# Some New Explicit Values of Quotients of Ramanujan’s Theta Functions and Continued Fractions

Authors: Nipen Saikia
Journal: International Journal of Mathematics and Mathematical Sciences
 (2014)

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## Abstract

We evaluate some new explicit values of quotients of Ramanujan’s theta functions and use them to find explicit values of Ramanujan’s continued fractions.


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## Body

## 1. Introduction

Ramanujan’s general theta functionf(a,b) is defined by
(1)f(a,b):=∑k=-∞∞‍ak(k+1)/2bk(k-1)/2,|ab|<1.
Two important special cases of f(a,b) are the theta functions ϕ(q) and ψ(q) [1, page 36, Entry 22] defined by, for |q|<1,
(2)ϕ(q):=f(q,q)=∑n=-∞∞‍qn2=(-q;-q)∞(q;-q)∞,ψ(q):=f(q,q3)=∑n=0∞‍qn(n+1)/2=(q2;q2)∞(q;q2)∞,
where (a;q)∞=∏n=0∞(1-aqn).In his notebooks [2], Ramanujan recorded many explicit values of theta functions ϕ(q) and ψ(q). All these values were proved by Berndt [3, page 325] and Berndt and Chan [4]. Yi [5] introduced the parameter hk,n for positive real numbers k and n defined by
(3)hk,n=ϕ(q)k1/4ϕ(qk);q=e-πn/k
and used the particular case hn,n to find explicit values ϕ(q). Baruah and Saikia [6] defined the parameter gk,n for positive real numbers k and n as
(4)gk,n=ψ(-q)k1/4q(k-1)/8ψ(-qk);q=e-πn/k
and used the particular case gn,n to find explicit values ψ(q). Saikia [7] also established some explicit values ψ(q).In this paper, we consider the particular casesh3,n and g3,n of the parameters hk,n and gk,n, respectively. By using theta function identities, we find some new explicit values of the parameters h3,n and g3,n. Particularly, we evaluate h3,n and g3,n for n= 3/2, 2/3, 6, 1/6, 5/2, 2/5, 10, and 1/10. Previously, Yi [5] evaluated h3,n for n= 1, 3, 1/3, 9, 1/9, 5, 1/5, 25, and 1/25. Saikia [8] evaluated h3,n for n= 2, 1/2, 4, 1/4, 7, 1/7, 49, and 1/49. Baruah and Saikia [6] evaluated g3,n for n= 1, 3, 1/3, 9, 1/9, 5, 1/5, 25, 1/25, 7, 1/7, 13, 1/13, 49, and 1/49. As an application to our new values, we evaluate some old and new explicit values of Ramanujan’s cubic continued fraction G(q) and a continued fraction of order twelve H(q) which are, respectively, defined by, for |q|<1,
(5)G(q)≔q1/31+q+q21+q2+q41+q3+q61+⋯,H(q)∶=qf(-q,-q11)f(-q5,-q7)=q(1-q)1-q3+q3(1-q2)(1-q4)(1-q3)(1+q6)+q3(1-q8)(1-q10)(1-q3)(1+q12)+⋯.
The continued fraction G(q) was recorded by Ramanujan on page 366 of lost notebook [9]. We refer to [10–14] for explicit evaluations of G(q). The continued fraction H(q) was introduced by Naika et al. [15]. We refer to [8, 15] for explicit evaluations of H(q).The presentation of the paper is as follows. In Section2 we record some preliminary results for ready references in this paper. Section 3 is devoted to explicit evaluations of the parameters h3,n and g3,n. In Section 4 we use new explicit values of h3,n and g3,n to evaluate some explicit values of the continued fractions G(q) and H(q).We end this introduction by noting the following remarks regardingg3,n and h3,n from [6, page 1764, Theorem 4.1] and [5, page 385, Remark 2.3], respectively.Remark 1.
The parameterg3,n has positive real value with g3,1=1 and g3,n increases as n increases.Remark 2.
The parameterh3,n has positive real value with h3,1=1 and the values of h3,n decrease as n increases.


## 2. Preliminary Results

Lemma 3 (see [16, Theorem 3.2]).
IfP:=qψ(-q)ψ(-q9)/ψ2(-q3) and Q:=qψ(-q2)ψ(-q18)/ψ2(-q6), then
(6)P2Q2+Q2P2+(P2Q2+Q2P2)2=(3PQ-1PQ)(3PQ-1PQ+3PQ+3QP+P3Q3+Q3P3)+14.Lemma 4 (see [16, Theorem 3.1]).
IfL:=ϕ(q)ϕ(q9)/ϕ2(q) and M:=qψ(-q)ψ(-q9)/ψ2(-q3), then
(7)L2=M2+11-3M2.Lemma 5 (see [16, Theorem 3.2]).
IfP:=ψ(-q3)ψ(-q5)/qψ(-q)ψ(-q15) and Q:=ψ(-q6)ψ(-q10)/q2ψ(-q2)ψ(-q30), then
(8)(P2Q2+Q2P2)+(PQ+QP)-(PQ+1PQ)+(PQ-1PQ)(P3Q3+Q3P3+PQ+QP)=0.Lemma 6 (see [16, Theorem 3.9]).
IfL:=ϕ(q3)ϕ(q5)/ϕ(q)ϕ(q15) and M:=ψ(-q3)ψ(-q5)/qψ(-q)ψ(-q15), then
(9)M=1+L1-L.Lemma 7 (see [5, page 385, Theorem 2.2]).
For all positive real numbersk and n, one has
(10)hk,1/n=1hk,n.Lemma 8 (see [6, page 1764, Theorem 4.1]).
For all positive real numbersk and n, one has
(11)gk,1/n=1gk,n.Lemma 9 (see [5, page 393, Theorem 4.9(i)]).
For any positive real numbern, one has
(12)3(h3,nh3,9n+1h3,nh9,n)=(h3,9nh3,n)2+3.Lemma 10 (see [5, page 394, Theorem 4.12(i)]).
For any positive real numbern, one has
(13)3{(h3,nh3,25n)2+1(h3,nh3,25n)2}={(h3,25nh3,n)3-(h3,nh3,25n)3}+5{(h3,25nh3,n)2+(h3,nh3,25n)2}+5{(h3,25nh3,n)-(h3,nh3,25n)}.Lemma 11 (see [6, page 1769, Theorem 5.1(i) and (iii)]).
For any positive real numbern, one has(i)
(1+3g3,ng3,9n)3=1+3g3,9n4,(ii)
3{(g3,ng3,25n)2+1/(g3,ng3,25n)2} + 5{(g3,25n/g3,n)2+(g3,n/g3,25n)2} = {(g3,25n/g3,n)3-(g3,n/g3,25n)3} + 5{(g3,25n/g3,n)-(g3,n/g3,25n)}.


## 3. Explicit Evaluations of the Parametersgk,n and hk,n

In this section, we prove some general theorems for the explicit evaluations of the parametersgk,n and hk,n and evaluate some new explicit values therefrom.Theorem 12.
One has(14)(g3,ng3,36ng3,4ng3,9n)2+(g3,4ng3,9ng3,ng3,36n)2+((g3,ng3,36ng3,4ng3,9n)2+(g3,4ng3,9ng3,ng3,36n)2)2-14={3(g3,ng3,4ng3,9ng3,36n)-(g3,9ng3,36ng3,ng3,4n)}×{3(g3,ng3,4ng3,9ng3,36n)-(g3,9ng3,36ng3,ng3,4n)+3(g3,ng3,36ng3,4ng3,9n)}+{3(g3,ng3,4ng3,9ng3,36n)-(g3,9ng3,36ng3,ng3,4n)}×{3(g3,4ng3,9ng3,ng3,36n)+(g3,ng3,36ng3,4ng3,9n)3+(g3,4ng3,9ng3,ng3,36n)3}.Proof.
We setq:=e-πn/3 in Lemma 3 and use the definition of g3,n.Corollary 13.
One has(15)g3,3/2=(32+22-6+62)2,g3,6=6(1+2)32+22-6+62,g3,2/3=2(32+22-6+62),g3,6=32+22-6+626(1+2).Proof.
Settingn=1/6 in Theorem 12 and simplifying using Lemma 8, we obtain
(16)(3(g3,3/2g3,6)2-(g3,3/2g3,6)2)×(3(g3,3/2g3,6)2-(g3,3/2g3,6)2+8)+8=0.
Solving (16) for g3,3/2g3,6 and noting the fact in Remark 1, we find that
(17)g3,3/2g3,6=3(1+2).
Again, setting n=1/6 in Lemma 11(i) and simplifying using Lemma 8, we deduce that
(18)(1+3(g3,3/2g3,6))3=1+3g3,3/24.
Eliminating g3,6 from (18) using (17) and simplifying, we obtain
(19)g3,3/24+3(1+2-(1+2)3/2)g3,3/22+(3+32)=0.
Solving (19) and noting the fact in Remark 1, we obtain
(20)g3,3/2=(32+22-6+62)2.
Employing (20) in (17) and simplifying, we obtain
(21)g3,6=6(1+2)32+22-6+62.
Now the values of g3,2/3 and g3,1/6 follow from the values of g3,3/2 and g3,6, respectively, and Lemma 8.Theorem 14.
One has(22)(h3,nh3,9n)2{1-3(g3,ng3,9n)2}-(g3,ng3,9n)2-1=0.Proof.
We setq:=e-πn/3 in Lemma 4 and use the definitions of h3,n and g3,n.Corollary 15.
One has(23)h3,3/2=2-5/8(32-4)3/4(6+62+6)1/2,h3,6=21/831/2(32+4)1/4(6+62+6)-1/2,h3,2/3=25/8(32-4)-3/4(6+62+6)-1/2,h3,1/6=2-1/83-1/2(32+4)-1/4(6+62+6)1/2.Proof.
Settingn=1/6 in Theorem 14 and simplifying using Lemma 7, we obtain
(24)1(h3,6h3,3/2)2(1-3(g3,6g3,3/2)2)-1(g3,6g3,3/2)2-1=0.
Employing (17) in (24), solving the resulting equation, and noting the fact in Remark 2, we deduce that
(25)h3,3/2h3,6=21/434+32.
Again setting n=1/6 in Lemma 9 and simplifying using Lemma 7, we obtain
(26)3(h3,3/2h3,6+h3,6h3,3/2)=(h3,3/2h3,6)2+3.
Employing (25) in (26), solving the resulting equation, and noting the fact in Remark 2, we deduce that
(27)h3,3/2h3,6=6+62+63(4+32).
Multiplying (25) and (27), we evaluate the value of h3,3/2, and dividing (25) by (27) and simplifying, we evaluate the value of h3,6. Now the values of h3,2/3 and h3,1/6 easily follow from the values of h3,3/2 and h3,6, respectively, and Lemma 7.Theorem 16.
One has(28)(g3,4ng3,25ng3,ng3,100n)2+(g3,ng3,100ng3,4ng3,25n)2+(g3,4ng3,25ng3,ng3,100n)+(g3,ng3,100ng3,4ng3,25n)-(g3,25ng3,100ng3,ng3,4n)-(g3,ng3,4ng3,25ng3,100n)+{(g3,25ng3,100ng3,ng3,4n)1/2-(g3,ng3,4ng3,25ng3,100n)1/2}×{(g3,4ng3,25ng3,ng3,100n)3/2+(g3,ng3,100ng3,4ng3,25n)3/2+(g3,4ng3,25ng3,ng3,100n)1/2+(g3,ng3,100ng3,4ng3,25n)1/2}=0.Proof.
We setq:=e-πn/3 in Lemma 5 and use the definition of g3,n.Corollary 17.
One has(29)g3,5/2=(1+3+2+3)1/2(a+5b)1/4c-1/4,g3,10=(1+3+2+3)1/2(a+5b)-1/4c1/4,g3,2/5=(1+3+2+3)-1/2(a+5b)-1/4c1/4,g3,1/10=(1+3+2+3)-1/2(a+5b)1/4c-1/4,
where
(30)a=63+353+302+3+206+33,b=2574+14853+13302+3+7706+33,c=25+153+142+3+76+33.Proof.
Settingn=1/10 in Theorem 16 and simplifying using Lemma 8, we obtain
(31)(g3,5/2g3,10)2+1(g3,5/2g3,10)2-4(g3,5/2g3,10-1g3,5/2g3,10)-4=0.
Solving (31) and noting the fact in Remark 2, we find that
(32)g3,5/2g3,10=1+3+2+3.
Again, setting n=1/10 in Lemma 11 (ii) and simplifying using Lemma 8, we obtain
(33)3{(g3,5/2g3,10)2+(g3,5/2g3,10)}+5((g3,5/2g3,10)2+1(g3,5/2g3,10)2)-((g3,5/2g3,10)3-1(g3,5/2g3,10)3)-5(g3,5/2g3,10-1g3,5/2g3,10)=0.
Employing (32) in (33), solving the resulting equation, and noting the fact in Remark 1, we obtain
(34)g3,10g3,5/2=(a+5bc)1/2,
where a=63+353+302+3+206+33, b=2574+14853+13302+3+7706+33, and c=25+153+142+3+76+33.
Combining (32) and (34), we calculate the values of g3,5/2 and g3,10. Then, the values of g3,2/5 and g3,1/10 follow from the values of g3,5/2 and g3,10, respectively, and Lemma 8.Theorem 18.
One has(35)(g3,25ng3,n){1-(h3,25nh3,n)}-1-(h3,25nh3,n)=0.Proof.
We setq:=e-πn/3 in Lemma 6 and use the definitions of h3,n and g3,n.Corollary 19.
One has(36)h3,5/2=(3+2+3)1/2(2+3+2+3)-1/2×(f-45b)1/4d-1/4,h3,10=(3+2+3)1/2(2+3+2+3)-1/2×(f-45b)-1/4d1/4,h3,2/5=(3+2+3)-1/2(2+3+2+3)1/2×(f-45b)-1/4d1/4,h3,1/10=(3+2+3)-1/2(2+3+2+3)1/2×(f-45b)1/4d-1/4,
where b is given in Corollary 17, f=1143+6553+5822+3+3466+33, and d=1117+6453 + 5782+3+3346+33.Proof.
Settingn=1/10 in Theorem 18 and simplifying using Lemma 7, we get
(37)g3,5/2g3,10(1-h3,5/2h3,10)-1-h3,5/2h3,10=0.
Employing (32) and solving the resulting equation, we obtain
(38)h3,5/2h3,10=3+2+32+3+2+3.
Again setting n=1/10 in Lemma 10 and simplifying using Lemma 7, we find that
(39)3{(h3,5/2h3,10)2+(h3,10h3,5/2)2}-((h3,5/2h3,10)3-1(h3,5/2h3,10)3)-5((h3,5/2h3,10)2+1(h3,5/2h3,10)2)-5(h3,5/2h3,10-1h3,5/2h3,10)=0.
Employing (38) in (39), solving the resulting equation, and noting the fact in Remark 2, we obtain
(40)h3,5/2h3,10=f-45bd,
where b is given in Corollary 17, f=1143+6553+5822+3+3466+33, and d=1117+6453 + 5782+3 + 3346+33.
Combining (38) and (40), we evaluate the values of h3,5/2 and h3,10. Then the values of h3,2/5 and h3,1/10 follow from h3,5/2 and h3,10, respectively, and Lemma 7.


## 4. Explicit Evaluations of Continued Fractions

This section is devoted to finding some explicit values of the continued fractionsG(q) and H(q) by using new values of g3,n and h3,n evaluated in Section 3.Lemma 20 (see [6, page 1788, Theorem 9.1(i)]).
One has(41)G3(e-πn/3)=-11+3g3,n4.Theorem 21.
One has(42)G3(-e-π/2)=2936+422-(38+362),G3(-e-π2)=(32+22-6+62)26((2+22)(6+62)-(58+402)),G3(-e-π2/3)=(32+22-6+62)26((2+22)(6+62)-(6+42)),G3(-e-π/32)=2(3+22)(36+242-(10+82)),G3(-e-π5/6)=-cc+3(a+5b)(1+3+2+3)2,G3(-e-π10/3)=-(a+5b)(a+5b)+3c(1+3+2+3)2,G3(-e-π2/15)=-(a+5b)(1+3+2+3)2(a+5b)(1+3+2+3)2+3c,G3(-e-π/30)=-c(1+3+2+3)2c(1+3+2+3)2+3(a+5b),
where a, b, and c are given in Corollary 17.Proof.
We setn= 3/2, 6, 2/3, 1/6, 5/2, 10, 2/5, and 1/10 in Lemma 20 and use the corresponding values of g3,n from Corollaries 13 and 17 to complete the proof.The explicit values ofG3(-e-π/32), G3(-e-π5/6), G3(-e-π10/3), G3(-e-π2/15), and G3(-e-π/30) are new.Lemma 22 (see [8, page 144, Theorem 5.1]).
One has(43)H(e-πn/3)=31/4h3,n-131/4h3,n+1.By settingn= 3/2, 6, 2/3, 1/6, 5/2, 10, 2/5, and 1/10 in Lemma 22 and employing the corresponding values of h3,n from Corollaries 15 and 19, we calculate the following new explicit values of the continued fraction H(q).Theorem 23.
One has(44)H(e-π/2)=-2+23/831/4(32-4)3/41+62+62+23/831/4(32-4)3/41+62+6,H(e-π2)=21/833/4(4+32)1/4-1+62+621/833/4(4+32)1/4+1+62+6,H(e-π2/3)=25/831/4-(32-4)3/41+62+625/831/4+(32-4)3/41+62+6,H(e-π/18)=27/81+62+6-2(12+92)1/427/81+62+6+2(12+92)1/4,H(e-π5/6)=(31/4(3+2+3)1/2(f-45b)1/4-(2+3+2+3)1/2d1/4)×(31/4(3+2+3)1/2(f-45b)1/4+(2+3+2+3)1/2d1/4)-1,H(e-π10/3)=(31/4(3+2+3)1/2d1/4-(2+3+2+3)1/2(f-45b)1/4)×(31/4(3+2+3)1/2d1/4+(2+3+2+3)1/2(f-45b)1/4)-1,H(e-π2/15)=(31/4(2+3+2+3)1/2d1/4-(3+2+3)1/2(f-45b)1/4)×(31/4(2+3+2+3)1/2d1/4+(3+2+3)1/2(f-45b)1/4)-1,H(e-π/30)=(31/4(2+3+2+3)1/2(f-45b)1/4-(3+2+3)1/2d1/4)×(31/4(2+3+2+3)1/2(f-45b)1/4+(3+2+3)1/2d1/4)-1,
where b is given in Corollary 17 and d and f are given in Corollary 19.



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