Research Article Some New Explicit Values of Quotients of Ramanujan s Theta Functions and Continued Fractions

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1 International Mathematics and Mathematical Sciences Article ID pages Research Article Some New Explicit Values of Quotients of Ramanujan s Theta Functions and Continued Fractions Nipen Saikia Department of Mathematics Rajiv Gandhi University Rono Hills Doimukh Arunachal Pradesh 79 India Correspondence should be addressed to Nipen Saikia; nipennak@yahoo.com Received 7 December 03; Accepted May 04; Published 5 May 04 Academic Editor: Seppo Hassi Copyright 04 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 evaluate some new explicit values of quotients of Ramanujan s theta functions and use them to find explicit values of Ramanujan s continued fractions.. Introduction Ramanujan s general theta function f(a b) is defined by f (a b) := a k(k+)/ b k(k )/ ab <. () k= Two important special cases of f(a b) are the theta functions φ(q) and ψ(q) [ page 36 Entry] defined by for q < φ (q) := f (q q) = q n = ( q; q) n= (q; q) () ψ (q) := f (q q 3 )= n=0 q n(n+)/ = (q ;q ) (q; q ) where (a; q) = n=0 ( aqn ). In his notebooks [] Ramanujan recorded many explicit values of theta functions φ(q) and ψ(q). All these values were proved by Berndt [3 page35]andberndtandchan[4]. Yi [5]introducedtheparameterh kn for positive real numbers k and n defined by φ(q) h kn = k /4 φ(q k ) ; q = e π n/k (3) and used the particular case h nn to find explicit values φ(q). Baruah and Saikia [6] defined the parameter g kn for positive real numbers k and n as ψ( q) g kn = k /4 q (k )/8 ψ( q k ) ; q = e π n/k (4) andusedtheparticularcaseg nn to find explicit values ψ(q). Saikia [7] also established some explicit values ψ(q). In this paper we consider the particular cases h 3n and g 3n of the parameters h kn and g kn respectively.byusingtheta function identities we find some new explicit values of the parameters h 3n and g 3n.Particularlyweevaluateh 3n and g 3n for n=3/ /3 6 /6 5/ /5 0 and /0. Previously Yi [5] evaluatedh 3n for n = 3 /3 9 /9 5 /5 5 and /5. Saikia [8] evaluatedh 3n for n = / 4 /4 7 /7 49 and /49. Baruah and Saikia [6] evaluatedg 3n for n= 3 /3 9 /9 5 /5 5 /5 7 /7 3 /3 49 and /49. As an application to our new values we evaluate some old and new explicit values of Ramanujan s cubic continued fraction G(q) and a continued fraction of order twelve H(q) which are respectively defined by for q < G(q):= q/3 q+q q +q 4 q 3 +q H(q) := qf ( q q ) f( q 5 q 7 ) = q( q) q 3 ( q )( q 4 ) q 3 ( q 8 )( q 0 ) q 3 + ( q 3 )( + q 6 ) + ( q 3 )( + q. ) + The continued fraction G(q) was recorded by Ramanujan on page 366 of lost notebook [9]. We refer to [0 4] for explicit evaluations of G(q). The continuedfraction H(q) was (5)

2 International Mathematics and Mathematical Sciences introducedby Naika et al. [5]. We refer to [8 5] for explicit evaluations of H(q). The presentation of the paper is as follows. In Section we record some preliminary results for ready references in this paper. Section 3 is devoted to explicit evaluations of the parameters h 3n and g 3n.InSection4 we use new explicit values of h 3n and g 3n to evaluate some explicit values of the continued fractions G(q) and H(q). We end this introduction by noting the following remarks regarding g 3n and h 3n from [6 page 764 Theorem 4.] and [5page385Remark.3]respectively. Remark. The parameter g 3n has positive real value with g 3 =and g 3n increases as n increases. Remark. The parameter h 3n has positive real value with h 3 =and the values of h 3n decrease as n increases.. Preliminary Results Lemma 3 (see [6 Theorem 3.]). If P:= qψ( q)ψ( q 9 )/ ψ ( q 3 ) and Q:=qψ( q )ψ( q 8 )/ψ ( q 6 )then P Q + Q P +(P Q + Q P ) =(3PQ PQ )(3PQ PQ + 3P Q + 3Q P + P3 Q 3 + Q3 P 3 ) Lemma 7 (see [5page385Theorem.]). For all positive real numbers k and nonehas h k/n = h kn. (0) Lemma 8 (see [6 page 764 Theorem 4.]). For all positive real numbers k and nonehas g k/n = g kn. () Lemma 9 (see [5 page 393 Theorem 4.9(i)]). For any positive real number none has 3(h 3n h 39n + )=( h 39n ) +3. () h 3n h 9n h 3n Lemma 0 (see [5 page394theorem4.(i)]).for any positive real number none has 3{(h 3n h 35n ) + (h 3n h 35n ) } + 4. Lemma 4 (see [6 Theorem 3.]). If L := φ(q)φ(q 9 )/φ (q) and M:= qψ( q)ψ( q 9 )/ψ ( q 3 )then (6) L = M + 3M. (7) ={( h 3 35n ) ( h 3 3n ) } h 3n h 35n +5{( h 35n ) +( h 3n ) } h 3n h 35n +5{( h 35n h 3n ) ( h 3n h 35n )}. (3) Lemma 5 (see [6 Theorem 3.]). If P := ψ( q 3 )ψ( q 5 )/ qψ( q)ψ( q 5 ) and Q := ψ( q 6 )ψ( q 0 )/q ψ( q )ψ( q 30 ) then ( P Q + Q P )+(P Q + Q P ) (PQ+ PQ ) +( PQ PQ )( P3 Q 3 + Q3 P 3 + P Q + Q P )=0. Lemma 6 (see [6 Theorem 3.9]). If L := φ(q 3 )φ(q 5 )/ φ(q)φ(q 5 ) and M := ψ( q 3 )ψ( q 5 )/qψ( q)ψ( q 5 )then (8) M= +L L. (9) Lemma (see [6 page 769 Theorem 5.(i) and (iii)]). For any positive real number none has (i) ( + 3g 3n g 39n ) 3 =+3g 4 39n (ii) 3{(g 3n g 35n ) +/(g 3n g 35n ) } + 5{(g 35n /g 3n ) + (g 3n /g 35n ) } = {(g 35n /g 3n ) 3 (g 3n /g 35n ) 3 } + 5{(g 35n /g 3n ) (g 3n /g 35n )}. 3. Explicit Evaluations of the Parameters g kn and h kn In this section we prove some general theorems for the explicit evaluations of the parameters g kn and h kn and evaluate some new explicit values therefrom.

3 International Mathematics and Mathematical Sciences 3 Theorem. One has ( g 3ng 336n ) +( g 34ng 39n ) g 34n g 39n g 3n g 336n +(( g 3ng 336n ) +( g 34ng 39n ) ) 4 g 34n g 39n g 3n g 336n ={3( g 3ng 34n g 39n g 336n ) ( g 39ng 336n g 3n g 34n )} {3( g 3ng 34n g 39n g 336n ) ( g 39ng 336n g 3n g 34n )+3( g 3ng 336n g 34n g 39n )} +{3( g 3ng 34n g 39n g 336n ) ( g 39ng 336n g 3n g 34n )} {3( g 34ng 39n )+( g 3 3ng 336n ) +( g 3 34ng 39n ) }. g 3n g 336n g 34n g 39n g 3n g 336n (4) Proof. We set q:=e π n/3 in Lemma 3 andusethedefinition of g 3n. Corollary 3. One has Again setting n=/6in Lemma (i) and simplifying using Lemma 8wededucethat ( + 3( g 3 33/ )) =+3g 4 33/ g. (8) 36 Eliminating g 36 from (8) using(7) andsimplifyingwe obtain g 4 33/ +3( + (+ ) 3/ )g 33/ +(3+3 ) = 0. (9) Solving (9) and noting the fact in Remark weobtain g 33/ = ( ) Employing (0) in(7) and simplifying we obtain. (0) 6(+ ) g 36 =. () Now the values of g 3/3 and g 3/6 follow from the values of g 33/ and g 36 respectively and Lemma 8. g 33/ = g 36 = g 3/3 = ( ) 6(+ ) ( ) (5) Theorem 4. One has ( h 3n ) { 3( g 3n ) } ( g 3n ) =0. () h 39n g 39n g 39n Proof. We set q:=e π n/3 in Lemma 4 andusethedefinitions of h 3n and g 3n. Corollary 5. One has h 33/ = 5/8 (3 4) 3/4 ( ) / g 36 = (+ ) Proof. Setting n=/6in Theorem and simplifying using Lemma 8weobtain 3 ( (g 33/ g 36 ) (g 33/g 36 ) ) 3 ( (g 33/ g 36 ) (g 33/g 36 ) +8) +8=0. (6) Solving (6)forg 33/ g 36 and noting the fact in Remark we find that g 33/ g 36 = 3(+ ). (7) h 36 = /8 3 / (3 +4) /4 ( ) / h 3/3 = 5/8 (3 4) 3/4 ( ) / h 3/6 = /8 3 / (3 +4) /4 ( ) /. (3) Proof. Setting n=/6in Theorem 4 and simplifying using Lemma 7weobtain (h 36 h 33/ ) ( 3 (g 36 g 33/ ) ) (g 36 g 33/ ) =0. (4) Employing (7) in(4) solving the resulting equation and noting the fact in Remark wededucethat h 33/ h 36 = / (5)

4 4 International Mathematics and Mathematical Sciences Again setting n = /6 in Lemma 9 and simplifying using Lemma 7weobtain 3( h 33/ + h 36 ) = (h h 36 h 33/ h 36 ) +3. (6) 33/ Employing (5) in(6) solving the resulting equation and noting the fact in Remark wededucethat h 33/ h 36 = (4+3 ). (7) Multiplying (5)and(7) we evaluate the value of h 33/ and dividing (5) by(7) and simplifying we evaluate the value of h 36.Nowthevaluesofh 3/3 and h 3/6 easily follow from the values of h 33/ and h 36 respectively and Lemma 7. Theorem 6. One has ( g 34ng 35n ) +( g 3ng 300n ) +( g 34ng 35n ) g 3n g 300n g 34n g 35n g 3n g 300n +( g 3ng 300n g 34n g 35n ) ( g 35ng 300n g 3n g 34n ) ( g 3ng 34n g 35n g 300n ) Proof. Setting n = /0 in Theorem 6 and simplifying using Lemma 8weobtain (g 35/ g 30 ) + (g 35/ g 30 ) 4(g 35/ g 30 g 35/ g 30 ) 4=0. Solving (3) and noting the fact in Remark wefindthat (3) g 35/ g 30 = (3) Again setting n = /0 in Lemma (ii) and simplifying using Lemma 8we obtain 3{( g 35/ ) +( g 35/ )} g 30 g 30 +{( g / 35ng 300n ) ( g / 3ng 34n ) } g 3n g 34n g 35n g 300n {( g 3/ 34ng 35n ) +( g 3/ 3ng 300n ) g 3n g 300n g 34n g 35n +5((g 35/ g 30 ) + ((g 35/ g 30 ) 3 (g 35/ g 30 ) ) (g 35/ g 30 ) 3 ) (33) +( g / 34ng 35n ) +( g / 3ng 300n ) }=0. g 3n g 300n g 34n g 35n (8) Proof. We set q:=e π n/3 in Lemma 5 andusethedefinition of g 3n. Corollary 7. One has g 35/ =(+ 3+ / + 3) (a + 5b) /4 c /4 5(g 35/ g 30 g 35/ g 30 )=0. Employing (3) in(33) solving the resulting equation and noting the fact in Remark weobtain g 30 =( a+ 5b g 35/ c / ) (34) g 30 =(+ 3+ / + 3) (a + 5b) /4 c /4 g 3/5 =(+ 3+ / + 3) (a + 5b) /4 c /4 g 3/0 =(+ 3+ / + 3) (a + 5b) /4 c /4 where a= b = c= (9) (30) where a= b = andc = Combining (3)and(34) we calculate the values of g 35/ and g 30.Thenthevaluesofg 3/5 and g 3/0 follow from the values of g 35/ and g 30 respectively and Lemma 8. Theorem 8. One has ( g 35n g 3n ){ ( h 35n h 3n )} ( h 35n h 3n )=0. (35) Proof. We set q:=e π n/3 in Lemma 6 andusethedefinitions of h 3n and g 3n.

5 International Mathematics and Mathematical Sciences 5 Corollary 9. One has / / h 35/ =( ) ( ) (f 4 5b) /4 d /4 / / h 30 =( ) ( ) (f 4 5b) /4 d /4 / / h 3/5 =( ) ( ) (f 4 5b) /4 d /4 / / h 3/0 =( ) ( ) (f 4 5b) /4 d /4 (36) where b is given in Corollary 7 f = andd = Proof. Setting n = /0 in Theorem 8 and simplifying using Lemma 7weget g 35/ g 30 ( h 35/ h 30 ) h 35/ h 30 =0. (37) Employing (3) and solving the resulting equation we obtain h 35/ h 30 = (38) Again setting n = /0 in Lemma 0 and simplifying using Lemma 7wefindthat 3{( h 35/ ) +( h 30 ) } h 30 h 35/ ((h 35/ h 30 ) 3 5((h 35/ h 30 ) + 5(h 35/ h 30 (h 35/ h 30 ) 3 ) (h 35/ h 30 ) ) h 35/ h 30 )=0. (39) Employing (38) in(39) solving the resulting equation and noting the fact in Remark weobtain h 35/ h 30 = f 4 5b (40) d where b isgivenincorollary7 f = andd = Combining (38)and(40) we evaluate the values of h 35/ and h 30.Thenthevaluesofh 3/5 and h 3/0 follow from h 35/ and h 30 respectively and Lemma Explicit Evaluations of Continued Fractions This section is devoted to finding some explicit values of the continued fractions G(q) and H(q) by using new values of g 3n and h 3n evaluatedinsection3. Lemma 0 (see [6 page 788 Theorem 9.(i)]). One has G 3 (e π n/3 ) = +3g 4 3n. (4) Theorem. One has G 3 ( e π/ )= ( ) G 3 ( e π ) ( ) = 6( ( + ) (6 + 6 ) ( )) G 3 ( e π /3 ) ( ) = 6( ( + ) (6 + 6 ) (6 + 4 )) G 3 ( e π/3 )= (3+ ) ( (0 + 8 )) G 3 ( e π 5/6 )= G 3 ( e π 0/3 )= G 3 ( e π /5 )= G 3 ( e π/ 30 )= c c+3(a+ 5b) ( ) (a+ 5b) (a + 5b) + 3c( ) (a+ 5b) ( ) (a + 5b) ( ) +3c c( ) c( ) +3(a+ 5b) where a bandc are given in Corollary 7. (4)

6 6 International Mathematics and Mathematical Sciences Proof. We set n = 3/ 6 /3 /6 5/ 0 /5 and /0 in Lemma 0 and use the corresponding values of g 3n from Corollaries 3 and 7 to complete the proof. H(e π 0/3 ) =(3 /4 (3 + / + 3) d /4 The explicit values of G 3 ( e π/3 ) G 3 ( e π 5/6 ) G 3 ( e π 0/3 ) G 3 ( e π /5 )andg 3 ( e π/ 30 ) are new. Lemma (see [8page44Theorem5.]). One has H(e π n/3 )= 3/4 h 3n 3 /4 h 3n +. (43) By setting n = 3/ 6 /3 /6 5/ 0 /5 and /0 in Lemma and employing the corresponding values of h 3n from Corollaries 5 and 9 wecalculatethefollowingnew explicit values of the continued fraction H(q). Theorem 3. One has (+ 3+ / + 3) (f 4 5b) /4 ) (3 /4 (3 + / + 3) d /4 H(e π /5 ) / +( ) (f 4 5b) /4 ) =(3 /4 ( + 3+ / + 3) d /4 (3+ / + 3) (f 4 5b) /4 ) H(e π/ )= + 3/8 3 /4 (3 4) 3/ /8 3 /4 (3 4) 3/ H(e π )= /8 3 3/4 (4 + 3 ) / /8 3 3/4 (4 + 3 ) / H(e π /3 )= 5/8 3 /4 (3 4) 3/ /8 3 /4 +(3 4) 3/ H(e π/ 8 )= 7/ ( + 9 ) /4 H(e π 5/6 ) 7/ ( + 9 ) /4 =(3 /4 (3 + / + 3) (f 4 5b) /4 (+ 3+ / + 3) d /4 ) (3 /4 (3 + / + 3) (f 4 5b) /4 / +( ) d /4 ) H(e π/ 30 ) (3 /4 ( + 3+ / + 3) d /4 +(3+ / + 3) (f 4 5b) /4 ) =(3 /4 ( + 3+ / + 3) (f 4 5b) /4 (3+ / + 3) d /4 ) (3 /4 ( + 3+ / + 3) (f 4 5b) /4 +(3+ / + 3) d /4 ) (44) wherebisgivenincorollary7 and d and f are given in Corollary 9. Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this paper.

7 International Mathematics and Mathematical Sciences 7 Acknowledgment The author is thankful to the University Grants Commission NewDelhiIndiaforpartiallysupportingtheresearchwork under Grant F. no /0(SR). References [] B. C. Berndt Ramanujan s Notebooks Springer New York NY USA 99. [] S. Ramanujan Notebooks Tata Institute of Fundamental Research Bombay India 957. [3] B. C. Berndt Ramanujan s Notebooks Springer New York NY USA 998. [4] B.C.BerndtandH.H.Chan Ramanujan sexplicitvaluesfor the classical theta-functions Mathematika vol.4no.pp [5] J. Yi Theta-function identities and the explicity formulas for theta-function and their applications Mathematical Analysis and Applicationsvol.9no.pp [6] N. D. Baruah and N. Saikia Two parameters for Ramanujan s theta-functions and their explicit values Rocky Mountain Mathematicsvol.37no.6pp [7] N. Saikia A new parameter for Ramanujan s theta-functions and explicit values Arab Mathematical Sciencesvol. 8no.pp [8] N. Saikia Modular identities and explicit values of a continued fraction of order twelve JPJournalofAlgebraNumberTheory and Applications vol. no. pp [9] S. Ramanujan The Lost Notebook and Other Unpublished Papers Narosa New Delhi India 988. [0] C. Adiga M. S. M. Naika and K. R. Vasuki Some new explicit evaluations for Ramanujan s cubic continued fraction New Zealand Mathematicsvol.3no.pp [] C. Adiga T. Kim M. S. M. Naika and H. S. Madhusudhan On Ramanujan s cubic continued fraction and explicit evaluations of theta-functions Indian Pure and Applied Mathematicsvol.35no.9pp [] N. D. Baruah and N. Saikia Some general theorems on the explicit evaluations of Ramanujan s cubic continued fraction Computational and Applied Mathematicsvol.60no. -pp [3] B.C.BerndtH.H.ChanandL.-C.Zhang Ramanujan sclass invariants and cubic continued fraction Acta Arithmeticavol. 73 pp [4] H. H. Chan On Ramanujan s cubic continued fraction Acta Arithmetica vol. 73 pp [5] M.S.M.NaikaB.N.DharmendraandK.Shivashankara A continued fraction of order twelve Central European Journal of Mathematicsvol.6no.3pp [6] M. S. M. Naika S. Chandankumar and M. Harish On some new mixed modular equations involving Ramanujan s thetafunctions Matematicki Vesnikvol.66no.3pp

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