Journal of Mathematical Analysis and Applications

Transcription

1 J Math Anal Appl 396 (0) 3 0 Contents lists available at SciVerse ScienceDirect Journal of Mathematical Analysis Applications journal homepage: wwwelseviercom/locate/jmaa On Ramanujan s modular equations of degree 35 KR Vasuki G Sharath Department of Studies in Mathematics University of Mysore Manasagangotri Mysore India a r t i c l e i n f o a b s t r a c t Article history: Received October 0 Available online 7 June 0 Submitted by BC Berndt In this paper we give an alternative proof of four of Ramanujan s modular equations of degree 35 which have been proved by BC Berndt using the theory of modular forms Our proofs involve only the identities stated by Ramanujan 0 Elsevier Inc All rights reserved Keywords: Theta functions Modular equations Introduction Ramanujan [ pp 49 50] has recorded eight beautiful modular equations of mixed degrees Berndt has proved all these modular equations in [ pp 43 46] Out of eight modular equations Berndt proved two of them by employing only the tools known to Ramanujan proved the rest by using the theory of modular forms Berndt [ pp 36] expressed that the primary disadvantage of proof by employing the modular form method is that the modular equation must be known in advance Thus the proofs are perhaps more aptly called verifications Motivated by these in this paper we prove four of the remaining six identities which have been proved by Berndt by employing the modular form Now we shall recall the definition of the modular equation before stating the aftermost mentioned modular equation of Ramanujan As usual for any complex number a (a) k = a(a )(a ) (a k ) k = 3 (a) 0 = The Gauss hypergeometric series is defined by a b F c ; z (a) k (b) k = z n z < (c) k k! Let F(x) = k=0 F ; x F ; x Corresponding author addresses: vasuki_kr@hotmailcom (KR Vasuki) sharath_gns@rediffmailcom (G Sharath) 00-47X/$ see front matter 0 Elsevier Inc All rights reserved doi:006/jjmaa00508

2 4 KR Vasuki G Sharath / J Math Anal Appl 396 (0) 3 0 Suppose that the equality nf(α) = F(β) holds for some positive integer n Then a modular equation of degree n is the relation between α β which is induced by the equality () We say β has degree n over α We also define the multiplier m with respect to α β by m = Suppose that F ; α F ; β n F(α) = F(β) n F(α) = F(γ ) n 3 F(α) = F(δ) () hold Then a mixed modular equation is a relation between α β γ δ induced by () In such an instant we say that β γ δ are of degree n n n 3 respectively over α or α β γ δ have degree n n n 3 respectively Ramanujan [] without proof obtained the following modular equations Theorem Let β γ δ have degree respectively over α Let m m denote the multipliers connecting α β γ δ respectively Then () (i) (ii) (iii) (iv) (αδ) /4 {( α)( δ)} /4 4/3 {αδ( α)( δ)} / (βγ ) /4 {( β)( γ )} /4 4/3 {βγ ( β)( γ )} / = { 4/3 {αβγ δ( α)( β)( γ )( δ)} /4 } (3) {(αδ) /4 {( α)( δ)} /4 4/3 {αδ( α)( δ)} / } {(βγ ) /4 {( β)( γ )} /4 4/3 {βγ ( β)( γ )} / } = 7/3 {αβγ δ( α)( β)( γ )( δ)} /4 {(αβγ δ) /8 {( α)( β)( γ )( δ)} /8 } (4) m / (αδ) /4 {( α)( δ)} /4 4/3 {βγ ( β)( γ )} / = (5) m (βγ ) /4 {( β)( γ )} /4 4/3 {αδ( α)( δ)} / m m / = (6) (v) (vi) (vii) {6βγ ( β)( γ )} /4 {6αδ( α)( δ)} /8 {6βγ ( β)( γ )} /4 {6βγ ( β)( γ )} /8 = {6αδ( α)( δ)} /8 {6αδ( α)( δ)} /4 {6βγ ( β)( γ )} /8 {6αδ( α)( δ)} /4 = m / (7) m m / (8) αδ /8 ( α)( δ) /8 αδ( α)( δ) /8 βγ ( β)( γ ) βγ ( β)( γ ) αδ( α)( δ) / m / = (9) βγ ( β)( γ ) m m

3 KR Vasuki G Sharath / J Math Anal Appl 396 (0) (viii) βγ /8 ( β)( γ ) /8 αδ ( α)( δ) βγ ( β)( γ ) αδ( α)( δ) βγ ( β)( γ ) /8 αδ( α)( δ) / m / = (0) m Berndt has proved (5) (6) by employing one of the Schröter [3] formulas as he has proved the rest by the theory of modular forms In this paper we derive (7) (0) by using the tools known to Ramanujan We are unable to extend the method of proof of (7) (0) adopted to prove the remaining two modular equations (3) (4) Using Entries 0(i) (iii) (i) (iii) (i) (iii) in [ Chapter 7] Berndt translates the modular equations (7) (0) to the following theta-function identities respectively: with ϕ(q 5 )ϕ(q 7 ){qψ( q 5 )ψ( q 7 ) q 4 ψ( q)ψ( q 35 ) f (q 5 )f (q 7 )} = ϕ(q)ϕ(q 35 )f (q 5 )f (q 7 ) () ϕ(q)ϕ(q 35 ){ψ( q 5 )ψ( q 7 ) q 3 ψ( q)ψ( q 35 ) f (q)f (q 35 )} = ϕ(q 5 )ϕ(q 7 )f (q)f (q 35 ) () q 3 ψ(q)ψ(q 35 ) ψ(q 5 )ψ(q 7 ) ϕ( q )ϕ( q 70 ) ϕ( q 0 )ϕ( q 4 ) q 3 ψ( q)ψ( q 35 ) q f ( q )f ( q 70 ) = (3) ψ( q 5 )ψ( q 7 ) f ( q 0 )f ( q 4 ) ψ(q 5 )ψ(q 7 ) q 3 ψ(q)ψ(q 35 ) ϕ( q 0 )ϕ( q 4 ) ϕ( q )ϕ( q 70 ) ϕ(q) := ψ(q) := f ( q) := n= n=0 (a; q) 0 = (a; q) = q n = ( q; q ) (q ; q ) (q; q ) ( q ; q ) q n(n) = (q ; q ) (q; q ) ( ) n q n(3n ) = (q; q) n= ( aq n ) n=0 One can easily show that ϕ(q) = f 5 ψ(q) = f f f f 4 f n := f ( q n ) For convenience set = qf f 35 := q f f 70 f 5 f 7 f 0 f 4 ψ( q 5 )ψ( q 7 ) q 3 ψ( q)ψ( q 35 ) 0 f ( q )f ( q 4 ) = (4) q f ( q )f ( q 70 ) ϕ( q) = f ψ( q) = f f 4 (5) f f R := q 4 f 4 f 40 f 0 f 8 S = f f 35 q 3/ f f 70 T = f 5f 7 q / f 0 f 4 K = f 5f 7 q 5/ f f 70 (6) Changing q to q in () then using (5) the above () is equivalent to ST = Changing q to q in () then using (5) (6) () is equivalent to K = (7) (8)

4 6 KR Vasuki G Sharath / J Math Anal Appl 396 (0) 3 0 Using (5) (6) in (3) (4) then (3) (4) respectively translate into 3 3 R R R = R R R 3 R = R Thus (7) (0) are respectively equivalent to (7) (0) In Section of this paper we prove (9) (0) in Section 3 we prove (7) (8) roof of (9) (0) (9) (0) Lemma We have 3 3 = 4 8 () roof We have from [4 p 55] [5] if then A = f q /4 f B = f 5 q 5/4 f 0 (AB) 4 (AB) = 3 3 B A A B () Changing q to q 7 in the above we obtain 3 (A B ) 4 (A B ) = B A A B A = f 7 q 7/4 f 4 B = f 35 q 35/4 f 70 3 (3) Multiplying () (3) we obtain (AA BB ) 6 AB (AA BB ) A 4 B BB AA = (4) A B AB AA BB We have from [ p 35] [6 p 06] if then C = f q /6 f 5 D = f q /3 f 0 CD 5 CD = 3 3 D C C D (5) Changing q to q 7 in the above we obtain C D 5 C D = D C 3 C D 3 (6) C = f 7 q 7/6 f 35 D = f 4 q 7/3 f 70 Multiplying (5) (6) then using the fact that C D = A B C D = A B CD C D = CD C D =

5 KR Vasuki G Sharath / J Math Anal Appl 396 (0) we obtain CC DD 5 CC DD 5 3 AA BB = BB AA Subtracting (4) from (7) (AA BB ) 6 AB (AA BB ) A 4 B A B AB 3 3 = We have from [ p 37] [6 p 07] if then E = f q /8 f 5 F = f q 3/8 f (7) CC DD 5 5 CC DD (8) EF 5 EF = F E 4 E F (9) Changing q to q 7 in the above we obtain E F 5 E F = F E 4 E F (0) E = f 4 q 7/8 f 35 F = f 7 q /8 f 70 Multiplying (9) (0) then using the fact that EE FF = CC DD EF FE = A B AB FF EE = AA BB EF E F = we obtain AB A B 4 5 = (AA BB ) 6 A B AB (AA BB ) CC DD 5 CC DD Using this in (8) we obtain AB A B 4 = 5 A B AB 3 3 () We have from [ p 86] [7 p 38] if α β γ δ have degree respectively then R 4 = 6 R () R := := 6 γ δ( γ )( δ) /48 αβ( α)( β) αδ( α)( δ) /48 βγ ( β)( γ )

6 8 KR Vasuki G Sharath / J Math Anal Appl 396 (0) 3 0 Recently Baruah [8] has deduced () from the tools known to Ramanujan Using Entries 0(i) (i) in Chapter 7 [ pp 4] convert () into the theta function identity then changing q to q we obtain AB A B = A B AB From () (3) we obtain () This completes the proof of Lemma The identity () can be written as This implies = 3/ ± 3/ = 3/ 3/ By definition of one can see that 3/ 3/ = This implies is positive < Thus the above implies 5 (3) ( ) = 3 3 (4) The identity (4) is due to Bhargava et al [9] they employ (9) (0) to prove (4) roof of (8) Changing q to q in (4) then multiplying throughout by we obtain R R = 3 R 3 R R Multiplying (4) by R then subtracting the resulting identity from the above we obtain (8) roof of (9) Changing q to q in (4) then multiplying throughout by we obtain R R = 3 R 3 R R Multiplying (4) by R then subtracting the resulting identity from the above we obtain (9) 3 roof of (7) (8) We have from [ p 86] [7 p 379] if α β γ δ have degree respectively then 4 4 = (3) as in () = [56αβγ δ( α)( β)( γ )( δ)] /48 Recently Baruah [8] has deduced (3) from the tools known to Ramanujan Using Entries 0(i) (i) of Chapter 7 [ pp 4] convert (3) into the theta function identity then changing q to q we obtain ST ST = S T S T S T T (3) S S T are as in (6) Lemma 3 We have 4 (ST) = ( 4 ) ( 4 ) (33)

7 KR Vasuki G Sharath / J Math Anal Appl 396 (0) roof It is easy to see that S = Using this in (3) we obtain T ST ST = Squaring this on both sides then setting 4 = x we obtain (ST) x x = t (34) t = 4 Solving (34) we obtain Consider x = t ± t 4 t 4 = [ ] 6( ) 8 ( ) ( ) 4 ( ) By employing () to the first factor of the number of the right h side we obtain (35) t 4 = ( ) ( ) ( ) ( ) 4 6( ) 6 Using this in (35) choosing positive sign we obtain x = 4t( )3 ( )( )( )( ) 8( ) 3 Multiplying this by ( 4 ) we obtain ( 4 )x = ( 4 ) ( )( ) ( 3 3 )( ) Since the third factor in the second term of the above is zero by (4) we obtain x = ( 4 ) ( 4 ) Since (34) is the reciprocal equation we have x = ( 4 ) ( 4 ) From the above two identities we can rewrite (34) as [( 4 )x ( 4 )][ ( 4 )x ( 4 )] = 0 (36) One can see that series expansion of x as follows = q q q 3 q 6 q 7 q 8 q 9 q 0 3q q q 4 q 5 = q q 4 q 6 q q 4 q 6 x = 4q 4 4q 5 4q 8 8q 9 8q 0 8q 8q 4q 4 8q 5 Using these in the first the second factor of (36) we see that First factor = G(q) = 4q 8 q 0 q q 8q 3 4q 5 Second factor = H(q) = q 3 3q 4 5q 6 3q 8 7q 9 4q 0 5q 4q 9q 3 3q 4 8q 5

8 0 KR Vasuki G Sharath / J Math Anal Appl 396 (0) 3 0 Clearly q 3 H(q) does not tend to zero as q 0 as q 3 G(q) tends to zero as q 0 Hence by analytic continuation G(q) = 0 Thus we have x = ( 4 ) ( 4 ) This completes the proof of the lemma Lemma 3 We have = ( )( ) ( )( ) (37) roof Multiplying (4) by ( ) we obtain This implies 4 4 ( ) ( ) ( ) ( ) = 0 [ 3 3 ] = [ 3 3 ] From this (37) follows roof of (7) We have ST = K Multiplying (33) (37) then using the above we obtain 4 ± (ST) = ( ) ( ) F ; x For q = exp π ;x F one can see that q is real 0 < q < Hence ST z = z if Re(z) is positive thus we choose positive sign in (39) This proves (7) roof of (8) Dividing (33) by (37) then using (38) we obtain (8) Acknowledgments (38) (39) is always positive We know that The authors thank the referee for his valuable comments The authors are grateful to DST New Delhi for awarding research project [No SR/S4/MS:57/08] under which this work has been done References [] S Ramanujan Notebooks ( Volumes) Tata Institute of Fundamental Research Bombay 957 [] BC Berndt Ramanujan s Notebooks art III Springer-Verlag New York 99 [3] H Schröter De aequationibus modularibus Dissertatio Inauguralis Albertina Litterarum Universitate Regiomonti 854 [4] S Ramanujan The Lost Notebook Other Unpublished apers Narosa New Delhi 988 [5] BC Berndt Modular equations in Ramanujan s lost notebook in: R Bambah VC Dumir RJ Hans-Gill (Eds) Number Theory Hindustan Book Agency New Delhi 000 pp [6] BC Berndt Ramanujan s Notebooks art IV Springer-Verlag New York 994 [7] BC Berndt Ramanujan s Notebooks art V Springer-Verlag New York 998 [8] ND Baruah On some of Ramanujan s Schlafli-type mixed modular equations J Number Theory 00 (003) [9] S Bhargava C Adiga MSM Naika A new class of modular equations in Ramanujan s alternative theory of elliptic functions of signature 4 some new eta-function identities Indian J Math 45 () (003) 3 39