# Parametric Evaluations of the Rogers-Ramanujan Continued Fraction

Authors: Nikos Bagis
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

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


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## Body

## 1. Introductory Definitions and Formulas

For|q|<1, the Rogers-Ramanujan continued fraction (RRCF) (see [1]) is defined as (1.1)R(q):=q1/51+q11+q21+q31+⋯. We also define (1.2)(a;q)n:=∏k=0n-1(1-aqk),f(-q):=∏n=1∞(1-qn)=(q;q)∞.
Ramanujan give the following relations which are very useful: (1.3)1R(q)-1-R(q)=f(-q1/5)q1/5f(-q5),(1.4)1R5(q)-11-R5(q)=f6(-q)qf6(-q5).
From the theory of elliptic functions (see [1–3]), (1.5)K(x)=∫0π/211-x2sin⁡(t)2dt,
is the elliptic integral of the first kind. It is known that the inverse elliptic nome k=kr, k′r2=1-kr2 is the solution of the equation (1.6)K(kr′)K(k)=r,
where r∈R+*. When r is rational then the kr are algebraic numbers.We can also write the functionf using elliptic functions. It holds (see [3]) (1.7)f(-q)8=28/3π4q-1/3(kr)2/3(kr′)8/3K(kr)4,
and also holds (1.8)f(-q2)6=2krkr′K(kr)3π3q1/2.
From [4] it is known that (1.9)R′(q)=15q-5/6f(-q)4R(q)R(q)-5-11-R(q)56.
Consider now for every 0<x<1 the equation (1.10)x=kr,
which has solution (1.11)r=k(-1)(x).
Hence for example (1.12)k(-1)(12)=1.
With the help of k(-1) function we evaluate the Rogers Ramanujan continued fraction.


## 2. Propositions

The relation betweenk25r and kr is (see [1] page 280)(2.1)krk25r+kr′k25r′+2⋅41/3(krk25rkr′k25r′)1/3=1.
For to solve (2.1) we give the following.Proposition 2.1.
The solution of the equation(2.2)x6+x3(-16+10x2)w+15x4w2-20x3w3+15x2w4+x(10-16x2)w5+w6=0.
when one knows w is given by
(2.3)y1/2w1/2=w1/2x1/2=124+23(L1/6M1/6-4M1/6L1/6)2+1223(L1/6M1/6-4M1/6L1/6),
where
(2.4)w=L(18+L)6(64+3L)<1,(2.5)M=18+L64+3L.
If it happens that x=kr and y=k25r, then r=k(-1)(x) and w2=k25rkr, (w′)2=k25r′kr′.Proof.
The relation (2.3) can be found using Mathematica. See also [5].Proposition 2.2.
Ifq=e-πr and
(2.6)a=ar=(kr′k25r′)2krk25rM5(r)-3,
then
(2.7)ar=R(q)-5-11-R5(q),
where M5(r) is root of (5x-1)5(1-x)=256(kr)2(kr′)2x.Proof.
Suppose thatN=n2μ, where n is positive integer and μ is positive real then it holds that
(2.8)K[n2μ]=Mn(μ)K[μ],
where K[μ]=K(kμ)
The following formula forM5(r) is known:
(2.9)(5M5(r)-1)5(1-M5(r))=256(kr)2(kr′)2M5(r).
Thus, if we use (1.4) and (1.7) and the above consequence of the theory of elliptic functions, we get:
(2.10)R-5(q)-11-R5(q)=f6(-q)qf6(-q5)=a=ar=(kr′k25r′)2krk25rM5(r)-3.
See also [4, 5].


## 3. The Main Theorem

From Proposition2.2 and relation w2=k25rkr we get (3.1)w5-kr2w=kr3(kr2-1)arM5(r)3.
Combining (2.2) and (3.1), we get (3.2)[-10kr4+26kr6+arM5(r)3kr6-16kr8]+[-kr3-6arM5(r)3kr3+kr5-6arM5(r)3kr5]w+[arM5(r)3kr2+15arM5(r)kr4]w2-20arM5(r)3kr3w3+15arM5(r)3kr2w4=0.
Solving with respect to arM5(r)3, we get (3.3)arM5(r)3=16kr6-26kr4-wkr3+10kr2+wkrkr4-6kr3w-20kr3w+15w2kr2-6krw+15w4+w2.
Also we have (3.4)K(k25r)K(kr)=M5(r)=1m=(k25rkr+k25r′kr′-k25rk25r′krkr′)-1=(wkr+w′kr′-ww′krkr′)-1.
The above equalities follow from [1] page 280 Entry 13-xii and the definition of w. Note that m is the multiplier.Hence for given0<w<1, we find L∈R and we get the following parametric evaluation for the Rogers Ramanujan continued fraction (3.5)R(e-πr(L))-5-11-R(e-πr(L))5=ar=16kr6-26kr4-wkr3+10kr2+wkrkr4-6kr3w-20krw3+15w2kr2-6krw+15w4+w2(wkr+w′kr′-ww′krkr′)3.
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 R(e-πr). The clearer result is as follows.Main Theorem 3.
Whenw is a given real number, one can find x from (2.3). Then for the Rogers-Ramanujan continued fraction the following holds:
(3.6)R(e-πk(-1)(x))-5-11-R(e-πk(-1)(x))5=ar=16x6-26x4-wx3+10x2+wxx4-6x3w-20xw3+15w2x2-6xw+15w4+w2×(wx+w′1-x2-ww′x1-x2)3.Theorem 3.1.
(the first derivative). One has(3.7)R′(e-πk(-1)(x))=24/3x1/2(1-x2)5w1/6w′2/3(wx+w′1-x2-ww′x1-x2)1/2×R(e-πk(-1)(x))K2(x)eπk(-1)(x)π2.Proof.
Combining (1.7) and (1.9) and Proposition 2.2 we get the proof.We will see now how the functionk(-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)G(x)=x2x-3x+2x3/2-2x1-3x+4x-3x3/2+x2.
Set for a given 0<w3<1(3.9)x=G(w3).
Then as in Main Theorem, for the Cubic continued fraction V(q), the following holds (see [5]): (3.10)t=V(e-πk(-1)(x))=(1-x2)1/3w31/421/3x1/3(1-w3).
Observe here that again we only have to know k(-1)(x).Ifx=kr, for a certain r, then (3.11)k9r=w3kr,
and if we set (3.12)T=1-8V(q)3,
then the follwing holds: (3.13)(kr)2=x2=(1-T)(3+T)3(1+T)(3-T)3,
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 knowt=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)krkr′+k9rk9r′=1.
Let now (3.15)V(q)=z⟺q=V(-1)(z),
if (3.16)Vi(t):=1-1-8t31+1-8t3(3+1-8t33-1-8t3)3,
then (3.17)Vi(V(e-πx))=kx,
or (3.18)V(e-πr)=Vi(-1)(kr),V(e-πk(-1)(x))=Vi(-1)(x),
or (3.19)e-πk(-1)(x)=V(-1)(Vi(-1)(x))=(Vi∘V)(-1)(x),k(-1)(Vi(V(q)))=1π2log⁡(q)2=r.
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 ifx=kr, r∈Q+*, then we have the classical evaluations with kr and k25r.Evaluations
(1) We have(4.1)R(e-2π)=-12-52+5+52,R′(e-2π)=825(9+55-250+225)e2ππ3Γ(54)4.
(2) Assume that x=1/2, hence k(-1)(1/2)=1. From (2.5) which for this x can be solved in radicals, with respect to w, we find
(4.2)w=24(5-1)-1275-15.
Hence from
(4.3)w′=1-w4x21-x2,
we get
(4.4)w′=(1+21-30+145-9-150+7052)1/4.
Setting these values to (3.6) we get the value of ar and then R(q) in radicals. The result is
(4.5)R(e-π)-5-11-R(e-π)5=-18(3+5--30+145)×[(22)221-5+-30+145+23/8(-3+5--30+145)×(1+21-30+145-9-150+705)1/4(22)22]×[-1574+7045-655-30+145+293-150+705]-1.
(3) Set w=1/64 and a=1359863889, b=36855, then
(4.6)x=9(a+b)5/6[49152⋅61/3(a+b)1/6-960(a+b)5/6+2⋅62/3(a+b)3/2+384⋅22/3⋅31/6√[-64(a+b)+31/6⋅22/3453287963⋅(b+a)2/3+8192⋅61/3(a+b)1/3+12285⋅62/3(a+b)2/3]-2⋅65/6(a+b)1/6√[4096.21/3⋅35/6453287963(a+b)1/3+36855⋅22/331/6453287963(a+b)2/3+150958080⋅61/3(a+b)1/3453025819⋅62/3(a+b)2/3-192(453025819+12285a)4096.21/3]]-1.
(4) For
(4.7)w=277108+13385108,
we get
(4.8)x=277/12+(13385)/124+7.
Hence
(4.9)R(exp⁡[-π⋅k(-1)(277/12+(13385)/124+7)1/2])=(--807118+10755518+1185(25740148-347053055))1/5.
(5) Set q=e-πr0, then from
(4.10)V(e-πr0)=Vi(-1)(kr0)=V0,V(q1/3)=V(q)1-V(q)+V(q)21+2V(q)+4V(q)23.
We can evaluate all
(4.11)V(q0(n))=b0(n)=Algebraicfunctionofr0,
where
(4.12)q0(n)=e-πr0/3n,Vi(V(q0(n)))=Vi(b0(n))=kr0/9n,
hence
(4.13)k(-1)(Vi(b0(n)))=r09n.
An example for r0=2 is
(4.14)V(e-π2)=-1+32,V(e-π2/3)=121/3(-1+32)1/3,V(e-π2/9)=ρ31/3,
where ρ3 can be evaluated in radicals but for simplicity we give the polynomial form
(4.15)-1-72x-6408x2+50048x3+51264x4-4608x5+512x6=0⋯.
Then, respectively, we get the values
(4.16)k(-1)(-49+352+43(99-702))=29,k(-1)(Vi(ρ31/3))=281,⋯.
Hence
(4.17)k(-1)(Vi(b0(n)))=r09n.
Also it holds that
(4.18)R(e-πr0/3n)-5-11-R(e-πr0/3n)5=16xn6-26xn4-wnxn3+10xn2+wnxnxn4-6xn3wn-20xnwn3+15wn2xn2-6xnwn+15wn4+wn2×(wnxn+wn′1-xn2-wnwn′xn1-xn2)3,
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)w=108+144a4+24a8+a12--11664+(108+144a4+24a8+a12)263,
then
(4.20)x=108+a4(12+a4)2-a4(6+a4)(12+a4)2(36+18a4+a8)23(3+a4-a26+a4),R(e-πk(-1)(x))-5-11-R(e-πk(-1)(x))5=A(a),
where
(4.21)w′=1-w4x21-x2.
The A(a) is a known algebraic function of a and can calculated from the Main Theorem.



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