Research Article A Parameter for Ramanujan s Function χ(q): Its Explicit Values and Applications

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1 International Scholarly Research Network ISRN Computational Mathematics Volume 01 Article ID pages doi:1050/01/ Research Article A Parameter for Ramanujan s Function χq: Its Explicit Values and Applications Nipen Saikia Department of Mathematics Rajiv Gandhi University Rono Hills Doimukh India Correspondence should be addressed to Nipen Saikia nipennak@yahoocom Received 3 May 01; Accepted 8 June 01 Academic Editors: L Hajdu L S Heath and H J Ruskin Copyright 01 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 I kn involving quotient of Ramanujan s function χq for positive real numbers k and n and study its several properties We prove some general theorems for the explicit evaluations of the parameter I kn and find many explicit values Some values of I kn are then used to find some new and known values of Ramanujan s class invariant G n 1 Introduction In Chapter 16 of his second notebook [1] Ramanujan develops the theory of theta-function Ramanujan s general theta-function is defined by f a b = n= After Ramanujan for q < 1 we define a nn1/ b nn 1/ ab < 1 1 q := e πn and n is a positive rational number Since from [ page 1 Entry 1v & vi] χ q { 1/ α1 α = 1/6 q 5 it follows from that χ q = 1/6 1 α 1/1 α q 1/ f q := f q q = 1 n q n3n 1/ = q; q n= G n ={α1 α 1/ g n = 1/1 1 α 1/1 α 1/ 6 a; q := n=0 1 aq n If q = e πiz with Imz > 0 then f q = q 1/ ηz ηz denotes the classical Dedekind eta function Ramanujan s function χqisdefinedby χ q := f q f q = q; q 3 The function χq is intimately connected to Ramanujan s class invariants G n and g n which are defined by G n = 1/ q 1/ χ q g n = 1/ q 1/ χ q Also if β has degree r over α then G r n = { β 1 β 1/ g r n = 1/1 1 β 1/1 β 1/ 7 In his notebooks [1] and paper [3]Ramanujanrecordeda total of 116 class invariants or monic polynomials satisfied by them The table at the end of Weber s book [ page 71 76] 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

2 ISRN Computational Mathematics of Ramanujan s class invariants and applications can be found in Berndt s book [5] For further references see [6 1] Ramanujan and Weber independently and many others in the literature calculated class invariants G n for odd values of n and g n for even values of n For the first time i [13] calculated some values of g n for odd values of n by finding explicit values of the parameter r kn see [13 page 11 11] or [1 page 111] defined by r kn := f q k 1/ q k 1/ f q k q = n/k e π 8 In particular she established the result [13 page 18 Theorem 3] g n = r n/ 9 However the values of G n for even values of n have not been calculated The main objective of this paper is to evaluate some new values of G n for even values of n We also prove some known values of G n For evaluation of class invariant G n in this paper we introduce the parameter I kn whichis defined as I kn := χ q q k1/ χ q k q = n/k e π 10 k and n are positive real numbers In Section 3 westudysomepropertiesofi kn and also establish its relations with Ramanujan s class invariant G n In Section by employing Ramanujan s modular equations we present some general theorems for the explicit evaluations of I kn and find several explicit values of I kn InSection 5 we establish some general theorems connecting the parameter I kn and the class invariant G n We also evaluate some explicit values of the product G nk G n/k by employing some values of I kn evaluated in Section Finally in Section 6 wecalculate new and known values of class invariant G n by combining the explicit values of I kn and the product G nk G n/k evaluated in Sections and 5 respectivelysection is devoted to record some preliminary results Since Ramanujan s modular equations are key in our evaluations of I kn and G n we complete this introduction by defining Ramanujan s modular equation from Berndt s book [] The complete elliptic integral of the first kind Kk is defined by π/ Kk : = 0 dφ 1 k sin φ = π n=0 = π F ;1;k 1/ n n! kn 11 0 < k < 1 F 1 denotes the ordinary or Gaussian hypergeometric function and a n = aa 1a a n 1 1 The number k is called the modulus of Kandk := 1 k is called the complementary modulus Let K K L and L denote the complete elliptic integrals of the first kind associated with the moduli k k l and l respectively Suppose that the equality n K K = L 13 L holds for some positive integer n Then a modular equation of degree n is a relation between the moduli k and lwhichis implied by 13 If we set q = exp π K q = exp K π L L 1 we see that 13 is equivalent to the relation q n = q Thus a modular equation can be viewed as an identity involving theta-functions at the arguments q and q n Ramanujan recorded his modular equations in terms of α and β α = k and β = l 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 [ page 35] Let K K L 1 L 1 L L L 3 andl 3 denote complete elliptic integrals of the first kind corresponding in pairs to the moduli α β γand δ and their complementary moduli respectively Let n 1 n andn 3 be positive integers such that n 3 = n 1 n Suppose that the equalities n 1 K K = L 1 L 1 n K K = L L n 3 K K = L 3 L 3 16 hold Then a mixed modular equation is a relation between the moduli α β γand δ that is induced by 16 In such an instance we say that β γandδ are of degrees n 1 n andn 3 respectivelyoverα or α β γ andδ have degrees 1 n 1 n andn 3 respectively Denoting z r = φ q r q = exp πk φ q = f q q K q < 1 17 the multipliers m and m associated with α β andγ δ respectively are defined by m = z 1 /z n1 and m = z n /z n3 Preliminary Results Lemma 1 see [ page 3 Entry 7v] If α and β are such that the modulus of each exponential argument is less than 1 and αβ = π then e α/ χe α = e β/ χ e β 18

3 ISRN Computational Mathematics 3 Lemma see [15 page 1 Lemma 3] Let := q 1/1 χq/χq 3 and := q 1/6 χq /χq 6 ; then 1 1 [ 6 6 ] [ { { ] = { { Lemma 3 see [15 page 1 Lemma 8] Let := q 1/6 χq/χq 5 and := q 1/3 χq /χq 10 ; then [ 3 3 ] [ ] 6 6 = Lemma see [15 page 5 Lemma 13] Let := q 1/ χq/χq 7 and := q 1/ χq /χq 1 ; then [ 8 [ [ [ [ ] [ ] 8 8 ] [7 { 19 ] 6 6 ] [ { { ] ] [16 { { 35 ] ] [96 { 56 { { 10 10] 176 { = [{ 8 8] 96 [{ ] 1 Lemma 5 see [ page 31 Entry 5xii] Let P = {16αβ1 α1 β 1/8 and Q = β1 β/α1 α 1/ ; then β has degree 3 over α Q Q 1 P P 1 = 0 Lemma 6 see [ page 8 Entry 13xiv] Let P = {16αβ1 α1 β 1/1 and Q = β1 β/α1 α 1/8 ; then β has degree 5 over α Q Q 1 P P 1 = 0 3 Lemma 7 see [ page 315 Entry 13xiv] Let P = {16αβ1 α1 β 1/8 and Q = β1 β/α1 α 1/6 ; then Q Q 1 7= P P 1 = 0 β has degree 7 over α Lemma 8 see [5 page 378 Entry 1] Let P = 1/6 {αβ1 α1 β 1/ and Q = β1 β/α1 α 1/ ; then Q 7 Q 7 13 Q 5 Q 5 5 Q 3 Q 3 78 Q Q 1 8 P 6 P 6 = 0 β has degree 13 over α For Lemmas 9 to 15weset P := 56αβγδ1 α 1 β 1 γ 1 δ 1/8 Q := R := T := 1/8 αδ1 α1 δ βγ 1 β 1 γ 1/8 γδ 1 γ 1 δ αβ1 α 1 β 1/8 βδ 1 β 1 δ αγ1 α 1 γ 5 6 Lemma 9 see [5 page 381 Entry 50] If α β γandδ have degrees and 35 respectively then R R Q 6 Q 6 5 Q Q 10 Q Q 15 = 0 7 Lemma 10 see [5 page 381 Entry 51] If α β γandδ have degrees and 39 respectively then Q Q 3 Q Q T T 3= 0 8

4 ISRN Computational Mathematics Lemma 11 see [5 page 381 Entry 5] If α β γandδ have degrees and 65 respectively then Q 6 Q 6 5 Q Q 1 T T 1 T T = 0 9 Lemma 1 see [16 page 77 Lemma 31] If α β γ andδ have degrees and 1 respectively then R R Q Q 3= 0 30 Lemma 13 see [15 page 3 Theorem 5] If α β γ and δ have degrees 1 3 and 6 respectively then R R P P = 0 31 Lemma 1 see [15 page 8 Theorem 10] If α β γ and δ have degrees 1 5 and 10 respectively then R 6 R 6 5 R R = P P 3 Lemma 15 see [15 page 5 Theorem 1] If α β γ and δ have degrees 1 7 and 1 respectively then T 3 T 3 = P 1 P P 3 P Properties of I kn In this section we study some properties of I kn We also establish a relation connecting I kn and Ramanujan s class invariants G n Theorem 16 For all positive real numbers k and n one has i I k1 = 1 ii I 1 = 1 iii I kn = I nk k I kn 3 n Proof Using the definition of I kn and Lemma 1 weeasily arrive at i Replacing n by 1/n in I kn and using Lemma 1 we find that I kn I k1/n = 1 which completes the proof of ii To prove iii we use Lemma 1 in the definition of I kn to arrive at I kn /I nk =1 Remark 17 By using the definitions of χq andi kn itcan be seen that I kn has positive real value less than 1 and that the values of I kn decrease as n increases when k>1 Thus by Theorem 16i I kn < 1foralln>1ifk>1 Theorem 18 For all positive real numbers k m and n one has I kn/m = I mkn Inkm 1 35 Proof Using the definition of I kn weobtain I mkn χ e π n/mk = I nkm e π m/nk n/mk/ χ 36 e π m/nk Using Lemma 1 in the denominator of the right-hand side of 36 and simplifying we complete the proof Corollary 19 For all positive real numbers k and n one has I k n = I nkn I kn/k 37 Proof Setting k = n in Theorem 18 and simplifying using Theorem 16ii we obtain I k m = I mkk I km/k 38 Replacing m by n we complete the proof Theorem 0 Letkabcanddbepositiverealnumberssuch that ab=cd Then I ab I kckd = I kakb I cd 39 Proof From the definition of I kn we deduce that for positive real numbers k a b candd I kakb I 1 ab = I kckd I 1 cd = χ e π ab e πk ab ab/ χ e kπ ab χ e π cd e πk cd cd/ χ e kπ cd 0 Now the result follows readily from 0 and the hypothesis that ab = cd Corollary 1 For any positive real numbers n and pwehave I npnp = I np ni pp 1 Proof The result follows immediately from Theorem 0 with a = p b = 1 c = d = pandk = n Now we give some relations connecting the parameter I kn and Ramanujan s class invariants G n Theorem Let k and n beanypositiverealnumbersthen i I kn = G n/k G 1 nk ii G 1/n = G n Proof Proof of i follows easily from the definitions of I kn and G n from 10 and respectively To prove ii we set k = 1 in part i and use Theorem 16i and iii General Theorems and Explicit Evaluations of I kn In this section we prove some general theorems for the explicit evaluations of I kn and find its explicit values

5 ISRN Computational Mathematics 5 Theorem 3 One has I3n I3n I3n I 3n I 3n I 3n 6 I3n I 3n [ I3n 10 I 3n 10 I 3n I 3n 16 6 I3n I 3n I 3n I 3n 6 71 I3n I 3n I 3n I 3n ] = I 3n I 3n 1 I 3n I 3n 1 5 I3n I 3n 8 I 3n I 3n 8 00 I3n I 3n I 3n I 3n Proof The proof follows easily from the definition of I kn and Lemma Corollary One has i I 3 = /1 3 1/ ii I 3 = iii I 31/ = /1 3 1/ iv I 31/ = Proof Setting n = 1/ in Theorem 3 and using Theorem 16ii we obtain I3 I3 176 I3 1 I = 0 5 Equivalently B 176B 100 = 0 6 B = I3 1 I3 1 7 Solving 6 and using the fact in Remark 17weobtain B = Employing 8 in7 solving the resulting equation for I 3 and noting that I 3 < 1 we arrive at I 3 = /1 3 9 This completes the proof of i Again setting n = 1 in Theorem 3 and using Theorem 16i we obtain I 6 3 I 6 3 [ I 10 3 I ] I3 6 I I 3 I3 50 = 5 I3 8 I I 3 I3 550 Equivalently D 1 D D 50 = 0 51 D = I 3 I 3 5 Since the first factor of 51 is nonzero solving the second factor we deduce that D = 3 6 1/ 53 Employing 53 in5 solving the resulting equation and using the fact that I 3 < 1 we obtain 1/ I 3 = 5 This completes the proof of ii Now iii and iv follow from i and ii respectively and Theorem 16ii Theorem 5 One has I5n I5n I5n I 5n I 5n I 5n 3 I5n I 5n [ I5n 5 I 5n 5 I 5n I 5n 8 3 I5n I 5n 3 I 5n I 5n 19 I 5n I 5n I5n I 5n ] = I 5n I 5n 6 I 5n I 5n 6 13 I5n I 5n I 5n I 5n 5 I5n I 5n I 5n I 5n 8 55 Proof The proof follows from Lemma 3 and the definition of I kn Corollary 6 One has i I 5 = /6 10 ii I 5 = 11 1/ iii I 51/ = /6 10 iv I 51/ = 11 1/ Proof Setting n = 1/ in Theorem 5 and using Theorem 16ii we obtain C 56C 16 = 0 57

6 6 ISRN Computational Mathematics C = I 6 5 I Solving 57 and noting the fact in Remark 17we obtain C = Employing 59 in58 solving the resulting equation and noting that I 5 < 1 we obtain I 5 = / This completes the proof of i Again setting n = 1 in Theorem 5 and using Theorem 16i we obtain Solving 61 we obtain B 8 B = 0 61 B = I 5 I B = / 63 Using 63 in6 solving the resulting equation and noting that I 5 < 1 we arrive at [ I7n 10 I 7n 10 I 7n I 7n 3 { I7n I 7n 6 I 7n I 7n 6 81 { I7n I 7n I 7n I 7n ] I7n I7n I 7n I 7n [ { I7n 8 16 I 7n 8 I 7n I 7n 88 { I7n I 7n I 7n I 7n 35 ] I7n I7n I 7n I 7n [ { I7n 96 I 7n I 7n I 7n 56 { I7n I 7n 6 I 7n I 7n 6 8 { I7n I 7n 10 I 7n I 7n 10 ] 176 { I7n 1 = I 7n 1 I 7n I 7n 15 { I7n I 7n 8 I 7n I 7n 8 96 { I7n I 7n I 7n I 7n Corollary 8 One has 65 I 5 = 11 1/ This completes the proof of ii Now iii and iv follow from i and ii respectively and Theorem 16ii Theorem 7 One has 1 1 I7n I7n I 7n I 7n I7n I7n I 7n I 7n I7n I7n I 7n I 7n [ { I7n 7 I 7n I 7n I 7n 19 ] 6 6 I7n I7n I 7n I 7n [ I7n I 7n I 7n I 7n ] i I 7 = 1/ ii I 7 = iii I 71/ = 1/ iv I 71/ = / 1/ 66

7 ISRN Computational Mathematics 7 Proof Setting n = 1/ and simplifying using Theorem 16 ii we obtain I 7 I7 3 I 0 7 I7 0 0 I7 16 I I 1 7I I 8 7I I 7 I 7 6 = 0 67 Equivalently A A 3A 3 A 18A 191 = 0 68 A = I 7 I 7 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 A = Combining 69and70 and noting that I 7 < 1 we obtain I 7 = 1/ / Theorem 9 One has I7n I 75n 3 3 I75n I75n I 7n I 75n I 7n I 7n I75n I75n 5 I 7n I 7n 10 { I75n I 7n I75n I 7n 15= 0 Proof Using 5inLemma 9 we find that 76 Q = qχ q 5 χ q 7 χ q χ q 35 R = q3/ χ q χ q 5 χ q 7 χ q Setting q = e πn/7 and using the definition of I kn in 77 we get 1/ I75n Q = R = 1/ I 7nI 75n 78 I 7n Employing 78in7 we complete the proof Corollary 30 One has i I 75 = h 36 h 6 This completes the proof of i 71 To prove ii setting n = 1 and simplifying using Theorem 16i we arrive at E E E 108E 113 = 0 7 ii I 71/5 = iii I 75 = iv I 71/5 = h 36 h 6 d 1 d 1 d 1 d 1 79 E = I 7 I 7 73 Using the fact in Remark 17 it is seen that the first two factors of 7 are nonzero and so solving the third factor we obtain E = Combining 73 and7 and noting that I 7 < 1 we deduce that I 7 = 75 So the proof of ii is complete Now iii and iv follow from i and ii respectively and Theorem 16ii h = / /3 and d = 1 / /3 / /3 Proof Setting n = 1/5 intheorem9 and simplifying using Theorem 16ii we obtain I75 6 I I75 I75 10 I 75 I75 17 = 0 80 Equivalently H 3 5H 7H 7 = 0 81 H = I 75 I 75 8 Solving 81 and noting the fact in Remark 17weobtain 1/3 1/ H = 3 83

8 8 ISRN Computational Mathematics Combining 8and83 and noting that I 75 < 1 we deduce that I 75 = h 36 h 6 8 h = / /3 85 This completes the proof of i Again setting n = 1 and simplifying using Theorem 16i we arrive at U 3 6U 7U 3 = 0 86 Solving 86 and using Remark 17weget U = U = I 75 I / 1/ / 1/ Combining 87and88 and noting that I 75 < 1 we obtain I 75 = d 1 d 1 89 d = 1 / /3 / /3 90 This completes the proof of ii Now ii and iv easily follow from i and ii respectively and Theorem 16ii Theorem 31 One has I13n I13n I13n 3 I 139n I 139n I 139n 1 I13n I 139n 91 { I13n 1 I 139n I13n I 139n 3= 0 Proof Proceeding as in the proof of Theorem 9 using 5 in Lemma 10 setting q := e π n/13 and using the definition of I kn we arrive at Q = I139n I 13n 1/ T = 1/ I 13nI 139n 9 Employing 9in8 we complete the proof Corollary 3 One has i I 133 = ii I 139 = iii I 131/3 = iv I 131/9 = 93 Proof Setting n = 1/3 intheorem31 and simplifying using Theorem 16ii we obtain V 3V 1 = 0 9 V = I133 I Solving 9 and using Remark 17weget V = Combining 95and96 and noting that I 133 < 1 we obtain I 133 = 97 So we complete the proof of i Again setting n = 1 and using Theorem 16i we obtain J J 1= 0 98 J = I 139 I Solving 98 and using Remark 17weget J = Combing 99and100 and noting that I 139 < 1 we deduce that I 139 = 101 So the proofs of ii is complete Now the proof of iii and iv follow from i and ii respectively and Theorem 16ii Theorem 33 One has 3 3 I135n I135n I135n 5 I 13n I 13n I 13n { I13 1 5n I 13n I135n I 13n I135n I 13n { I135n I 13n I 135n I 13n =

9 ISRN Computational Mathematics 9 Proof Using 5 inlemma 11 setting q := e π n/13 and using the definition of I kn we arrive at Q = I135n I 13n 1/ T = 1/ I 13nI 135n 103 Employing 103 in9 we complete the proof Corollary 3 One has i I 135 = c 36 c ii I 135 = iii I 131/5 = c 36 c iv I 131/5 = c = / / Proof Setting n = 1/5 and simplifying using Theorem 16ii we arrive at L 3 3L = L = I 135 I Solving 105 and using the fact in Remark 17weobtain L = Employing 107 in106 solving the resulting equation and noting that I 135 < 1 we obtain I 135 = This completes the proof of i To prove ii setting n = 1 and simplifying using Theorem 16i we arrive at A 3 6A 3A 18 = A = I 135 I Solving 109 and using the fact in Remark 17weobtain A = / / Employing 111 and110 solving the resulting equation and noting that I 135 < 1 we obtain I 135 = c 36 c 6 11 c = / /3 This completes the proof of ii Now the proofs of iii and iv follow from i and ii respectively and Theorem 16ii Theorem 35 One has 1 I79n I7n I 79n I7n I 79n I 7n I79n I 7n 3= Proof Using 5 inlemma 1 setting q := e π n/7 and using the definition of I kn we arrive at Q = I79n I 7n 1/ R = 1/ I 7nI 79n 11 Employing 11 in30 we complete the proof Corollary 36 One has 1/ 5 1 i I 73 = ii I 79 = 1/ 5 1 iii I 71/3 = iv I 71/9 = 115 Proof Setting n = 1/3 and simplifying using Theorem 16ii we arrive at I73 I73 5 = Solving 116 and noting the factin Remark 17we obtain I 73 = 1/ This completes the proof of i To prove ii setting n = 1 and simplifying using Theorem 16i we arrive at D D 5 = D = I 79 I

10 10 ISRN Computational Mathematics Solving 118 and using the fact in Remark 17weobtain 1 1 D = 10 Employing 10 in119 solving the resulting equation and noting that I 79 < 1 we deduce that I 79 = 11 This completes the proof of ii Now the proofs of iii and iv follow from i and ii respectively and Theorem 16ii 5 General Theorems and Explicit Evaluations of G nk G n/k In this section we evaluate some explicit values of the product G nk G n/k by establishing some general theorems and employing the values of I kn obtained in Section Werecall from Theorem ii that G 1/n = G n for ready references in this section Theorem 37 One has i I 3n I 3n I 3n I 3n G n/3 G n/3 G 9n/3 G 36n/3 1 G n/3 G n/3 G 9n/3 G 36n/3 = 0 Solving 15 and noting that G 6 G 3/ > 1 we complete the proof of i To prove ii setting n = 1 in Theorem 37i; using Theorem 16i and noting that G 1/n = G n weobtain I3 I3 G 1 3G /3 G 1 G 3 G /3 G 1 = 0 16 Employing 53 in16 solving the resulting equation and noting that G 3G /3 G 1 > 1 we obtain G 3G /3 G 1 = 17 Using the value G 3 = 1/1 from [5 p 189] in 17 we complete the proof of ii To prove iii setting n = 13 in Theorem 37ii we obtain I I { G 3/13 G 39 G 3/13 G 39 = 0 18 Cubing 96 and then employing in 18 and solving the resulting equation we complete the proof Theorem 39 One has i 3 I 5n I 5n 3 I 5n I 5n 5 { 1 I5n I 5n I5n I 5n = G n/5 G n/5 G 5n/5 G 100n/5 G n/5 G n/5 G 5n/5 G 100n/5 ii I 6 3n I 6 3n { G n/3 G 3n 3 G n/3 G 3n 3 = 0 1 ii I 3 5n I 3 5n { G n/5 G 5n G n/5 G 5n = 0 19 Proof To prove i using 5 inlemma 13 setting q := e πn/3 and employing the definitions of I kn and G n we obtain R = q1/ χ q χ q χ q 3 χ q 6 = 1/ I 3n I 3n 13 P = G n/3 G n/3 G 9n/3 G 36n/3 1/ Employing 13in31 we complete the proof ii follows similarly from Lemma 5 and the definition of I kn and G n with q := e π n/3 Corollary 38 One has 1 3 i G 6 G 3/ = ii G 1 G /3 = 13/ iii G 39 G 13/3 = 1/ / Proof Setting n = 1/ in Theorem 37i and simplifying using Theorem 16ii and the result G 1/n = G n weobtain G 6 G 3/ G 6 G 3/ = 0 15 Proof Using 5 inlemma 1 setting q := e πn/5 and employing the definitions of I kn and G n weobtain R = q1/ χ q χ q χ q 5 χ q 10 = 1/ I 5n I 5n 130 P = G n/5 G n/5 G 5n/5 G 100n/5 1/ Employing 130 in 3 we complete the proof of i Similarly ii follows from Lemma 6 and the definition of I kn and G n with q := e π n/5 Corollary 0 One has 1/ 3 10 i G 10 G 5/ = ii G 0 G 5/ = / 3/M M denotes Proof Setting n = 1/ in Theorem 39i and simplifying using Theorem 16ii and the result G 1/n = G n weobtain G 10 G 5/ G 10 G 5/ 1 = 0 13

11 ISRN Computational Mathematics 11 Solving 13 and noting that G 10 G 5/ > 1 we complete the proofofi For proof of ii setting n = 1inTheorem 39ii and simplifying using Theorem 18i and the result G 1/n = G n we obtain I5 3 I5 3 { G /5 G 0 G /5 G 0 = Employing the value of I 5 I5 1 from 63in133 solving the resulting equation and noting that G /5 G 0 > 1 we get G /5 G 0 = / 3/ M 13 So the proof is complete Theorem 1 One has { In I 9n 3/ I n I 9n 3/ = G n/ G n/ G 7n/ G 196n/ 1/ G n/ G n/ G 7n/ G 196n/ 1/ G n/ G n/ G 7n/ G 196n/ 3/ G n/ G n/ G 7n/ G 196n/ 5/ 135 Proof Using 5 inlemma 15 setting q := e πn/ and employing the definitions of I kn and G n we obtain R = q1/3 χ q χ q 7 χ q χ q 1 = 1/ I n I 9n 136 P = G n/ G n/ G 7n/ G 196n/ 1/ Employing 136 in33 we complete the proof Corollary One has 8 G 1 G 7/ = 137 Proof Setting n = 1/7 and simplifying using Theorem 16ii and the result G 1/n = G n weget G 1 G 7/ 6 G 1 G 7/ 5 G 1 G 7/ 138 G 1 G 7/ 1 = 0 Solving 138 and noting that G 1 G 7/ > 1 we complete the proof Theorem 3 One has I7n I7n 7= { G 7/n G 7n 3 G 7/n G 7n Proof Using 5 inlemma 7 setting q := e πn/7 and employing definitions of I kn and G n we arrive at P = G n/7 G 7n 3 q 1/ χ q 10 Q = χ q 7 = I7n Using 10 in we complete the proof Corollary One has i G 7/ G 8 = ii G 7/3 G 1 = 1/ /3 iii G 7/9 G 63 = 1/ /3 Proof Setting n = intheorem 3weget 11 I 7 I 7 7= { G 7/ G 8 3 G 7/ G Squaring 7 and simplifying we obtain I 7 I 7 = Employing 13 in1 solving the resulting equation and noting that G 7/ G 8 > 1 we complete the proof of i Toproveiiwesetn = 3inTheorem 3 and employing 116 we obatin G 1 G 7/3 3 G 1 G 7/3 3 3 = 0 1 Solving 1 and noting that G 1 G 7/3 > 1 we complete the proof To prove iii setting n = 9inTheorem 3weget I 79 I 79 7= { G7/9 G 63 3 G 7/9 G Squaring 10 twice and simplifying we obtain I79 I79 = 16 Employing 16 in15 solving the resulting equation and noting that G 7/9 G 63 > 1 we complete the proof Theorem 5 One has I13n 7 I 13n 7 13 { I13n 5 I 13n 5 5 { I13n 3 I 13n 3 78 { I13n I13n 1 8 { G n/13 G 13n 6 G n G 13n 6 = 0 17 Proof Using 5 inlemma 8 setting q := e π n/13 and employing definitions of I kn and G n we arrive at P = G n/713 G 13n 1 Q = q 1/ χ q = I 13n 18 χ q 13 Using 18 in5 we complete the proof Corollary 6 One has G 117 G 13/9 = /6 3 19

12 1 ISRN Computational Mathematics Proof Setting n = 9inTheorem 5weget 7 I139 7 I { 5 I139 5 I { 3 I139 { 3 I139 1 I I { G 9/13 G G 9/13 G = 0 Employing 100in150 and simplifying we obtain G 9/13 G G 9/13 G = Theorem 8 One has i G /3 = 3 / / / 6 ii G 1 = 63 / / / Solving the resulting equation 151 and noting that G 9/13 G 117 > 1 we complete the proof 6 New Values of Class Invariant G n In this section we find some new values of Ramanujan s class invariant G n by using explicit values of I kn and G nk G n/k evaluated in Sections 3 and respectively For ready references in this section we recall from Theorem that I kn = G n/k /G nk We also recall from Theorem ii that G 1/n = G n Theorem 7 One has i G 3/ = 1/ 1 3 1/ / 3 ii G 6 = 1/ 1 3 1/ / 3 Proof From Corollary i we have 15 G3/ I 3 = = /1 3 G Also from Corollary 38i we have G 6 G 3/ = 1/ Multiplying 153 and15 and simplifying we complete the proof of i Dividing 15 by153 and simplifying we complete the proof of ii The proofs of the Theorems 8 56 are identical to the proof of Theorem 7 So we give the references of the required results only Proof We use Corollary ii and Corollary 38ii and proceed as in Theorem 7 Theorem 9 One has i G 5/ = 1/ / /1 10 ii G 10 = 1/ / / Proof We employ Corollary 6i and Corollary 0i and proceed as in Theorem 7 Theorem 50 One has i G 5/ = / / 3/ / ii G 0 = / / 3/ / 1/ / 157 Proof We employ Corollary 6ii and Corollary 0ii and proceed as in Theorem 7

13 ISRN Computational Mathematics 13 Theorem 51 One has i G 7/ = 9/8 8 1/ ii G 1 = 9/8 8 1/ /8 1/8 158 Proof We employ Corollary 8i and Corollary i and proceed as in Theorem 7 Theorem 5 One has i G 7/ = 3/ 1/ / ii G 8 = 3/ 1/ / 159 Proof We employ Corollary 8ii and Corollary i and proceed as in Theorem 7 Theorem 53 One has i G 7/3 = 5/ 3 7 1/ /8 ii G 1 = 5/ 3 7 1/ /8 160 The values G 7/3 and G 1 can also be found in [5 6] Proof We employ Corollary 36i and Corollary ii and proceed as in Theorem 7 Theorem 5 One has i G 9/7 = 11/ / / 1 ii G 63 = 11/ / / The value G 63 can also be found in [5 page 19] Proof We employ Corollary 36ii and Corollary ii and proceed as in Theorem 7 Theorem 55 One has i G 13/3 = 11/ / / 13 ii G 39 = 11/ / / The values G 13/3 and G 39 can also be found in [5 6] Proof We use Corollary 3i and Corollary 38iii and proceed as in Theorem 7 Theorem 56 One has i G 13/9 = 50 1/ / / 3 ii G 117 = 50 1/ / / 3 The values G 39 and G 117 can also be found in [5 page 193] 163 Proof We employ Corollary 3ii and Corollary 6 and proceed as in Theorem 7 References [1] S Ramanujan Notebooks vol 1- Tata Institute of Fundamental Research Bombay India 1957 [] B C Berndt Ramanujan s Notebooks Part III Springer New ork N USA 1991 [3] S Ramanujan Modular equations and approximations to π Quarterly Mathematics vol 5 pp [] H Weber Lehrburg Der Algebra II Chelsea New ork N USA 1961 [5] B C Berndt Ramanujan s Notebooks Part V Springer New ork N USA 1998 [6] N D Baruah On some class invariants of Ramanujan The the Indian Mathematical Society vol 68 no 1 pp [7] B C Berndt and H H Chan Some values for the Rogers- Ramanujan continued fraction Canadian Mathematics vol 7 no 5 pp [8] B C Berndt H H Chan S Kang and L C Zhang A certain quotient of eta-functions found in Ramanujan s lost notebook Pacific Mathematics vol 0 no pp [9] B C Berndt H H Chan and L-C Zhang Ramanujan s class invariants with applications to the values of q-continued fractions and theta functions in Special Functions q-series and Related Topics MIsmailDMassonandMRahman Eds vol 1 of Fields Institute Communications Series pp American Mathematical Society Providence RI USA 1997

14 1 ISRN Computational Mathematics [10] B C Berndt H H Chan and L C Zhang Ramanujan s class invariants and cubic continued fraction Acta Arithmetica vol 73 no 1 pp [11]BCBerndtHHChanandLCZhang Ramanujan s remarkable product of theta-functions Proceedings of the Edinburgh Mathematical Society vol 0 no 3 pp [1] N Saikia Ramanujan s modular equations and Weber-Ramanujan s class invariants Gn and gn Bulletin of Mathematical Sciences vol no 1 pp [13] J i Construction and application of modular equation [PhD thesis] University of Illionis 001 [1] J i Theta-function identities and the explicity formulas for theta-function and their applications Mathematical Analysis and Applications vol 9 no pp [15] K R Vasuki and T G Sreeramamurthy Certain new Ramanujan s Schläfli-type mixed modular equations Mathematical Analysis and Applications vol 309 no 1 pp [16] N D Baruah On some of Ramanujan s Schläfli-type mixed modular equations Number Theory vol 100 no pp

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