# Modular Identities and Explicit Evaluations of a Continued Fraction of Ramanujan

Authors: Nipen Saikia
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We study a new continued fraction of Ramanujan. We prove its modular identities and give some explicit evaluations.


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## Body

## 1. Introduction

Throughout the paper, we assume|q|<1. As usual, for positive integers n and any complex number a, we write
(1.1)(a)n:=(a;q)n:=∏j=0n-1‍(1-aqj),(a)∞:=(a;q)∞:=∏n=0∞‍(1-aqn).Ramanujan's general theta-functionf(a,b) is defined by
(1.2)f(a,b):=∑k=-∞∞‍ak(k+1)/2bk(k-1)/2,
where |ab|<1. After Ramanujan, we define
(1.3)ϕ(q):=f(q,q)=1+2∑k=1∞‍qk2=(-q;q2)∞(q2;q2)∞(q;q2)∞(-q2;q2)∞,(1.4)ψ(q):=f(q,q3)=∑k=0∞‍qk(k+1)/2=(q2;q2)∞(q;q2)∞,(1.5)f(-q):=f(-q,-q2)=∑k=0∞‍(-1)kqk(3k-1)/2+∑k=1∞‍(-1)kqk(3k+1)/2=(q;q)∞,(1.6)χ(q)=f(q)f(-q2)=(-q;q2)∞.
Ramanujan recorded many q-continued fractions and some of their explicit values in his second notebook [1] and in his lost notebook [2]. The following beautiful continued fraction identity was recorded by Ramanujan in his second notebook and can be found in [3, p. 11, Entry 11]:
(1.7)(-a)∞(b)∞-(a)∞(-b)∞(-a)∞(b)∞+(a)∞(-b)∞=a-b1-q+(a-bq)(aq-b)1-q3+q(a-bq2)(aq2-b)1-q5+⋯,
where either q, a, and b are complex numbers with |q|<1, or q, a, and b are complex numbers with a=bqm for some integer m. Several elegant q-continued fractions have representations as q-products and some of them can be expressed in terms of Ramanujan’s theta-functions. An account of this can be found in in Chapter 32 of Berndt's book [4] (also see [5]). The most famous one, of course, is the Rogers-Ramanujan continued fraction R(q) defined by
(1.8)R(q):=q1/51+q1+q21+q31+⋯.
The continued fraction R(q) has a very beautiful and extensive theory almost all of which was developed by Ramanujan. In particular, his lost notebook [2] contains several results on the Rogers-Ramanujan continued fraction. We refer to the paper by Berndt et al. [6], Kang [7, 8] for proofs of many of these theorems.In this paper, we examine another continued fractionT(q) of Ramanujan arising from (1.7) and is defined by
(1.9)T(q):=q1-q2+q41-q6+q81-q10+⋯.
Note that, replacing q by q2 and then setting a=q and b=0 in (1.7), we obtain (1.9).In Section2, we record some preliminary results. Section 3 is devoted to prove some modular identities for the continued fraction T(q). Finally, in Section 4, we give some explicit evaluations of T(q).We complete this introduction by defining Ramanujan’s modular equation from Berndt’s book [3]. The complete elliptic integral of the first kind K(k) is defined by
(1.10)K(k):=∫0π/2‍dϕ1-k2sin2ϕ=π22F1(12,12;1;k2),
where 0<k<1,2F1 denotes the ordinary or Gaussian hypergeometric function. The number k is called the modulus of K, and k′:=1-k2 is called the complementary modulus. Let K,K′,L, and L′ denote the complete elliptic integrals of the first kind associated with the moduli k, k′, l, and l′, respectively. Suppose that the equality
(1.11)nK′K=L′L
holds for some positive integer n. Then, a modular equation of degree n is a relation between the moduli k and l which is implied by (1.11). If we set
(1.12)q=exp⁡(-πK′K),q′=exp⁡(-πL′L),
we see that (1.11) is equivalent to the relation qn=q′. Thus, a modular equation can be viewed as an identity involving theta-functions at the arguments q and qn. Ramanujan recorded his modular equations in terms of α and β, where α=k2 and β=l2. We say that β has degree n over α. The multiplier m connecting α and β is defined by
(1.13)m=KL,
where zr=ϕ2(qr).


## 2. Preliminary Results

In this section, we record some results that will be used in the subsequent sections.Lemma 2.1 (see [3, p. 124, Entry 12(i) and (ii)]).
One has(2.1)f(q)=z12-1/6{α(1-α)}1/24q-1/24,f(-q)=z12-1/6(1-α)1/6α1/24q-1/24.Lemma 2.2 (see [3, p. 214, Entry 24(iii)]).
Ifβ has degree 2 over α, then
(2.2)mα-1+β=1,m2α-1+β=1.Lemma 2.3 (see [3, p. 230, Entry 5(ii)]).
Ifβ has degree 3 over α, then
(2.3)(αβ)1/4+{(1-α)(1-β)}1/4=1.Lemma 2.4 (see [3, p. 215, (24.22)]).
Ifβ has degree 4 over α, then
(2.4)β=(1-(1-α)1/41+(1-α)1/4)2.Lemma 2.5 (see [3, p. 280-281, Entry 13(v) and (vi)]).
Ifβ has degree 5 over α, then
(2.5)m=1+((1-β)5/(1-α))1/81+{(1-α)3(1-β)}1/8,5m=1-((1-α)5/(1-β))1/81-{(1-α)(1-β)3}1/8.Lemma 2.6 (see [3, p. 314, Entry 19(i)]).
Ifβ has degree 7 over α, then
(2.6)(αβ)1/8+{(1-α)(1-β)}1/8=1.


## 3. Modular Identitites forT(q)

In this section, we use Ramanujan's modular equations to prove certain modular identities forT(q).Theorem 3.1.
One has(3.1)T(q)=f(q)-f(-q)f(q)+f(-q).Proof.
Replacingq by q2 and the setting a=q and b=0 in (1.7) and simplifying, we obtain
(3.2)(-q;q2)∞-(q;q2)∞(-q;q2)∞+(q;q2)∞=q1-q2+q41-q6+q81-q10+⋯.
Employing (1.6) and (1.9) in (3.2) and simplifying, we complete the proof.Corollary 3.2.
One has(3.3)1+T(q)1-T(q)=f(q)f(-q).Proof.
Dividing numerator and denominator on right-hand side of the identity in Theorem3.1 by f(-q) and simplifying, we complete the proof.Theorem 3.3.
One has(3.4)(i)α=1-(1-T(q)1+T(q))8,(ii)β=1-(1-T(qn)1+T(qn))8,
where β has degree n over α.Proof.
We employ Lemma2.1 in Corollary 3.2 to complete the proof.Theorem 3.4.
Letu=T(q) and v=T(-q). Then,
(3.5)u+v=0.Proof.
Replacingq by -q in Corollary 3.2, we obtain
(3.6)1+T(-q)1-T(-q)=f(-q)f(q).
Now, eliminating f(q)/f(-q) between (3.6) and Corollary 3.2 and simplifying, we complete the proof.Theorem 3.5.
Letu=T(q) and v=T(q2). Then,
(3.7)u2-v-2u2v-u4v+6u2v2-v3-2u2v3-u4v3+u2v4=0.Proof.
Eliminatingm in (2.2) and then simplifying, we deduce that
(3.8)(1+β+(β-1)1-α)2-4β=0.
From Theorem 3.3(i), we have
(3.9)1-α=(1-T(q)1+T(q))4.
Now, employing Theorem 3.3(ii) with n=2 and (3.9) in (3.8) and factorizing using Mathematica, we obtain
(3.10)(-1+v)8(u2-v-2u2v-u4v+6u2v2-v3-2u2v3-u4v3+u2v4)×(1+2u2+u4-16u2v+6v2+12u2v2+6u4v2-16u2v3+v4+2u2v4+u4v4)=0.
It can be seen that the first and the last factors in (3.10) do not vanish for |q|→0. So, by identity theorem, we have
(3.11)u2-v-2u2v-u4v+6u2v2-v3-2u2v3-u4v3+u2v4=0.Theorem 3.6.
Letu=T(q) and v=T(q3). Then,
(3.12)u3-v-3u2v+3uv2+3u3v2-3u2v3-u4v3+uv4=0.Proof.
From Lemma2.3, we obtain
(3.13)αβ-(1-(1-α)1/4(1-β)1/4)4=0.
From Theorem 3.3, we deduce that
(3.14)α=1-(1-u1+u)8,β=1-(1-v1+v)8,(1-α)1/4=(1-u1+u)2,(1-β)1/4=(1-v1+v)2,
where β has degree 3 over α.
Employing (3.14) in (3.13) and factorizing using Mathematica, we arrive at
(3.15)(-u3+v+3u2v-3uv2-3u3v2+3u2v3+u4v3-uv4)×(-u+3u2v+u4v-3uv2-3u3v2+v3+3u2v3-u3v4)=0.
It can be seen that the second factor of (3.15) does not vanish for |q|→0, so by identity theorem, we have
(3.16)u3-v-3u2v+3uv2+3u3v2-3u2v3-u4v3+uv4=0.Theorem 3.7.
Letu=T(q) and v=T(q4). Then,
(3.17)u4-v-4u2v+2u4v-4u6v-u8v+28u4v2-7v3-28u2v3+14u4v3-28u6v3-7u8v3+70u4v4-7v5-28u2v5+14u4v5-28u6v5-7u8v5+28u4v6-v7-4u2v7+2u4v7-4u6v7-u8v7+u4v8=0.Proof.
Squaring the modular equation in Lemma2.4 and simplifying, we obtain
(3.18)β-(1-(1-α)1/41+(1-α)1/4)4=0.
From Theorem 3.3(i), we have
(3.19)(1-α)1/4=(1-T(q)1+T(q))2.
Now, employing Theorem 3.3(ii) with n=4 and (3.19) in (3.18) and simplifying, we complete the proof.Theorem 3.8.
Letu=T(q) and v=T(q5). Then,
(3.20)u5-v-5u2v+10u3v2+5u5v2-10u2v3-10u4v3+5uv4+10u3v4-5u4v5-u6v5+uv6=0.Proof.
From Theorem3.3, we obtain
(3.21)c:=(1-α)1/8=(1-u1+u),d:=(1-β)1/8=(1-v1+v),
where β has degree 5 over α.
Employing (3.21) in (2.5), we find that
(3.22)m=c+d5c(1+c3d),(3.23)5m=d-c5d(1-cd3),
respectively.
Eliminatingm between (3.22) and (3.23) and simplifying, we deduce that
(3.24)5cd(1+c3d)(1-cd3)-(c+d5)(d-c5)=0.
Substituting for c and d from (3.21) in (3.24) and simplifying, we arrive at
(3.25)u5-v-5u2v+10u3v2+5u5v2-10u2v3-10u4v3+5uv4+10u3v4-5u4v5-u6v5+uv6=0.Theorem 3.9.
Letu=T(q) and v=T(q7). Then,
(3.26)u8-uv-7u3v-7u5v+7u7v+28u6v2-7uv3-49u3v3+7u5v3-7u7v3+70u4v4-7uv5+7u3v5-49u5v5-7u7v5+28u2v6+7uv7-7u3v7-7u5v7-u7v7+v8=0.Proof.
From Lemma2.6, we obtain
(3.27)αβ-(1-(1-α)1/4(1-β)1/4)8=0.
Again, from Theorem 3.3, we deduce that
(3.28)α=1-(1-u1+u)8,β=1-(1-v1+v)8,(1-α)1/8=(1-u1+u),(1-β)1/8=(1-v1+v),
where β has degree 7 over α.
Employing (3.28) in (3.27) and simplifying using Mathematica, we arrive at
(3.29)u8-uv-7u3v-7u5v+7u7v+28u6v2-7uv3-49u3v3+7u5v3-7u7v3+70u4v4-7uv5+7u3v5-49u5v5-7u7v5+28u2v6+7uv7-7u3v7-7u5v7-u7v7+v8=0.


## 4. Explicit Evaluations ofT(q)

In this section, we establish some general theorems for the explicit evaluations of the continued fractionT(q) and give examples.Forq:=e-πn, Ramanujan's two class invariants Gn and gn are defined by
(4.1)Gn=2-1/4q-1/24χ(q),gn=2-1/4q-1/24χ(-q).
The class invariants Gn and gn are connected by the relation [4, p. 187, Entry 2.1]:
(4.2)g4n=21/4gnGn.The singular modulusαn is defined by αn:=α(e-πn), where n is a positive integer and unique positive number between 0 and 1 satisfying
(4.3)n=F21(1/2,1/2;1;1-αn)F21(1/2,1/2;1;αn).Class invariants and singular moduli are intimately related by the equalities [4, p. 185, (1.6)]:
(4.4)Gn=(4αn(1-αn))-1/24,gn=(4αn(1-αn)-2)-1/24.
An account of Ramanujan's class invariants and singular moduli can be found in Chapter 34 of Berndt's book [4].Theorem 4.1.
One has(4.5)T(e-πn)=1-(1-αn)1/81+(1-αn)1/8.Proof.
We setq:=e-πn in Theorem 3.3(i) and use the definition of singular moduli αn and simplifying, we complete the proof.In the scattered places of his first notebook [1], Ramanujan calculated over 30 singular moduli αn. See Chapter 34 of Berndt's book [4] for details. Thus, one can use Theorem 4.1 to find the values of T(e-πn) if the corresponding values of αn are known. For example, from [4, p. 281, Theorem 9.2], we note that
(4.6)α2=(2-1)2.
Employing (4.6) in Theorem 4.1, we calculate
(4.7)T(e-π2)=1-(-2+22)1/81+(-2+22)1/8.
Many other values of T(e-πn) can be computed by using the known values of αn.Theorem 4.2.
One has(4.8)T(e-πn)=(g4n-21/4gn2g4n-21/4gn2).Proof.
Dividing numerator and denominator of right-hand side of Theorem3.1 and employing (1.6), we obtain
(4.9)T(q)=χ(q)-χ(-q)χ(q)+χ(-q).
Setting q:=e-πn, employing the definitions of Gn and gn from (4.1) in (4.9) and simplifying, we obtain
(4.10)T(e-πn)=Gn-gnGn+gn.
Substituting for Gn from (4.2) in (4.10) and simplifying, we complete the proof.Theorem4.2 implies that if we know the values of gn and g4n for any positive number n, then corresponding values of T(e-πn) can easily be calculated. Saikia [9] evaluated several values of gn and g4n for positive number n. For example, noting from [9, Theorem 3.5], we have
(4.11)g3=2-1/6(2+3)1/8,g12=21/6(2+3)1/8.
Employing (4.11) in Theorem 4.2, we obtain
(4.12)T(e-π3)=2-23/4(2+3)1/82+23/4(2+3)1/8.
Many other values of T(e-πn) can be determined by using the values of gn and g4n evaluated in [9].



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