2011 Boonrod Yuttanan

Transcription

1 0 Boonrod Yuttanan

2 MODULAR EQUATIONS AND RAMANUJAN S CUBIC AND QUARTIC THEORIES OF THETA FUNCTIONS BY BOONROD YUTTANAN DISSERTATION Submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics in the Graduate College of the University of Illinois at Urbana-Champaign, 0 Urbana, Illinois Doctoral Committee: Professor A. J. Hildebrand, Chair Professor Kenneth B. Stolarsky, Co-Chair Professor Bruce C. Berndt, Director of Research Professor Alexandru Zaharescu

3 Abstract In this thesis, we prove several identities involving Ramanujan s general theta function. In Chapter, we give proofs for new Ramanujan type modular equations discovered by Somos and establish applications of some of them. In Chapter 3, we will give proofs for several Dedekind eta product identities which Somos discovered through computational searches and which Choi discovered in his work on basic bilateral hypergeometric series and mock theta functions. In Chapter, we derive new identities related to the Ramanujan-Göllnitz-Gordon continued fraction that are similar to those for the famous Rogers-Ramanujan continued fraction. We give a new proof of the 8-dissection of the Ramanujan-Göllnitz-Gordon continued fraction and also show that the signs of the coefficients of power series associated with this continued fraction are periodic with period 8. In Chapter 5, we prove several infinite series identities involving hyperbolic functions and hypergeometric functions by using the classical and quartic theories of theta functions. In Chapter 6, we study a new function called a quartic analogue of Jacobian theta functions. Finally, Chapter 7 is devoted to establishing new identities related to the Borweins cubic theta functions and Ramanujan s general theta function. We also give equivalent combinatorial interpretations of such identities. ii

4 To my parents and my advisor. iii

5 Acknowledgments I would like to give special thanks to my thesis advisor, Professor Bruce C. Berndt, for his very helpful comments and encouragement. He read my thesis carefully and corrected the errors. I have learned much from him throughout the years that I worked with him. I offer my thanks to my committe members, Professors A. J. Hildebrand, Kenneth Stolarsky, and Alexandru Zaharescu, who offered guidance and support. I also thank Michael Somos and Daniel Schultz for useful communications. I am grateful to the Thai government providing me financial support. Finally, I would like to thank my family and my friends for their love and support. iv

6 Table of Contents Chapter Introduction Chapter P Q Modular Equations Main Theorems Applications Chapter 3 Dedekind Eta Product Identities Somos s Identities Choi s Identities Chapter The Ramanujan-Göllnitz-Gordon Continued Fraction Introduction New Identities for the Ramanujan-Göllnitz-Gordon Continued Fraction....3 The Expansion of the Ramanujan-Göllnitz-Gordon Continued Fraction Chapter 5 Hyperbolic Infinite Series Connected with Theta Functions Introduction Results Explicit Evaluations of Infinite Series Chapter 6 Ramanujan s Alternative Quartic Theory of Theta Functions 7 6. Introduction Results Chapter 7 The Borweins Cubic Theta Functions Introduction Results Applications References v

7 Chapter Introduction In his notebooks and his lost notebook, Srinivasa Ramanujan developed numerous mathematical results involving theta-functions. Among those results, many of them are related to modular equations, the Ramanujan-Göllnitz-Gordon continued fraction, and infinite series. We now provide some definitions and preliminary results. As customary and throughout this thesis, we assume that q < and use the standard notation n a; q n := aq k, a; q := aq n, k=0 and a, a,..., a k ; q := a ; q a ; q a k ; q. For ab <, Ramanujan s general theta-function fa, b is given by fa, b := a nn+/ b nn /..0. n= Jacobi s triple product identity [9, p. 35] is given by fa, b = a; ab b; ab ab; ab..0. The three most important special cases of fa, b [9, p. 36] are ϕq := fq, q = q n = q; q q ; q,.0.3 n=

8 ψq := fq, q 3 = q nn+/ = q ; q q; q,.0. and f q := f q, q = n q n3n / = q; q..0.5 n= where the product representations in follow from.0.. After Ramanujan, we define χq := q; q..0.6 Ramanujan recorded several identities for fa, b, ϕq, ψq, and f q. The following lemma provides such identities. Lemma.0. [9, p. 8]. Let U n = a nn+/ b nn / and V n = a nn / b nn+/ for each integer n. Then k fu, V = U r f r=0 Uk+r U r, V k r,.0.7 U r for every positive integer k. Lemma.0. [9, p. 3]. We have fa, b = fb, a,.0.8 f, a = fa, a 3,.0.9 f, a = 0,.0.0 and if n is an integer, then fa, b = a nn+/ b nn / faab n, bab n..0.

9 Lemma.0.3 [9, pp. 39 0]. We have ϕqψ q = fqf q,.0. ϕqϕ q = ϕ q,.0.3 ψqψ q = ψq ϕ q,.0. ϕqψq = ψ q,.0.5 ϕq + ϕ q = ϕq,.0.6 ϕq ϕ q = qψq The complete elliptic integral of the first kind Kk is defined by K := Kk := π/ 0 dθ k sin θ, where 0 < k < is called the modulus of K, and the complementary modulus k is defined by k := k. Let K, K, L, and L denote complete elliptic integrals of the first kind associated with the moduli k, k, l, and l, respectively. Suppose that n K K = L L.0.8 holds for some positive integer n. Then a modular equation of degree n is a relation between the moduli k and l which is induced by.0.8. Ramanujan expressed his modular equations in terms of α and β, where α = k and β = l. We say that β has degree n. If q = exp πk /K, then one of the fundamental properties of elliptic functions affirms that [9, p. 0] ϕ q = π Kk = F, ; ; k,.0.9 where F, ; ; x denotes an ordinary hypergeometric function with x <. Ramanujan also recorded several fomulas for ϕ, ψ, f, and χ at different arguments in 3

10 terms of α, q, and z := F, ; ; α by using.0.9. The following lemmas provide such formulas. First, we give evaluations for ϕ. Lemma.0. [9, p. ]. If α, q, and z are defined as above, then ϕq = z,.0.0 ϕq = { } / z + α,.0. ϕq = z + α /,.0. ϕq / = z + α /,.0.3 ϕ q = z α /,.0. ϕ q = z α /8,.0.5 ϕ q / = z α /..0.6 Next, the following are evaluations for ψ. Lemma.0.5 [9, p. 3]. In the notation above, we have ψq = z αq /8,.0.7 ψq = z αq /,.0.8 ψq 8 = { } z α / q,.0.9 ψq / = { } / z + α αq /6,.0.30 ψ q / = { } / z α αq / Finally, we give formulas for f and χ. Lemma.0.6 [9, p. ]. We have fq = z /6 { α αq } /,.0.3

11 f q = z /6 α /6 αq /,.0.33 f q = z { /3 α αq } /,.0.3 f q = z /3 α / αq /6,.0.35 χq = { /6 α αq } /,.0.36 χ q = /6 α / αq /,.0.37 χ q = /3 α / αq / Suppose β is of degree n over α. If q is replaced by q n above, then the same formulas hold with β in place of α and z n := F, ; ; β in place of z. In Chapter, our purpose is to prove new Ramanujan type modular equations which we call P Q modular equations. In his notebook [33], [0, pp. 0 37], Ramanujan stated 3 beautiful modular equations of a type that we now describe. These modular equations involve quotients of the function f q at certain arguments. For example [0, p. 06], let P = f q q /6 f q 5 and Q = f q q /3 f q 0. Then P Q + 5 P Q = 3 3 Q P P Q Note that these modular equations are also called Schläfli-type. Since the publication of [0], several authors, including N. D. Baruah [5, 6], M. S. Mahadeva Naika [9, 30], and K. R. Vasuki [37], have found additional modular equations of the type In Chapter 3, we prove several new Dedekind eta product identities which Michael Somos discovered through computational searches; for example, ηq ηq 6 ηq 0 ηq 30 = ηqηq ηq 5 ηq 0 + ηq 3 ηq ηq 5 ηq 60, 5

12 where ηq := q / f q. Furthermore, we prove identities of a similar type discovered by Youn-Seo Choi in his work on basic bilateral hypergeometric series and mock theta functions. In Chapter, new properties for the Ramanujan-Göllnitz-Gordon continued fraction are presented, in particular, fractorizations. These are similar to the fractorization of the famous Rogers-Ramanujan continued fraction. We also provide a new proof of the 8-dissection of the Ramanujan-Göllnitz-Gordon continued fraction by using the technique that Andrews and Berndt use to prove the 5-dissection of the Rogers-Ramanujan continued fraction, see [, p. 09]. Moreover, we show that this 8-dissection yields immediately the sign of the coefficients of power series associated with this continued fraction being periodic with period 8. In Chapter 5, we derive new infinite series identities which are similar to those of Ramanujan recorded in his notebooks, for instance [9, p. 3] k= k 3 sinhky = z x 8, where z := zx := F, ; ; x, and y := π z x. zx We can also prove these kinds of identities where F, ; ; x is replaced by F, 3 ; ; x. In Chapter 6, we study a new function, namely, ϑq, q, x, x, x 3 := a,b,c= q a+c q a+b + b+c x a x b x c 3, 6

13 where and q := exp π F, ; ; α F, ; ; α q := exp π F, 3; ; α F, 3; ; α. Some propertities analogous to those of classical Jacobian theta functions are established here. In Chapter 7, we prove several identities related to one of the three most special cases of Ramanujan s general theta function, f q, and one of the Borweins s cubic theta functions, namely, aq := q m +mn+n. m,n= We also establish combinatorial interpretations of these identities. 7

14 Chapter P Q Modular Equations M. Somos recently used a computer to discover new elegant modular equations in the spirit of the 3 original modular equations found by Ramanujan. He has a large list of eta-product identities, and he runs PARI/GP scripts to look at each identity to see if it is equivalent to an identity in P Q forms. The purpose of this chapter is to prove these modular equations found by Somos. Some of Somos s equations are simple consequences of others, and we do not record them here. Moreover, some of our proofs are similar to each other, and so, in the interest of brevity and avoiding the repetition of ideas, we provide proofs for only six of the equations and simply indicate the basic ideas for remaining proofs. In Section., we establish combinatorial interpretations of some of our identities.. Main Theorems In this section, we will employ the theory of modular equations to prove some of Somos s discoveries. The modular equations here appear in terms of Ramanujan s general theta functions instead of moduli via the relation.0.9, Lemma.0., Lemma.0.5 and Lemma.0.6. First, we establish some elementary modular equations. Theorem... i If P = f q f q and Q = f q qf q 6, 8

15 then P Q + P Q = Q P P. Q ii If P = ϕ q ϕq and Q = ϕq qψq 8, then P Q + 8 P Q = Q P P +. Q Proof of i. Recall from Entry ii in Chapter 6 of Ramanujan s second notebook [33], [9, p. 39] that f 3 q = ϕ qψq... By.0.3,.0.6,.0.30,.0.3, and.., f q / = z α /3 { + α } / αq /8,.. and fq / = z + α /3 { α } / αq /8...3 Next, by.0.3,.0.36,.., and..3, P = { } / α + and Q = α /. α Finally, P Q + = α + α α / + = α α + α = Q P, which completes the proof. Proof of ii. The proof is similar to i, but we use.0.,.0.,.0.5, and

16 Next, we establish some of Somos s modular equations of degree 3. Theorem... i Let P = χ q χ q and Q = χ q χ q 3. Then P Q q P Q = Q P + q. P Q ii Let P = χ 3 q and Q = χ 3 q 3. Then P Q 8q P Q = Q P + q. P Q iii Let P = f q f q and Q = f q3 f q. Then P Q + q P Q = Q P + q. P Q iv Let P = fq f q and Q = fq3 f q 3. Then P Q Q P = +. P Q P Q v Let P = ϕq ϕq 3 and Q = ψq ψq 6. Then P Q 3q P Q = Q P P + q. Q 0

17 vi Let P = ϕ q ϕ q 3 and Q = ϕq ϕq 3. Then Q P + P Q = P Q + 3 P Q. Proof of i. First, rewrite the desired equation in the form P 3 Q 3 qp Q = Q + qp. Then P 6 Q 6 qp Q + q P Q = Q 8 + qp Q + q P 8, P 6 Q 6 + q P Q = Q 8 + 6qP Q + q P 8, P Q + 6q P Q = Q 6 + qp Q + 30q P 8 Q 8 + q 3 P Q + q P 6... Let R = P and S = Q. Then.. becomes R 3 S 3 + 6q RS = S + qrs q R S + q 3 R 3 S + q R...5 From Entry 5iii in Chapter 6 [9, p. 0],.0. and.0.5, ϕ q = ϕ q ϕq = z α /8 + α /, and from Entry iii in Chapter 6 [9, p. 39], χ q = ϕ q f q

18 = z α /6 { } / + α z /3 α / αq /6 = 5/ α /8 + α / αq /6. Then R = P = q / + α α / α / and S = Q = q 3/ + β. β / β/..6 Now from 5. and 5.3 in [9, p. 33], m 3 + m3 α =, β = m m, 6m 3 6m m + 3 m3 α =, and β = m m, 6m 3 6m..7 where m = z /z 3. For convenience, set x = {mm + 3 m} /, y = m + m 3 + m3 m, r = m + 3 mx, s = m + m + x,..8 t = 6m + 3 mx + mm + x + m + 3 m. Then by..6..8, we have R = S = 6mqr x3 mm 3 + m 3, R = 6q 3 s xm + m m, S = 6 m q r my3 m m 3 + m 5, 6 q 6 s mym + m m, and RS = 6mq t ym 3 + m.

19 Then the right side of..5 becomes Factoring out F R, S := S + qrs q R S + q 3 R 3 S + q R 6 q 6 s = mym + m m + 6 mq 6 ts xym + m 3 + m 6 30m q 6 t + y m 3 + m + 6 m q 6 tr xy3 mm 3 + m 6 m q 6 r + my3 m m 3 + m q 6 on the right side of..9, we have mxy m m 3 F R, S = =: 6 q 6 mxy m m 3 { x3 m 3 + m s + m ty3 m3 + m s + 30m 3 t xym 3 + m + m 3 tym + m r + m xm + m r } 6 q 6 U...0 mxy m m 3 Next, expanding U in terms of m and x, we find that U = 6m 3 796m + 588m m m m m m m 833m m 6888m m 6 7m 7 8m 8 m m x 583m x + 778m 5 x + 589m 6 x m 7 x m 8 x + 650m 9 x + 838m 0 x 5500m x + 96m x 696m 3 x + 633m x 896m 5 x 8m 6 x + m 8 x = 6m t 3 x + m txy m 3 + m. 3

20 Using the calculation above in..0, we conclude that F R, S = = 6 q 6 mxy m m 3 6m t 3 x + m txy m 3 + m 6mq t 3 y 3 m m + 6 mq 6 t 3 ym 3 + m = R 3 S 3 + 6q RS... Recall that F R, S is defined in..9. Then, by.., this establishes the equation..5. To obtain the desired result, we need to reverse the argument in.. and take square roots. To resolve the sign of the square root in each case, we examine the series expansions for P and Q. We then see that we must take the positive root in each instance, and so we complete the proof of i. Proof of ii. By.0.36, P = {α αe y } /8 and Q = { β βe 3y} /8, where α and β have degrees and 3, respectively. Let m be the multiplier. Now from 5. in [9, p. 3], β 3 /8 = m +, α β 3 /8 = m and α α 3 /8 = 3 m β m, α 3 /8 = 3 + m β m... It follows that, by.., and P 3 Q = 8m 9 m..3 P Q = m q

21 From..3, we have m = 9P 3 P 3 + 8Q and from.., we have m = 8qP Q 3 9P 3 P 3 + 8Q = 8qP Q +, 3 +. Then or P 3 Q 3 8qP Q = Q + qp...5 This is equivalent to the identity claimed in the theorem. Proof of iii. By.0.33 and.0.35, P = / q /8 α α /8 /8 β and Q = / q 3/8, β where α and β have degrees and 3, respectively. Then P 3 Q 3 + qp Q = 8q 3/ { α β αβ } /8 { } / α β +. αβ By Entry 5ii in Chapter 9 of Ramanujan s second notebook [9, p. 30], we obtain { } /8 α β P 3 Q 3 + qp Q = 8q 3/. α 3 β 3 Finally, by Entry 5ix in Chapter 9 [9, p. 3], P 3 Q 3 + qp Q = q 3/ {α β}/ + {β α} / = Q + qp. αβ / This completes the proof. 5

22 Proof of iv. The proof is similar to iii. We employ.0.3,.0.33 and... Proof of v. The proof is similar to iii. We employ.0.0,.0.8 and Entry 5x in Chapter 9 [9, p. 3], which is mα / β / = 3 m β/ + α /, where α, β and m are as in the proof of iii. Proof of vi. The proof is similar to iii. We employ.0.0,.0. and Entry 5x in Chapter 9 [9, p. 3], which is m α / + β / = 3 m β/ α /, where α, β and m are as in the proof of iii. We next establish a P Q modular equation of degree. This modular equation also lacks symmetry. Theorem..3. If P = ψq ϕ q 8 and Q = χ qχ q 8, then P Q q P Q = Q P P q. Q Proof. By.0.5,.0.7,.0.37, and.0.38, P = { } /8 { } α α m / q /8 and Q = / / β q 3/8, β αβ where α and β have degrees and, respectively, and m = z /z is the multiplier. Then P Q q P Q 3 + qp 3 Q 6

23 { = q α m β +m 3/ α α / β / β / /3 m / α + }...6 β By.8 and.9 in Chapter 8 [9, pp. 5], m = + β / and m α / + β / =...7 Then by..6 and..7, P Q q P Q 3 + qp 3 Q { = q /3 /3 } m α β + m3/ β/ β / β m β / = q m + α + α / α / β / m mβ / + β /3...8 By.0 in Chapter 8 [9, p. 5], + α / = m and + α + β = m...9 Then by..8 and..9, P Q q P Q 3 + qp 3 Q = q m β / m α / β / /3...0 Employing..7 in..0, we arrive at P Q q P Q 3 + qp 3 Q = q m β / + β / β / β / /3 7

24 = 0. This yields the desired identity. The last theorem for this section provides a P Q modular equation of degree 5. Theorem... If P = fq f q and Q = fq5 f q 0, then P Q q P Q = 3 3 Q P + q. P Q Proof. By.0.3 and.0.3, P = /6 q / {α α} / and Q = /6 q 5/ {β β} /, where β has degree 5 over α. Let m = z /z 5 be the multiplier. Then P 5 Q 5 qp Q = /3 q 5/ { {6αβ α β} /6}... {αβ α β} 5/ By Entries 3i, 3xiii, and 3iv in Chapter 9 of Ramanujan s second notebook [9, pp. 80 8], {6αβ α β} /6 = + αβ / + { α β} / = m + 5 m = /3 { α 5 α 5 β β } / { } β + /3 5 β 5 /. α α Using the last identity in.., we conclude that P 5 Q 5 qp Q = /3 q 5/ {αβ α β} 5/ /3 { α 5 α 5 β β } / 8

25 { } β + /3 5 β 5 / α α = Q 6 + qp 6... Hence we obtain the desired result.. Applications The purpose of this section is to establish some applications of new P Q modular equations in partition theory. Theorem... Let S k denote the set consisting of 3k copies of the odd positive integers for k =,, 3,. Let AN be the number of partitions of N into distinct parts, which are elements in S 3 that are not multiples of 3, and elements in S that are multiples of 3. Let BN be the number of partitions of N into distinct parts, which are elements in S that are not multiples of 3. Let CN be the number of partitions of N into distinct parts, which are elements in S that are not multiples of 3, and elements in S that are multiples of 3. Let DN be the number of partitions of N into distinct parts, which are elements in S that are multiples of 3. Then AN = 8BN + CN + DN. Proof. By.0.6, the equation..5 in Theorem..ii can be written as q; q 9 q 3 ; q qq; q 3 q 3 ; q qq; q = q 3 ; q 6, that is, q; q 9 q 3 ; q qq, q 5 ; q 6 3 q 3 ; q qq, q 3, q 5 ; q 6 = q 3 ; q 6, 9

26 or q, q 5 ; q 6 9 q 3 ; q 6 + 8qq, q 5 ; q qq, q 5 ; q 6 q 3 ; q 6 6 = q 3 ; q The partition identity in the theorem is obtained directly from the last equality by q replaced by q. Example. Let N =. Then A = 3, B =, C = 6, and D = 0. The representations are given by = 3 α + β, where α {,,..., } and β {,,..., 9} = ω + ɛ + γ + δ, where ω, ɛ, γ, δ {,,..., 9}, and ω, ɛ, γ, δ are distinct, 3 = + + 3, 3 = 3 = 3 = 3 3 = 3 = 3 5 = 3 6 = α + β + γ, where α, β, γ {,,..., }, and α, β, γ are distinct. Theorem... Let S k denote the set consisting of 3k copies of the odd positive integers for k =,, 3,. Let AN be the number of partitions of N into elements in S that are not multiples of 3. Let BN be the number of partitions of N into elements in S 3 that are not multiples of 3, and elements in S that are multiples of 3. Let CN be the number of partitions of N into elements in S that are multiples of 3. Let DN be the number of partitions of N into elements in S that are not multiples of 3, and elements in S that are multiples of 3. Then AN + 8BN + CN = DN. Proof. We divide.. by q, q 5 ; q 6 q 3 ; q 6. We find that q, q 5 ; q 6 3 8q + q, q 5 ; q 6 9 q 3 ; q 6 q + q 3 ; q 6 6 =. q, q 5 ; q 6 q 3 ; q 6 6 The proposed result follows readily from the above equation. 0

27 Example. Let N =. Then A = 5, B = 77, C = 6, and D = 37. The representations are given by = α + β + γ + δ, where α, β, γ, δ {,, 3}, 3 = α + β + γ, where α, β, γ {,,..., 9}, = 3 = 3 = = 3, 3 = 3 = 3 = 3 3 = 3 = 3 5 = 3 6, = 3 α + β, where α {,,..., 6} and β {,,..., }, = ω + ɛ + γ + δ, where ω, ɛ, γ, δ {,,..., }. Theorem..3. Let AN denote the number of partitions of N + into parts that are congruent to ±3 or 6 modulo with two distinct colors, say blue and green, and parts that are congruent to ±, ± or ±5 modulo. Let BN denote the number of partitions of N into parts that are congruent to ±3 or 6 modulo with six distinct colors, say blue, green, red, silver, indigo and orange, and parts that are congruent to ±, ± or ±5 modulo with three distinct colors, say yellow, violet and pink. Let CN denote the number of partitions of N + into parts that are not congruent to 0 modulo with four distinct colors, say blue, green, red and silver. Let DN denote the number of partitions of N into parts that are congruent to ±3 or 6 modulo with four distinct colors, say blue, green, red and silver. Then AN + BN = CN + DN. Proof. The identity in Theorem..iii can be expressed via.0.5 as q; q 3 q 3 ; q 3 3 q ; q 3 q ; q 3 q; q q 3 ; q 3 + q = q3 ; q 3 q ; q q ; q q ; q + q q; q... q ; q

28 Multiplying both sides of.. by q ; q q ; q, q 3 ; q 3 q; q we have q ; q q ; q q 3 ; q 3 q; q + q q ; q 3 q ; q 3 q 3 ; q 3 3 q; q 3 = q ; q q; q + q q ; q. q 3 ; q 3 After simplification, we arrive at q + q 3, q 6, q 9 ; q q, q, q 5, q 7, q 0, q ; q q 3, q 6, q 9 ; q 6 q, q, q 5, q 7, q 0, q ; q 3 q = +, q, q, q 3 ; q q 3, q 6, q 9 ; q which gives the desired partition identity. Example. Let N = 3. Then A3 = 5, B3 = 5, C3 = 0, and D3 =. The representations are given by = 3 b + = 3 g + = + = + + = + + +, 3 = 3 b = 3 g, = 3 r = 3 s = 3 i = 3 o = α + β, where α, β {y, v, p} = α + β + γ, where α, β, γ {y, v, p}, = 3 α + β = α + β, where α, β {b, g, r, s} = α + β + γ, where α, β, γ {b, g, r, s} = α + β + γ + δ, where α, β, γ, δ {b, g, r, s}, 3 = 3 b = 3 g = 3 r = 3 s. Theorem... Let S denote the set consisting of six copies, say in colors blue, green,

29 white, silver, cyan and red, of the odd positive integers and, let S denote the set consisting of six copies, say in colors yellow, pink, orange, indigo, violet and navy, of the odd positive integers that are multiples of 5. Let AN be the number of partitions of N+ into distinct parts, which are elements in S S, except red and navy colors. Let BN be the number of partitions of N into distinct parts, which are black and yellow elements in S S. Let CN be the number of partitions of N + into distinct parts, which are elements in S. Let DN be the number of partitions of N into distinct parts, which are elements in S. Then AN = BN + CN + DN. Proof. We transcribe P and Q in Theorem.. by using.0.5 to find that P = q; q q ; q = q; q and Q = q5 ; q 5 q 0 ; q 0 = q 5 ; q 0. Hence.. becomes q; q 5 q 5 ; q qq; q q 5 ; q 0 = q 5 ; q 0 6 qq; q This immediately yeilds the partition identity in the theorem by q replaced by q. Example. Let N =. Then A = 30, B =, C = 0, and D = 6. The representations are given by = b + g + w + s = b + g + w + c = b + g + s + c = b + w + s + c = g + w + s + c = 3 α + β, where α, β {b, g, w, s, c}, 3 = 3 b, 3 = 3 b = 3 g = 3 w = 3 s = 3 c = 3 r 3

30 = α + β + γ, where α, β, γ {b, g, w, s, c} and α, β, γ are distinct. Theorem..5. Let S denote the set consisting of six copies, say in colors blue, green, white, silver, cyan and red, of the odd positive integers and let S denote the set consisting of six copies, say in colors yellow, pink, orange, indigo, violet and navy, of the odd positive integers that are multiples of 5. Let AN be the number of partitions of N into black and yellow elements in S S. Let BN be the number of partitions of N into elements in S S, except red and navy colors. Let CN be the number of partitions of N into elements in S. Let DN be the number of partitions of N into elements in S. Then AN + BN + CN = DN. Proof. Divide..3 by q; q 6 q 5 ; q 0 6 and obtain q; q q 5 ; q 0 + q q; q 5 q 5 ; q 0 5 = q; q 6 q. q 5 ; q 0 6 We achieve the proposed formula in the theorem. Example. Let N =. Then A =, B = 0, C = 0, and D = 6. The representations are given by = 3 b + b = b + b + b + b, 3 = 3 b = 3 g = 3 w = 3 s = 3 c = 3 r = α + β + γ, where α, β, γ {b, g, w, s, c}, = 3 α + β, where α, β {b, g, w, s, c, r}, = ω + ɛ + γ + δ, where ω, ɛ, γ, δ {b, g, w, s, c, r}.

31 Chapter 3 Dedekind Eta Product Identities In this chapter, we will give proofs of Dedekind eta product identities discovered by M. Somos in the first section and discovered by Y.-S. Choi in the second section. The Dedekind eta function is defined by ηq := q / f q. For our convenience, we write η n := ηq n Furthermore, for any positive integers n, m, we define η n,m as η n,m := η n,m q := q n P m n f qm, q n m, 3.0. f q n where P t = {t} {t} + 6 is called the second Bernoulli function, and {t} := t [t] is the fractional part of t. 3. Somos s Identities Theorem 3... The following identities hold: i η η 6 η 0 η 30 = η η η 5 η 0 + η 3 η η 5 η 60, 3.. ii η η 3 η 5 η 5 + η η 6 η 0 η 30 = η η η 5 η 30 + η 3 η 5 η 6 η 0, 3.. 5

32 iii η η 7 η 9 η 8 η 36 η 6 = η η η η 8 η 63 η 5 + η η 6 η η 8 η η 6, 3..3 iv η η η 7η 8 + η η η 7 η 8 = η 3 η 3, 3.. and v η η 8 η 3 + η η 6η 3 = η η η 3 + η η 8 η Proof of i. We rewrite the identity 3.. in the form f q f q 6 f q 0 f q 30 = f qf q f q 5 f q 0 + qf q 3 f q f q 5 f q Let α, β, γ, δ be of the first, third, fifth, and fifteenth degrees, respectively. Let m denote the multiplier connecting α and β, and let m be the multiplier relating γ and δ. Then applying to the terms of 3..6, we find that f q f q 6 f q 0 f q 30 = /3 z z 3 z 5 z 5 {αβγδ α β γ δ} /, f qf q f q 5 f q 0 = 5/3 z z 3 z 5 z 5 {αδ β γ} / {βγ α δ} /6, and qf q 3 f q f q 5 f q 60 = 5/3 z z 3 z 5 z 5 {αδ β γ} /6 {βγ α δ} /. 6

33 Recall from Entries i and ii in [9, p. 383], respectively that {αδ} /8 + { α δ} /8 = m m and {βγ} /8 + { β γ} /8 = m m. We find that {αδ} /8 + { α δ} /8 {βγ} /8 + { β γ} /8 =. If we expand and rearrange this modular equation, then we arrive at {αβγδ} /8 { α β γ δ} /8 = {βγ α δ} /8 + {αδ β γ} / Note that Entry xiv in [9, p. 385] is {αβγδ} /8 { α β γ δ} /8 = /3 {αβγδ α β γ δ} / Then combine 3..7 and 3..8 to obtain /3 {αβγδ α β γ δ} / = {βγ α δ} /8 + {αδ β γ} /8, 3..9 which is equivalent to We complete the proof. Proof of ii. Let α, β, γ, δ be of the first, third, fifth, and fifteenth degrees, respectively. Let 7

34 m denote the multiplier connecting α and β, and let m be the multiplier relating γ and δ, i.e., m = z /z 3 and m = z 5 /z 5. Then by.0.33 and.0.3, the left-hand side of 3.. becomes η η 3 η 5 η 5 + η η 6 η 0 η 30 = z z 3 z 5 z 5 /3 { α β γ δ} /6 αβγδ / + /3 {αβγδ α β γ δ} /. We factor the term /3 { α β γ δ} /6 αβγδ / out and then use Entry xiv in [9, p. 385] to obtain η η 3 η 5 η 5 + η η 6 η 0 η 30 = z z 3 z 5 z 5 /3 { α β γ δ} /6 αβγδ / { α β γ δ} /8 + /3 {αβγδ α β γ δ} / = z z 3 z 5 z 5 αβγδ /8 { α β γ δ} /8 αβγδ / For the right-hand side of 3.., by.0.33 and.0.3, we find that η η η 5 η 30 + η 3 η 5 η 6 η 0 = m z z 3 z 5 z 5 m αδ/8 { α δ} / + m m βγ/8 { β γ} /, 3.. for which we use the fact that m = z /z 3 and m = z 5 /z 5. We will use the following shorthand notation for our convenience. Let A = αδ /8, A = { α δ} /8, B = βγ /8, B = { β γ} /8, and M = m/m /. We 8

35 deduce that 3..0 and 3.. become η η 3 η 5 η 5 + η η 6 η 0 η 30 = z z 3 z 5 z 5 AB A B AB 3.. and η η η 5 η 30 + η 3 η 5 η 6 η 0 = z z 3 z 5 z 5 MAA + BB, 3..3 M respectively. Note that we can rewrite the equations. in [9, p. 385] in our notation as AB + AB + A B + A B =, B A M = BB = B A M, and A + A = B + B = M. Finally, by using the equality above and elementary manipulations, we can show that MAA + BB B M = AB BB + B M A = AB ABB + BB A + A A B = AB + A B A M A BB + A M = AB + A B A BB AA AM + AA M = AB + A B A BB AA AB + AB + AA M. By Entry ix in [9, p. 38], we have A B BB + AB + AA M = 0. 9

36 Then MAA + BB M = AB + A B A BB AA AB + BB A = AB + A B A BB AA AB + BB A = AB + A B AB = AB A B AB., 3.. By 3.., 3..3 and 3.., we complete the proof. Proof of iii. From Entry 9ii in [9, p. 6], we have η 6 η = q ψq 7 ψq 9 q 8 ψqψq 63. Then 3..3 can be written as η η 7 η 9 η 8 η 36 η 6 + q 8 η η η 8 η 6 ψqψq 63 = η η η η 8 η 63 η 5 + q η η η 8 η 6 ψq 7 ψq Let α, β, γ, δ have degrees, 7, 9, 63, respectively. Let m denote the multiplier connecting α and β, and let m be the multiplier relating γ and δ. Then by.0.7,.0.33,.0.3 and.0.35, the left-hand side of 3..5 becomes η η 7 η 9 η 8 η 36 η 6 + q 8 η η η 8 η 6 ψqψq 63 = z z 63 z7 z 9 7/3 {αβγδ α β γ δ} / m m {βγ β γ}/8 + αδ /8,

37 and the right-hand side of 3..5 becomes η η η η 8 η 63 η 5 + q η η η 8 η 6 ψq 7 ψq 9 = z z 63 z7 z 9 7/3 {αβγδ α β γ δ} / m {αδ α δ} /8 + m βγ/ By Entry 9iii in [9, p. 6], we have m m = αδ/8 {αδ α δ} / βγ /8 /8 {βγ β γ} Employing 3..8 in 3..6 and 3..7, we find that the right sides of 3..6 and 3..7 are seen to be equal and we achieve the proposed identity Proof of iv. If β has degree seven over α, then by.0.33 and.0.35, the left-hand side of 3.. reduces to η η η 7η 8 + η η η 7 η 8 = z z 7 3/ {αβ α β} / { { α β} /8 + αβ /8}. Recall from Entry 9i in [9, p. 3] that { α β} /8 + αβ /8 =. Then η η η 7η 8 + η η η 7 η 8 = z z 7 3/ {αβ α β} / = η 3 η 3, 3

38 which completes the proof by.0.3. Proof of v. By Entry iii in [9, p. 39], we have ϕ q = η η and q /8 ψq = η η Now, consider the left-hand side of We have η ηη 8 η 3 + η η6η 3 = η η 8 η 3 + η 6 η η 8 = η η 8 η 3 ϕ q + qψq 8. Let β be of the fourth degree over α and let m denote the associated multiplier. By.0. and.0.8, we deduce that η η 8 η 3 + η η 6η 3 = η η 8 η 3 z α / + q = η η 8 η 3 z m α / + β /. { } z βq / Note from Entry iii in [9, p. ] that m α / + β / =. Then η η 8 η 3 + η η 6η 3 = η η 8 η 3 z Next, consider the right-hand side of By 3..9, we find that η η ηη 3 + η η 8 η6 = η η 8 η 3 + q η 6 η 8 η 3 3

39 = η η 8 η 3 ϕ q + q ψq 3. By.0. and.0.9, it follows that { η ηη 3 + η η 8 η6 z = η η 8 η 3 β / + q { } } z β / q = η η 8 η 3 z. 3.. We complete the proof by 3..0 and 3... Next, we will establish an identity between multipliers which follows from the above theorem. Corollary 3... We have = / / / X Y + + XY /, Y X XY where X = 3 mm + 3 m m+, Y = m m, m = z z 3 and m = z 5 z 5. Proof. If we multiply both sides of 3..9 by /3, and then cancel the modular parameters from the left-hand side of the formula, we obtain the identity = /3 { β γ β γ αδ α δ + /3 { α δ α δ βγ β γ } / { α δ β γ } / { β γ α δ } /8 } /8. By Entries iv and v in [9, pp ], our formula becomes = } /8 + βγ /8 + { β γ} /8 { α δ β γ + αδ /8 { α δ} /8 { } /8 β γ, α δ 33

40 and finally by Entries i, ii in [9, p. 383] and the equation.0 in [9, p. 388], we find that { m m 3 m m + = + m m m + 3 m m { m m + 3 m + m m 3 m m + = + / X Y + Y X Y X / + XY / = / X Y + Y X } /8 XY } /8 / /, which completes the poof. 3. Choi s Identities Theorem 3... We have η η 3η 8 η 6η 3,η 6, η η η 6 6η η,η 6 6,η, = 3η 8 η η 6η 3.. and η 6 η 8 3η η 3,η, η 6 η η 6η = η 0 η η 3η 6 η 6η Proof. Note that P = + 6 =, P = = 8, and P = = 36. Let α and β have degrees and 3, respectively and let m = z /z 3 be a multiplier. By the 3

41 definitions 3.0. of η n and 3.0. of η n,m, the first term of 3.. becomes η η 3η 8 η 6η 3,η 6, = q 3/6 f qf 8 q f q, q f q, q 5. Note that by Examplev in [9, p. 5], we have fq, q 5 = ψ q 3 χq Then by.0.5 and 3..3, we see that η η 3η 8 η 6η 3,η 6, = q 3/6 f qf 8 q ψ q 3 χ q. By.0.7,.0.33,.0.35, and.0.37, we obtain η η 3η 8 η 6η 3,η 6, = z 6 z 3 /3 α 7/ α 7/6 β /. 3.. Next for the second term of 3.., we find that η η η 6 6η η,η 6 6,η, = q 3/6 f 0 q f q, q f q, q 0. Then by using.0.3, 3..3,.0.5,.0.8,.0.3,.0.38, respectively, we obtain η η η 6 6η η,η 6 6,η, = q 3/6 f 0 q ϕ q ψ q 6 χ q = z 6 z 3 /3 α /3 α 7/6 β / Similarly, for the last term of 3.., by using , we deduce that 3η 8 η η 6η = 3q 3/6 f 8 qf q f q 6 f q = 3z 5 z 3 /3 α /3 α 7/ β / β /

42 We denote F q to be the left-hand side of 3... We see that by 3.. and 3..5, F q = zz 6 3 /3 α 7/ α 7/6 β / zz 6 3 /3 α /3 α 7/6 β / α = zz 6 3 /3 α /3 α 7/6 β / 3 /. β By.., the right side of the above equation becomes 3 + m F q = zz 6 3 /3 α /3 α 7/6 β / m m + 3 m = 3zz 5 3 /3 α /3 α 7/6 β / 3 / m + 3 / 3 m. 6m 3 6m 3..7 For the last equality, we use the fact that m = z /z 3. Note that the equation 5.5 in [9, p. 33] is α = m + 3 m3 and β = m m m 3 6m We substitute 3..8 into 3..7 to obtain F q = 3z 5 z 3 /3 α /3 α 7/6 β / α / β / = 3z 5 z 3 /3 α /3 α 7/ β / β /, 3..9 By 3..6 and 3..9, the proof of the first identity is complete. To prove 3.., we note that P = = 8, and P = =

43 For the first term of 3.., by using.0.8,.0.33,.0.3,.0.35, and.0.38, we obtain η 6 η 8 3η η 3,η, = q 3/3 f qf 6 q f q 3 f q ψ q 6 χ q = z 0 z 3 8/3 α /3 α 5/ β β 3/ Next by.0.33,.0.3, and.0.35, the second term of 3.. becomes η 6 η η 6η = q 3/3 f 6 qf q f q 6 f q = z 0 z 3 8/3 α /3 α 7/6 β β /. 3.. For the last term of 3.., by using , we find that η 0 η η 3η 6 η 6η 6 = q 6/3 f 0 qf q f q 3 f 6 q f q 6 f 6 q = 3z 9 z 5 3 8/3 α 9/ α 5/ β 5/ β 3/. 3.. We set Gq to be the left side of 3... We see that by 3..0 and 3.., Gq = z 0 z 3 8/3 α /3 α 5/ β β 3/ z 0 z 3 8/3 α /3 α 7/6 β β / = z 0 z 3 8/3 α /3 α 5/ β β 3/ α 3/ β /. By.., the right side of the above equation becomes Gq = z 0 z 3 8/3 α /3 α 5/ β β 3/ 3 m = 3zz /3 α /3 α 5/ β β 3/ m m / m 3 / m + 3, m 3 6m m 37

44 where for the last equality, we use the fact that m = z /z 3. Note that the equation 5. in [9, p. 33] is α = m m + 33 and β = m 3 m m 3 6m Substituting 3.. into 3..3 yields Gq = 3z 9 z 5 3 8/3 α /3 α 5/ β β 3/ α / β / = 3z 9 z 5 3 8/3 α 9/ α 5/ β 5/ β 3/ By 3.. and 3..5, we complete the proof of the second identity. 38

45 Chapter The Ramanujan-Göllnitz-Gordon Continued Fraction. Introduction The Rogers-Ramanujan continued fraction is defined by [, p. 9], [3, p. xxviii], Rq := q/5 q + + q q 3 + +, q <. The famous Rogers-Ramanujan functions Gq and Hq are defined by Gq := q n q; q n = q; q 5 q ; q 5.. and Hq := q nn+ q; q n = q ; q 5 q 3 ; q 5,.. where the two equalities on the right sides of.. and.. are the celebrated Rogers- Ramanujan identities [, p. 87], [3, p. 37]. Using.. and.., Rogers [36] proved that /5 Gq Rq = q Hq. In his notebooks, Ramanujan recorded many identities involving Rq which can be found in [], [9], [33], [3]. For example, two of the most important formulas for Rq [, p. ], [3, pp. 35, 38] are Rq Rq = q/5 ; q /5 q /5 q 5 ; q

46 and R 5 q R5 q = q; q6... qq 5 ; q 5 6 Furthermore, he established factorizations of..3 and.. [, pp. ], [3, p. 06], namely, γ Rq = Rq Rq 5 q /0 δ Rq = Rq Rq 5 q /0 γ 5 Rq = q / δ 5 Rq = q / q; q q 5 ; q 5 n= q; q q 5 ; q 5 n= q; q q 5 ; q 5 n= q; q q 5 ; q 5 n= + γq n/5 + q n/5, + δq n/5 + q n/5, + γq n/5 + q n/5 5, + δq n/5 + q n/5 5,..5 where γ = 5/ and δ = + 5/. On page 9 of his second notebook [33], [9, p. ], Ramanujan defined another continued fraction vq that has properties similar to those of Rq, namely, q q vq := q/ + q + + q q q 7 +,..6 q 6 later called the Ramanujan-Göllnitz-Gordon continued fraction. In addition, Ramanujan recorded an identity for vq [9, p. ], which is vq = q / f q, q7 f q 3, q H. Göllnitz [] and B. Gordon [3] rediscovered and proved..7 independently. A proof of..7 can also be found in [9, p. ]. Moreover, Ramanujan established two further 0

47 identities for vq, [9, p. ], [33, p. 9], namely, vq vq = ϕq q / ψq..8 and vq + vq = ϕq q / ψq...9 The Göllnitz-Gordon functions Sq and T q are defined as Sq := q; q n q ; q n q n and T q := q; q n q ; q n q n +n...0 Then the Göllnitz-Gordon identities are given by Sq = q; q 8 q ; q 8 q 7 ; q 8.. and We see that T q = q 3 ; q 8 q ; q 8 q 5 ; q 8... / T q vq = q Sq. Recently, several authors, including N. D. Baruah and N. Saikia [7], H.-H. Chan and S.-S. Huang [7], S.-D. Chen and S.-S. Huang [8], S. Cooper [0] and M. D. Hirschhorn [], have found additional identities and properties for vq. In this chapter, we will establish several new identities and properties for vq motivated by identities involving the Rogers- Ramanujan continued fraction. In [6], T. Horie and N. Kanou found that v q 6 + v q = q; q q ; q,..3 qq ; q q 8 ; q 8

48 by employing modular forms; also see [0]. Here in Section., we will present a shorter proof of..3 by utilizing factorizations of vq that are similar to..5. Here in Section., we will present factorizations for vq in Theorem.. and Theorem.. which are similar to..5. In the sequel, we prove that vq + vq = q/ ; q q; q q ; q 8 q / q 8 ; q 8 and vq vq = q/ ; q q; q q ; q 8 q / q 8 ; q 8, which are employed to establish a shorter proof of..3. Hirschhorn [] showed that when infinite products associated with vq are expanded as power series, the sign of the coefficients is periodic with period 8, and he also derived some identities for such coefficients. In Section.3, we will provide a new proof of Hirschhorn s results. The forms given in Theorems.3.3 and.3. are the same as the ones given by Hirschhorn, but are given here in a more compact form. Both Hirschhorn s form as well as the formulas in Theorems.3.3 and.3. immediately imply the sign of the coefficients. In the same section, some new identities for vq are found here, namely Theorem New Identities for the Ramanujan-Göllnitz-Gordon Continued Fraction In this section, we shall provide identities for vq which resemble..3 and..5. Define v := q / f q, q7 f q 3, q 5...

49 Theorem... We have and v + i v = v i v = f q q / f iq / f q 8 χ q,.. f q q / fiq / f q 8 χ q,..3 n + q n/,.. + v = χq χ q v q / χ q v v = χq χ q q / χ q n= n= n q n/,..5 where n is the Kronecker symbol. Proof. By.., v + i v = f q3, q 5 + iq / f q, q 7 q / f q, q 7 f q 3, q By Jacobi s triple product identity.0., f q, q 7 f q 3, q 5 = q; q 8 q 7 ; q 8 q 3 ; q 8 q 5 ; q 8 q 8 ; q 8 = q; q q 8 ; q 8 = χ qf q Take k =, a = iq /, and b = iq 3/ in.0.7 to obtain fiq /, iq 3/ = f q 3, q 5 + iq / f q, q 7. By.0., fiq /, iq 3/ = iq / ; q iq 3/ ; q q ; q 3

50 = f q f iq /...8 Substituting..7 and..8 in..6, we deduce that v + i v = f q q / f iq / f q 8 χ q. To prove..3..5, take k, a, b =, iq /, iq 3/,, q /, q 3/, and, q /, q 3/, respectively, in.0.7. We notice that Theorem.. provides a factorization of..8 and..9. Before proceeding further, we define α :=, β := + and ζ := e πi/. Theorem... We have v α v = v + α v = v β v = + α n q n/ αq n n q 3n/ n= q / f q 8 χ q + αq n/ αq n q 3n/ n=,..9 q / f q 8,..0 χ q βq n/ + βq n q 3n/ n= q / f q 8 χ q,.. and v + β v = β n q n/ + βq n n q 3n/ n= q / f q 8 χ q... Proof. By.., α v = f q3, q 5 αq / f q, q 7 v q / f q, q 7 f q 3, q 5...3

51 Take k =, a = ζ, and b = ζ 3 q / in.0.7 to obtain fζ, ζ 3 q / = f q 3, q 5 +ζf q 5, q 3 +ζ 6 q / f q 7, q+ζ 7 q 3/ f q 9, q... Take n =, a = q, and b = q 9 in.0.. This yeilds f q 9, q = q f q 7, q...5 Substituting..5 in.., we obtain fζ, ζ 3 q / = + ζf q 3, q 5 + ζ 6 q / f q 7, q ζ 7 q / f q 7, q = + ζf q 3, q 5 + ζ 6 ζ 7 q / f q, q 7, by.0.8. Note that α = ζ + ζ 7, so ζ 6 ζ 7 = α + ζ. It follows that fζ, ζ 3 q / + ζ = f q 3, q 5 αq / f q, q Substituting..7 and..6 in..3, we deduce that v α v = fζ, ζ 3 q / + ζ q / χ qf q By Jacobi s triple product identity.0., fζ, ζ 3 q / + ζ = ζ; q/ ζ 3 q / ; q / q / ; q / + ζ = ζq / ; q / ζ 3 q / ; q / q / ; q / = + ζ q / n ζ 3 q / n q / n. n= 5

52 Note that ζ ζ 3 = and ζ =. Then fζ, ζ 3 q / + ζ = = = + q / n + q / n q / n n= + n q n/ q n n q 3n/ n= + α n q n/ αq n n q 3n/...8 n= Substituting..8 in..7, we complete the proof of..9. To prove..0.., take k, a, b =, ζ 3, ζ 5 q /,, ζ 3, ζ 5 q /, and, ζ, ζ 3 q /, respectively, in.0.7 and use the fact that ζ 6 + ζ = α ζ 3, ζ 6 ζ = β + ζ 3, and ζ 6 + ζ 7 = β ζ, respectively. Corollary..3. We have v + v = ϕq/ q / ψq..9 and v v = ϕ q/ q / ψq...0 6

53 Proof. By..9 and.., observe that v + v = v α v v + β v + α n q n/ αq n n q 3n/ β n q n/ + βq n n q 3n/ = n= q / χ qf q 8... We see that + α n q n/ αq n n q 3n/ β n q n/ + βq n n q 3n/ = n q n/ + q n + q n n q 5n/ + q 3n = + q n n q n/ + q n + q n q k + q k, if n = k, = + q n + q k / + q k, if n = k... Substituting.. in.., we deduce that v + v = + q n q n + q n / n= q / χ qf q 8 = q ; q q; q q / ; q q / q; q q 8 ; q 8 = q ; q q ; q q; q q / ; q q / q 8 ; q 8 = q ; q 8 q; q q / ; q q / q 8 ; q 8 = ϕq/ q / ψq, 7

54 by.0.3 and.0., which completes the proof of..9. The proof of..0 follows precisely along the same lines. Corollary... If ζ = e πi/, then v + + v = f qζ 3 q / ; q ζ 5 q / ; q, q / f q f q 8..3 v + v = f qζq / ; q ζ 7 q / ; q, q / f q f q 8.. v α v = f qχq ζ 3 q; q ζ 5 q; q,..5 q / f q 8 and v β v = f qχq ζq; q ζ 7 q; q...6 q / f q 8 Proof. By..0 and.., we see that v + + v = v + α v v + β v + αq n/ αq n q 3n/ β n q n/ + βq n n q 3n/ = n= q / χ qf q We observe that + αq n/ αq n q 3n/ β n q n/ + βq n n q 3n/ = q n/ + + αq n/ + q n n q n/ + β n q n/ + q n 8

55 q k + q k, if n = k, =..8 q k ζ 3 q k / ζ 5 q k /, if n = k. Substituting..8 in..7, we deduce that v + + v = q n + q n q n ζ 3 q n / ζ 5 q n / n= q / χ qf q 8 = q; q q ; q q; q ζ 3 q / ; q ζ 5 q / ; q q / q; q q 8 ; q 8 = q; q ζ 3 q / ; q ζ 5 q / ; q q / q ; q q 8 ; q 8 = f qζ 3 q / ; q ζ 5 q / ; q, q / f q f q 8 which completes the proof of..3. The proofs of....6 proceed along similar lines. The following corollary is..3, and originally was proved by using modular forms. Corollary..5. We have v 6 + v = f qf q qf q f q 8 = ϕ q qψ q...9 Proof. By..9 and..0, we deduce that v 6 + v = v + v v v ϕq / ϕ q / = q / ψq q / ψq = ϕ q qψ q. The last equation is obtained by.0.3. By employing the product representations in 9

56 , we easily obtain the last equality of The Expansion of the Ramanujan-Göllnitz-Gordon Continued Fraction In [8], Chen and Huang established the -dissection of the Ramanujan-Göllnitz-Gordon Continued Fraction. In this section, we shall provide a new proof the 8-dissection identities of this continued fraction found by Hirschhorn []. Define Dq := f q3, q 5 f q, q 7. By..8 and..9, we find that If fq = Dq = ϕq + ϕq ψq = q n + q n ψq n= n= =: u n q n..3. a n q n, we define the operator U 8 operating on fq by [, p. 6] U 8 fq := a 8n q n = 8 7 fζ j q /8, j=0 where ζ = e πi/. Then apply U 8 to.3. and deduce that u 8n+a q n = U 8 q a Dq 50

57 = 7 ζ aj q a/8 Dζ j q /8 8 j=0 = 7 ζ aj q a/8 ζ nj q n / + ζ nj q n /8 6 ψζ j q / j=0 n= n= = 7 ζ n aj q n / a/8 + ζ n aj q n /8 a/8 6 ψ j q / j=0 n= n= 3 = ζ n aj q n / a/8 + ζ n aj q n /8 a/8 6ψq / j=0 n= n= 3 + ζ n aj+ q n / a/8 + ζ n aj+ q n /8 a/8. 6ψ q / j=0 n= n=.3. Lemma.3.. If ζ = e πi/, then we have 3, if r 0 mod, ζ rj = 0, otherwise, j=0.3.3 and Lemma.3.. We have, if r 0 mod 8, 3 ζ rj+ =, if r mod 8, j=0 0, otherwise..3. ψqψ q = f q f q,.3.5 ψq + ψ q = fq 6, q and ψq ψ q = qfq, q..3.7 Proof. The identity.3.5 follows immediately from.0. and.0.. 5

58 To prove.3.6, by.0., we have ψq + ψ q = = q nn+/ + q nn+ + + n q nn+ + = q nn+ + = n= By.0. and.0.8, we obtain.3.6. Similarly, by.0., we deduce that ψq ψ q = = q nn+. q nn+/ q nn+ + q nn+/ q n+n+ n q n+n+ q n+n+3 q nn+/ q n+n+ n q nn+ n q n+n+ = q n+n+3 + q n+n+ = n= q n+n+, so the proof of.3.7 is complete by employing.0. and.0.8. Theorem.3.3. We have u 8n q n = ϕq fq 3, q 5 f qf q, 5

59 u 8n+ q n = ψqfq3, q 5 f qf q, u 8n+ q n = ψq fq 3, q 5 f qf q, u 8n+3 q n = 0, u 8n+ q n = qψq8 fq, q 7 f qf q, u 8n+5 q n = ψqfq, q7 f qf q, u 8n+6 q n = ψq fq, q 7 f qf q, u 8n+7 q n = 0. Proof. It is obvious from.3.,.3.3 and.3. that the fourth and the last identities hold. To prove the first equality, by.3. with a = 0,.3.3 and.3., we have u 8n q n = ψq / + ψ q / n= q n + n= n= q n + q n / n= n q n / = ϕq + ϕq / + ϕq + ϕ q / ψq / ψ q / = ϕq ψq / + ϕq / + ψ q / ψq / + ϕ q/..3.8 ψ q / Employing.0.5 and.0., respectively, we see that ϕq / ψq / = ψq/ ψq = ϕ q ψ q /

60 Similarly, ϕ q / ψ q / = ψ q/ ψq = ϕ q ψq /..3.0 Putting.3.9 and.3.0 in.3.8 yields By.0.6,.3.5 and.3.6, u 8n q n = ϕq + ϕ q ψq / +. ψ q / which completes the proof of the first identity. u 8n q n = ϕq fq 3, q 5 f qf q, Now by.3. with a =,.3.3 and.3., we find that u 8n+ q n = = ψq / n= q nn+/ + ψq / + ψ q / ψ q / n= q nn+/ ψq..3. Utilize.3.5 and.3.6 to deduce that u 8n+ q n = ψqfq3, q 5 f qf q. This completes the proof of the second identity. Again by.3. with a =,.3.3 and.3., we have u 8n+ q n = = ψq / n= ψq / + ψ q / q nn+ + ψ q / n= q nn+ ψq..3. so the proof of the third identity is complete after employing.3.5 and.3.6 in.3.. 5

61 As before, by.3. with a =,.3.3 and.3., we have u 8n+ q n = ψq / ψ q / n= q n / + n= n= q n / + q n / n= n q n / = q / ϕq + ϕq / q / ϕq + ϕ q / ψq / ψ q / = q / ϕq ψq / ϕq / + ψ q / ψq / ϕ q/..3.3 ψ q / Putting.3.9 and.3.0 in.3.3 yields u 8n+ q n = q / By.0.7,.3.5 and.3.7 above, The proof of the fifth identity is complete. ϕq ϕ q ψq /. ψ q / u 8n+ q n = qψq8 fq, q 7 f qf q. By.3. with a = 5,.3.3 and.3., we have u 8n+5 q n = ψq / = q / n= q n +n / ψ q / ψq. ψq / ψ q / n= q n +n / Using.3.5 and.3.7 above, we find that the sixth identity holds. Again by.3. with a = 6,.3.3 and.3. above, we see that u 8n+6 q n = ψq / n= q nn+ / ψ q / n= q nn+ / 55

62 = q / ψq / ψq. ψ q / After utilizing.3.5 and.3.7, the proof of the seventh identity is complete. By..8 and..9, we see that Dq = ϕq ϕq qψq = q n q n ψq n= n= =: v n q n..3. Then apply U 8 to.3. and deduce that v 8n+a q n = U 8 q a Dq = 7 ζ aj q a/8 8 Dζ j q /8 j=0 = 7 ζ aj q a/8 ζ n j q n /8 ζ n j q n /8 6 ψζ j q / j=0 n= n= = 7 ζ n a j q n a /8 ζ n a j q n a /8 6 ψ j q / j=0 n= n= 3 = ζ n a j q n a /8 ζ n a j q n a /8 6ψq / j=0 n= n= 3 + ζ n a j q n a /8 ζ n a j q n a /8. 6ψ q / j=0 n= n=

63 Theorem.3.. We have v 8n q n = ψqfq3, q 5 f qf q, v 8n+ q n = ψq fq 3, q 5 f qf q, v 8n+ q n = 0, v 8n+3 q n = ϕq fq, q 7 f qf q, v 8n+ q n = ψqfq, q7 f qf q, v 8n+5 q n = ψq fq, q 7 f qf q, v 8n+6 q n = 0, Proof. We observe that by.3. and.3.5, and v 8n+a q n = v 8n+7 q n = ψq8 fq 3, q 5 f qf q. u 8n+a+ q n, for a = 0,, 5, 6, v 8n+a q n = u 8n+a+ q n, for a =,. Then by Theorem.3.3, we can deduce all identities except the fourth and the last one. To prove the fourth equality, by.3.5 with a = 3,.3.3 and.3., we have v 8n+3 q n = ψq / + ψ q / n= q n / n= n= q n / n+ q n / + 57 n= q n /

64 = q / ϕq / ϕq + q / ϕq ϕ q / ψq / ψ q / = q / ϕq ψ q / ϕq / + ψq / ψq / ϕ q/..3.6 ψ q / Putting.3.9 and.3.0 in.3.6 yields v 8n+3 q n = q / ϕq + ϕ q ψ q / ψq / After using.0.6,.3.5 and.3.7 above, we complete the proof of the fourth identity. To prove the last equality, by.3.5 with a = 7,.3.3 and.3., we have v 8n+7 q n = ψq / + ψ q / n= q n / n= n= q n n+ q n / n= q n = q ϕq / ϕq q + ϕ q / ϕq ψq / ψ q / = ϕq / q ψq / + ϕ q/ ϕq ψ q / ψq / ψ q /. Putting.3.9 and.3.0 in.3.7 yields v 8n+7 q n = q ϕq ϕ q ψq / +, ψ q / so the proof is complete after employing.0.7,.3.5 and.3.6 above. Corollary.3.5. With u n and v n defined by.3. and.3., we have, for n 0, u 8n > 0, u 8n+ > 0, u 8n+ > 0, u 8n+3 = 0, u 8n+ < 0, u 8n+5 < 0, u 8n+6 < 0, u 8n+7 = 0, u = 0, v 8n > 0, v 8n+ < 0, v 8n+ = 0, v 8n+3 > 0, v 8n+ < 0, v 8n+5 > 0, v 8n+6 = 0, v 8n+7 < 0. 58

65 Proof. These results follow immediately from Theorem.3.3 and Theorem.3.. The last corollary was proved by Chen and Huang [8] and Hirschhorn [] by different methods. Moreover, the formulas of Theorem.3.3 and Theorem.3. are in more compact forms than those of Hirschhorn. We observe that by the two identities.. and.., Dq = Sq T q..3.8 Chen and Huang [8] showed that with u n and v n defined by.3. and.3., u n q n = Dq ψq ψq,.3.9 v n+ q n = ψq Dq ψq.3.0 and Theorem.3.6. We have u n+ q n = v n q n = ψq ψq..3. Dq = qψq 8 + q ψq 6 + q ψq 3 + q 8 ψq 6 + ψq and Dq = ψq 8 qψq 6 + q ψq 3 q ψq 6 +. ψq Proof. By.3.9 and.3., it follows that Dq = u n+ q n+ + u n q n = q ψq8 ψq + Dq ψq8 ψq 59

66 = q ψq8 ψq + ψq6 q ψq + ψq3 q ψq + ψq6 q8 ψq + = qψq 8 + q ψq 6 + q ψq 3 + q 8 ψq 6 +, ψq which completes the proof of the first result. Similarly, by.3.0 and.3., we find that Dq = v n q n + v n+ q n+ = ψq8 ψq q ψq8 ψq Dq = ψq8 ψq q ψq6 ψq + ψq3 q ψq ψq6 q ψq + = ψq 8 qψq 6 + q ψq 3 q ψq 6 +, ψq as desired. 60

67 Chapter 5 Hyperbolic Infinite Series Connected with Theta Functions 5. Introduction In his notebooks [33], [9, pp. 3 38], Ramanujan records several formulas for infinite series involving hyperbolic functions, and B. C. Berndt offers proofs in [9, pp. 3 38] using the classical theory of elliptic functions. In this chapter, we establish new identities of these types. For 0 < x <, we let z := zx := F, ; ; x, and y := yx := π F, ; ; x F,, 5.. ; ; x q := e y. We now define the Eisenstein series for n, E n q := n B n k= k n q k q k, where B n is the nth Bernoulli number. Recall that E n can be expressed as a polynomial in E and E 6 [3, p. 8], [35, p. 99]. Say, E n q = f n E q, E 6 q, 6

68 where f n is a certain polynomial. Lemma 5.. [3, p. 8], [3], [3, p. ], [35, pp. 95, 97]. We have f E, E 6 = E, 5.. f 6 E, E 6 = E 6, 5..3 f 8 E, E 6 = E, 5.. and f 0 E, E 6 = E E 6, 5..5 f E, E 6 = 69 E E 6, 5..6 f E, E 6 = E E Lemma 5.. [9, p. 7]. For 0 < x <, we have E q = z + x + x, 5..8 E 6 q = z 6 + x 3x + x, 5..9 E q = z x + 6 x, 5..0 and E 6 q = z 6 x x 3 x

69 5. Results Theorem 5... Suppose x, y and z are related as in the previous section. Then k= k= k= k= k= k= coshky + sinhky k 3 = z 6 + x x, coshky + k 5 = z x x x, sinhky 8 coshky + k 7 = z8 sinhky 6 + x3 + 08x + 7x x, coshky + k 9 = z0 sinhky x + 53x + 590x 3 3x x, coshky + sinhky coshky + sinhky k = z x x x 3 + 0x + 69x 5 x, k 3 = z x x x x x 5 56x 6 x. Note that one can obtain further evaluations involving summand with k n+ for n 7, but the computation will be very complicated because of the function f n. Proof. By elementary manipulations, we find that E n q E n q = n B n = n B n = n B n q k q qk k q k q k + q k + q 3k k= q k coshky + sinhky k= k= k n k n k n. 63

70 Then for n, k= coshky + k n = B n En q E n q sinhky n = B n n fn E q, E 6 q f n E q, E 6 q. 5.. To prove the first identity, set n = in 5... We know that B =. Then by and 5..0, it follows that k= coshky + k 3 = B E q E q sinhky = 0 = z 6 + x x, 7 z x + 6 x z + x + x which is the first identity. For the second equality, let n = 3 in 5.. and note that B 6 =. By 5..9 and 5.., we find that k= coshky + k 5 = B 6 E6 q E 6 q sinhky 6 = 5 z 6 x x 3 x = z x x x, z 6 + x 3x + x which completes the proof of the second equality. Now, by 5..8, 5..0, and 5.. with n = and B 8 =, we deduce that 30 k= coshky + k 7 = B 8 E sinhky 8 q Eq 6