# Some New Explicit Values of Quotients of Theta-Functionϕ(q) and Applications to Ramanujan's Continued Fractions

Authors: Nipen Saikia
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

We find some new explicit values of the parameterhk,n for positive real numbers k and n involving Ramanujan's theta-function ϕ(q) and give some applications of these new values for the explicit evaluations of Ramanujan's continued fractions. In the process, we also establish two new identities for ϕ(q) by using modular equations.


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## Body

## 1. Introduction

Forq:=e2πiz, Im(z)>0, define Ramanujan's theta-function ϕ(q) as
(1)ϕ(q):=∑n=-∞∞‍qn2=ϑ3(0,2z),
where ϑ3 [1, page 464] is one of the classical theta-functions.In his notebook [2, volume I, page 248], Ramanujan recorded several explicit values of theta-functions ϕ(q) and its quotients which are proved by Berndt and Chan [3]. They also found some new explicit values. An account of these can also be found in Berndt's book [4]. Yi [5] also evaluated many new values of ϕ(q) by finding explicit values of the parameters hk,n and hk,n′ for positive real numbers k and n which are defined by
(2)hk,n:=ϕ(q)k1/4ϕ(qk),q=e-πn/k,hk,n′:=ϕ(-q)k1/4ϕ(-qk),q=e-2πn/k.
Yi [5] established several properties of these parameters and found their explicit values by appealing to transformation formulas and theta-function identities for ϕ(q). Recently, Saikia [6] found many new explicit values of quotients of ϕ(q) by finding explicit values of the parameter An which is a particular case of the parameter hk,n where k=4. Saikia [6] also established some new theta-function identities for ϕ(q).In the sequel of the previous work, in this paper we find some new explicit values of the parametersh4,n and h2,n which are particular cases of the parameter hk,n by using some properties of hk,n established in [5] and two new theta-function identities for ϕ(q). In addition, we give some applications of these new values of hk,n for the explicit evaluations of Ramanujan's continued fractions c(q) and K(q) defined by
(3)c(q):=11+2q1-q2+q2(1+q2)21-q6+q4(1+q4)21-q10+⋯,|q|<1,(4)K(q):=q1/21+q+q21+q3+q41+q5+⋯,|q|<1.
The continued fraction c(q) is studied by Adiga and Anitha [7]. For explicit evaluations of c(q), see [6]. The continued fraction K(q) is called Ramanujan-Go¨llnitz-Gordon continued fraction [4, page 50]. For further references on K(q), see [8, 9].In Section2, we record some preliminary results. Section 3 is devoted to prove two new identities for theta-function ϕ(q). In Section 4, we find some new explicit values of the parameter h4,n. In Section 5, we evaluate some new values of the parameter h2,n. Finally in Section 6, we give applications of these values of h4,n and h2,n for the explicit evaluations of Ramanujan's continued fractions c(q) and K(q).We end the introduction by defining Ramanujan's modular equation. The complete elliptic integral of the first kindK(k) is defined by
(5)K(k):=∫0π/2dϕ1-k2sin2ϕ=π2∑n=0∞‍(1/2)n2(n!)2k2n=π22F1(12,12;1;k2),
where 0<k<1, 2F1 denotes the ordinary or Gaussian hypergeometric function, and
(6)(a)n=a(a+1)(a+2)⋯(a+n-1).
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
(7)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 (7).If we set(8)q=exp(-πK′K),q′=exp(-πL′L),
we see that (7) 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 m=z1/zn where zr=ϕ2(qr).


## 2. Preliminary Results

Lemma 1 (see [5, page 385, Theorem 2.2]).
For positive real numbersk and n, (i)
hk,1=1,(ii)
hk,1/n=1/hk,n.Lemma 2 (see [5, page 387, Corollary 2.6]).
For positive real numbersk and n,
(9)hk2,n=hk,nkhk,n/k.Lemma 3 (see [5, page 392, Theorem 4.6]).
One has(10)2(h2,2nh2,4n+1h2,2nh2,4n)=h2,4nh2,n+2,
for any positive real number n.Lemma 4 (see [10, page 122, Entry 10 (i) and (v)]).
One has(i)
ϕ(q)=z,(ii)
ϕ(q4)=z(1+(1-α)1/4))/2.Lemma 5 (see [10, page 280-281, Entry 13(vii)]).
Ifβ has degree 5 over α, then
(11)(αβ3)1/8+{(1-α)(1-β)3}1/8=(α3β)1/8+{(1-α)3(1-β)}1/8.Lemma 6 (see [10, page 314, Entry 19(i)]).
Ifβ has degree 7 over α, then
(12)(αβ)1/8+{(1-α)(1-β)}1/8=1.


## 3. Two New Identities forϕ(q)

In this section, we prove two new identities for theta-functionϕ(q) by using Ramanujan's modular equations and transformation formulas.Theorem 7.
IfP=ϕ(q)/ϕ(q4), Q=ϕ(q5)/ϕ(q20), then
(13)P6-256PQ+640P2Q-640P3Q+320P4Q-70P5Q+640PQ2-1600P2Q2+1600P3Q2-785P4Q2+160P5Q2-640PQ3+1600P2Q3-1620P3Q3+800P4Q3-160P5Q3+320PQ4-785P2Q4+800P3Q4-400P4Q4+80P5Q4-70PQ5+160P2Q5-160P3Q5+80P4Q5-16P5Q5+Q6=0.Proof.
TranscribingP and Q using Lemma 4(i) and (iv) and then simplifying, we get
(14)(1-α)1/4=2P-1,(1-β)1/4=(2Q-1),
where β has degree 5 over α.
Equivalently,(15)α=1-(2P-1)4,β=1-(2Q-1)4.
Now by Lemma 5, we have
(16)(αβ3)1/8-(α3β)1/8={(1-α)3(1-β)}1/8-{(1-α)(1-β)3}1/8.
Squaring (16) and simplifying, we arrive at
(17)(αβ3)1/4+(α3β)1/4=x+2(αβ)1/2,
where(18)x={(1-α)3(1-β)}1/4+{(1-α)(1-β)3}1/4-2{(1-α)(1-β)}1/2.
Squaring (17) and simplifying, we obtain
(19)(αβ3)1/2+(α3β)1/2=y+4x(αβ)1/2,
where y=x2+2αβ.
Squaring (19) and simplifying, we obtain
(20)αβ3+α3β+2α2β2-y2-16x2αβ=8xy(αβ)1/2.
Again squaring (20), we obtain
(21)(αβ3+α3β+2α2β2-y2-16x2αβ)2-64x2y2αβ=0.
Now employing (14) and (15) and factorizing using Mathematica, we deduce that
(22)f(P,Q)g(P,Q)h(P,Q)=0,
where
(23)f(P,Q)=(P-Q)4,g(P,Q)=P6-256PQ+640P2Q-640P3Q+320P4Q-70P5Q+640PQ2-1600P2Q2+1600P3Q2-785P4Q2+160P5Q2-640PQ3+1600P2Q3-1620P3Q3+800P4Q3-160P5Q3+320PQ4-785P2Q4+800P3Q4-400P4Q4+80P5Q4-70PQ5+160P2Q5-160P3Q5+80P4Q5-16P5Q5+Q6,h(P,Q)=16P10-96P9Q-32P10Q+240P8Q2+192P9Q2+24P10Q2-320P7Q3-480P8Q3-144P9Q3-8P10Q3+256P6Q4+608P7Q4+384P8Q4+40P9Q4+P10Q4-448P5Q5+352P6Q5-1264P7Q5+248P8Q5-70P9Q5+256P4Q6+352P5Q6-880P6Q6+1640P7Q6-785P8Q6+160P9Q6-320P3Q7+608P4Q7-1264P5Q7+1640P6Q7-1620P7Q7+800P8Q7-160P9Q7+240P2Q8-480P3Q8+384P4Q8+248P5Q8-785P6Q8+800P7Q8-400P8Q8+80P9Q8-96PQ9+192P2Q9-144P3Q9+40P4Q9-70P5Q9+160P6Q9-160P7Q9+80P8Q9-16P9Q9+16Q10-32PQ10+24P2Q10-8P3Q10+P4Q10.
By examining the behavior of the first factor f(P,Q) and the last factor h(P,Q) of the left-hand side of (22) near q=0, it can be seen that there is a neighborhood about the origin, where these factors are not zero. Then the second factor g(P,Q) is zero in this neighborhood. By the identity theorem, this factor is identically zero. Hence, we complete the proof.Theorem 8.
IfP=ϕ(q)/ϕ(q4), Q=ϕ(q7)/ϕ(q28), then
(24)P8-4096PQ+14336P2Q-21504P3Q+17920P4Q-8736P5Q+2352P6Q-280P7Q+14336PQ2-51968P2Q2+80640P3Q2-69440P4Q2+35056P5Q2-9772P6Q2+1176P7Q2-21504PQ3+80640P2Q3-129472P3Q3+115360P4Q3-60424P5Q3+17528P6Q3-2184P7Q3+17920PQ4-69440P2Q4+115360P3Q4-106330P4Q4+57680P5Q4-17360P6Q4+2240P7Q4-8736PQ5+35056P2Q5-60424P3Q5+57680P4Q5-32368P5Q5+10080P6Q5-1344P7Q5+2352PQ6-9772P2Q6+17528P3Q6-17360P4Q6+10080P5Q6-3248P6Q6+448P7Q6-280PQ7+1176P2Q7-2184P3Q7+2240P4Q7-1344P5Q7+448P6Q7-64P7Q7+Q8=0.Proof.
TranscribingP and Q using Lemma 4(i) and (iv) and then simplifying, we obtain
(25)(1-α)1/4=2P-1,(1-β)1/4=(2Q-1),
where β has degree 7 over α.
Equivalently,(26)α=1-(2P-1)4,β=1-(2Q-1)4.
Now by Lemma 6, we have
(27){(1-α)(1-β)}1/8=1-(αβ)1/8.
Squaring (27) and simplifying, we obtain
(28)x-(αβ)1/4=-2(αβ)1/8,
where x={(1-α)(1-β)}1/4-1.
Squaring (28) and simplifying, we obtain
(29)x2+(αβ)1/2=(4+2x)(αβ)1/4.
Squaring (29) and simplifying, we obtain
(30)x4+αβ=((4+2x)2-2x2)(αβ)1/2.
Again, squaring (30), we arrive at
(31)(x4+αβ)2=((4+2x)2-2x2)2(αβ).
Employing (25) and (26) in (31) and simplifying with the help of Mathematica, we complete the proof.


## 4. New Values ofh4,n

In this section, we find some new values ofh4,n and h2,n by using theta-function identities proved in Section 3 and the properties of hk,n listed in Lemmas 1 and 2. We begin with following remarks.Remark 9.
The values ofhk,n are real and hk,n<1 for all n>1 if k>1. We also note that the values of hk,n decrease as n increases when k>1. In view of this, in the following theorem we have h4,4>h4,5>h4,7>h4,8>h4,25>h4,49 where h4,4 and h4,8 are evaluated in [6, Theorem 4.3]. Yi [5] also evaluated the value of h4,4.Theorem 10.
One has(i)
h4,5=(22+105-2-10-(2+10-22+105)2-4)/2,(ii)
h4,1/5=(22+105-2-10+(2+10-22+105)2-4)/2,(iii)
h4,25=114-802+25965-183602-(114-802+25965-183602)2-1,(iv)
h4,1/25=114-802+25965-183602+(114-802+25965-183602)2-1,(v)
h4,7=(12-72-7(34-242))/2,(vi)
h4,1/7=(12-72+7(34-242))/2,(vii)
h4,49=a+b--1+(a+b)2,(viii)
h4,1/49=a+b+-1+(a+b)2,
wherea=1317-9312+3470684-24541442 and b = 6940535 − 49077002 + 13061331967/(123953-876482) − 92357564014/(123953-876482).Proof.
For (i) and (ii), settingk=4 and employing the definition of hk,n in Theorem 7, we get
(32)P=2h4,n,Q=2h4,25n.
Setting n=1/5 in (32), applying to (13), and then simplifying using Lemma 1(ii), we obtain
(33)(h4,56+h4,5-6)-70(h4,54+h4,5-4)+3202(h4,53+h4,5-3)-1425(h4,52+h4,5-2)+19202(h4,5+h4,5-1)-3348=0.
Equivalently,
(34)A6-76A4+3202A3-1136A2+9602A-640=0,
where
(35)A:=h4,5+h4,5-1.
Solving (34) for A and noting that A has positive real value greater than 1, we obtain
(36)A=-2-10+22+105.
Invoking (36) in (35), solving for h4,5, and using the fact in Remark 9, we complete the proof of (i). Noting h4,1/5=1/h4,5 from Lemma 1(ii), we arrive at (ii).
For proofs of (iii) and (iv), we setn=1 in (32), applying to (13) and then simplifying using Lemma 1(i), we arrive at
(37)(h4,252+h4,25-2)-(456-3202)(h4,25+h4,25-1)-(674-8402)=0.
Equivalently,(38)B2+8(-57+402)B+(-674+8402)=0,
where
(39)B:=h4,25+h4,25-1.
Solving (38) for B and noting that B has positive real value greater than 1, we obtain
(40)B=2(114-802+5(5193-36722)).
Invoking (40) in (39), solving for h4,25, and using the fact in Remark 9, we complete the proof of (iii). Noting h4,1/25=1/h4,25 from Lemma 1(iv), we prove (ii).
To prove (v) and (vi), applying the definition ofh4,n in Theorem 8, we get
(41)P=2h4,n,Q=2h4,49n.
Setting n=1/7 in (41), applying in (24), and then simplifying using Lemma 1(ii), we arrive at
(42)h4,716-280h4,714+23522h4,713-18508h4,712+440162h4,711-140616h4,710+1592642h4,79-262810h4,78+1592642h4,77-140616h4,76+440162h4,75-18508h4,74+23522h4,73-280h4,72+1=0.
Dividing (42) by h4,78 and simplifying, we get
(43)(h4,78+h4,7-8)-280(h4,76+h4,7-6)+23522(h4,75+h4,7-5)-18508(h4,74+h4,7-4)+440162(h4,73+h4,7-3)-140616(h4,72+h4,7-2)+1592642(h4,7+h4,7-1)-262810=0.
Equivalently,
(44)L8-288L6+23522L5-16808L4+322562L3-69120L2+389762L-18032=0,
where
(45)L:=h4,7+h4,7-1.
Solving (44) by using Mathematica and noting that L has positive real value greater that 1 satisfying the fact in Remark 9, we obtain
(46)L=12-72.
Employing (46) in (45), solving for h4,7, and using the fact in Remark 9, we complete the proof of (v). Noting h4,1/7=1/h4,7, we arrive at (vi).
For proofs of (vii) and (viii), settingn=1 in (41), applying in (24), and simplifying using Lemma 1(ii), we arrive at
(47)h4,494+h4,49-4+4(-1317+9312)(h4,493+h4,49-3)+28(-2053+14522)(h4,492+h4,49-2)+56(-2409+17032)(h4,49+h4,49-1)+210(-837+5922)=0.
Equivalently,(48)D4+8(-1317+9312)D3+16(-3593+25412)D2+64(-1614+11412)D+128(-475+3362)=0,
where
(49)D=h4,49+h4,49-1.
Solving (48) for D and noting that D has positive real value greater than 1, we obtain
(50)D=2(a+b),
where a=1317-9312+3470684-24541442 and b = 6940535 − 49077002 + 13061331967/(123953-876482) − 92357564014/(123953-876482).
Invoking (50) in (49), solving for h4,49, and using the fact in Remark 9, we arrive at (vi). Noting h4,1/49=1/h4,49 from Lemma 1(ii), we complete the proof of (vii).


## 5. New Values ofh2,n

In this section, we find some new values of the parameterh2,n by using the values of h4,n evaluated in Section 4 and in [6].Theorem 11.
One has(i)
h2,6=82-10+(2-2)18-122,(ii)
h2,3/2=(2-2+18-122)/(282-10+(2-2)18-122),(iii)
h1/6=1/82-10+(2-2)18-122,(iv)
h2,2/3=282-10+(2-2)18-122/(2-2+18-122).Proof.
Settingk=2 and n=3 in Lemma 2, we deduce that
(51)h4,3=h2,6h2,3/2.
From [6, page 174, Theorem 4.3(vii)], we have
(52)h4,3=(2-2+18-122)2.
Combining (51) and (52), we obtain
(53)h2,6h2,3/2=(2-2+18-122)2.
Next, setting n=1/6 in Lemma 3 and simplifying using Lemma 1(ii), we obtain
(54)2(h2,6h2,3/2+(h2,6h2,3/2)-1)=(h2,6h2,3/2)+2.
Invoking (53) in (54) and simplifying, we deduce that
(55)(h2,6h2,3/2)=2(82-10+(2-2)18-122)2-2+18-122.
Multiplying (53) and (55) and simplifying, we complete the proof of (i). Dividing (53) by (55) and simplifying, we arrive at (ii). (iii) and (iv) follow from (i) and (ii), respectively, and Lemma 1(ii).The proofs of Theorems12–15 are identical to the proof of Theorem 11. So we omit details and give only references of the required results to prove them.Theorem 12.
One has(i)
h2,10=(4+c2)2-4c/2,(ii)
h2,5/2=c/(4+c2)2-4c,(iii)
h2,1/10=2/(4+c2)2-4c,(iv)
h2,2/5=(4+c2)2-4c/c,
wherec=22+105-2-10-(2+10-22+105)2-4.Proof of Theorem12 follows from Theorem 10(i), Lemma 2 with k=2 and n=5, Lemma 1(ii), and Lemma 3 with n=1/10.Theorem 13.
One has(i)
h2,50=-2d+2(d2+1),(ii)
h2,25/2=d/-2d+2(d2+1),(iii)
h2,1/50=1/-2d+2(d2+1),(iv)
h2,2/25=-2d+2(d2+1)/d,
whered=114-802+25965-183602-(114-802+25965-183602)2-1.To prove Theorem13, we use Theorem 10(iii), Lemma 2 with k=2 and n=25, Lemma 1(ii), and Lemma 3 with n=1/50.Theorem 14.
One has(i)
h2,14=(-8+62)(12-72-238-1682),(ii)
h2,7/2=12-72-238-1682/2-8+62,(iii)
h2,1/14=1/(-8+62)(12-72-238-1682),(iv)
h2,2/7=2-8+62/12-72-238-1682.We employ Theorem10(v), Lemma 2 with k=2 and n=7, Lemma 1(ii), and Lemma 3 with n=1/14 to prove Theorem 14.Theorem 15.
One has(i)
h2,98=2(a+b-(a+b)2-1)(2a+2b-1),(ii)
h2,49/2=(a+b-(a+b)2-1)/2(2a+2b-1),(iii)
h2,1/98=1/2(a+b-(a+b)2-1)(2a+2b-1),(iv)
h2,2/49=2(2a+2b-1)/(a+b-(a+b)2-1),
where a and b are given in Theorem10(viii).Proof follows from Theorem10(vii), Lemma 2 with k=2 and n=49, Lemma 1(ii), and Lemma 3 with n=1/98.


## 6. Applications ofh4,n and h2,n

In this section, we use the new values of the parametersh4,n and h2,n to find explicit values of continued fractions c(q) and K(q) defined in (3) and (4), respectively.The parameterh4,n is useful in finding explicit values of the continued fraction c(q). If we know values of the parameter h4,n for any positive real number n, then explicit values of c(e-πn/2) can be calculated by appealing to the following theorem.Theorem 16 (see [6, page 177, Theorem 5.1]).
One has(56)c(e-πn/2)=12h4,n,
where n is any positive real number.For example, employing the value ofh4,5 from Theorem 10(i) in Theorem 16, we obtain
(57)c(e-π5/2)=2×((2+10-22+105)2-422+105-2-10-(2+10-22+105)2-4)-1.
Similarly, we can find new values of c(e-π/25), c(e-5π/2), c(e-π/10), c(e-π7/2), c(e-π/27),  c(e-7π/2), and c(e-π/14) by employing the values of h4,n from Theorem 10(ii)–(viii), respectively, in Theorem 16. Since it is a routine calculation, we omit details.Next, the parameterh2,n is connected to continued fraction K(q) by the following theorem.Theorem 17 (see [8, page 281, Theorem 4.1]).
For any positive real number n, one has(58)K2(e-πn/2)=21/4h2,n-121/4h2,n+1.From Theorem17, we note that if the values of h2,n are known, then the values of K(e-πn/2) can easily be evaluated. For example, using the value of h2,6 from Theorem 11(i) in Theorem 17, we evaluate
(59)K(e-π6/2)=21/482-10+(2-2)18-122-121/482-10+(2-2)18-122+1.Similarly, we can evaluate new values ofK(e-πn/2) for n = 3/2, 1/6, 2/3, 10, 5/2, 1/10, 2/5, 50, 25/2, 1/50, 2/25, 14, 7/2, 1/14, 2/7, 98, 49/2, and 2/49 by using Theorem 17 and the values of h2,n evaluated in Theorems 11–15.



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