A quasi-theta product in Ramanujan s lost notebook

Transcription

1 Math. Proc. Camb. Phil. Soc. 2003, 35, c 2003 Cambridge Philosophical Society DOI: 0.07/S Printed in the United Kingdom A quasi-theta product in Ramanujan s lost notebook By BRUCE C. BERNDT Department of Mathematics, University of Illinois, 409 West Green Street, Urbana, IL 680, U.S.A. berndt@math.uiuc.edu HENG HUAT CHAN Department of Mathematics, National University of Singapore, 2 Science Dr. 2, Singapore 7543, Republic of Singapore. chanhh@math.nus.edu.sg and ALEXANDRU ZAHARESCU Department of Mathematics, University of Illinois, 409 West Green Street, Urbana, IL 680, U.S.A. zaharesc@math.uiuc.edu Received September 200; revised 8 January 2002 Abstract On page 209 of his lost notebook, Ramanujan records an unusual product formula, reminiscent of a product of theta functions. The formula involves hypergeometric functions and has a connection with elliptic functions. In this paper, we prove the formula, offer some generalizations, and indicate some further connections with Ramanujan s work.. Introduction At the top of page 209 in the Narosa edition of his lost notebook [5], Ramanujan recorded the enigmatic formula, { n q 2n+/2 2n+ } log q { + n iq n n } 2πi where + n q 2n+/2 n iq n n= π 2 =exp 4 k 3F 2,, ; 3, 3; k F, ;;k2, 2 2 q =exp πk /K, q =exp πk/k and 0 <k<. 2 Because of poor photocopying, is very difficult to read in [5]. If the powers 2n + and n on the two pairs of large parentheses were absent, the products could Research partially supported by grant MDA from the National Security Agency.

2 2 B. C. Berndt, H. H. Chan and A. Zaharescu be expressed in terms of theta functions. Ramanujan did not use the notation 3 F 2 and 2 F for hypergeometric functions, but instead only recorded the first three terms of each series. Also, Ramanujan did not divulge the meaning of the notations K and K.However, from considerable work in both the ordinary notebooks [4]andthelost notebook [5], we can easily deduce that K denotes the complete elliptic integral of the first kind defined by π/2 dϕ K Kk 0 k2 sin 2 ϕ, where k, 0 <k<, denotes the modulus. Furthermore, K = Kk, where k k2 is the complementary modulus. There are no other formulas like in Ramanujan s work, and apparently there are none like it in the literature either. The purpose of this paper is to prove. As will be seen in our proof, the unique character of derives from a single, almost miraculous, connection with the theory of elliptic functions given in the identity 2n + 2 cosh{2n +α/2} = k 2z 3 F 2,, ; 3 2, 3 2 ; k2, 3 where α = πk /K and z = 2 F 2, 2 ;;k2. The identity 3 is found in entry 6 of chapter 8 in Ramanujan s second notebook [4], [, p. 53]. Like many of Ramanujan s discoveries, 3 is not fully understood. Is this connection between hypergeometric series and elliptic functions a singular accident, or are there deeper, still to be recognized connections? In his notebooks [4, p. 280], Ramanujan also attempted to find a formula similar to 3, but with 2n + 2 replaced by 2n + 4. In fact, Ramanujan struck out his imprecisely stated formula by putting two lines through it. See [2, pp ] for Berndt s failed attempt to find a correct version. In Section2, we first establish intheorem 2 an equivalent formulation of as an identity amongst infinite series of hyperbolic trigonometric functions. Secondly, we prove this identity. In Section 3, we briefly indicate generalizations of and Theorem 2 andoffer some related hyperbolic series of Ramanujan. 2. An equivalent formulation of in terms of hyperbolic series Theorem 2. Let α and β be any complex numbers with nonzero real parts and with αβ = π 2.Then is equivalent to the identity α sinh{2n +α/2} 2n +cosh 2 {2n +α/2} + π n 2n +cosh 2 {2n +β/2} = π n + 2 cosh{2n +α/2}. 2 Proof. We assume that α and β are positive real numbers. The general result will then follow by analytic continuation. Taking logarithms on both sides of, we

3 Quasi-theta product 3 find that { log n q 2n+/2 2n+ } log q { +log + n iq n n } 2πi + n q 2n+/2 n iq n n= = π2 4 k 3F 2,, ; 3, 3 ; k F,. 2 2 ;;k2 2 2 Here and in the following step, we have ignored branches of the logarithm. The justification lies in our eventual proof of 2. For brevity, let L and R denote, respectively, the left-hand and right-hand sides of 2 2. Then L =logq 2n +{ log n q 2n+/2 log + n q 2n+/2 } +2πi n{log + n iq n log n iq n } n= log qs S 2 +2πiS 3 S Recall that q and q are defined in 2. Set α = πk /K and β = πk/k,sothat αβ = π 2.Wenow proceed to show that S,...,S 4 can be expressed as sums of hyperbolic functions. Using the Taylor series of log + z about z =0and recalling the definition of β, we find that S 3 n m+mn i m e βmn m n= m= i m n{ e β m } n m m= n= ie β m m e β m By a similar calculation, m= S 4 m= Combining 2 4 and 2 5, we find that ie β m m e β m ie β m + ie β m S 3 S 4 = m e β m 2 m= m e 2m+β 2i 2m + + e 2m+β 2 i 2 m 2m +cosh 2 {2m +β/2}. 2 6 Next, again using the Taylor series of log + z about z = 0 and recalling the

4 4 B. C. Berndt, H. H. Chan and A. Zaharescu definition of α, wefind that S 2n + mn e α2n+m/2 m m= 2n mn e α2n+m/2 + mn e α2n+m/2 m m= 2 m e 3αm/2 m e α m + e αm/2 2 e α m m= m e 3αm/2 + e αm/ m e α m 2 m= By an analogous argument, S 2 m= Thus, combining 2 7 and 2 8, we deduce that S S 2 = e 3αm/2 + m e αm/2 m e α m m e 3αm/2 e αm/2 + e 3αm/2 + m e αm/2 m= =2 m e α m 2 e 32m+α/2 e 2m+α/2 2m + + e 2m+α 2 e 2m+α/2 e 2m+α/2 =2 2m +e 2m+α/2 + e 2m+α/2 2 sinh{2m +α/2} 2m +cosh 2 {2m +α/2}. 2 9 If we use 2 6 and 2 9 in 2 3 and recall that log q α, wededuce that sinh{2m +α/2} α 2m +cosh 2 {2m +α/2} + π m 2m +cosh 2 {2m +β/2} m =0 m =0 = π2 4 k 3F 2,, ; 3, 3 ; k F,. 2 0 ;;k2 2 2 We now invoke 3. If we substitute 3 into 2 0, we deduce 2 to complete the proof. It should be emphasized that the only time we used the definitions 2 of q and q in our proof is in the application of 3. Thus, it would seem that is a very special result in that there are likely very few if any other results like it. We now prove 2. Proof of 2. Our first main idea is to introduce the functions F and G in 2 and 2 3, respectively, and use them to find a simpler identity which is equivalent

5 to 2. Define Then Set Gβ 2β F α α Quasi-theta product 5 F α α 2α n + 2 cosh{2n +α/2}. 2 sinh{2n +α/2} 2n +cosh 2 {2n +α/2} n + 2 cosh{2n +α/2} 2n + 2 cosh{2n +π 2 /2β} =2π2 F π 2 β, 2 3 by 2 and the fact that αβ = π 2.Thus, by 2 2, π G β =2π 2 F 2 π2 sinh{2n +π 2 /2β} = α β β 2 2n +cosh 2 {2n +π 2 /2β} +2 2n + 2 cosh{2n +π 2 /2β}. 2 4 If we define then Hβ βπ2 4 2π n tanh{2n +β/2} 2n + 2, 2 5 H β= π2 4 π n 2n +cosh 2 {2n +β/2}. 2 6 In view of 2 4 and 2 6, we see that 2 is equivalent to It follows that for some constant c, G β =H β. Gβ =Hβ+c. 2 7 Clearly, from the definitions of Gβ andhβ in2 3 and 2 5, respectively, both Gβ andhβ tendto0asβ 0. Thus, in 2 7, c =0. Hence, it now suffices to prove that 2β 2n + 2 cosh{2n +π 2 /2β} = βπ2 4 2π n tanh{2n +β/2} 2n

6 6 B. C. Berndt, H. H. Chan and A. Zaharescu It is easily seen that 2 8 is equivalent to β 2n + 2 cosh{2n +π 2 /2β} + π n= n= n tanh{2n +β/2} 2n + 2 βπ2 4 = The second primary idea is to introduce a function f of a complex variable and use contour integration to prove 2 9. To that end, define, for fixed η>0, fz tanηz z 2 cosh z The function fz ismeromorphic in the entire complex plane with a simple pole at z =0andsimplepoles at z =2n +πi/2 andz =2n +π/2η for each integer n. Let γ Rm be a sequence of positively oriented circles centred at the origin and with radii R m tending to as m,wheretheradiir m are chosen so that the circles remain at a bounded distance from all the poles of fz. From the definition 2 20 of f, itistheneasy to see that fzdz η γ Rm R m, 2 2 as R m. For brevity, let Radenote the residue of fzatapolea.then, brief calculations show that R0 = η, n +π 4η R 2η π 2 2n + 2 cosh{2n +π/2η}, n +πi R 4 n tanh{2n +πη/2}, π 2 2n + 2 for each integer n. Hence, using and the residue theorem, we deduce that 4η fzdz = η 2πi γ Rm π 2 2n + 2 cosh{2n +π/2η} 2n+ <2ηR m/π 2n+ <2R m/π 4 n tanh{2n +πη/2} π 2 2n Letting R m tend to in 2 25 and employing 2 2, we conclude that 0=η 4η π 2 4 π 2 n= n= 2n + 2 cosh{2n +π/2η} n tanh{2n +πη/2} 2n Now set η = β/π in Then multiply both sides by π 3 /4. We then readily obtain 2 9, and so this completes the proof.

7 Quasi-theta product 7 3. Concluding remarks Theorem 2 can easily be generalized in at least two directions. First, in the proof of 2, we could replace fz by f n z tanηz z n cosh z, where η is a positive integer exceeding. The generalization of 2 would then involve Bernoulli numbers arising from the Taylor expansion of tan z about z =0 and Euler numbers arising from the expansion of /cosh z about z =0. Second, in the proof of 2, we could replace fz by fz,θ coshθztanηz, z 2 cosh z where <θ<. Then by a proof analogous to that given above, we can deduce that, for any complex numbers α and β with Re α, Reβ 0andαβ = π 2,andfor any real number θ with θ <, sinh{2n +α/2} cosh{2n +θα/2} α 2n +cosh 2 θα {2n +α/2} n cos{2n +πθ/2} + π 2n +cosh 2 {2n +β/2} = π2 4 2 sinh{2n +θα/2} 2n +cosh{2n +α/2} cosh{2n +θα/2} 2n + 2 cosh{2n +α/2}. The identity 3 is equivalent to n q 2n+ θ/2 n q 2n++θ/2 2n+ + n q 2n+ θ/2 + n q 2n++θ/2 { { n= log q/2 n q 2n+ θ/2 + n q 2n++θ/2 } θlog q/2 + n q 2n+ θ/2 n q 2n++θ/2 + n ie θπi/2 q n + n ie θπi/2 q n n } πi n ie θπi/2 q n n ie θπi/2 q n 3 π 2 =exp 4 2 cosh{2n +θα/2} n + 2 cosh{2n +α/2} When θ =0,3 and 3 2 reduce to 2 and, respectively. If α, β > 0and θ u + iv, whereu and v are real, then 3 2 can be analytically continued to the rectangle <u<, 2π/α < v < 2π/α. If we differentiate 3 2k times with respect to θ and then setθ =0,wededuce that 2n + 2k sinh{2n +α/2} α cosh 2 + k β 2k π 2k n 2n + 2k {2n +α/2} cosh 2 {2n +β/2} =4k 2 2n + 2k 2 cosh{2n +α/2}, 3 3

8 8 B. C. Berndt, H. H. Chan and A. Zaharescu which is valid for any integer k andanycomplex numbers α and β with Re α, Re β 0andαβ = π 2. If we let α or β 0 in 2, we deduce Leibniz s well-known evaluation n 2n + = π 4, while if we let α 0orβ in2, we deduce Euler s well-known evaluation 2n + = π We remark that care must be taken when taking certain limits inside summation signs above. Ramanujan examined several other infinite series of hyperbolic functions in [4] and [5]. We cite two examples giving evaluations of series involving cosh z which are very similar to those arising above. First, in entry 6x of chapter 7 in his second notebook [4], [, p. 34], Ramanujan asserted that 2n + 2 cosh{2n +π/2} = π 3/2 2 2Γ In fact, it is shown in [, pp ] that one can also evaluate in closed form the more general sum 2n + 2m cosh{2n +α/2}, 3 5 where m is a positive integer. However, the evaluations are in terms of z 2F, 2 2 ;;k2 see[, p.0, eq. 6 3] for the relation between α and k, where in [], y = α. Note that the sums 3 5 appear on the right-hand side of 3 3, and so these evaluations also automatically yield evaluations for the left-hand side of 3 3. Second, the evaluation 2n + 2 cosh 2 {2n +π/2} = π 2 2Γ arises in Ramanujan s formulas for the power series coefficients of the reciprocals, or, more generally, quotients, of certain Eisenstein series [3, cor. 3 9]. In a future paper, the authors will study a multi-variable generalization of the products in and derive a transformation formula for them. REFERENCES [] B. C. Berndt. Ramanujan s notebooks, Part III Springer-Verlag, 99. [2] B. C. Berndt. Ramanujan s notebooks, Part V Springer-Verlag, 998. [3] B. C.Berndt, P.R.Bialek and A. J.Yee. Formulas of Ramanujan for the power series coefficients of certain quotients of Eisenstein series. Internat. Math. Res. Not , [4] S. Ramanujan. Notebooks 2 volumes Tata Institute of Fundamental Research, 957. [5] S. Ramanujan. The lost notebook and other unpublished papers Narosa, 988.