a parameter for ramanujan s function q its explicit values and applications

International Scholarly Research Network ISRN Computational Mathematics Volume 2012, Article ID 169050, 14 pages doi:10.5402/2012/169050 Research Article A Parameter for Ramanujan’s Function χ(q): Its Explicit Values and Applications Nipen Saikia Department of Mathematics, Rajiv Gandhi University, Rono Hills, Doimukh 791112, India Correspondence should be addressed to Nipen Saikia, nipennak@yahoo.com Received May 2012; Accepted 28 June 2012 Academic Editors: L Hajdu, L S Heath, and H J Ruskin Copyright © 2012 Nipen Saikia This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited We define a new parameter Ik,n involving quotient of Ramanujan’s function χ(q) for positive real numbers k and n and study its several properties We prove some general theorems for the explicit evaluations of the parameter Ik,n and find many explicit values Some values of Ik,n are then used to find some new and known values of Ramanujan’s class invariant Gn √ Introduction In Chapter 16 of his second notebook [1], Ramanujan develops the theory of theta-function Ramanujan’s general theta-function is defined by ∞ n(n+1)/2 n(n−1)/2 f (a, b) = a b , |ab| < (1) n=−∞ ∞ f −q := f −q, −q n n(3n−1)/2 = χ q = 21/6 α(1 − α) q χ −q = 21/6 (1 − α)1/12 −1/24 , α q (5) −1/24 , it follows from (4) that After Ramanujan, for |q| < 1, we define where q := e−π n and n is a positive rational number Since from [2, page 124, Entry 12(v) & (vi)] (−1) q Gn = {4α(1 − α)}−1/24 , = q; q n=−∞ ∞, (2) (6) gn = 2−1/12 (1 − α)1/12 α−1/24 Also, if β has degree r over α, then ∞ n=0 (1 where (a; q)∞ := If q = with Im(z) > 0, then f (−q) = q−1/24 η(z), where η(z) denotes the classical Dedekind eta function Ramanujan’s function χ(q) is defined by χ q := − aqn ) e2πiz f q = −q; q2 f −q ∞ (3) The function χ(q) is intimately connected to Ramanujan’s class invariants Gn and gn , which are defined by Gn = 2−1/4 q−1/24 χ q , gn = 2−1/4 q−1/24 χ −q , (4) Gr n = 4β − β −1/24 , gr n = 2−1/12 − β 1/12 −1/24 β (7) In his notebooks [1] and paper [3], Ramanujan recorded a total of 116 class invariants or monic polynomials satisfied by them The table at the end of Weber’s book [4, page 721– 726] also contains the values of 107 class invariants Weber primarily was motivated to calculate class invariants so that he could construct Hilbert class fields On the other hand Ramanujan calculated class invariants to approximate π and probably for finding explicit values of Rogers-Ramanujan continued fractions, theta-functions, and so on An account ISRN Computational Mathematics of Ramanujan’s class invariants and applications can be found in Berndt’s book [5] For further references, see [6– 12] Ramanujan and Weber independently and many others in the literature calculated class invariants Gn for odd values of n and gn for even values of n For the first time, Yi [13] calculated some values of gn for odd values of n by finding explicit values of the parameter rk,n (see [13, page 11, (2.1.1)] or [14, page 4, (1.11)]) defined by rk,n := f −q k1/4 q(k−1)/24 f q = e−2π , −q k √ n/k (8) In particular, she established the result [13, page 18, Theorem 2.2.3] gn = r2,n/2 (9) However, the values of Gn for even values of n have not been calculated The main objective of this paper is to evaluate some new values of Gn for even values of n We also prove some known values of Gn For evaluation of class invariant Gn in this paper, we introduce the parameter Ik,n , which is defined as Ik,n := χ q q(−k+1)/24 χ qk q = e−π , √ n/k , (10) where k and n are positive real numbers In Section 3, we study some properties of Ik,n and also establish its relations with Ramanujan’s class invariant Gn In Section 4, by employing Ramanujan’s modular equations, we present some general theorems for the explicit evaluations of Ik,n and find several explicit values of Ik,n In Section 5, we establish some general theorems connecting the parameter Ik,n and the class invariant Gn We also evaluate some explicit values of the product Gnk Gn/k by employing some values of Ik,n evaluated in Section Finally, in Section 6, we calculate new and known values of class invariant Gn by combining the explicit values of Ik,n and the product Gnk Gn/k evaluated in Sections and 5, respectively Section is devoted to record some preliminary results Since Ramanujan’s modular equations are key in our evaluations of Ik,n and Gn , we complete this introduction by defining Ramanujan’s modular equation from Berndt’s book [2] The complete elliptic integral of the first kind K(k) is defined by K(k) : = π/2 dφ − k2 sin2 φ ∞ = (1/2)2n 2n k π n=0 (n!) (11) π 1 = F1 , ; 1; k2 , 2 L denote the complete elliptic integrals of the first kind associated with the moduli k, k , l, and l , respectively Suppose that the equality n L K = K L (13) 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 (13) If we set q = exp −π K , K q = exp −π L , L (14) we see that (13) is equivalent to the relation qn = q Thus, a modular equation can be viewed as an identity involving theta-functions at the arguments q and qn Ramanujan recorded his modular equations in terms of α and β, where α = k2 and β = l2 We say that β has degree n over α The multiplier m connecting α and β is defined by m= K L (15) Ramanujan also established many “mixed” modular equations in which four distinct moduli appear, which we define from Berndt’s book [2, page 325] Let K, K , L1 , L1 , L2 , L2 , L3 , and L3 denote complete elliptic integrals of the first kind corresponding, in pairs, to √ √ √ the moduli α, β, γ, and δ and their complementary moduli, respectively Let n1 , n2 , and n3 be positive integers such that n3 = n1 n2 Suppose that the equalities n1 K L = 1, K L1 n2 K L = 2, K L2 n3 K L = K L3 (16) hold Then, a “mixed” modular equation is a relation √ √ √ between the moduli α, β, γ, and δ that is induced by (16) In such an instance, we say that β, γ, and δ are of degrees n1 , n2 , and n3 , respectively, over α or α, β, γ, and δ have degrees 1, n1 , n2 , and n3 , respectively Denoting zr = φ2 (qr ), where q = exp − πK K , φ q = f q, q , q Thus, by Theorem 16(i), Ik,n < for all n > if k > Theorem 18 For all positive real numbers k, m, and n, one has −1 Ik,n/m = Imk,n Ink,m (35) Proof Using the definition of Ik,n , we obtain √ χ e−π n/mk Imk,n = √ √ √ Ink,m eπ( m/nk− n/mk)/24 χ e−π m/nk (36) Using Lemma in the denominator of the right-hand side of (36) and simplifying, we complete the proof Inp,np = Inp2 ,n I p,p (41) Proof The result follows immediately from Theorem 20 with a = p2 , b = 1, c = d = p, and k = n Now, we give some relations connecting the parameter Ik,n and Ramanujan’s class invariants Gn Theorem 22 Let k and n be any positive real numbers Then −1 (i) Ik,n = Gn/k Gnk , (ii) G1/n = Gn (42) Proof Proof of (i) follows easily from the definitions of Ik,n and Gn from (10) and (4), respectively To prove (ii), we set k = in part (i) and use Theorem 16(i) and (iii) General Theorems and Explicit Evaluations of Ik,n In this section, we prove some general theorems for the explicit evaluations of Ik,n and find its explicit values ISRN Computational Mathematics Theorem 23 One has I3,n I3,4n × 12 I3,n I3,4n + 10 I3,n I3,4n 12 ⎡ −10 −12 + −6 I3,n I3,4n Equivalently, ⎤ ⎦ +25 I3,n I3,4n + I3,n I3,4n −4 + I3,n I3,4n −8 D = −2 + ⎛ , This completes the proof of (ii) Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii) √ (44) 1/12 (iii) I3,1/2 = −44 + 27 + 458 − 264 ⎛ × Proof Setting n = 1/2 in Theorem 23 and using Theorem 16(ii), we obtain 24 I3,2 −24 + I3,2 + 176 12 I3,2 −12 + I3,2 − 1002 = Theorem 25 One has I5,n I5,4n , √ √ ⎞1/2 −6 + + −2 + ⎠ (iv) I3,1/4 = ⎝ −6 I + 5,n I5,4n ⎡ I + ⎣ 5,n I5,4n −5 I5,n I5,4n + I5,n I5,4n −3 I + 5,n I5,4n + I5,n I5,4n ⎤ ⎦ + I5,n I5,4n −3 +19 I5,n I5,4n + I5,n I5,4n = I5,n I5,4n + I5,n I5,4n (45) + 52 I5,n I5,4n Equivalently, −6 + 13 I5,n I5,4n + I5,n I5,4n −2 + I5,n I5,4n −4 + 82 (55) B2 + 176B − 1004 = 0, (46) 12 −12 + I3,2 B = I3,2 (47) where Proof The proof follows from Lemma and the definition of Ik,n Corollary 26 One has Solving (46) and using the fact in Remark 17, we obtain √ B = 54 − 88 (48) Employing (48) in (47), solving the resulting equation for I3,2 , and noting that I3,2 < 1, we arrive at √ √ 1/12 I3,2 = −44 + 27 − 458 − 264 −10 I10 3,4 + I3,4 −6 −2 +16 I3,4 + I3,4 + 71 I3,4 + I3,4 −8 −4 = 25 I3,4 + I3,4 + 200 I3,4 + I3,4 + 550 √ (50) √ 1/6 (i) I5,2 = −14 + 10 − 445 − 140 10 √ 11 + 5 1/4 (ii) I5,4 = (49) This completes the proof of (i) Again setting n = in Theorem 23 and using Theorem 16(i), we obtain −6 I3,4 + I3,4 (54) 1/12 √ √ ⎞1/2 −2 + − −6 + ⎠ , √ (53) √ √ ⎞1/2 −2 + − −6 + ⎠ I3,4 = ⎝ (i) I3,2 = −44 + 27 − 458 − 264 (ii) I3,4 1/2 Employing (53) in (52), solving the resulting equation, and using the fact that I3,4 < 1, we obtain Corollary 24 One has √ (52) Since the first factor of (51) is nonzero, solving the second factor, we deduce that √ √ (51) −2 + I3,4 D = I3,4 + 550 Proof The proof follows easily from the definition of Ik,n and Lemma =⎝ D4 + 4D2 − 50 = 0, where −2 (43) ⎛ D2 + +16 I3,n I3,4n +(I3,n I3,4n )−6 + I3,n I3,4n + I3,n I3,4n + 200 I3,n I3,4n I3,n I3,4n +⎣ + I3,n I3,4n +71 I3,n I3,4n = I3,n I3,4n −12 , √ − −4 + 11 + 5 , √ √ (iii) I5,1/2 = −14 + 10 + 445 − 140 10 √ 11 + 5 (iv) I5,1/4 = 1/4 (56) 1/6 , √ + −4 + 11 + 5 Proof Setting n = 1/2 in Theorem 25 and using Theorem 16(ii), we obtain C + 56C − 216 = 0, (57) ISRN Computational Mathematics × where −6 C = I5,2 + I5,2 (58) Solving (57) and noting the fact in Remark 17, we obtain √ C = −28 + 10 10 (59) Employing (59) in (58), solving the resulting equation, and noting that I5,2 < 1, we obtain √ √ + I7,n I7,4n −81 + I7,n I7,4n I7,n I7,4n × 16 B8 − 22B4 − = 0, × 296 −8 where −1 B = I5,4 + I4,5 (62) = I7,n I7,4n I7,n I7,4n I7,n I7,4n 12 10 B = 11 + 5 1/4 −8 √ √ − −4 + 11 + 5 1/4 I5,4 = −2 ⎦ + I7,n I7,4n I7,n I4,7n + ⎡ + 16⎣ 10 I7,n I4,7n ⎡ + + 8⎣ I7,n I4,7n (64) + I7,n I7,4n × I7,n I7,4n ⎡ +⎣ I7,n I4,7n + −10 + I7,n I7,4n + 496 I7,n I7,4n + I7,n I7,4n −8 −4 Corollary 28 One has (i) I7,2 = 2−1/4 ⎛ − √ √ −4+ −8 − 2+ 163 + 116 −8 ⎞1/4 ⎠ I7,n I7,4n + I7,n I7,4n √ ⎛ −2 √ √ × ⎝ − − + 163 + 116 ⎦ ⎤ ⎦ , −6 + − −58 + 45 √ , ⎤ + I7,n I7,4n −6 (iii) I7,1/2 = 2−1/4 , ⎤ ⎦ (65) (ii) I7,4 = −10 + 1746 −12 √ I7,n I4,7n I7,n I4,7n −6 √ √ × ⎝ − − + 163 + 116 −12 I + 7,n I4,7n −2 + I7,n I7,4n + I7,n I7,4n Theorem 27 One has I7,n I4,7n + 352 ⎤ + 145 I7,n I7,4n This completes the proof of (ii) Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii) 12 −4 + I7,n I7,4n (63) Using (63) in (62), solving the resulting equation, and noting that I5,4 < 1, we arrive at 11 + 5 −2 ⎦ Solving (61), we obtain √ −6 ⎤ + I7,n I7,4n I + 7,n I4,7n I7,n I7,4n − 256 (61) I7,n I7,4n I7,n I4,7n −4 I + 7,n I4,7n I7,n I7,4n ⎡ +⎣ This completes the proof of (i) Again, setting n = in Theorem 25 and using Theorem 16(i), we obtain I + ⎣ 7,n I4,7n (60) −10 −288 + I7,n I7,4n + 32 I7,n I7,4n ⎡ 1/6 I5,2 = −14 + 10 − 445 − 140 10 10 I7,n I7,4n √ √ ⎞1/4 + −4+ −8 − 2+ 163+116 ⎠ −4 − 19 √ (iv) I7,1/4 = √ −6 + + −58 + 45 √ (66) ISRN Computational Mathematics Proof Setting n = 1/2 and simplifying using Theorem 16 (ii), we obtain I7,n I7,25n 24 −24 20 −20 16 −16 − 40 I7,2 I7,2 + I7,2 + 32 I7,2 + I7,2 + I7,2 12 −12 −8 −4 − 192 I7,2 − 96 I7,2 +I7,2 +I7,2 +64 I7,2 +I7,2 +462 = (67) Equivalently, Theorem 29 One has A − A + 32A − 42A − 128A − 191 = 0, (68) where + I7,n I7,25n ⎧ ⎨ I + 5⎩ 7,25n I7,n − 10 2 I7,25n I7,n −2 ⎧ ⎨ I 7,25n − ⎩ I7,n ⎫ I + 7,25n I7,n −3 ⎬ ⎭ ⎫ −2 ⎬ I + 7,25n I7,n ⎭ I7,25n I7,n + + 15 = (76) Proof Using (5) in Lemma 9, we find that −4 + I7,2 A = I7,2 (69) By using the fact in Remark 17, it is seen that the first factor of (68) is nonzero, and so from the second factor, we deduce that √ √ Q= qχ q5 χ q7 , χ q χ q35 Setting q = e−π we get (70) A = −8 − + 163 + 116 Q= Combining (69) and (70) and noting that I7,2 < 1, we obtain ⎛ −1/4 ⎝ I7,2 = √ n/7 √ √ −4 + −8 − 2+ 163 + 116 and using the definition of Ik,n in (77), 1/2 I7,25n I7,n R = I7,n I7,25n , 1/2 (78) √ ⎞1/4 ⎠ h − −36 + h2 √ , (i) I7,5 = √ h + −36 + h2 √ , (ii) I7,1/5 = (79) √ This completes the proof of (i) E2 E2 − E4 + 108E2 − 1134 = 0, d − −144 + d2 (iii) I7,25 = To prove (ii), setting n = and simplifying using Theorem 16(i), we arrive at , 12 √ (iv) I7,1/25 = (72) d + −144 + d2 √ 12 , √ 1/3 1/3 where h = + (62 − 105) + (62 + 105) and d = √ 1/3 √ 1/3 12 + 22/3 (135 − 15 21) + 22/3 (135 + 15 21) where −2 + I7,4 E = I7,4 (73) Using the fact in Remark 17 it is seen that the first two factors of (72) are nonzero, and so solving the third factor, we obtain √ (74) E = −6 + Combining (73) and (74) and noting that I7,4 < 1, we deduce that √ (77) Corollary 30 One has (71) I7,4 = q3/2 χ q χ q5 χ q7 χ q35 Employing (78) in (27), we complete the proof √ − − + 163 + 116 − √ R= Proof Setting n = 1/5 in Theorem 29 and simplifying using Theorem 16(ii), we obtain −6 −4 −2 − 17 = + I7,5 − I7,5 + I7,5 + 10 I7,5 + I7,5 I7,5 Equivalently, (75) So the proof of (ii) is complete Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii) H − 5H + 7H − = 0, (81) −2 + I7,5 H = I7,5 (82) where √ −6 + − −58 + 45 √ (80) Solving (81) and noting the fact in Remark 17, we obtain √ H= + 62 − 105 1/3 √ + 62 + 105 1/3 (83) ISRN Computational Mathematics Combining (82) and (83) and noting that I7,5 < 1, we deduce that Corollary 32 One has √ √ (i) I13,3 = h − −36 + h2 √ , I7,5 = (84) √ h = + 62 − 105 1/3 √ + 62 + 105 1/3 , √ (85) This completes the proof of (i) Again setting n = and simplifying using Theorem 16(i), we arrive at U − 6U + 7U − = 0, (86) , √ 2+ 3− 3+4 (ii) I13,9 = where √ √ + 13 − −2 + 13 (93) √ + 13 + −2 + 13 (iii) I13,1/3 = √ √ 2+ 3+ 3+4 (iv) I13,1/9 = , Proof Setting n = 1/3 in Theorem 31 and simplifying using Theorem 16(ii), we obtain where V − 3V − = 0, (94) −2 + I13,3 V = I13,3 (95) where −1 U = I7,25 + I7,25 (87) Solving (94) and using Remark 17, we get Solving (86) and using Remark 17, we get √ U= + 15 − 21 /2 1/3 √ √ + 15 + 21 /2 + 13 V= 1/3 (88) Combining (95) and (96) and noting that I13,3 < 1, we obtain √ √ I7,25 = 12 (89) , where 2/3 d = 12 + √ 135 − 15 21 1/3 2/3 +2 I13,n I13,9n − + I13,n I13,9n ⎧ ⎨ 135 + 15 21 (90) I13,n I13,n I13,9n + I13,n I13,9n −1 + I13,n I13,9n I13,9n I13,n (98) −1 J = I13,9 + I13,9 (99) √ J = + ⎭ (100) Combing (99) and (100) and noting that I13,9 < 1, we deduce that √ −1 ⎬ (91) + = √ 2+ 3− 3+4 I13,9 = Proof Proceeding as in the proof of Theorem 29, using (5) √ in Lemma 10, setting q := e−π n/13 , and using the definition of Ik,n , we arrive at Q= J − 4J + = 0, Solving (98) and using Remark 17, we get ⎫ −3 ⎩ I13,9n (97) So we complete the proof of (i) Again setting n = and using Theorem 16(i), we obtain 1/3 Theorem 31 One has −2 where √ This completes the proof of (ii) Now (ii) and (iv) easily follow from (i) and (ii), respectively, and Theorem 16(ii) √ + 13 − −2 + 13 I13,3 = Combining (87) and (88) and noting that I7,25 < 1, we obtain d − −144 + d2 (96) (101) So the proofs of (ii) is complete Now the proof of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii) Theorem 33 One has I13,25n I13,n I13,25n I13,n −3 ⎧ ⎨ I 13,25n −5 ⎩ I13,n I + 13,25n I13,n −1 ⎫ ⎬ + 2⎭ 1/2 , T = I13,n I13,9n Employing (92) in (28), we complete the proof 1/2 (92) × I13, 25n I13,n + I13,25n I13,n − I13,25n I13,n + I13,25n I13,n −1 −2 +2 = (102) ISRN Computational Mathematics Proof Using (5) in Lemma 11, setting q := e−π using the definition of Ik,n , we arrive at Q= I13,25n I13,n √ n/13 , and √ 1/2 1/2 T = I13,n I13,25n , (103) √ + 65 − 58 + 65 (ii) I13,25 = √ c − −36 + c2 I7,n I7,9n + I7,n I7,9n , √ + 65 + 58 + 65 √ (iv) I13,1/25 = √ where c = + (1080 − 15 39) 1/3 Q= √ and ⎧ ⎨ I −1 7,9n − ⎩ I7,n ⎫ R = I7,n I7,9n , simplifying using √ (i) I7,3 = + I13,5 √ + 65 − 58 + 65 A3 − 6A2 − 23A − 18 = 0, −1 A = I13,25 + I13,25 (iv) I7,1/9 = (110) √ + 1080 + 15 39 (114) , (115) 1/4 , √ + 21 + + 21 = 1/3 Proof Setting n Theorem 16(ii), we arrive at and simplifying −4 + I7,3 − = I7,3 using (116) Solving (116) and noting the fact in Remark 17, we obtain √ I7,3 = − 21 1/4 (117) This completes the proof of (i) To prove (ii), setting n = and simplifying using Theorem 16(i), we arrive at Solving (109) and using the fact in Remark 17, we obtain + 21 √ (109) where 1/3 (iii) I7,1/3 = (108) This completes the proof of (i) To prove (ii), setting n = and simplifying using Theorem 16(i), we arrive at √ 1/2 √ + 21 − + 21 (107) Employing (107) in (106), solving the resulting equation, and noting that I13,5 < 1, we obtain + 1080 − 15 39 and , √ √ + 65 L= 2 n/7 , 1/4 − 21 (ii) I7,9 = (106) Solving (105) and using the fact in Remark 17, we obtain I13,5 = √ Corollary 36 One has (105) −2 √ + = Employing (114) in (30), we complete the proof √ L= ⎭ 1/2 I7,9n I7,n where I13,5 −2 ⎬ I + 7,9n I7,n 1/3 L3 − 23L − 42 = 0, A= (113) , + (1080 + 15 39) Proof Setting n = 1/5 Theorem 16(ii), we arrive at 1/3 Proof Using (5) in Lemma 12, setting q := e−π using the definition of Ik,n , we arrive at , c + −36 + c2 √ 1/3 (104) √ (iii) I13,1/5 = (112) , Theorem 35 One has , I13,25 = where c = + (1080 − 15 39) + (1080 + 15 39) This completes the proof of (ii) Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii) Corollary 34 One has √ c − −36 + c2 √ Employing (103) in (29), we complete the proof (i) I13,5 = Employing (111) and (110), solving the resulting equation, and noting that I13,25 < 1, we obtain D2 − D − = 0, (118) −1 D = I7,9 + I7,9 (119) where 1/3 (111) 10 ISRN Computational Mathematics Solving (118) and using the fact in Remark 17, we obtain √ + 21 D= (120) Employing (120) in (119), solving the resulting equation, and noting that I7,9 < 1, we deduce that √ √ + 21 − + 21 I7,9 = (121) This completes the proof of (ii) Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii) General Theorems and Explicit Evaluations of Gnk Gn/k In this section we evaluate some explicit values of the product Gnk Gn/k by establishing some general theorems and employing the values of Ik,n obtained in Section We recall from Theorem 22(ii) that G1/n = Gn for ready references in this section + I3,n I3,4n −2 + (Gn/3 G4n/3 G9n/3 G36n/3 )−1 √ Proof To prove (i), using (5) in Lemma 13, setting q := √ e−π n/3 , and employing the definitions of Ik,n and Gn , we obtain q2 q χ χ q3 χ q6 = I3,n I3,4n −1/2 P = (Gn/3 G4n/3 G9n/3 G36n/3 ) √ G23 G4/3 G12 1/2 = √ Cubing (96) and then employing in (128) and solving the resulting equation, we complete the proof Theorem 39 One has + I5,n I5,4n (123) 1+ , √ + + −2 + , 1/3 + I5,n I5,4n + I5,n I5,4n −1 −3 (ii) I5,n + I5,n + (Gn/5 G5n )−2 − (Gn/5 G5n )2 = (129) Proof Using (5) in Lemma 14, setting q := e−π employing the definitions of Ik,n and Gn , we obtain q1/2 χ q χ q2 = I5,n I5,4n χ q5 χ q10 (124) 1/2 √ n/5 , and , (130) P = (Gn/5 G4n/5 G25n/5 G100n/5 )−1/2 Employing (130) in (32), we complete the proof of (i) Similarly, (ii) follows √from Lemma and the definition of Ik,n and Gn with q := e−π n/5 Corollary 40 One has √ √ −3 = 4(Gn/5 G4n/5 G25n/5 G100n/5 ) (i) G10 G5/2 = (iii) G39 G13/3 = 21/6 + 13 (127) −6 + I3,13 + 2 (G3/13 G39 )−2 − (G3/13 G39 )2 = (128) I3,13 , √ (ii) G12 G4/3 = 2−13/6 √ + + −2 + Using the value G3 = 21/12 from [5, p 189] in (127), we complete the proof of (ii) To prove (iii), setting n = 13 in Theorem 37(ii), we obtain Corollary 38 One has √ (126) Employing (53) in (126), solving the resulting equation, and noting that G23 G4/3 G12 > 1, we obtain R= Employing (123) in (31), we complete the proof (ii) follows similarly from Lemma and the definition of Ik,n and Gn √ with q := e−π n/3 (i) G6 G3/2 = − G23 G4/3 G12 = −2 −6 (ii) I3,n + I3,n + 2 (Gn/3 G3n )−3 − (Gn/3 G3n )3 = (122) q1/4 χ −1 − (Gn/5 G4n/5 G25n/5 G100n/5 ) , − 2(Gn/3 G4n/3 G9n/3 G36n/3 ) = 0, R= −2 I3,4 + I3,4 + G23 G4/3 G12 (i) I5,n I5,4n Theorem 37 One has (i) I3,n I3,4n Solving (125) and noting that G6 G3/2 > 1, we complete the proof of (i) To prove (ii), setting n = in Theorem 37(i); using Theorem 16(i), and noting that G1/n = Gn , we obtain (ii) G20 G5/4 = + 10 1/4 , √ −3 11+5 1/4 √ + 11 + 5 √ 3/4 +M, (131) √ where M denotes 16 + 11 + 5(−3 + 11 + 5) Proof Setting n = 1/2 in Theorem 37(i) and simplifying using Theorem 16(ii) and the result G1/n = Gn , we obtain Proof Setting n = 1/2 in Theorem 39(i) and simplifying using Theorem 16(ii) and the result G1/n = Gn , we obtain 2(G6 G3/2 )2 − (G6 G3/2 )−2 − = 4(G10 G5/2 )4 − (G10 G5/2 )−4 − 12 = (125) (132) ISRN Computational Mathematics 11 Solving (132) and noting that G10 G5/2 > 1, we complete the proof of (i) For proof of (ii), setting n = in Theorem 39(ii) and simplifying using Theorem 18(i) and the result G1/n = Gn , we obtain −3 I5,4 + I5,4 + (G4/5 G20 )−2 − (G4/5 G20 )2 = 1/4 √ 3/4 + 11 + 5 √ 2+ (i) G7/4 G28 = , √ (ii) G7/3 G21 = 2−1/6 + (133) −1 Employing the value of I5,4 + I5,4 from (63) in (133), solving the resulting equation, and noting that G4/5 G20 > 1, we get √ G4/5 G20 = −3 11 + 5 Corollary 44 One has + M (134) So the proof is complete √ (iii) G7/9 G63 = 21/6 + 21 √ −4 + I7,4 + = 2 (G7/4 G28 )3 + (G7/4 G28 )−3 I7,4 √ −4 + I7,4 = −56 + 45 I7,4 3/2 2 I2,n I2,49n + I2,n I2,49n = 4(Gn/2 G4n/2 G47n/2 G196n/2 ) (142) 1/2 −1/2 (135) Employing (143) in (142), solving the resulting equation, and noting that G7/4 G28 > 1, we complete the proof of (i) To prove (ii), we set n = in Theorem 43, and employing (116), we obatin √ (G21 G7/3 )3 + (G21 G7/3 )−3 − = + 4(Gn/2 G4n/2 G47n/2 G196n/2 )−3/2 −5/2 − (Gn/2 G4n/2 G47n/2 G196n/2 ) √ Proof Using (5) in Lemma 15, setting q := e−π employing the definitions of Ik,n and Gn , we obtain, q1/3 χ q χ q7 = I2,n I2,49n χ q2 χ q14 1/2 n/2 , and √ −4 I7,9 + I7,9 + = 2 (G7/9 G63 )3 + (G7/9 G63 )−3 (145) , (136) Squaring (120) twice and simplifying, we obtain √ −4 + I7,9 = I7,9 Employing (136) in (33), we complete the proof √ 2+ 4+8 (137) Proof Setting n = 1/7 and simplifying using Theorem 16(ii) and the result G1/n = Gn , we get 4(G14 G7/2 ) − 2(G14 G7/2 ) − 2(G14 G7/2 ) (138) + 4(G14 G7/2 )2 − = Theorem 43 One has √ −4 I7,n + I7,n + = 2 (G7/n G7n )−3 + (G7/n G7n )3 Proof Using (5) in Lemma 7, setting q := e−π employing definitions of Ik,n and Gn , we arrive at (139) √ n/7 , χ I13,n −7 + I13,n + 52 I13,n + 13 I13,n + I13,n −3 + I13,n + 78 I13,n + I13,n −6 −5 −8 (Gn/13 G13n ) − (Gn G13n ) −1 = = I7,n Using (140) in (24), we complete the proof P = (Gn/713 G13n )−1 , Q= q1/2 χ q χ q13 n/13 , and = I13,n (148) Corollary 46 One has (140) Proof Using (5) in Lemma 8, setting q := e−π employing definitions of Ik,n and Gn , we arrive at √ Using (148) in (25), we complete the proof P = (Gn/7 G7n ) , q Theorem 45 One has and −3 q7 (146) (147) Solving (138) and noting that G14 G7/2 > 1, we complete the proof q1/4 χ 31 + 21 Employing (146) in (145), solving the resulting equation, and noting that G7/9 G63 > 1, we complete the proof Corollary 42 One has √ (144) Solving (144) and noting that G21 G7/3 > 1, we complete the proof To prove (iii), setting n = in Theorem 43, we get P = (Gn/2 G4n/2 G47n/2 G196n/2 )−1/2 G14 G7/2 = (143) −3/2 − 2(Gn/2 G4n/2 G47n/2 G196n/2 ) Q= 1/3 Squaring (74) and simplifying, we obtain √ (141) , Proof Setting n = in Theorem 43, we get Theorem 41 One has R= 1/3 √ √ G117 G13/9 = 450 + 260 + 405299 + 23400 1/6 (149) 12 ISRN Computational Mathematics Proof Setting n = in Theorem 45, we get I13,9 + I13,9 + 52 I13,9 −7 + 13 I13,9 −3 + I13,9 + I13,9 Theorem 48 One has −5 + 78 I13,9 + I13,9 −6 −8 (G9/13 G117 ) − (G9/13 G117 ) −1 (150) √ × Solving the resulting equation (151) and noting that G9/13 G117 > 1, we complete the proof √ 1/4 1/2 , √ + + −2 + √ √ −2 + + −6 + 1/4 Proof We use Corollary 24(ii) and Corollary 38(ii) and proceed as in Theorem 47 √ In this section we find some new values of Ramanujan’s class invariant Gn by using explicit values of Ik,n and Gnk Gn/k evaluated in Sections and 4, respectively For ready references in this section, we recall from Theorem 22 that Ik,n = Gn/k /Gnk We also recall from Theorem 22(ii) that G1/n = Gn (i) G5/2 = 2−1/8 + 10 1/8 √ √ × −14 + 10 − 445 − 140 10 −1/8 (ii) G10 = √ + 10 √ √ × −44 + 27 − 458 − 264 (ii) G6 = 2−1/4 + , (156) 1/12 1/4 (i) G3/2 = 2−1/4 + √ 1/12 1/8 √ √ × −14 + 10 + 445 − 140 10 Theorem 47 One has √ (155) 1/2 Theorem 49 One has New Values of Class Invariant Gn Proof We employ Corollary 26(i) and Corollary 40(i) and proceed as in Theorem 47 1/24 , (152) 1/4 Theorem 50 One has ⎛ √ √ × −44 + 27 + 458 − 264 √ (i) G5/4 = ⎝ 11 + 5 1/24 1/4 √ √ = −44 + 27 − 458 − 264 −4 + 11 + 5⎠ − √ × ⎝ − 11 + 5 1/12 1/4 √ + 11 + 5 √ ⎞1/2 √ ⎛ Proof From Corollary 24(i), we have G3/2 G6 √ + + −2 + (ii) G12 = 2−4/3 (G9/13 G117 ) − 20 45 + 26 (G9/13 G117 ) + = (151) I3,2 = √ × = Employing (100) in (150) and simplifying, we obtain 12 √ √ −2 + − −6 + (i) G4/3 = 2−4/3 3/4 √ ⎞1/4 + 16+ 11+5 −3+ 11 + 5 ⎠ (153) ⎛ Also, from Corollary 38(i), we have √ G6 G3/2 = 1+ √ (ii) G20 = ⎝ 11 + 5 1/2 (154) Multiplying (153) and (154) and simplifying, we complete the proof of (i) Dividing (154) by (153) and simplifying, we complete the proof of (ii) 1/4 − √ −4 + 11 + 5⎠ ⎛ √ × ⎝ − 11 + 5 ⎞−1/2 1/4 √ √ + 11 + 5 3/4 √ ⎞1/4 + 16 + 11 + 5 −3 + 11 + 5 ⎠ (157) The proofs of the Theorems 48–56 are identical to the proof of Theorem 47 So we give the references of the required results only Proof We employ Corollary 26(ii) and Corollary 40(ii) and proceed as in Theorem 47 ISRN Computational Mathematics 13 Theorem 51 One has Theorem 55 One has √ (i) G7/2 = 2−9/8 + + 1/2 ⎛ √ √ × ⎝ − − + 163 + 116 − (ii) G14 = 2−9/8 + + 1/6 √ 1/6 (ii) G39 = 2−11/12 3+ 13 √ √ −4+ −8 − 2+ 163+116 √ √ (i) G13/3 = 2−11/12 3+ 13 √ √ 1/2 √ √ 1/2 1+ 13 − −2+ 13 1+ 13 + −2 +2 13 (162) ⎞1/8 ⎠ , The values G13/3 and G39 can also be found in [5, 6] , Proof We use Corollary 32(i) and Corollary 38(iii) and proceed as in Theorem 47 1/2 Theorem 56 One has ⎛ √ √ × ⎝ − − + 163 + 116 √ √ √ ⎞1/8 + −4 + −8 − + 163+116 ⎠ × 2+ (158) Proof We employ Corollary 28(i) and Corollary 42(i) and proceed as in Theorem 47 √ √ √ 1/12 1/2 3− 3+4 , √ × 2+ √ √ 3+ 3+4 1/2 (163) √ 1/2 √ 1/2 (ii) G28 = 2−3/4 2+ 1/12 (ii) G117 = 2−1/2 450 + 260 + 405299 + 23400 Theorem 52 One has (i) G7/4 = 2−3/4 2+ √ (i) G13/9 = 2−1/2 450 + 260 + 405299 + 23400 √ √ 1/4 √ √ 1/4 −6+5 − −58+45 −6+5 2+ −58+45 , The values G39 and G117 can also be found in [5, page 193] Proof We employ Corollary 32(ii) and Corollary 46 and proceed as in Theorem 47 (159) Proof We employ Corollary 28(ii) and Corollary 44(i) and proceed as in Theorem 47 Theorem 53 One has √ 1/6 √ 1/6 (i) G7/3 = 2−5/24 + (ii) G21 = 2−5/24 + √ 1/8 √ 1/8 − 21 + 21 , (160) The values G7/3 and G21 can also be found in [5, 6] Proof We employ Corollary 36(i) and Corollary 44(ii) and proceed as in Theorem 47 Theorem 54 One has √ 1/6 √ 1/6 (i) G9/7 = 2−11/12 + 21 (ii) G63 = 2−11/12 + 21 √ √ 1/2 √ √ 1/2 + 21 − + 21 + 21 + + 21 , (161) The value G63 can also be found in [5, page 192] Proof We employ Corollary 36(ii) and Corollary 44(ii) and proceed as in Theorem 47 References [1] S Ramanujan, Notebooks, vol 1-2, Tata Institute of Fundamental Research, Bombay, India, 1957 [2] B C Berndt, Ramanujan’s Notebooks Part III, Springer, New York, NY, USA, 1991 [3] S Ramanujan, “Modular equations and approximations to π,” Quarterly Journal of Mathematics, vol 45, pp 350–372, 1914 [4] H Weber, Lehrburg Der Algebra II,, Chelsea, New York, NY, USA, 1961 [5] B C Berndt, Ramanujan’s Notebooks Part V, Springer, New York, NY, USA, 1998 [6] N D Baruah, “On some class 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Zhang, “Ramanujan’s remarkable product of theta-functions,” Proceedings of the Edinburgh Mathematical Society, vol 40, no 3, pp 583–612, 1997 [12] N Saikia, “Ramanujan’s modular equations and Weber-Ramanujan’s class invariants Gn and gn,” Bulletin of Mathematical Sciences, vol 2, no 1, pp 205–223, 2012 [13] J Yi, Construction and application of modular equation [Ph.D thesis], University of Illionis, 2001 [14] J Yi, “Theta-function identities and the explicity formulas for theta-function and their applications,” Journal of Mathematical Analysis and Applications, vol 292, no 2, pp 381–400, 2004 [15] K R Vasuki and T G Sreeramamurthy, “Certain new Ramanujans Schlăafli-type mixed modular equations, Journal of Mathematical Analysis and Applications, vol 309, no 1, pp 238–255, 2005 [16] N D Baruah, On some of Ramanujans Schlăafli-type mixed modular equations,” Journal of Number Theory, vol 100, no 2, pp 270–294, 2003 ISRN Computational Mathematics Copyright of ISRN Computational Mathematics 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