
(4.13)
An example for
r0 = 2 is
(4.14)
where
ρ3 can be evaluated in radicals but for simplicity we give the polynomial form
(4.15)
Then, respectively, we get the values
(4.16)
Hence
(4.17)
Also it holds that
(4.18)
where
xn =
Vi(
b0(
n)) = known. The
wn are given from (
2.2) (in this case we do not find a way to evaluate
wn in radicals).
Theorem 4.1. Set
(4.19)
then
(4.20)
where
(4.21)
The
A(
a) is a known algebraic function of
a and can calculated from the Main Theorem.
