RAMANUJAN S MOST BEAUTIFUL IDENTITY

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1 RAMANUJAN S MOST BEAUTIFUL IDENTITY MICHAEL D. HIRSCHHORN Abstract We give a simple proof of the identity which for Hardy represented the best of Ramanujan. On the way, we give a new proof of an important identity that Ramanujan stated but did not prove. Of all the 4000 or so identities Ramanujan presented, Hardy chose one which for him represented the best of Ramanujan. I would like to show you this identity, and prove it. Following Euler, we define a partition of the positive integer n as a representation of n as a sum of positive integers, in which order is unimportant. The partitions of 4 are 4 = 3+1 = 2+2 = = The number of partitions of n is denoted by p(n); thus, p(4) = 5. For convenience, we define p(0) = 1. Euler showed that the partition generating function P(q) = p(n)q n = 1 + 1q + 2q 2 + 3q 3 + 5q 4 + satisfies where P(q) = 1 (q;q), (a;q) = (1 aq n ). He also showed that (q;q) = 1 q q 2 + q 5 + q 7 q 12 q In the series on the right, the terms occur in pairs, alternately with coefficients 1 and +1. The powers, 1, 2, 5, 7, 12, 15,..., are known as the 1

2 2 MICHAEL D. HIRSCHHORN pentagonal numbers, and Euler s expansion above as the pentagonal number theorem. The easiest way of generating the pentagonal numbers is to proceed as follows. The triangular numbers 1, 1 + 2, and so on are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78,.... Two of every three are divisible by 3. If we divide these by 3 we obtain the pentagonal numbers! A beautiful combinatorial proof of Euler s pentagonal number theorem was given by F. Franklin in 1881, and is reproduced in Hardy and Wright [3]. Euler s pentagonal number theorem is the special case a = 1 of Jacobi s triple product identity when written (a 1 q;q 3 ) (aq 2 ;q 3 ) (q 3 ;q 3 ) = 1 a 1 q aq 2 + a 2 q 5 + a 2 q 7 a 3 q 12 a 3 q A proof of the triple product identity is given in the Appendix. Also, it is important that I mention that Ramanujan found a marvellous and powerful extension of the triple product identity. Back to the main story: We have that the partition generating function is the reciprocal of the series 1 q q 2 + q 5 + q This implies p(0) = 1, p(1) p(0) = 0, p(2) p(1) p(0) = 0, p(3) p(2) p(1) = 0, and, more generally, for n > 0, p(n) p(n 1) p(n 2) + p(n 5) + p(n 7) + + = 0.

3 RAMANUJAN S MOST BEAUTIFUL IDENTITY 3 Here, you recognise the pattern of + and signs, and the numbers 1, 2, 5, 7,.... For each n, the sum on the left terminates, since all terms with negative argument are zero. P. MacMahon, who was in Cambridge with Hardy and Ramanujan, used the above recurrence to calculate p(n) for n 200, and serendipitously listed the values in groups of five thus: Ramanujan observed that the numbers at the bottom of each group are divisible by 5, that is, 5 p(5n + 4). He also observed that 7 p(7n + 5), 11 p(11n + 6), and, on the basis of the very small amount of evidence provided by MacMahon s table, formulated a very general conjecture, which was essentially correct; the proof was completed by Oliver Atkin in Ramanujan did much more than prove 5 p(5n + 4). We can write down the generating function of the p(5n + 4), p(5n + 4)q n = q + 135q q q q 5 +. Ramanujan [4, p.213] makes the claim that this series can be written as a neat product, (1) p(5n + 4)q n = 5 (q5 ;q 5 ) 5 (q;q) 6. Hardy [3, p. xxxv] says of (1): It would be difficult to find more beautiful formulae than the Rogers-Ramanujan identities, but here Ramanujan must take second place to Rogers; and, if I had to select one formula from all

4 4 MICHAEL D. HIRSCHHORN Ramanujan s work, I would agree with Major MacMahon in selecting [(1)]. Hence, I refer to (1) as Ramanujan s most beautiful identity. In short (1) says, if 1 (q;q) = p(n)q n then p(5n + 4)q n = 5 (q5 ;q 5 ) 5. (q;q) 6 Note that this formulation does not require that we have a combinatorial interpretation of the p(n); the statement can be regarded as purely algebraic. (2) My aim is to sketch a proof of the wonderful identity (1). We can write, with ω 1 a fifth root of unity, 1 (q;q) = (ωq;ωq) (ω 2 q;ω 2 q) (ω 3 q;ω 3 q) (ω 4 q;ω 4 q) (q;q) (ωq;ωq) (ω 2 q;ω 2 q) (ω 3 q;ω 3 q) (ω 4 q;ω 4 q). The denominator of the right-hand side of (2) is (1 q n )(1 ω n q n )(1 ω 2n q 2n )(1 ω 3n q 3n )(1 ω 4n q 4n ) = (1 q n ) 5 (1 q n )(1 ωq n )(1 ω 2 q n )(1 ω 3 q n )(1 ω 4 q n ) 5 n 5 n = (1 q 5n ) 5 (1 q 5n ) 5 n = (1 q 5n ) 5 (1 q 5n ) / (1 q 5n ) = (q 5 ;q 5 ) 6 /(q 25 ;q 25 ), 5 n and (2) becomes (3) 1 (q;q) = (ωq;ωq) (ω 2 q;ω 2 q) (ω 3 q;ω 3 q) (ω 4 q;ω 4 q) (q 5 ;q 5 ) 6 /(q 25 ;q 25 ). Now we deal with the numerator of the right-hand side of (3). The pentagonal numbers are congruent to 0, 1, or 2 modulo 5, and we can write (q;q) = ( 1 + q 5 q 15 q 35 q 40 q 70 ) q ( 1 q 25 q 50 + q 125 ) q 2 ( 1 q 5 + q 10 q 20 q 55 + q 75 ).

5 RAMANUJAN S MOST BEAUTIFUL IDENTITY 5 Ramanujan [4, p. 212] simply states It can be shewn that (4) (q;q) = (q10 ;q 25 ) (q 15 ;q 25 ) (q 25 ;q 25 ) (q 5 ;q 25 ) (q 20 ;q 25 ) q(q 25 ;q 25 ) q 2(q5 ;q 25 ) (q 20 ;q 25 ) (q 25 ;q 25 ) (q 10 ;q 25 ) (q 15 ;q 25 ). Hardy regarded this as a gap in Ramanujan s proof of (1), and said [2, p. 90] Ramanujan never gave a complete proof. Here is a simple proof of (4) that I found recently. The triple product identity can be written (a 1 ;q) (aq;q) (q;q) = ( 1) n a n q (n2 +n)/2, or n= (1 a 1 )(a 1 q;q) (aq;q) (q;q) = ( 1) n (a n a n 1 )q (n2 +n)/2. If we suppose a 1 and divide by 1 a 1, we obtain (a 1 q;q) (aq;q) (q;q) = ( a ( 1) n n a n 1 ) 1 a 1 q (n2 +n)/2 = ( ) ( 1) n a n+1 2 a n 1 2 q (n2 +n)/2. a 1 2 a 1 2 If we now set a = exp 2iθ, we obtain 2cos 2θ q (1 n + q 2n )(1 q n ) = ( 1) nsin (2n + 1)θ q (n2 +n)/2. sin θ In particular, if θ = 2π 5, (5) (1 + αq n + q 2n )(1 q n ) = (q 10 ;q 25 ) (q 15 ;q 25 ) (q 25 ;q 25 ) βq(q 5 ;q 25 ) (q 20 ;q 25 ) (q 25 ;q 25 ),

6 6 MICHAEL D. HIRSCHHORN while if θ = π 5, (6) (1 + βq n + q 2n )(1 q n ) = (q 10 ;q 25 ) (q 15 ;q 25 ) (q 25 ;q 25 ) αq(q 5 ;q 25 ) (q 20 ;q 25 ) (q 25 ;q 25 ), where α = and β = 1 5 2, and where we have used the triple product identity to sum the series that arise. By way of explanation, when we substitute θ = 2π 5, the terms of the sum in which n 0 or 1 (mod 5) are ( 1) 5m sin (10m + 1)2π 5 m 0 sin 2π 5 q (25m2 +5m)/2 + m 1( 1) 5m 1 sin (10m 1)2π 5 sin 2π 5 q ((5m 1)2 +(5m 1))/2 = ( 1) m q (25m2 +5m)/2 + m 0 m 1 = m= ( 1) m q (25m2 +5m)/2 = (q 10 ;q 25 ) (q 15 ;q 25 ) (q 25 ;q 25 ) ( 1) m q (25m2 5m)/2 by the triple product identity. In the same way, we can sum the terms corresponding to n 1 and n 2 (mod 5), and the terms corresponding to n 2 (mod 5) vanish. Thus we obtain (5). We can handle (6) similarly. If we multiply (5) and (6) together, we obtain (q;q) (q 5 ;q 5 ) = (q 10 ;q 25 ) 2 (q 15 ;q 25 ) 2 (q 25 ;q 25 ) 2 q(q 5 ;q 5 ) (q 25 ;q 25 ) q 2 (q 5 ;q 25 ) 2 (q 20 ;q 25 ) 2 (q 25 ;q 25 ) 2, and if we now divide by (q 5 ;q 5 ) we obtain Ramanujan s result (4).

7 RAMANUJAN S MOST BEAUTIFUL IDENTITY 7 We now complete the proof of (1). If we write a = (q10 ;q 25 ) (q 15 ;q 25 ) (q 5 ;q 25 ) (q 20 ;q 25 ) and b = (q5 ;q 25 ) (q 20 ;q 25 ) (q 10 ;q 25 ) (q 15 ;q 25 ) = a 1, then (4) becomes (q;q) = (q 25 ;q 25 ) (a q q 2 b) and the numerator of the right-hand side of (3) is (ωq;ωq) (ω 2 q;ω 2 q) (ω 3 q;ω 3 q) (ω 4 q;ω 4 q) = (q 25 ;q 25 ) 4 (a ωq ω 2 q 2 b)(a ω 2 q ω 4 q 2 b)(a ω 3 q ω 6 q 2 b)(a ω 4 q ω 8 q 2 b) = (q 25 ;q 25 ) 4 ( (a 4 q 5 (2ab 2 + b)) + q(a 3 + q 5 (ab 3 + b 2 )) + q 2 ((a 3 b + a 2 ) q 5 b 3 ) ) +q 3 ((2a 2 b + a) + q 5 b 4 ) + q 4 (a 2 b 2 + 3ab + 1) = (q 25 ;q 25 ) 4 ( (a 4 3q 5 b) + q(a 3 + 2q 5 b 2 ) + q 2 (2a 2 q 5 b 3 ) + q 3 (3a + q 5 b 4 ) + 5q 4). So (3) becomes p(n)q n = (q25 ;q 25 ) 5 (q 5 ;q 5 ) 6 ( (a 4 3q 5 b) + q(a 3 + 2q 5 b 2 ) + q 2 (2a 2 q 5 b 3 ) + q 3 (3a + q 5 b 4 ) + 5q 4). If we extract those terms in which the powers are 4 modulo 5, we obtain or which is (1). p(5n + 4)q 5n+4 = 5q 4(q25 ;q 25 ) 5 (q 5 ;q 5 ) 6, p(5n + 4)q n = 5 (q5 ;q 5 ) 5 (q;q) 6,

8 8 MICHAEL D. HIRSCHHORN Appendix: The triple product identity We have ( aq;q 2 ) = (1 + aq)( aq 3 ;q 2 ). If we write ( aq;q 2 ) = a k c k (q) k 0 then c 0 (q) = 1 and It follows that for k 1, k 0 a k c k (q) = (1 + aq) k 0 a k q 2k c k (q). c k (q) = q 2k c k (q) + q 2k 1 c k 1 (q), or c k (q) = q2k 1 1 q 2k c k 1(q). It follows that, if we write (a;q) k = (1 a)(1 aq) (1 aq k 1 ) for k 1, (a;q) 0 = 1, c k (q) = qk2 (q 2 ;q 2 ) k and ( aq;q 2 ) = k 0 a k q k2 (q 2 ;q 2 ) k. This is an identity of Euler. It is now an easy induction to prove that ( a 1 q;q 2 ) n ( aq;q 2 ) = If we now let n, we obtain k= n a k q k2 (q 2 ;q 2 ) k+n. or ( a 1 q;q 2 ) ( aq;q 2 ) = k= ( a 1 q;q 2 ) ( aq;q 2 ) (q 2 ;q 2 ) = a k q k2 (q 2 ;q 2 ), k= a k q k2,

9 RAMANUJAN S MOST BEAUTIFUL IDENTITY 9 and this is the triple product identity. Note that the triple product can be written in the equivalent forms (a 1 q;q 3 ) (aq 2 ;q 3 ) (q 3 ;q 3 ) = ( 1) k a k q (3k2 +k)/2 k= and (a 1 ;q) (aq;q) (q;q) = ( 1) k a k q (k2 +k)/2 k= under the substitutions (a,q) ( aq 1 2,q 3 2) and (a,q) ( aq 1 2,q 1 2) respectively. Postscript The marvellous and powerful extension of the triple product identity that I alluded to earlier is known technically as Ramanujan s 1 ψ 1 - summation. I recommend that the reader interested in learning more about Ramanujan s work might begin by consulting Bruce Berndt s book [1]. References [1]. B. C. Berndt, Number Theory in the Spirit of Ramanujan, Student Mathematical Library, vol. 34, American Mathematical Society, Providence, RI, [2] G. H. Hardy, Ramanujan, Twelve Lectures on Subjects Suggested by His Life and Work, AMS Chelsea, Providence, RI, [3] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, Oxford, [4] S. Ramanujan, Collected Papers, G. H. Hardy, P. V. Seshu Aiyar, B. M. Wilson, eds., AMS Chelsea, Providence, RI, 2000.

10 10 MICHAEL D. HIRSCHHORN School of Mathematics and Statistics, University of New South Wales, Sydney, Australia 2052