In this paper, we establish some new modular relations connecting Ramanujan's cubic continued fraction V (q) with

V(qn), for n= 4, 6, 8,10, 12, 14, 16 and 22

Chandankumar, S.

Mahadeva Naika, M.S.

Sushan Bairy, K.

ePrints@Bangalore University

Int. J. Contemp. Math. Sciences, Vol. 7, 2012, no. 20, 953 - 962On Some New Identities for Ramanujan’sCubic Continued FractionM. S. Mahadeva NaikaDepartment of MathematicsBangalore UniversityCentral College CampusBangalore-560 001, Indiamsmnaika@rediffmail.comS. ChandankumarDepartment of MathematicsBangalore UniversityCentral College CampusBangalore-560 001, Indiachandan.s17@gmail.comK. Sushan BairyPG Department of MathematicsVijaya CollegeR. V. Road, BasavanagudiBangalore-560 004, Indiaksbairy@rediffmail.comAbstractIn this paper, we establish some new modular relations connectingRamanujan’s cubic continued fraction V (q) with V (qn), for n = 4, 6, 8,10, 12, 14, 16 and 22.Mathematics Subject Classification: 33D10, 11A55, 11F27Keywords: Cubic continued fraction, Modular equations, Theta-function954 M. S. Mahadeva Naika, S. Chandankumar and K. Sushan Bairy1 IntroductionOn page 366 of his ‘lost’ notebook [11], S. Ramanujan has recorded cubiccontinued fractionV (q) :=q1/31 +q + q21 +q2 + q41 +q3 + q61 +. . . , |q| < 1 (1)and other identities related to V (q). H. H. Chan [4] has established theseidentities. Subsequently, many mathematicians contributed to the theory ofRamanujan’s cubic continued fraction. Some of them are N. D. Baruah [2], C.Adiga, T. Kim, M. S. Mahadeva Naika and H. S. Madhusudhan [1], MahadevaNaika [6], Mahadeva Naika, M. C. Maheshkumar and K. Sushan Bairy [9], B.Cho, J. K. Koo and Y. K. Park [5], Mahadeva Naika, S. Chandankumar andBairy [7], [8].In this paper, we establish several new modular identities connecting V (q)with V (qn), for n = 4, 6, 8, 10, 12, 14, 16 and 22.2 Preliminary resultsIn Chapter 16, of his second notebook [10], [3, pp.257-262], Ramanujan devel-ops the theory of theta-function and his theta-function is defined byf(a, b) :=∞∑n=−∞an(n+1)/2bn(n−1)/2, |ab| < 1,= (−a; ab)∞(−b; ab)∞(ab; ab)∞,where (a; q)∞ :=∞∏n=1(1− aqn−1), |q| < 1.Following Ramanujan, we defineϕ(q) := f(q, q) =∞∑n=−∞qn2= (−q; q2)2∞(q2; q2)∞, (2)ψ(q) := f(q, q3) =∞∑n=−∞qn(n+1)/2 =(q2; q2)∞(q; q2)∞. (3)Now, we define modular equation in brief. The complete elliptic integral ofthe first kind K(k) is defined byK(k) :=∫ pi20dφ√1− k2 sin2 φ=pi2∞∑n=0(12)2n(n!)2k2n =pi22F1(12,12; 1; k2), (4)On some new identities for Ramanujan’s cubic continued fraction 955where 0 < k < 1 and 2F1 is the ordinary or Gaussian hypergeometric functiondefined by2F1(a, b; c; z) :=∞∑n=0(a)n (b)n(c)n n!zn, 0 ≤ |z| < 1,where (a)0 = 1, (a)n = a(a + 1) · · · (a + n − 1) for n a positive integer anda, b and c are complex numbers such that c 6= 0,−1,−2, . . .. The number kis called the modulus of K, and k′ :=√1− k2 is called the complementarymodulus. Let K, K ′, L and L′ denote the complete elliptic integrals of thefirst kind associated with the moduli k, k′, l and l′, respectively. Suppose thatthe equalitynK ′K=L′L(5)holds for some positive integer n. Then a modular equation of degree n isa relation between the moduli k and l which is induced by (5). FollowingRamanujan, set α = k2 and β = l2. Then we say β is of degree n over α. Themultiplier m is defined bym =KL. (6)Let K, K ′, L1, L′1, L2, L′2, L3 and L′3 denote complete elliptic integrals ofthe first kind corresponding, in pairs, to the moduli√α,√β,√γ and√δ,and their complementary moduli, respectively. Let n1, n2 and n3 be positiveintegers such that n3 = n1n2. Suppose that the equalitiesn1K ′K=L′1L1, n2K ′K=L′2L2and n3K ′K=L′3L3(7)hold. Then a “mixed” modular equation is a relation between the moduli√α,√β,√γ and√δ that is induced by (7). We say that β, γ and δ are of degreesn1, n2 and n3, respectively over α. The multipliers m and m′ are associatedwith α, β and γ, δ respectively.We end this section by listing some relevant identities that are useful inproving our main results.Lemma 2.1. [3, Ch. 20, Entry 3 (xii), pp. 352–353] Let α, β and γ beof the first, third and ninth degrees respectively. Let m denote the multiplierconnecting α, β and m′ be the multiplier relating γ, δ, then(β2αγ)1/4+((1− β)2(1− α)(1− γ))1/4−(β2(1− β)2αγ(1− α)(1− γ))1/4= −3mm′, (8)(αγβ2)1/4+((1− α)(1− γ)(1− β)2)1/4−(αγ(1− α)(1− γ)β2(1− β)2)1/4=m′m. (9)956 M. S. Mahadeva Naika, S. Chandankumar and K. Sushan BairyLemma 2.2. [3, Ch. 20, Entry 9 (vii), p. 377] We have{ψ(q3)ψ(q5)− qψ(q)ψ(q15)}ϕ(−q3)ϕ(−q5)={ψ(q3)ψ(q5) + qψ(q)ψ(q15)}ϕ(−q)ϕ(−q15). (10)Lemma 2.3. [3, Ch. 20, Entry 13 (i) and (ii), p. 401] Let α, β, γ and δ beof the first, third, seventh and twenty first degrees respectively. Let m denotethe multiplier connecting α and β and m′ be the multiplier relating γ and δ.Then (βγαδ)1/4+((1− β)(1− γ)(1− α)(1− δ))1/4−(βγ(1− β)(1− γ)αδ(1− α)(1− δ))1/4+ 4(βγ(1− β)(1− γ)αδ(1− α)(1− δ))1/6=mm′,(11)(αδβγ)1/4+((1− α)(1− δ)(1− β)(1− γ))1/4−(αδ(1− α)(1− δ)βγ(1− β)(1− γ))1/4+ 4(αδ(1− α)(1− δ)βγ(1− β)(1− γ))1/6=m′m.(12)Lemma 2.4. [3, Ch. 20, Entry 14 (i) and (ii), p. 408] Let α, β, γ and δ beof the first, third, eleventh and thirty third degrees respectively. Let m denotethe multiplier connecting α and β and m′ be the multiplier relating γ and δ.Then (βδαγ)1/8+((1− β)(1− δ)(1− α)(1− γ))1/8−(βδ(1− β)(1− δ)αγ(1− α)(1− γ))1/8− 2(βδ(1− β)(1− δ)αγ(1− α)(1− γ))1/12=√mm′,(13)(αγβδ)1/8+((1− α)(1− γ)(1− β)(1− δ))1/8−(αγ(1− α)(1− γ)βδ(1− β)(1− δ))1/8− 4(αγ(1− α)(1− γ)βδ(1− β)(1− δ))1/12=3√mm′.(14)Lemma 2.5. [3, Ch. 17, Entry 10 (i), Entry 11 (ii), pp. 122–123] For0 < α < 1,ϕ(q) =√z, (15)√2q1/8ψ(−q) = √z{α(1− α)}1/8, (16)where z := 2F1(12,12; 1;α).On some new identities for Ramanujan’s cubic continued fraction 957Lemma 2.6. [4] If u = V (q) and v = V (q2), thenu2 + 2v2u− v = 0. (17)Lemma 2.7. [6] We haveϕ2(q)− ϕ2(q3) = 4qχ2(q)ψ(q6)f(−q,−q5), (18)ϕ2(q) + ϕ2(q3) = 2χ2(q)ϕ(−q3)f(q2, q4). (19)Lemma 2.8. [1, Theorem 5.1] If P =ψ(−q)q1/4ψ(−q3) and Q =ϕ(q)ϕ(q3), thenQ4 + P 4Q4 = 9 + P 4. (20)3 Modular identities for Ramanujan’s cubiccontinued fractionIn this section, we establish several new modular relations connecting V (q)with V (qn), for n = 4, 6, 8, 10, 12, 14, 16 and 22.Theorem 3.1. If a := V (q) and b := V (q6), then(16b4 + 1 + 12b2 + 4b+ 16b3)a6 + 2b4 − 3b3 − b− b2 − 4b5+ (6b+ 14b2 + 12b4 + 2b3 + 8b6 + 8b5)a3 = 0.(21)Proof. Using the equations (8), (9), (15) and (16), we deduce thatd2b21 + a21d2 + 3c2a21 − b21c2 = 0, (22)wherec =ϕ(q)ϕ(q3), d =ϕ(q3)ϕ(q9), a1 =ψ(−q)q1/4ψ(−q3) , b1 =ψ(−q3)q3/4ψ(−q9) .Using the equations (18) and (19), we deduce thatϕ2(q)ϕ2(q3)=1 + 2V (q)V (q2)1− 2V (q)V (q2) . (23)Using the equation (17), we deduce thatV (q2) =3 + r1− r , (24)V (q) =1 + 2V (q2){t− V 2(q2)}1− 2V (q2){t− V 2(q2)} , (25)958 M. S. Mahadeva Naika, S. Chandankumar and K. Sushan Bairywhere r := ±√1− 8V 3(q) and t := ±√V 4(q2) + V (q2).Collecting the terms containing a21 on one side of the equation (22), squaringand then employing the equations (20), (23), (24) and (25), we find that1− 8bs− 24b4s− 96b7s− 32b3r − 56b6r + 96b9r − 128b10s+ 128b12r − 6a3 + 56b4sr − 32b7sr − 128b10sr + 128b12 + r+ 56b6 + 160b9 + 128b7sra3 + 176b4sra3 + 24bsra3 + 512b9a3+ 464b6a3 + 1024b9a6 + 1024b6a6 + 144b3a6 − 240b6ra3− 128b9ra3 − 96b3ra3 − 208b4sa3 − 512b7sa3 − 1024b7sa6+ 16b3 − 2ra3 − 16bsa6 + 8bsr + 40bsa3 − 512b4sa6 = 0,(26)where s := ±√V 4(q6) + V (q6).Eliminating r and s from the equation (26), we obtain(2b4 − b2 − b− 4b5 + 12b4a3 + 6ba3 + 16b4a6 + 4ba6 + 2b3a3 + 8b6a3+ 16b3a6 + a6 + 8b5a3 + 12b2a6 + 14b2a3 − 3b3)(b2 − 2b4 + 7b6 − 7b5+ 14b7 − 8b8 + 8b9 − 8b4a3 + 186b4a6 + ba6 + 20b3a3 − 16b9a3 + 4b6a3− 88b7a3 + 72b7a6 + a12 + 44b5a3 + 88b8a3 + 29b2a6 − 12b2a3 + 144b8a6+ 156b5a6 + 32b11a3 − b3 − 64b11a6 + 32b5a9 + 160b8a9 + 4b2a12 + 34b2a9+ 256b8a12 − 64b5a12 + 64b12a6 + 128b9a9 − 320b6a9 + 64b6a12 − 16b3a12− 124b3a9 − 80b10a3 − 32a6b10 + 220a9b4 − 6a9b− 64a9b7 − 128a9b10+ 64a12b4 − 4a12b− 256a12b7 + 16b10 − 300b6a6 − 74b3a6)(1 + 2ba)2(1− 2ba+ 4b2a2)2 = 0.(27)As q → 0, the first factor vanishes faster than the second factor, whereas otherfactors does not vanish. This completes the proof.Theorem 3.2. If a := V (q) and b := V (q10), then64b12a6 + 32(5a4 − a)b11 + 16(5a2 + 20a8 − 10a5 + 32a11)b10+ 64(20a9 + 10a6 − 5a3)b9 + 20(34a4 − 4a7 − a− 48a10)b8+ 4(5a2 + 160a11 − 238a5 − 20a8)b7 + 20(67a6 − 4a3 + 64a9)b6+ (10a4 − 896a10 − 760a7)b5 + (80a8 + 5a2 + 120a11 + 10a5)b4+ 20(a3 − 8a6 − 11a9)b3 + (112a7 + 130a10 − 15a4)b2+ (10a5 − a2 − 20a11 − 15a8)b+ a12 + b6 = 0.(28)Proof of the identity (28) is similar to the proof of the identity (21), exceptthat in place of results (8) and (9), result (10) is used.On some new identities for Ramanujan’s cubic continued fraction 959Theorem 3.3. If a := V (q) and b := V (q14), then256b16a8 + 128(7a4 − a)b15 + 256(32a15 + 28a12 + 7a6)b14 + 1792(12a8 − 3a5+ 16a11)b13 + 112(464a10 + 512a13 − a+ 17a4 − 8a7)b12 + 448(32a15 − 3a3− 100a12 − 112a9 + 20a6)b11 + 112(512a8 − 40a5 − 256a14 + a2 + 272a11)b10+ 16(4480a13 − 1240a7 + 3192a10 + 7a4)b9 + 112(56a15 − 453a12 + 128a6− 6a3 − 259a9)b8 + 8(2513a8 − 3200a14 − 336a11 − 392a5)b7 + 28(200a10− 228a7 + 432a13 + 29a4)b6 + 56(2a3 − 17a6 − 12a9 + 4a12)b5 + 7(a2 − 16a11− 453a8 + 50a5 + 464a14)b4 + 56(20a7 − 27a10 − 50a13 − 2a4)b3 + 4(203a12+ 200a9 − 28a6 − 28a15)b2 + (56a14 − a2 − 49a8 + 14a5)b+ a16 + b8 = 0.(29)Proof of the identity (29) is similar to the proof of the identity (21), exceptthat in place of results (8) and (9), results (11) and (12) are used.Theorem 3.4. If a := V (q) and b := V (q22), thenb12 + a24 + 8(11b+ 110b4 + 225280b13 + 675840b16 + 1408b10 + 720896b19+ 262144b22 − 6688b7)a23 + 4(5046272b20 + 5406720b17 − 29744b5 − 4302898b14− 3906048b11 + 726704b8 + 869b2)a22 + 88(84800b9 + 106496b18 + 425600b12+ 525b3 − 22812b6 + 686080b15 − 262144b21)a21 + 22(131072b22 − 1080896b10− 7456768b13 + 10606b4 + 176464b7 − 5b+ 1277952b19 − 5062656b16)a20+ 88(1449984b20 − 25502b5 + 169b2 + 2756608b17 + 913056b11 + 39612b8+ 2154880b14)a19 + 44(5703b3 − 212080b9 − 3180544b18 − 4571648b15− 1245184b21 − 2875232b12 + 35626b6)a18 + 44(614656b16 + 16384b22 − 20204b7− 19b+ 1863680b19 + 905680b10 − 325632b13 − 11029b4)a17 + 11(2205888b11+ 23824384b17 + 10297344b14 + 16191488b20 + 4129b2 − 39612b5 − 435440b8)a16+ 88(1325b3 − 1825024b18 − 1195552b12 + 13255b6 − 2708288b15 − 319488b21+ 4272b9)a15 + 22(6697408b13 − 16889b4 + 6679680b16 − 4096b22 + 1948416b10− b+ 2046976b19 − 226420b7)a14 + 4(7879b2 − 17135712b14 + 15851264b17− 16050496b11 − 758274b8 + 313863b5 + 18625024b20)a13 + 2(7904512b18− 36575b3 + 6575536b9 + 21602944b15 − 988361b6 + 26518184b12 + 3942400b21+ 2048b24)a12 + 8(5632b22 − 2302234b10 − 27984b7 + 451616b16 + 40051b4+ 55b− 574016b13 − 1515008b19)a11 + 22(80448b8 − 8192b20 + 670976b17+ 389448b11 − 191b2 − 16835b5 − 2048b23 + 1058624b14)a10 + 44(8929b6 + 335b3− 84634b9 + 530432b18 + 391840b15 + 79360b21 − 122744b12)a9 + 11(205824b19− 4802b7 − 2472b4 − 15b− 41056b13 + 208740b10 + 3136b16 + 17408b22)a8960 M. S. Mahadeva Naika, S. Chandankumar and K. Sushan Bairy+ 44(1346b5 − 11633b8 − 165056b17 + 22516b11 − 15b2 − 80768b20 − 41936b14− 1024b23)a7 + 22(11228b12 − 157248b18 − 35328b21 − 132608b15 − 1553b6− 13b3 + 14258b9)a6 + 22(910b7 − 8608b13 − 12864b16 − 39b4 − 3998b10 + b+ 3072b22 − 8064b19)a5 + 11(7b2 − 354b5 + 2048b23 + 3953b8 + 26112b20+ 40384b17 − 13228b11 − 256b14)a4 + 88(1104b18 − 78b9 + 19b6 − 175b12+ 384b21 + 620b15 − b3)a3 + 11(b4 − 2b10 − 2b7 − 1024b22 − 8b13 + 272b16− 768b19)a2 − ba2 + 8(−256b23 − 88b17 − 352b20 + 11b14)a = 0.(30)Proof of the identity (30) is similar to the proof of the identity (21), exceptthat in place of results (8) and (9), results (13) and (14) are used.Theorem 3.5. If a := V (q) and b := V (q4), then4ab3 + b− a4 − 8a3b4 − 4a3b− 6a2b2 = 0. (31)Proof. Employing the equation (17), we arrive at the equation (31).Theorem 3.6. If a := V (q) and b := V (q8), thena8 − b− 28a2b3 − 64a3b7 − 16ab5 + 8a3b− 8ab2 + 70a4b2 + 224a5b6+ 168a5b3 + 256a6b7 + 280a6b4 − 4a6b+ 128a7b8 + 128a7b5 + 8a7b2 = 0.(32)Proof. Employing the equations (17) and (31), we arrive at the equation (32).Theorem 3.7. If a := V (q) and b := V (q12), then512b12a9 + 256(4a9 + 6a6 − a3)b11 + 128(10a9 − a3 + 15a6)b10 + 64(5a3 − 6a6− 1 + 20a9)b9 + 32(8a3 − 1 + 38a9 + 8a12 + 57a6)b8 + 32(16a12 + 87a6 − 2a3+ 58a9 − 2)b7 + 8(39a6 − 7 + 62a3 + 218a9 + 80a12)b6 + 16(21a6 − 4 + 26a3+ 32a12 + 14a9)b5 + 2(332a9 − 19 + 92a3 + 152a12 + 498a6)b4 + (68a3 − 19+ 128a12 + 868a9 + 534a6)b3 + (40a12 − 5 + 210a6 + 12a3 + 140a9)b2+ (12a3 − 20a9 − 1 + 8a12 − 30a6)b+ a12 = 0.(33)Proof. Employing the equations (17) and (21), we arrive at the equation (33).Theorem 3.8. If a := V (q) and b := V (q16), then32768a15b16 + 16384(8a10 − 4a13 − a7)b15 + 8192(71a11 + 4a14 − 8a8)b14+ 4096(184a12 − a3 − 64a9 + 16a15 + 12a6)b13 + 2048(199a13 + 88a7 − 8a4− 160a10)b12 + 7168(24a14 − 5a5 + 22a11 + 32a8)b11 + 512(1032a12 + 59a9On some new identities for Ramanujan’s cubic continued fraction 961+ 88a6 + 72a15 − 12a3)b10 + 256(836a7 − 1096a10 − a+ 1556a13 − 88a4)b9+ 128(931a14 − 2576a11 − 224a5 − 8a2 + 2712a8)b8 + 64(48a15 − 3052a12− 59a6 − 64a3 + 4770a9)b7 + 32(1092a7 − 2166a13 + 6089a10 − 160a4 − 8a)b6+ 16(5768a11 − 71a2 + 154a5 − 1224a14 + 2576a8)b5 + 8(3987a12 − 184a3+ 1032a6 − 124a15 + 3052a9)b4 + 4(2166a10 − 199a4 + 1556a7 + 1220a13− 4a)b3 + (2448a11 + 632a14 + 1862a8 + 8a2 − 336a5)b2 + (16a15 + 48a9− 1 + 16a3 − 72a6 + 124a12)b+ a16 = 0.(34)Proof. Employing the equations (17) and (32), we arrive at the equation (34).ACKNOWLEDGEMENTS. The authors would like to thank Univer-sity Grants Commission, New Delhi for their support under the Major ResearchProject No.F.No.34–140\2008 (SR).References[1] C. Adiga, Taekyun Kim, M.S. Mahadeva Naika and H.S. Madhusudhan,On Ramnujan’s cubic continued fraction and explicit evaluations of theta-functions, Indian J. Pure Appl. Math., 35(9) (2004), 1047 - 1062.[2] N.D. Baruah, Modular equations for Ramanujan’s cubic continued frac-tion, J. Math. Anal. Appl., 268 (2002), 244 - 255.[3] B.C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, NewYork, 1991.[4] H.H. Chan, On Ramanujan’s cubic continued fraction, Acta Arith., 73(1995), 343 - 355.[5] B. Cho, J.K. Koo and Y.K. Park, On the Ramanujan’s cubic continuedfraction as modular function, Tohoku Math. J., 62(4) (2010), 579 - 603.[6] M.S. Mahadeva Naika, Some theorems on Ramanujan’s cubic continuedfraction and related identities. Tamsui Oxf. J. Math. Sci., 24(3) (2008),243 - 256.[7] M.S. Mahadeva Naika, S. Chandankumar and K. Sushan Bairy, New iden-tities for Ramanujan’s cubic continued fraction, Funct. Approx. Comment.Math., (to appear).[8] M.S. Mahadeva Naika, S. Chandankumar and K. Sushan Bairy, On someparameter involving Ramanujan’s cubic continued fraction, (communi-cated).962 M. S. Mahadeva Naika, S. Chandankumar and K. Sushan Bairy[9] M.S. Mahadeva Naika, M.C. Maheshkumar and K. Sushan Bairy, Generalformulas for explicit evaluations of Ramanujan’s cubic continued fraction,Kyungpook Math. J., 49(3) (2009), 435 - 450.[10] S. Ramanujan, Notebooks (2 volumes), Tata Institute of FundamentalResearch, Bombay, 1957.[11] S. Ramanujan, The lost notebook and other unpublished papers, Narosa,New Delhi, 1988.Received: November, 2011

On some new identities for Ramanujan's cubic continued fraction.

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On some new identities for Ramanujan's cubic continued fraction.

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