# A Product of Theta-Functions Analogous to Ramanujan's Remarkable Product of Theta-Functions and Applications

Authors: Nipen Saikia
Journal: Journal of Mathematics
 (2013)

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## Abstract

We define a productlk,n for any positive real numbers k and n involving
Ramanujan's theta-functions ϕ(q) and ψ(q) which is analogous to Ramanujan's remarkable product of
theta-functions recorded by Ramanujan (1957) and study its several properties. We prove general theorems for the explicit evaluations of lk,n and find some explicit values.
As application of the product lk,n, we also offer explicit formulas for explicit values of Ramanujan's
continued fraction V(q) in terms of lk,n and give examples.


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## Body

## 1. Introduction

Ramanujan’s theta-functionsϕ(q), ψ(q), and f(-q) are defined as
(1)ϕ(q):=∑n=-∞∞‍qn2=(-q;-q)∞(q;-q)∞,ψ(q):=∑n=0∞‍qn(n+1)/2=(q2;q2)∞(q;q2)∞,(2)f(-q):=f(-q,-q2)=∑n=-∞∞‍(-1)nqn(3n-1)/2=(q;q)∞,
where (a;q)∞:=∏k=0∞‍(1-aqk).On page 338 of his first notebook, Ramanujan [1] defined the remarkable product of theta-functions as
(3)am,n=ne-(π/4)(n-1)m/nψ2(e-πmn)ϕ2(-e-2πmn)ψ2(e-πm/n)ϕ2(-e-2πm/n),
where m and n are positive real numbers. He then, on pages 338 and 339, offered a list of eighteen particular values of the product am,n. All these eighteen values are proved by Berndt et al. [2]. An account of these can be found in Berndt’s book [3]. Naika and Dharmendra [4] also established some general theorems for explicit evaluations of the product am,n and found some new explicit values therefrom. Further results on am,n can be found in [5, 6].In [7], Mahadeva Naika et al. defined the product
(4)bm,n=ne-(π/4)(n-1)m/nψ2(-e-πmn)ϕ2(-e-2πmn)ψ2(-e-πm/n)ϕ2(-e-2πm/n).
They established general theorems for explicit evaluations of bm,n and obtained some particular values. Mahadeva Naika et al. [8] established general formulas for explicit values of Ramanujan’s cubic continued fraction V(q) in terms of the products am,n and bm,n defined above, where
(5)V(q):=q1/31+q+q21+q2+q41+q3+q61+⋯,|q|<1.
and found some particular values of V(q).Motivated by the above work, in this paper, we define the product of theta-functionslk,n as
(6)lk,n=ψ(-q)ϕ(qk)q(k-1)/8ψ(-qk)ϕ(q);q:=e-πn/k,
where n and k are positive real numbers. We establish several properties of the product lk,n. We prove general formulas for explicit evaluations of lk,n and find its explicit values. As application of the product lk,n, we also offer explicit formulas for explicit values of Ramanujan’s cubic continued fraction V(q) in terms of lk,n and give examples.In Section2, we collect some preliminary results. In Section 3, we prove several properties of the product lk,n. Section 4, is devoted to find explicit values of lk,n. Finally in Section 5, we offer explicit formulas for explicit evaluations of continued fraction V(q) in terms of lk,n with examples.To end this introduction, we define Ramanujan’s modular equation. LetK,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). 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. By denoting zr=ϕ2(qr), where q=exp(-πK′/K),  |q|<1, the multiplier m connecting α and β is defined by m=z1/zn.


## 2. Preliminary Results

Lemma 1 (see [9, p. 43, Entry 27(i)]).
Ifαβ=π then
(8)αϕ(e-α2)=βϕ(e-β2).Lemma 2 (see [10, p. 1049, (1.13)]).
Ifαβ=π2 then
(9)e-α/8α4ψ(-e-α)=e-β/8β4ψ(-e-β).Lemma 3 (see [9, p. 122, Entry 10(i)]).
One has(10)ϕ(q)=z1.Lemma 4 (see [9, p. 123, Entry 11(ii)]).
One has(11)ψ(-q)=z12{α(1-α)}1/8q-1/8.Lemma 5 (see [9, p. 124, Entry 12(v), (vi)]).
One has(12)χ(q)=21/6{α(1-α)q}-1/24,χ(-q)=21/6(1-α)1/12(αq)-1/24.
One also notes that if we replace q by qn in the Lemmas 3 and 4, then z1 and α will be replaced by zn and β, respectively, where β has degree n over α.Lemma 6 (see [9, p. 345, Entry 1(i), (ii)]).
IfV(q) is as defined in (5), then
(13)1+1V(q)=ψ(q1/3)q1/3ψ(q3),(14)1+1V3(q)=ψ4(q)qψ4(q3),(15)2V(q)=1-ϕ(-q1/3)ϕ(-q3).Lemma 7 (see [9, p. 347]).
IfV(q) is as defined in (5), then
(16)1-8V3(q)=ϕ4(-q)ϕ4(-q3).Lemma 8 (see [9, p. 231, Entry 5(xii)]).
LetP={16αβ(1-α)(1-β)}1/8 and Q=(β(1-β)/α(1-α))1/4, then
(17)Q+Q-1+22(P-P-1)=0,
where β has degree 3 over α.Lemma 9 (see [9, p. 282, Entry 13(xiv)]).
LetP={16αβ(1-α)(1-β)}1/12 and Q=(β(1-β)/α(1-α))1/8, then
(18)Q+Q-1+2(P-P-1)=0,
where β has degree 5 over α.


## 3. Some Properties oflk,n

In this section, we study some properties of the productlk,n.Theorem 10.
For all positive real numbersk and n, one has (i)
lk,1=1,(ii)
lk,1/n=1/lk,n,(iii)
lk,n=ln,k.Proof.
Using the definition oflk,n and Lemmas 1 and 2, we easily arrive at (i). Replacing n by 1/n in lk,n and using Lemmas 1 and 2, we find that lk,nlk,1/n=1 which completes the proof of (ii). To prove (iii), we interchange n and k in lk,n.Remark 11.
By using the definitions ofϕ(q), ψ(q), and lk,n, it can be seen that lk,n has positive real value and that the values of lk,n increases as n increases when k>1. Thus, by Theorem 10(i), lk,n>1 for all n>1 if k>1.Theorem 12.
For all positive real numbers k, m, and n, one has(19)lk,n/m=lmk,nlnk,m.Proof.
Using the definition oflk,n, we find that
(20)lmk,nlnk,m=e-π(n/mk-m/nk)/8ψ(-e-πn/mk)ϕ(e-πm/nk)ψ(-e-πm/nk)ϕ(e-πn/mk).
Applying Lemmas 1 and 2 in the denominator of right hand side of (20) and simplifying using Theorem 10(ii) and (iii), we complete the proof.Corollary 13.
For all positive real numbers k and n, one has(21)lk2,n=lnk,nlk,n/k.Proof.
Settingk=n in Theorem 12 and simplifying using Theorem 10(ii), we obtain
(22)lk2,m=lmk,klk,m/k.
Replacing m by n, we complete the proof.Theorem 14.
Let k, a, b, c, and d be positive real numbers such thatab=cd. Then
(23)la,blkc,kd=lka,kblc,d.Proof.
From the definition oflk,n and using ab=cd for positive real numbers k,a,b,c, and d, we deduce that
(24)lka,kbla,b-1=lkc,kdlc,d-1.
Rearranging the terms in (24) we arrive at the desired result.Corollary 15.
For any positive real numbersn and p, one has
(25)lnp,np=lnp2,nlp,p.Proof.
The result follows immediately from Theorem14 with a=p2,b=1,c=d=p and k=n.Theorem 16.
For all positive real numbersa, b, c, and d, one has
(26)la/b,c/d=lad,bclac,bd.Proof.
Applying Theorem10(iii) in Theorem 12, we deduce that, for all positive real numbers a, b, and n(27)la/b,n=la,bnlb,an-1.
Now setting n=c/d and again employing Theorems 12 and 10(iii) in (27), we complete the proof.Theorem 17.
For all positive real numbersa and b, one has (i)
la/b,a/b=lb,bla,a/b2,(ii)
la,ala,b2/a=lb,blb,a2/b,(iii)
la,alb,a2b=lb,ba,ab2.Proof.
By using Theorems10(ii) and 16, we find that
(28)la/b,a/b=lb,bla,a/b2.
So we complete the proof of (i). Setting n=a/b in (27), we find that
(29)la/b,a/b=la,alb,a2/b-1.
Now (ii) follows from (28) and (29) and Theorem 10(ii).
By using Theorems10(ii) and 12, we find that
(30)la/b,a/b=lb,blab2,ala2,b2-1.
Similarly, we find that
(31)lb/a,b/a=la,ala2b,blb2,a2-1.
From (30), (31), and Theorem 10(ii) and (iii), we complete the proof of (iii).


## 4. Explicit Values oflk,n

In this section, we prove general theorems for explicit evaluations oflk,n and find some explicit values. First we define Ramanujan’s class invariants. Ramanujan’s two class invariants Gn and gn which are defined by
(32)Gn=2-1/4q-1/24χ(q),gn=2-1/4q-1/24χ(-q),q:=e-πn,
where χ(q)=(-q;q2)∞ and n is a positive rational number. Employing Lemma 5 in (32) it follows that
(33)Gn={4α(1-α)}-1/24,gn=2-1/12(1-α)1/12α-1/24.
Also, if β has degree r over α, then
(34)Gr2n={4β(1-β)}-1/24,gr2n=2-1/12(1-β)1/12β-1/24.
In his notebooks [1] and paper [11], Ramanujan recorded a total of 116 class invariants or monic polynomials satisfied by them. An account of these can also be found in Berndt’s book [3]. For further references, see [2, 12–17].Theorem 18.
One has(35)lk,n=(GnkGn/k)3.Proof.
Employing Lemmas3 and 4 in the definition of lk,n, we find that
(36)lk,n=(α(1-α)β(1-β))1/8.
Again from (32) and (33) with q:=e-πn/k, we find that
(37)GnkGn/k=(α(1-α)β(1-β))1/24.
Combining (36) and (37), we complete the proof.From Theorem18, it is obvious that if we know (Gnk/Gn/k) or (GnkandGn/k) then explicit values of lk,n can easily be determined. For example, we find some values of lk,n in next theorem by employing the values of (Gnk/Gn/k) or (GnkandGn/k) from literature.Theorem 19.
One has(i)
l3,5=(G15G5/3)3=(3+52),(ii)
l3,7=(G21/G7/3)3=(33+27)1/2,(iii)
l3,11=(G33/G11/3)3=(3+11/2),(iv)
l3,13=(G39/G13/3)3=1/23/2((3+13)/2+(3+313)/23/2),(v)
l5,11=(G55/G11/5)3=1/2((9+55)/2+(95+455)/2),(vi)
l5,13=(G65/G13/5)3=((7+65)/8+(65-1)/8)3,(vii)
l3,23=(G69/G23/3)3=(24+523)1/4(25+369/2)3/8,(viii)
l3,47=(G141/G47/3)3=(48+747)3/12(95+8141)3/8,(ix)
l3,71=(G213/G71/3)3=(3480+41371)1/4((73+5213)/2)3/8.Proof.
For (i)–(vi), the corresponding values of(Gnk/Gn/k) can be found in equations (4.10), (4.20), (4.29), (4.37), (4.51), and (4.60), respectively of [12]. For (vii) to (ix), the corresponding values of can be found in equations [p.227,(5.8)], [p.229,(5.15)], and [p.233,(5.25)], respectively of [3].Corollary 20.
One has(38)ln,n=(Gn2)3.Proof.
We setk=n in Theorem 18 and use the result G1=1, to complete the proof.From Corollary20, it is obvious explicit values of ln,n can easily be determined if the corresponding values of the class invariants Gn2 are known. We give some examples in the next theorem.Theorem 21.
One has(i)
l2,2=G43=(1+2)3/4/29/16,(ii)
l4,4=G163=2-3/4(16+1521/4+122+923/4)3,(iii)
l3,3=G93=(1+3)/2,(iv)
l5,5=G253=((71/4+4+7)/(2))3,(v)
l7,7=G493=((1+5)/(2))3,(vi)
l9,9=G813=(2(3+1)3+1)/(2(3-1)3+1).Proof.
For (i) and (ii), we use the values ofG4 and G16 from [18, p. 114-115, Theorem 6.2.2(ii) (vi)]. For (iii)–(vi), we use the corresponding values of Gn from [3, p. 189-193].Ramanujan’s Schalfli type modular equations of prime degree can also be used to explicit values of the productlk,n. We offer two theorems as examples.Theorem 22.
One has(39)l3,n-2+l3,n2+22{(G3nGn/3)-3-(G3nGn/3)3}=0.Proof.
We use the definitions of class invariantslk,n and Gn in Lemma 8, to complete the proof.Theorem 23.
One has(40)l5,n+l5,n-1+2{(G5nGn/5)-2-(G5nGn/5)2}=0.Proof.
We use the definitions of class invariantslk,n and Gn in Lemma 9, to complete the proof.


## 5. Evaluations of Ramanujan’s Cubic Continued Fraction

In this section, we prove two formulas of explicit evaluations Ramanujan’s cubic continued fractionV(q) in terms of the product lk,n. We also give examples.Theorem 24.
One has(41)V3(-e-πn/3)=l3,n4+1-l3,n8+34l3,n4+116l3,n4.Proof.
Replacingq by -q in (14) and Lemma 7, we obtain
(42)-(1+1V3(q))=ψ4(-q)qψ4(-q3),(43)1-8V3(-q)=ϕ4(q)ϕ4(q3).
Dividing (42) by (43), setting q:=e-πn/3 and simplifying using the definition of lk,n, we obtain
(44)8l3,44V6(-e-πn/3)-(l3,44+1)V3(-e-πn/3)-1=0.
Solving (2) for V3(-e-πn/3) and noting that V(-e-πn/3)<0 and the fact in Remark 11, we arrive at the desired result.From Theorem24, it is clear that to find explicit values of V3(-e-πn/3) it is enough to know the explicit values of l3,n. For example, noting l3,1=1 from Theorem 10(i) we evaluate
(45)V3(-e-π/3)=-1/4.Theorem 25.
One has(46)V(-e-πn)=l9,n+1-l9,n2+10l9,n+14l9,n.Proof.
Replacingq by q3 and then replacing q by -q in (13) and (15), we obtain
(47)-(1+1V(-q3))=ψ(-q)qψ(-q9),(48)1-2V(-q3)=ϕ(q)ϕ(q9).
Dividing (47) by (48), setting q:=e-πn/9 and simplifying by using the definition of lk,n, we arrive at
(49)2l9,nV2(-e-πn)-(l9,n+1)V(-e-πn)-1=0.
Solving (49) for V(-e-πn) and noting V(-e-πn)<0 and the fact in Remark 11, we complete the proof.From Theorem25, it is obvious that if we know the values of l9,n the corresponding values of V(-e-πn) can easily be evaluated. For example, by noting l9,1=1 from Theorem 10(i) we calculate
(50)V(-e-π)=1-32.
Similarly, one can find explicit values of V(-e-3π) by using the value of l9,9 from Theorem 21(vi).



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*Source: 620756-2012-12-20.xml*