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Kyungpook Mathematical Journal -0001; 56(3): 715-726

Published online November 30, -0001

Copyright © Kyungpook Mathematical Journal.

On Some Modular Equations in the Spirit of Ramanujan

Belakavadi Radhakrishna Srivatsa Kumar

Department of Mathematics, Manipal Institute of Technology, Manipal University, Manipal 576 104 India.

Received: May 12, 2014; Accepted: April 7, 2016

Abstract

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In this paper, we establish some new P-Q type modular equations, by using the modular equations given by Srinivasa Ramanujan.

Keywords: Theta functions, Modular equations

1. Introduction

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In Chapter 16 of his second notebook [9], S. Ramanujan developed, theory of theta-function and his theta-function is defined by

f(a,b):=∑n=-∞∞an(n+1)/2bn(n-1)/2,         ∣ab∣ <1.

Note that, if we set a = q2iz, b = q−2iz, where z is complex and Im(τ ) > 0, then f(a, b) = ϑ3(z, τ), where ϑ3(z, τ) denotes one of the classical theta-functions in its standard notation [16, p. 464]. The three most important special cases of f(a, b) [4, p, 36] are

ϕ(q):=f(q,q)=∑n=-∞∞qn2=(-q;q2)2∞(q2;q2)∞=(-q;-q)∞(q;-q)∞,ψ(q):=f(q,q3)=∑n=0∞qn(n+1)2=(q2;q2)∞(q;q2)∞,f(-q):=f(-q,-q2)=∑n=-∞∞(-1)nqn(3n-1)/2=(q;q)∞.

After Ramanujan, we define

χ(q):=(-q;q2)∞,

where we employ the customary notation

(a;q)∞:=∏n=0∞(1-aqn),         ∣q∣ <1.

We now define a modular equation as given by Ramanujan. The complete elliptic integral of the first kind K(k) is defined by

K(k):=∫0π/2dφ1-k2Sin2φ=π2∑n=0∞(12)n(n!)2k2n=π2F21 (12,12;1;k2),

where 0 < k < 1. The series representation in (1.1) is found by expanding the integrand in a binomial series and integrating termwise and 2F1 is the ordinary or Gaussian hypergeometric function defined by

F21(a,b;c;z):=∑n=0∞(a)n(b)n(c)nn!zn,         ∣z∣ <1,

with

(a)k=Γ(a+k)Γ(a).

where a, b and c are complex numbers such that c is not a nonpositive integer. 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 moduli k, k′ l and l′ respectively. Suppose that the equality

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 (1.2). Ramanujan recorded his modular equations in terms of α and β, where α = k2 and β = l2. We often say that β has degree n over α. The multiplier m is defined by

m=KL.

Ramanujan [4, p. 122–124] recorded several formulae for ϕ, ψ, f and χ at different arguments of α q and z:=F21(12,12;1;α) by using

ϕ2(q)=2πK(k)=F21 (12,12;1;k2),         q=exp(-πK′/K).

Ramanujan’s modular equations involve quotients of function f(−q) at certain arguments. For example [5, p. 206], let

P:=f(-q)q1/6f(-q5)         and         Q:=f(-q2)q1/3f(-q10),

then

PQ+5PQ=(QP)3+(PQ)3.

These modular equations are also called Schläfli-type. Since the publication of [5], several authors, including N. D. Baruah [2], [3] M. S. M. Naikia [7], [8] K. R. Vasuki [12], [13] and K. R. Vasuki and B. R. Srivatsa Kumar [14] have found additional modular equations of the type (1.3). Recently C. Adiga, et. al. [1] have established several modular relations for the Rogers-Ramanujan type functions of order eleven which analogous to Ramanuja’s forty identities for Rogers-Ramanujan functions and also they established certain interesting partition-theoritic interpretation of some of the modular relations and H. M. Srivastava and M. P. Chaudhary [11] established a set of four new results which depicit the interrelationships between q-product identities, coninued fraction identities and combinatorial partition identities.

On page 366 of his ‘Lost’ notebook [10], Ramanujan has recorded a continued fraction

G(q):=q1/31+q+q21+q2+q41+…         ∣q∣ <1,

and claimed that there are many results of G(q) which are analogous to the famous Roger’s-Ramanujan continued fraction. Motivated by Ramanujan’s claim H. H. Chan [6], N. D. Baruah [2], K. R. Vasuki and B. R. Srivatsa Kumar [15] have established new identities providing the relations between G(q) and seven continued fractions G(−q), G(q2), G(q3), G(q5), G(q7), G(q11) and G(q13). We conclude this introduction by recalling certain results on G(q) stated by Ramanujan [4] and H. H. Chan [6].

G(-q):=q1/3χ(q)χ(q3)

where χ(q) is defined as χ(q) = (−q; q2)∞.

G(q)+G(-q)+2G2(-q)G2(q)=0

and

G2(q)+2G2(q2)G(q)-G(q2)=0.

For a proof of (1.5) and (1.6), see [6].

Motivated by the above works in this paper, we establish some new P-Q type modular equations, by employing Ramanujan’s modular equations.

2. Main Results

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Theorem 2.1

If

X:=q1/3χ(q)χ(q6)χ(q3)χ(q2)         and         Y:q2/3χ(q2)χ(q12)χ(q6)χ(q4)

then

2X2-22Y4X3-2Y+4Y2X-18X2Y3+17Y9X2-10Y8X+17Y10X3+Y11X+34Y5X+328Y7X3-160Y6X2-30Y7X6-30Y6X5+12Y5X4-371Y8X4+328Y9X5-10Y11X4-160Y10X6+34Y11X7-22Y9X8+12Y8X7+4Y11X10-18Y10X9-2Y12X11+10Y2X4+20Y4X6+20Y6X8+10X10Y8+2Y10X12=0.
Proof

From (1.4) and the definition of X and Y, it can be seen that

B-AX=0         and         C-BY=0.

where A = G(−q), B = G(−q2) and C = G(−q4). On changing q to q2 in (1.5), we have

G(q2)+G(-q2)+2G2(-q2)G2(q2)=0

and also change q to −q in (1.6), we have

G2(-q)+2G2(q2)G(-q)-G(q2)=0.

Eliminating G(q2) between (2.2) and (2.3) using Maple,

2(AB)4-4(AB)3+3(AB)2+AB+A3+B3=0.

Now on using first identity of (2.1) in (2.4), we obtain

2B6-4B4X+3B2X2+X3+BX+BX4=0.

On replacing q to q2 in (2.4) we see that

2(BC)4-4(BC)3+3(BC)2+BC+B3+C3=0.

Using second identity of (2.1) in the above, it is easy ton see that

2B6Y4-4B4Y3+3B2Y2+Y+B+BY3=0.

Finally, on eliminating B between (2.5) and the above, using Maple we obtain

P(X,Y)Q(X,Y)=0,

where

P(X,Y)=X-16Y4X2-6XY3-6Y5X3-2Y5+Y5X6+10Y3X4+10Y2X3+5YX2+5Y4X5-2Y6X

and

Q(X,Y)=-2Y+2X2-22Y4X3+4Y2X-18X2Y3+17Y9X2-10Y8X+17Y10X3+Y11X+328Y7X3+34Y5X-160Y6X2-30Y7X6-30Y6X5+12Y5X4-18Y10X9-371Y8X4+328Y9X5-10Y11X4-160Y10X6+34Y11X7-22Y9X8+12Y8X7+4Y11X10-2Y12X11+2Y10X12+10Y2X4+20Y6X8+20Y4X6+10X10Y8.

By examining the behavour of the first factor near q = 0, it can be seen that there is a neighbourhood about the origin, where P(X, Y ) ≠ 0 and Q(X, Y ) = 0 in this neighbourhood. Hence by the identity theorem, we have Q(X, Y ) = 0.

Theorem 2.2

If

X:=q1/6χ2(q3)χ(q)χ(q9)         and         Y:=q1/3χ2(q6)χ(q2)χ(q18)

then

(XY)3+(YX)3+{(XY)5/2+1(XY)5/2+11 ((XY)1/2+1(XY)1/2)} [(XY)3/2+(YX)3/2]=(XY)3+1(XY)3-11 ((XY)2+1(XY)2)+44 (XY+1XY)+8 (X3+1X3)+8 (Y3+1Y3)-86.
Proof

From Entry 12(v)of Chapter 17 [4, p. 124], we have

X={αγ(1-α)(1-γ)β2(1-β)2}1/24.

where β and γ be of the third and ninth degrees respectively, with respect to α. Let

B:=q1/3χ2(-q6)χ(-q2)χ(-q18).

Then from Entry 12(vii) of Chapter 17 [4, p. 124], we have

B={α2γ2(1-β)2β4(1-α)(1-γ)}1/24.

By (2.7) and (2.8), we deduce that

(αγβ2)1/8=XB         and         {(1-α)(1-γ)(1-β)2}1/8=X2B.

From Entry 3 (xii) and (xiii) of Chapter 20 [4, p. 352–358], we have

(β2αγ)1/4+((1-β)2(1-α)(1-γ))1/4-(β2(1-β)2αγ(1-α)(1-γ))1/4=-3mm′

and

(αγβ2)1/4+((1-α)(1-γ)(1-β)2)1/4-(αγ(1-α)(1-γ)β2(1-β)2)1/4=m′m.

where m = z1/z3 and m′ = z3/z9. Thus (2.9), (2.10) and (2.11) yields

M(X2B4+X4-B2X6)-B2=0         and         X4+X2B4-B2+3MX6B2=0.

where M = m/m′. Which implies

X6B6-6B4X4-B8X2+B6-X6+B2X2+X8B2=0.

Let

A:=q1/6χ2(-q3)χ(-q)χ(-q9)

Then, from Entry 12(vi) of Chapter 17 [4, p. 124], we have

A={αγ(1-β)4β2(1-α)2(1-γ)2}1/24.

From (2.7) and (2.13), we obtain

{(1-α)(1-γ)(1-β)2}1/8=XA         and         (αγβ2)1/8=AX2.

Using the above in (2.10) and (2.11), we deduce

(X4A4+X2-X6A2)M-A2=0         and         X2+X4A4-A2+3MX6A2=0.

From the above two identities, we obtain

X8A6-6X4A4-A8X6+A6X2-X2+A2+X6A2=0.

Changing q to q2 in the above, we have

Y8B6-6Y4B4-B8Y6+B6Y2-Y2+B2+Y6B2=0.

Now on eliminating B, between (2.12) and (2.14), using Maple we obtain

C(X,Y)D(X,Y)=0.

where

C(X,Y)=X4Y+X3+XY+6Y2X2+Y3X3+Y3+XY4

and

D(X,Y)=X8Y5-Y7X7-8Y4X7+X7Y+11X6Y6+11Y3X6+X5Y8-44Y5X5+11X5Y2-8X4Y7+86Y4X4-8X4Y+11Y6X3-44Y3X3+X3+11Y5X2+11Y2X2+XY7-8XY4-XY+Y3.

By examining the behavour of C(X, Y ) near q = 0, it can be seen that there is a neighbourhood about the origin, where this factor is not zero. Then the second factor D(X, Y ) = 0 in this neighbourhood. Hence by the identity theorem, we have

D(X,Y)=0.

On dividing the above throughout by (XY )4, we obtain the result.

Theorem 2.3

If

X:=q1/3χ(q3)χ(q5)χ(q)χ(q15)         and         Y:=q2/3χ(q6)χ(q10)χ(q2)χ(q30)

then

(XY)3+(YX)3+[(XY)5/2+1(XY)5/2+(XY)1/2+1(XY)1/2] ((XY)3/2+(YX)3/2)=(XY)3+1(XY)3-5 ((XY)2+1(XY)2)+10 (XY+1XY)+4 (X3+1X3+Y3+1Y3)-20.
Proof

Let

B:=q2/3χ(-q6)χ(-q10)χ(-q2)χ(-q30).

By Entry 12(v) and (vii) of Chapter 17 [4, p. 124], we have

X={αδ(1-α)(1-δ)βγ(1-β)(1-γ)}1/24         and         B={α2δ2(1-β)(1-γ)β2γ2(1-α)(1-δ)}1/24,

where α, β, γ and δ are of the first, third, fifth and fifteenth degrees respectively. From (2.16), we deduce that

(αδβγ)1/8=XB,         {(1-α)(1-δ)(1-β)(1-γ)}1/8=X2B.

From Entry 11(viii) and (ix) of Chapter 20 [4, p. 383–397], we have

(αδβγ)1/8+((1-α)(1-δ)(1-β)(1-γ))1/8-(αδ(1-α)(1-δ)βγ(1-β)(1-γ))1/8=m′m

and

(βγαδ)1/8+((1-β)(1-γ)(1-α)(1-δ))1/8-(βγ(1-β)(1-γ)αδ(1-α)(1-δ))1/8=-mm′.

Employing (2.17) in (2.18) and (2.19), we obtain

M(XB2+X2-X3B)-B=0   and   X2+B2X-B+MBX3=0,

where M=m/m′. Which implies

4X2B2+X3-X4B+XB4-X3B3-B3-BX=0.

Let

A:=q1/3χ(-q3)χ(-q5)χ(-q)χ(-q15).

Then, by employing Entry 12(vi) of Chapter 17 [4, p. 124] and (2.16) we deduce that

{(1-α)(1-δ)(1-β)(1-γ)}1/8=XA         and         (αδβγ)1/8=AX2.

Using these in (2.18) and (2.19), upon simplifying the resulting identities, and then replacing q by q2, we obtain

4B2Y2+Y-BY3+B4Y3-B3Y4-B3Y-B=0.

Eliminating B from (2.20) and (2.21), using Maple we obtain

C(X,Y)D(X,Y)=0.

where

C(X,Y)=X4Y+X3+XY+4Y2X2+Y3X3+Y3+XY4

and

D(X,Y)=X8Y5-X7Y7-4X7Y4+X7Y+5X6Y6+X6Y3+X5Y2+X5Y8-10Y5X5-4X4Y7+20X4Y4-4X4Y+X3Y6-10Y3X3+X3+X2Y5+5Y2X2+XY7-4XY4-XY+Y3.

It is same as discussed in Theorem 2.2, that C(X, Y ) ≠ 0 near q = 0 whereas D(X, Y ) = 0 in some neighbourhood q = 0. Hence by identity theorem, we have

D(X,Y)=0.

Finally, on dividing the above throughout by (XY )4, we obtain the result.

Theorem 2.4

If

X:=q2/3χ(q)χ(q7)χ(q3)χ(q21)         and         Y:=q4/3χ(q2)χ(q14)χ(q6)χ(q42)

then

p12+14p11+229p10+1328p9+1635p8-15550p7-8529p6-177572p5-37641p4+764070p3+2368728p2+4125694p1-2(2q23+24q21+158q19+586q17+663q15+13509q13+43169q11+36801q9-14490q7-612613q5-1259739q3-1742545q1)r3-2(2q21-6q19+51q17+208q15-111q13-2275q11-8880q9-22598q7-43267q5+65339q3-79989q1)r9-2(q15+2q13+4q11+20q9+78q7+88q5+38q3+155q1)r15+(6p11+60p10+162p9-560p8-5129p7-11254p6+10488p5+126726p4+406080p3+828738p2+1238441p1+1410116)s3+(p10-10p9+11p8+60p7+218p6+896p5+2022p4+3816p3+7277p2+111558p1+13838)s6+(p5-2p4-3p3+8p2+2p1-12)s9+4907562=0.

where

pn=(XY)n+1(XY)n,         qn=(XY)n/2+1(XY)n/2,rn=(XY)n/2+(YX)n/2,         sn=(XY)n+(YX)n.
Proof

Let

B:=q4/3χ(-q2)χ(-q14)χ(-q6)χ(-q42).

Then from Entry 12 (v) and (vii) of Chapter 17 [4, p. 124], we have

X={βδ(1-β)(1-δ)αγ(1-α)(1-γ)}1/24         and         B={β2δ2(1-α)(1-γ)α2γ2(1-β)(1-δ)}1/24.

where α, β, γ and δ are of the degrees 1, 3, 7 and 21 respectively. From (2.23), we deduce that

XB=(βδαγ)1/8,         X2B={(1-β)(1-δ)(1-α)(1-γ)}1/8.

From Entry 13 of Chapter 20 [4, p. 400–403], we have

(βδαγ)1/4+((1-β)(1-δ)(1-α)(1-γ))1/4+(βδ(1-β)(1-δ)αγ(1-α)(1-γ))1/4-2 (βδ(1-β)(1-δ)αγ(1-α)(1-γ))1/8{1+(βδαγ)1/8+((1-β)(1-δ)(1-α)(1-γ))1/8}=mm′

and

(αγβδ)1/4+((1-α)(1-γ)(1-β)(1-δ))1/4+(αγ(1-α)(1-γ)βδ(1-β)(1-δ))1/4-2 (αγ(1-α)(1-γ)βδ(1-β)(1-δ))1/8{1+(αγβδ)1/8+((1-α)(1-γ)(1-β)(1-δ))1/8}=9mm′.

Employing (2.24) in (2.25) and (2.26), we obtain

X2B4+X4+B2X6-2BX3(B+XB2+X2)-B2M=0

and

(X2B4+X4+B2-2BX(BX2+X+B2))M-9B2X6=0.

where M = mm′, which implies

X6+B6+6B4X4+X8B2-2X7B+X2B8+X6B6-2X4B7+B2X2-2B4X-2B4X7-2BX4-2B7X=0.

Let

A:=q2/3χ(-q)χ(-q7)χ(-q3)χ(-q21)

From Entry 12 (vi) of Chapter 17 [4, p. 124] and (2.23), we deduce that

XA={(1-β)(1-δ)(1-α)(1-γ)}1/8         and         AX2=(βδαγ)1/8.

Employing these in (2.25) and (2.26) up on simplifying, the resulting identities and then replacing q by q2, we obtain

B2Y6+B2-2BY+B8Y6+B6Y8-2B7Y7+B6Y2-2B4Y-2B4Y7-2B7Y4-2BY4+Y2+6B4Y4=0.

On eliminating B between (2.27) and (2.28), using Maple we obtain

C(X,Y)D(X,Y)E(X,Y)=0.

where

C(X,Y)=X6Y6-2X4Y7+X2Y8-2XY7-2X7Y-2X4Y+Y6+Y2X8+X2Y2+6Y4X4-2Y4X-2Y4X7+X6,D(X,Y)=Y6+256X6Y6+38X4Y7+2X2Y8-2X4Y+2Y2X8+X2Y2+29Y4X4-2Y4X+38Y4X7-16X14Y5+66X10Y7+14X8Y5-10X5Y11-20X3Y12-10X11Y5+X14Y2+72X7Y7-20Y13X4-35X4Y10+66Y9X6-16X5Y14-35X10Y4+X10Y16-2X6Y15-35X6Y12-4X7Y13+66X7Y10+2X8Y14+14X8Y11+466X8Y8+38X9Y12+72X9Y9+256X10Y10+12X11Y11+14X11Y8-2X12Y15+29X12Y12+38X12Y9-35X12Y6-14X10Y13-2X15Y6-2X13Y13-14X13Y10-4X13Y7+X14Y14+2X14Y8-2X15Y12+X16Y10+14X5Y8-4X3Y9+Y14X2-2XY10-2YX10-16X2Y11+X6-2X3Y3-20X13Y4+66X9Y6-16Y2X11-14X6Y3-20Y3X12-4Y3X9+12X5Y5-14X3Y6.

and E(X, Y ) is as in (2.22).

As discussed in Theorem 2.2, by examining the behaviour of C(X, Y) and D(X, Y ) near q = 0, it can be seen that there is a neighbourhood about the origin, where these factors are not zero. Then the third factor E(X, Y ) = 0 in this neighbourhood. Hence by identity theorem, we have E(X, Y ) = 0. Finally, on dividing E(X, Y ) throughout by (PQ)16 and then simplifying we have the result.

Acknowledgement

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The author would like to thank the anonymous referee for his/her valuable suggestions.

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