A FUNDAMENTAL MODULAR IDENTITY AND SOME APPLICATIONS · 2018-11-16 · mathematics of computation volume 61, number 203 july 1993, pages 83-95 A FUNDAMENTAL MODULAR IDENTITY AND SOME

mathematics of computationvolume 61, number 203july 1993, pages 83-95

A FUNDAMENTAL MODULAR IDENTITYAND SOME APPLICATIONS

RICHARD BLECKSMITH, JOHN BRILLHART, AND IRVING GERST

Dedicated to the memory ofD. H. Lehmer

Abstract. We prove a six-parameter identity whose terms have the form

xaT(kx, lx)T(k2 , l2), where T(k, I) = Y,-x xk"2+ln . This identity is then

used to give a new proof of the familiar Ramanujan identity H(x)G(x11) -

x2G(x)H{xu) = 1, where G(x) = Tl^oK1 " *5"+1)(l - x5"+4)]-' and

H(x) = n^oK1 - *5"+2)(l - x5n+3)]-' . Two other identities, called "bal-

anced Q2 identities", are also established through its use.

1. INTRODUCTION

It is our purpose in this paper to prove a fundamental identity, whose terms

are a certain type of modular form, and then to demonstrate its use in proving

other identities. This general identity, its proof, and the three applications of

the identity that are made here were discovered by a mixture of theoretical

studies and computer analyses of particular identities. This work is different

from that presented in [2-4], where the identity to be proved was first discovered

by a computer search and then was proved by a computer-assisted, theoretical

argument.

We will use the two single-variable T-functions 7b and Tx (cf. [4, equation

(2)]), defined byOO OO

T(k, I) = T0(k,l) dáfYíxkn2+ln = Y[(l -x2kn)(l + x2k"-k+,)(l + x2kn~k~l)

— oo n=l

andOO OO

Tx(k,l) dàiY/(-l)nxk"2+ln = Y[(l -x2kn)(l -x2k"-k+l)(l -x2k"-k-').

-oo «=1

We call an identity a " T2 identity" if each of its terms has the form

xaTtl(kx,lx)T(2(k2,l2),

where ex and e2 are 0 or 1. We also say that a T2 identity is "balanced" if

the first component pair (kx ,k2) in each of its terms is the same.

Received by the editor July 27, 1992.

1991 Mathematics Subject Classification. Primary 05A19.Key words and phrases. Triple and quintuple product, modular identity, balanced T2 and Q1

identity.

© 1993 American Mathematical Society

0025-5718/93 $1.00+ $.25 per page

83

License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use

84 RICHARD BLECKSMITH, JOHN BRILLHART, AND IRVING GERST

In §2 we will prove the fundamental identity (Theorem 1). It will then be

used in §3 to give a new proof of a familiar identity of Ramanujan (Theorem 2)

and in §4 to provide proofs for two interesting identities, called "balanced Q2

identities" (Theorems 3 and 4). In all of these proofs the fundamental identity

is specialized to produce a small family of identities by assigning sets of values

to its parameters. The identity in question is then established by showing it to

be a linear combination of the identities in this family.

These three proofs illustrate a tentative proof method of three steps: ( 1 ) If

possible, transform the given identity into a T2 identity; (2) Balance this iden-

tity using the expansion formula in [4, p. 779]; (3) Determine if and how the

fundamental formula can be used to generate a small family of identities in

terms of which the given identity can be expressed as a linear combination.

2. The fundamental identity

The following general identity, which is of interest in its own right, is also

important in generating special sets of identities from which proofs of other

identities can be made.

In working with T-functions, it is often important to re-index their sums.

This can be accomplished by simple transformation rules, which are also useful

in putting T in "reduced" form. (The function Te (k, I) is in "reduced form"

when 0<l<k (cf. [4, p. 780]).) Then, for e e {0, 1}, there is

the negative rule [4,(13)]:

(2.1) Te(k,-l) = T((k,l),

the single-step formula [4, (14)]:

(2.2) xaT((k, I) = (-l)exa-('-k)T€(k,2k-l),

and

the general transformation formula: If / = 2kq + r, where q e Z , then

(2.3) xaT((k,l) = (-iy"xa-('2k-i'rTf(k,r).

Proof of (2.3). We have thatOO

xaTt(k, I) = xtt^(-l)(nxknl+l2k<i+r)n

— OO

OO

— x<* y/ \y[n-q)xk(n-q)2M2kq+r)(n-q)

— oo

= (-l)(<lx"-q2k-'irT((k,r). D

Note, when we use (2.3) to put T(k, I) numerically into reduced form, we

take r to be in the interval -k < r < k, and then use (2.1) if necessary.

Throughout the rest of this paper we will give all T-functions with numerical

arguments in reduced form.

Theorem 1. Suppose that m, u, v e Z + , e, f e Q, and k e Q+ , where

uv < 2m . Then


A FUNDAMENTAL MODULAR IDENTITY AND SOME APPLICATIONS 85

(2.4) J2 xa"T(kx,lXn)T(k2,l2n)= £ xa»T(kx,l\n)T(k2,r2n),n€Rm n€R'm

where

2vk(2.5) a„ =-n +2en, kx = uk, k2 = (2m - uv)vk,

m

, 2uvk .hn =-n + ue + f,

m

(2.6)

■2vkI l2n = (2m-uv)(—n + e) -vf,

v m )

2uvkl\„ =-n + ue - f,'" m

l'2n = (2m - uv)(-n + ej +vf,<2vk

m

and Rm and R'm are any complete residue systems (mod m).

Proof. The proof rests on the following generalization of the expansion for-

mula in [4, p. 779]. (The notation in [4] is changed here to suit the present

proof. Note that the proof presented in [4] remains valid when we allow the

parameters k > 0 and / in T(k, I) = J^-^ xk"2+ln to be rational.)

Let a, b € Z, p e Z+, (kx , Àx), (k2 , X2) e Q+ x Q, and ex, e2 e {0, 1}.If the separability condition

(2.7) KXb = K2a(p - ab)

is satisfied, then

Tei(Kx,Àx)Tei(K2,À2)

(2.8) = ^{-^"x^^Ts^J^nWs^J^n)),n£Rf,

where

kx—KX+ K2a2, k2 = K2ß(p - ab),

lx(n) = ax - A2a - 2tc2an , l2(n) = (p.- ab)(2K2n + a2) + AXb,

¿i = (ei + é2#) (mod 2), 62 — (exb + e2(p - ab)) (mod 2),

and Ru is a complete residue system (mod p ).

In (2.8) substitute

(2m - uv)uk vkK\ =-»-— , k2 = —,

(2-9) if ; .v ; M=-f, h = e,

ex = u mod 2, e2 = 1,

and use the parameters a = u, b = v , and p = 2m . Note that (2.7) is satisfied,

since by (2.9)

kxv = K2u(p - uv).


86 RICHARD BLECKSMITH, JOHN BRILLHART. AND IRVING GERST

We then have that

, -, (2m - uv)uk vku2 ,kx =kx+ K2aL =---+ -z— = uk,

2m 2m

(2.10) k2 = K2p(p-ab) = (2m-uv)vk,

ôx = (ex + e2u) mod 2 = 0,

S2 = (exv + e2(2m - uv)) mod 2-v(ex+ e2u) mod 2 = 0.

Since T = To, the expansion is then

(2.11) T(l(Kx,AX)Te2(K2,A2)= J2 (-irnxa(n)T(kx,lx(n))T(k2,l2(n)),

neRim

where (using -/i(zz) by (2.1))

a(n) = K2n2 + l2n = -—n2 + en,

(2.12) /,(«) = 2K2an + À2a - ax =-n + ue + f,m

/ v k \l2(n) = (p - ab)(2K2n + X2) + Xxb = (2m - uv)(—n + e) - vf.

\ m )

Now separate the sum on the right side of (2.11 ) into the difference of two sums,

Tu(kx , Xx)T(1(k2 , k2) =S+-S-,

where

S+ = 53 xa^T(kx , lx(2n)) Tik2, l2i2n)) and

(2.13) "!!;S- = 51 xn{2n+l)T(kx,lx(2n+ l))T(k2,l2(2n+ 1)).

neRm

Next, consider the expansion of the product

Ttl(Ki,-Ai)Te2(K2,A2)=S'+-SL

= 53 x"^T(kx,l[(2n))T(k2,l2(2n))neR'„,

53 x^2n+]^T(kx, l[(2n + 1)) T(k2,l'2(2n + 1)),n£R'„

where /{ and 1'2 are obtained from lx and l2 by replacing / by -/ in (2.12).

Since rf|(zci, ax) = T(1(kx , -ax) , we have that

(2.14) S+-S- =5;-51.

Now repeat the preceding development but with ex — e2 = 0 in (2.8). Since

all the terms of the resulting expansion are the same as before, except that they



are now all positive, it follows that

T0(Kl,Al)To(K2,h) = S+ + S-

and

To(kx,-ax)T0(k2,a2)=S'+ + S'_,

so

(2.15) 5++ 5_ = 5;+51.

It then follows from (2.14) to (2.15) that 5+ = 5; and 5_ = 51. On the onehand, from (2.13) we have

5+= 53 xa"T(kx,lXn)T(k2,l2n),n€Rm

where from (2.12) we obtain

,_ , 2vk 7an = a(2n) =-n + 2en,

m, . ._ . 2uvk .h„ = h(2n) =-n + ue + f,

mjlvk

m

which is the left-hand side of (2.4). Similarly S'+ is the right-hand side of (2.4),so (2.4) is proved. D

¿2« = h(2n) = (2m - uv)(-n + e) - vf,\ m )

Remarks. 1. In (2.4) we can permit a„ to be negative, since negative powers of

x can be removed from the equation by multiplying through by an appropriate

power of x.

2. We do not obtain another identity from the equation 5_ = 51, since

this identity arises from (2.4) by replacing "e" by "e + ^ " in (2.5) and (2.6),

and multiplying the resulting identity by " xe+(vkl2m) ".

3. A NEW PROOF OF A RAMANUJAN IDENTITY

In this section and the next we will prove three identities using Theorem 1.

In each of these proofs a small family of identities is derived from (2.4) by

giving certain sets of values to its parameters. The identity in question is then

verified by showing it is a particular linear combination of these identities.

The first identity we will consider is a familiar result of Ramanujan.

gW = n(1_x5,+1)1(1_x5n+4) *«à H(x) = H 5n+2l){l_x5n+3),

Theorem 2 (Ramanujan [7, 9, 10, 1,6, 8]). If

1

n=0v" "' n=0

then

(3.1) H(x)G(xn)-x2G(x)H(xn)= 1.

Proof. The proof is in three steps.

(i) We begin by transforming (3.1) into a T2 identity. For simplicity, write

the infinite products in (3.1) in the abbreviated notation



(ri, r) d^ n ;i — jc")n=l

., r, (mod m)

and then clear the fractions, which gives the identity

(±l)5(±22)ss - x2(±2)5(±l 1)55 = (±1, ±2)s (±11, ±22)55.

Multiplying through by (0)5 (0)55 and rewriting, we obtain the identity

(0, ±22)55(0, ±1)5 -x2(0, ±11)55 (0, ±2)5(3.2)

(0),(0)„=(0, ±11)33(0, ±1)3.

Identifying each factor in (3.2) as a certain Tx(k, /), we obtain the desired T2identity, viz.

(3.3) r,(f, l±)Tx(\,\)-x2Tx(%,^)Tx(\,{) = T,(f , if)7i(f, \).

(ii) We next use the expansion formula (cf. [4, p. 779]) (here we use the

original notation) on each of the three terms in (3.3) to convert it into a bal-

anced T2 identity. Program Forward [4, p. 789] indicates that both (kx, k2) =

(■y , f ) and (kx, k2) = (",§) can be expanded to the same pair (Kx, K2) =

(30, 330). We therefore expand the two terms on the left side of (3.3) intosums that are balanced at (30, 330), using the parameters [a, b, m] = [1, 1, 12]

(cf. [4, p. 780 ]). This gives respectively the two groups of twelve terms in

Table 1. Next, the term on the right of (3.3) is expanded using the parameters

[3, 3, 20], giving the twenty terms listed in Table 2. (Note that the form of each

term in all these expansions is xa'T0(30, lXi)T0(330, l2i), while in Tables 1 and2 we have indicated only the signs and the values of a¡, lXi, l2i.)

Table 1. Expansions using [1, 1, 12]

Group 1 terms Group 2 terms

7i(f ,%)Txi\,\), [1,1,12]

1234

56789101112

sign

+

++

++

+

0147

13

18273446

557077

hi

491

14

619112416292126

hi

22*337788*

132*143187198*242*253297308*

5 1x'Tx(^,^)Tx(^,^), [1, 1,12]

234

5

67891011

12

sign

+

+

+

+

24

5111322263444

516773

162111266

291

244

199

14

hi

22337788

132143187198242253297308


A FUNDAMENTAL MODULAR IDENTITY AND SOME APPLICATIONS

Table 2. Expansion of Tx(^ , ^)TX(\ , \) using [3, 3, 20]

Group 3 positive terms Group 3 negative terms

89

1234

567

89

10

057

12232437466277

/,

42214202

2810168

26

hi

22*44

88*110176154220242*286308*

12

34

567

8910

12

1115162933486279

/,

135

2923111971

2517

hi

11

5577

121143187209253275319

Table 3. A¡(x) and B¡(x)

Ai(x)=xaiT(30,lXi)T(330,l2i) Bi(x)=xaiT(30,lXi)T(330,l2i)

1234

567

8910II1213141516

hi hi

oi34

1014152125283247

50616678

135

211129231129

1197

119259

17

1155337777

121143143187187209253253275297319

1234

567

89101112

131415

16

hi

01

34

1011

17

2223263643

54616972

9161

222620192

2811104

298

21

14

33227744

88110143176154187220242253286297308

Equation (3.3) now becomes

(3.4) Group 1 terms - Group 2 terms = Group 3 terms.



Four terms in Group 1 cancel across this equation with four positive terms in

Group 3 and two terms in Group 1 cancel the two negative terms in Group 2.

(These are marked with an asterisk in Tables 1 and 2.) This leaves 16 terms on

each side of (3.4). Transposing the negative terms that remain in (3.4) to the

other side of the equation and dividing out the factor x in all the terms, we

obtain the desired balanced T2 identity:

(3.5)

16 16

53^w = 53ß,wi=i r=l

where the A,(x) and B¡(x) are listed in Table 3. It is this equation we must

prove to establish (3.1).(iii) We now obtain three identities, given in Table 4 below, by choosing

three sets of values for the parameters m , k , u, v , e , and / in (2.4). The

first two identities have m = 6, £ = 30, u = v = 1, and (e, f) = (|, -y)

and (|, "), respectively. The third identity has m = 10, k = 10, u — v = 3,

and (e,f) = (\,xi).

Table 4. The three identities with kx = 30 and k2 = 330

Identity 1: m = 6, £ = 30, u = v = 1, e = \, /=-T

03

17

265467

212

8221828

4466

154176264286

137

101315

03

17

265469

91

1911

2921

3377

143187253297

Identity 2: m = 6, /: = 30, u = v 9 f.3> / '33T

34

891315

0 21

182247

63

291

3377

143187253297

9*

14

01

19205158

12

222

2818

6644

176154264286


A FUNDAMENTAL MODULAR IDENTITY AND SOME APPLICATIONS

Identity 3: w = 10, Â: = 10, u = v = 3, e = § , / = T

91

1

25

67

101112

1416

01

1014

15283247

6178

/,

135

2923111971

2517

h

11

5577

121143187209253275319

11

12*

16

1011

17

2636436772

/,

21626208

22104

2814

h

44

2288

110154176220242286308

Now form the following linear combination of the above three identities,

(3.6) Identity 1 + jc3 x Identity 2 + Identity 3,

which yields equation (3.5) when six terms on the left-hand side of Identity 1

are cancelled with six terms on the right-hand side of Identities 2 and 3. (The

terms that cancel are indicated in the tables by an asterisk. The term numbers

in the # columns for the left and right sides in Table 4 correspond respectively

to those in the "z" columns in Table 3. Note that the a-values in Identity 2

must be increased by 3 because of the multiplier x3.) □

4. TWO BALANCED Q2 IDENTITIES

Within the set of balanced T2 identities there is a special and interesting

subset—the balanced Q2 identities. (Here Q stands for the usual quintuple

product

Q(m,k)def }(1 -xn) = YJxm(i"2+n)l2(x-

n€S -oo

3kn ,lkn+k)•

0, ±k , ±(m - 2k), ±(m - k), m (mod 2m)}where 5 = {n e Z+ :

The two identities of this kind which we will prove in this section are tri-

nomials of similar form. They have similar proofs as well, but are of quite

different types, as we will show in another paper. Because these identities are

already balanced, there is no need to balance them as we did in the proof of

Theorem 2. At present, these are two of the three balanced Q2 identities we

know.

(a) The first of these identities was originally proved as part of a proof in

[4, equation (16)]. To our knowledge, this is the first example of a balanced Q2

identity to appear in the literature.

Theorem 3. We have that

(4.1) (2(8, 3)0(56, 7) + x3Q(S, 1)0(56, 21) = 0(8, 2)0(56, 14).



Proof. Since (4.1) is a Q2 identity (balanced at (8, 56)), we can routinely

rewrite it as a T2 identity (balanced at (12, 84)) by first transforming each Qin (4.1) into T terms by the following formula [4, equation (24)]:

(4.2) Q(m,k) = T(-j-, y - 3fcj - xkT[— , - + 3k).

Therefore, using (2.1) and (2.2), we have

ß(8,3) = 77(12,5)-x277(12, 11), ß(8, 1) = 77(12, l)-jrT(12, 7),

and

ß(8, 2) = 77(12, 2)-x2r( 12, 10).

Replacing x by x1 in these three results, we obtain

0(56,21) = 77(84, 35) - x1477(84, 77),

ß(56, 7) = 77(84, 7)-x777(84, 49),

and

ß(56, 14) = 77(84, 14)-x14r(84, 70).

Thus, the first term in (4.1) becomes

ß(8, 3)0(56, 7)= [77(12, 5)-x277(12, 11)][77(84, 7) - x777(84, 49)]

(4.3) =5(0, 5, 7)-5(2, 11, 7) - 5(7, 5, 49)

+ 5(9, 11,49),

where for simplicity we have written

5(a,/1,/2) = x«r(12,/1)77(84,/2).

If we expand the second and third terms in (4.1) in the same way, we obtain

x3ß(8, l)ß(56, 21) =5(3, 1, 35)-5(4, 7, 35)-5(17, 1, 77)+ 5(18, 7, 77)

and

ß(8, 2)ß(56, 14) = 5(0, 2, 14)-5(2, 10, 14)-5(14,2, 70)+5(16, 10, 70).

With the help of these three results, identity (4.1) becomes the balanced 772

identity

5(0, 5, 7)+ 5(2, 10, 14)+ 5(3, 1,35)+ 5(9, 11,49)

+ 5(14,2,70) + 5(18,7,77)

( ' ' =5(0,2, 14)+ 5(2, ll,7) + 5(4,7,35) + 5(7,5,49)

+ 5(16, 10, 70)+ 5(17, 1, 77).

To prove (4.4), we use the subsidiary identity derived from Theorem 1 by setting

m = 4, k = 12, and u = v = 1, viz.

3

535(6«2 + 2e«, 6n + e + f, 7(6«+ <?)-/)

(4.5) "=° 3

= ¿5(6«2 + 2é*«, 6n + e-f, 7(6« + <?)+ /).n=0



If we put (e, f) = (\, ]) and (e, f) = (-\, y) into (4.5), we obtain respec-tively the two identities

5(0,5,7) + 5(3, l,35) + 5(9, 11, 49)+5(18, 7, 77)[ ' ' =5(0,2, 14) + 5(3,8,28)+5(9,4,56) + 5(16, 10,70)

and

5(0, 10, 14)+5(l,8,28) + 5(7,4,56) + 5(12,2,70)

= 5(0, ll,7) + 5(2,7,35) + 5(5,5,49) + 5(15, 1,77).

Then the linear combination (4.6) + x2 x (4.7) of these equations produces

(4.4) when the common sum 5(3, 8, 28) +5(9, 4, 56) is cancelled from itstwo sides. (Note that multiplying an 5 by x" increases the first argument of

5 by n .) G

Remark. The proof of (4.1) provides a third proof of identity (9) in [4].

(b) In the summer of 1990 we sent identity (4.1) to Oliver Atkin. Within24 hours he sent us identity (4.8) below, another trinomial identity with a form

similar to that of (4.1). An impressive performance!

Theorem 4 [Atkin]. We have that

(4.8) ß(14,3)ß(70,5) + x3ß(14, l)ß(70, 25) = ß(14, 5)ß(70, 15).

Proof. As in the proof of (4.1), we first use (4.2) to transform (4.8) into a

T2 equation, balanced at (21, 105). We find that

ß(14,3) = 7(21,2)-x37(21, 16), ß(14, 1) = 77(21, 4) -xT(2l, 10),

and

ß(14,5) = 7(21,8)-x47(21,20),

and, replacing x by x5 in these results, we obtain the equations

ß(70, 15) = 77(105, 10)-x157(105, 80),

ß(70, 5) = 7(105, 20)-x577(105, 50),

and

ß(70,25) = 7(105, 40)-x207(105, 100).

Thus, the three terms in (4.8) transform into

ß(14, 3)ß(70, 5) = 5(0, 2, 20)-5(3, 16,20)-5(5,2, 50) + 5(8, 16, 50),(4.9)

(4.10)

and

(4.11)

x3Q(l4, l)ß(70, 25) = 5(3, 4, 40)-5(4, 10,40)-5(23,4, 100) + 5(24, 10, 100),

ß(14, 5)0(70, 15) = 5(0, 8, 10)-5(4, 20, 10)-5(15, 8, 80) + 5(19,20, 80),

where we have written

5(a,/,,/2) = x"7(21,/1)7(105,/2).



Thus, equation (4.8) becomes the balanced 72 equation

5(0, 2, 20) + 5(3, 4, 40) + 5(4, 20, 10) + 5(8, 16, 50)

+ 5(15, 8, 80)+ 5(24, 10, 100)(" ' =5(0, 8, 10)+ 5(3, 16,20) + 5(4, 10, 40)+ 5(5, 2, 50)

+ 5(19,20, 80)+ 5(23, 4, 100).

Again we derive a subsidiary identity from Theorem 1 by setting m = 3,k — 21, and u = v = I, viz.

2

535(14«2 + 2e«, 14n + e + f, 5(14« + e) - f)

(4.13) "=0

535(14«2 + 2e«, I4n + e-f, 5(\4n + e) + f).n=0

Putting (e, f) = (3, -5), (6, -10), (5, 15) into (4.13) gives respectively thefamily of three identities

(4.14)

(4.15)

and

(4.16)

5(0, 2, 20)+5(8, 16, 50) + 5(20, 12, 90)

= 5(0,8, 10) + 5(8,6,60) + 5(19,20,80),

5(0, 4, 40)+ 5(2, 18, 30)+ 5(21, 10, 100)

= 5(0, 16, 20) + 5(2, 2, 50) + 5(17, 12, 90),

5(0,20, 10) + 5(4,6,60) + 5(ll, 8, 80)

= 5(0, 10,40)+5(1, 18, 30) + 5(19,4, 100).

The linear combination (4.14) +x3 x (4.15) +x4 x (4.16) of these equations

gives an identity which becomes (4.12) when the common sum 5(5, 18, 30) +

5(8, 6, 60) + 5(20, 12, 90) is cancelled from its two sides. □

Acknowledgment

We would like to thank Greg Manning for his help in developing the algo-

rithms that were used to find the proofs in this paper.

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Department of Mathematical Sciences, Northern Illinois University, DeKalb, Illi-

nois 60115E-mail address : [email protected]

Department of Mathematics, University of Arizona, Tucson, Arizona 85721

E-mail address: [email protected]

Department of Applied Mathematics and Statistics, SUNY at Stony Brook, Stony

Brook, New York 11794


A FUNDAMENTAL MODULAR IDENTITY AND SOME APPLICATIONS · 2018-11-16 · mathematics of computation volume 61, number 203 july 1993, pages 83-95 A FUNDAMENTAL MODULAR IDENTITY AND SOME

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