Ramanujan's Lost Notebookin Five VolumesThoughts and

Transcription

1 Ramanujan's Lost Notebook in Five Volumes Thoughts and Comments SLC (KrattenthalerFest) 2018 September 2018 This talk is dedicated to my good friend Christian Krattenthaler

2 I. BACKGROUND

3

4 His memory, and his powers of calculation, were very unusual, but they could not be reasonably be called abnormal. If he had to multiply two very large numbers, he multiplied them in the ordinary way; he would do it with unusual rapidity and accuracy, but not more rapidly than any mathematician who is naturally quick and has the habit of computation. There is a table of partitions at the end of our paper.... This was, for the most part, calculated independently by Ramanujan and Major MacMahon; and Major MacMahon was, in general the slightly quicker and more accurate of the two.

5

6 THE PATH FROM THE LAST LETTER TO THE LOST NOTEBOOK

7 He returned from England only to die, as the saying goes. He lived for less than a year. Throughout this period, I lived with him without break. He was only skin and bones. He often complained of severe pain. In spite of it he was always busy doing his Mathematics. That, evidently helped him to forget the pain. I used to gather the sheets of papers which he lled up. I would also give the slate whenever he asked for it. He was uniformly kind to me. In his conversation he was full of wit and humour. Even while mortally ill, he used to crack jokes. One day he conded in me that he might not live beyond thirty-ve and asked me to meet the event with courage and fortitude. He was well looked after by his friends. He often used to repeat his gratitude to all those who had helped him in his life.

8 UNIVERSITY OF MADRAS, 12th January I am extremely sorry for not writing you a single letter up to now.... I discovered very interesting functions recently which I call Mock θ-functions. Unlike the False θ-functions (studied partially by Prof. Rogers in his interesting paper) they enter into mathematics as beautifully as the ordinary θ-functions. I am sending you with this letter some examples.... Mock θ-functions φ(q) = 1 + ψ(q) = q 1 + q 2 + q 4 (1 + q 2 )(1 + q 4 ) +, q 1 q + q 4 (1 q)(1 q 3 ) + q 9 (1 q)(1 q 3 )(1 q 5 ) +.

9

10

11 When the Royal Society asked me to write G. N. Watson's obituary memoir I wrote to his widow to ask if I could examine his papers. She kindly invited me to lunch and afterwards his son took me upstairs to see them. They covered the oor of a fair sized room to a depth of about a foot, all jumbled together, and were to be incinerated in a few days. One could make lucky dips and, as Watson never threw anything away, the result might be a sheet of mathematics but more probably a receipted bill or a draft of his income tax for By extraordinary stroke of luck one of my dips brought up the Ramanujan material which Hardy must have passed on to him when he proposed to edit the earlier notebooks.

12

13

14

15

16 THE MOCK THETA CONJECTURES

17 f 0(q) = 1 + q 1 + q + q 4 (1 + q)(1 + q 2 ) +, φ 0(q) = 1 + q(1 + q) + q 4 (1 + q)(1 + q 3 ) + q 9 (1 + q)(1 + q 3 )(1 + q 5 ) +, ψ 0(q) = q + q 3 (1 + q) + q 6 (1 + q)(1 + q 2 ) + q 10 (1 + q)(1 + q 2 )(1 + q 3 ) +, F 0(q) = 1 + χ 0(q) = 1 + q2 1 q + q 8 (1 q)(1 q 3 ) +, q 1 q 2 + q 2 (1 q 3 )(1 q 4 ) + q 3 (1 q 4 )(1 q 5 )(1 q 6 ) + = 1 + q 1 q + q 3 (1 q 2 )(1 q 3 ) + q 5 (1 q 3 )(1 q 4 )(1 q 5 ) +

18

19 φ 0( q) = +1 (1 q 5n+5 )(1 + q 5n+2 )(1 + q 5n+3 ) (1 q 10n+2 )(1 q 10n+8 ) q 5n2 (1 q)(1 q 6 ) (1 q 5n+1 )(1 q 4 )(1 q 9 ) (1 q 5n 1 ). φ 0( q) = +1 (1 q 5n+5 )(1 + q 5n+2 )(1 + q 5n+3 ) (1 q 10n+2 )(1 q 10n+8 ) { (1 q 5n+5 ) q + (1 q 1 ) n=1 } ( 1) n q n(15n+5)/2 (1 + q 5n ). (1 q 5n+1 )(1 q 5n 1 )

20 EQUIVALENTLY (1 q 10n q n2 ) (1 + q) (1 + q n ) n=1 = 2 n= n= ( 1) n q 2n 1 q 5n+1 ( 1) n q 15n2 +15n+2 1 q 10n+2

21 MOCK THETA CONJECTURES r a (n) = # OF PTNS OF n WITH RANK a( mod 5) rank = (largest part) (# of parts) The rank of is 5 6 = 1.

22 1ST MOCK CONJECTURE (Proved by Hickerson) r 1 (5n) r 0 (5n) EQUALS THE # OF PTNS OF n IN WHICH THE LARGEST PART IS ODD AND EACH PART IS 1 (LARGEST PART) 2

23 SUMS-OF-TAILS

24

25 OR IN STANDARD NOTATION IF S = n=1 (1 + qn ) = n=1 THEN 1 (1 q 2n 1 ) q n(n+1)/2 (1 + q)(1 + q 2 ) (1 + q n ) =1 = q ( q) n (1 q)(1 q 2 ) (1 q n ) =2{ S 2 + (S (1 + q)(1 + q 2 ) (1 + q n q n )) 2S 1 q n n=1 =2{ S 2 + (S n=1 1 (1 q)(1 q 3 ) (1 q 2n+1 ) ) 2S n=1 q 2n 1 q 2n

26 LET R(q) := q n(n+1)/2 (1 + q)(1 + q 2 ) (1 + q n ) =1 + q ( q) n (1 q)(1 q 2 ) (1 q n )

27 Dyson, Hickerson & I proved that if R(q) = c n q n, then asymptotically 100% of the c n are ZERO, and for any integer M between and, c n = M innitely many times.

28 ALSO IT IS INFORMATIVE TO CONTRAST R(q) = 1 + n=1 q n(n+1) 2 (1 + q)(1 + q 2 ) (1 + q n ) with 1 + n=1 q n(n+1) 2 (1 + q)(1 + q 2 ) (1 + q n )

29 ALSO IT IS INFORMATIVE TO CONTRAST R(q) = 1 + n=1 q n(n+1) 2 (1 + q)(1 + q 2 ) (1 + q n ) with 1 + n=1 q n(n+1) 2 (1 + q)(1 + q 2 ) (1 + q n ) = 2

30 NATHAN FINE, IN HIS WONDERFUL, RAMANUJANESQUE BOOK, BASIC HYPERGEOMETRIC SERIES AND APPLICATIONS, INDEPENDENTLY FOUND ONE RESULT OF THIS NATURE: LET THEN OR Φ(q) = n=1 q n 1 q n = ( 1) n 1 q n(n+1) 2 (1 q) (1 q n 1 )(1 q n ) Φ(q) = Φ(q) n=1 (1 qn ) = j=0 n=1 1 (1 q j ) ( j>n 1 k=1 (1 qk ) j k=1 ) 1 (1 q k )

31 SIMILARLY, D. ZAGIER HAS PROVED n (1 q j ) (1 q j ) where H(q) = n 0 j=1 j=1 = 1 2 H(q) + D(q) (1 q j ), n=1 j=1 nχ(n)q (n2 1)/24 χ the unique primitive character of conductor 12.

32 ZAGIER STARTS FROM (1 q j ) = j=1 n (1 q j ) j=1 (1 q)(1 q 2 ) (1 q n 1 )nq n n=1 WHICH MAY BE PROVED DIRECTLY BY EXAMINING THE PARTIAL SUMS ON EACH SIDE.

33 THIS IS EXACTLY ANALOGOUS TO THE TREATMENT OF RAMANUJAN'S FORMULA n S (1 + q j ) = (1 + q)(1 + q 2 ) (1 + q n 1 )nq n j=1 n=1 (Adv. in Math., 61(1986), p. 159). ALSO EASILY PROVED

34 KEN ONO, JORGE JIMENEZ URROZ and I EXAMINED THE FOLLOWING QUESTION: WHAT IS GOING ON HERE?

35 IN EACH CASE, THERE IS A MISSING VARIABLE WHICH HAS BEEN DIFFERENTIATED AND THEN SET = 1. IS THERE A GENERAL PRINCIPLE?

36 THEOREM. SUBJECT TO SOME TEDIOUS BUT NECESSARY CONVERGENCE CONDITIONS, IF lim n α n = α, AND THEN lim z 1 f (z) = z n α n, d dz (1 z)f (z) = (α α n ).

37 LET (a, n) = a(a + 1)(a + 2) (a + n 1) P. FREITAS AND I EXTENDED THIS RESULT TO { } 1 d p p lim (1 z)f (z) z 1 dzp = (n + 1, p 1)(α α n+p 1 )

38 DELIGHTFULLY, THIS GENERAL THEOREM (INSPIRED BY RAMANUJAN) BECAME ESSENTIAL IN JOINT WORK WITH GARVAN AND LIANG CONCERNING spt(n), THE NUMBER OF SMALLEST PARTS IN THE PARTITIONS ON n. FOR EXAMPLE: spt(4) = 10 4, 3+1, 2 + 2, ,

39 spt(n) CAME INTO PROMINENCE WHEN, IN THE SPIRIT OF RAMANUJAN'S CONGRUENCES FOR p(n), IT WAS FOUND THAT spt(5n + 4) 0( mod 5) spt(7n + 5) 0( mod 7) spt(13n + 6) 0( mod 13) THE LAST BEING A GREAT SURPRISE.

40 GARVAN, LIANG & I PROVIDE COMBINATORIAL EXPLANATIONS OF THE CONGRUENCES mod 5 AND 7 AS WELL AS A FULL ACCOUNT OF THE PARITY OF spt(n) (FIRST DONE BY FOLSOM & ONO). CENTRAL TO OUR DEVELOPMENT ARE RESULTS LIKE:

41 ((q) 2n (q) ) (q 2 ; q 2 ) n ( 1) n q n2 = (q; q 2 ) n=1 n ((q) n (q) ) = (q) 2 n n=1 nq n2, (q) 2 n where n 1 (A) n = (A; q) n = (1 Aq j ). j=0

42 THE 40 IDENTITIES FOR THE ROGERS-RAMANUJAN FUNCTIONS AND GENERALIZATIONS

43 LET AND G(q) := H(q) := n=1 n=1 1 (1 q 5n 4 )(1 q 5n 1 ), 1 (1 q 5n 3 )(1 q 5n 2 ). In 1975, Brian Birch uncovered Ramanujan's manuscript on 40 identities relating G(q) and H(q). A typical example is G(q)G(q 4 ) + qh(q)h(q 4 ) = φ(q) (q 2 ; q 2 ) where φ(q) = n= qn2.

44 PRIOR TO THE DISCOVERY OF THE LOST NOTEBOOK, EVERY KNOWN PROOF OF THESE FORMULAE RELIED HEAVILY ON THE MODULAR ASPECTS OF G(q) AND H(q). IN THE LOST NOTEBOOK, WE FIND A NON-MODULAR GENERALIZATION OF THE PREVIOUSLY STATED IDENTITY.

45 NAMELY, ( ) ( a n q n2 (q 4 ; q 4 ) n ( ) ( a n q (n+1)2 + (q 4 ; q 4 ) n 1 = a n q n2 (q 4 ; q 4 ) n= ) a 2n q 4n2 (q 4 ; q 4 ) n a 2n 1 q 4n2 +4n (q 4 ; q 4 ) n ) a = 1 YIELDS THE ORIGINAL FORMULA ONCE WE RECALL SOME RESULTS OF L. J. ROGERS

46 G(q) = q n2 (q; q) n = ( q 2 ; q 2 ) q n2 (q 4 ; q 4 ) n, AND H(q) = q n2 +n (q; q) n = ( q 2 ; q 2 ) q n2 +2n (q 4 ; q 4 ) n

47 SIMILAR GENERALIZATIONS HOLD FOR RELATED RESULTS OF G. N. WATSON G( q)φ(q) G(q)φ( q) = 2qH(q 4 )ψ(q 2 ) AND H( q)φ(q) + H(q)φ( q) = 2G(q 4 )ψ(q 2 ), WHERE φ(q) = ψ(q) = n= q n2 q n(n+1)/2

48 HOW RAMANUJAN PROVED THE RICHMOND-SZEKERES THEOREM

49 ROGERS-RAMANUJAN CONTINUED FRACTION = C(q) 1 + q q 2 q q = (1 q2 )(1 q 3 )(1 q 7 )(1 q 8 )(1 q 12 ) (1 q)(1 q 4 )(1 q 6 )(1 q 9 )(1 q 11 )

50 G. Szekeres & B. Richmond examined the power series of C(q) C(q) = 1 + q q 3 + q 5 + q 6 q 7 2q 8 + 2q q 11 q 12 3q 13 q q q 16 2q 17 5q 18 q q q 21 3q 22 8q 23 2q 24 +

51 LET C(q) = c nq n, R. and S. showed that eventually c n and c n+5 have the same sign by showing c n = { 2 4π e 25 5n (5n) 3/4 cos ( 2π 5 (n 2 ) 5 ) + O( 1 } ). n

52 IN THE LOST NOTEBOOK WE FIND C(q) = ( 1) n q (15n2 +n)/2 (1 + q 5n+1 ) n= (1 q 5n ) n=1 FROM THIS RAMANUJAN FINDS FORMULAE FOR c 5m+j q m j = 0, 1, 2, 3, 4. m=0

53 FOR EXAMPLE c 5m q m m=0 = n=1 n 0,±37( mod 75) + q 4 n=1 n 0,±13( mod 75) 1 1 q n 1 1 q n

54 HOWEVER c 5m+2 q m = m=0 q n=1 n 0,±23( mod 75) + q 8 n 0,±2( mod 75) 1 1 q n 1 1 q n

55 A PARTITION THEOREM OF B. GORDON ALLOWS US TO TREAT THE SIGN PROBLEM FULLY. A k,a (n) = B k,a (n) WHERE A k,a (n) =# OF PTNS OF n INTO PARTS 0, ±a( mod 2k + 1). B k,a (n) =# OF PTNS OF n INTO b 1 + b b j WITH b i b i+1, b i b i+k 1 2, & AT MOST a 1 OF THE b i ARE = 1.

56 FOR 1 < a k A k,a (n) is non-decreasing in n. B k,a (n) is non-decreasing in a. THUS c 5m+2 = B 37,23 (m 1) + B 37,2 (m 8) < 0 for m 1.

57 INNOCENTS ABROAD

58 HERE ARE TWO SEEMINGLY BENIGN IDENTITIES ( 1) n q n(n+1)/2 ( q; q) 2 n = ( q) n(n+1)/2 ( q 2 ; q 2 ) n 2 n=1 ( 1) n q 2n2 ( q; q 2 ) 2n, φ( q) q n(n+1)/2 (q 2 ; q 2 ) n = WHERE φ(q) = n= qn2 ( q) n(n+1)/2 ( q 2 ; q 2 ) n + 2 n=1 ( 1) n q 2n2 ( q; q 2 ) 2n.

59 THE PROOF REQUIRES TWO SEEMINGLY TRIVIAL FINITE IDENTITIES (q; q 2 ) m 2m 2m 1 (q; q 2 ) m ( 1) n (α; q) n (q; q) 2m n = m ( 1) n ( = 1 α ) m (q; q) n (α; q) 2m 1 n q n=1 ( 1) n q n2 (q 2 ; q 2 ) m n (αq; q 2 ) n. ( 1)q n2 (q 2 ; q 2 ) m n (α; q 2 ) n.

60 AND THREE q-hypergeometric SERIES IDENTITIES ( ) a0, a 1,..., a 2 ; q, t r+1φ r = b 1,..., b r (a 0, a 1,..., a r ; q) n t n (q, b 1,..., b r ; q) n

61 lim N 10 φ 9 ( a, q a, q a, b, r1, r 1, r 2, r 2, q N, q N ; q, a3 q 2N+3 ) a, a, aq b, aq r 1, aq r 1, aq r 2, aq r 2, aq N+1, aq N+1 = (a2 q 2 ; q 2 ) (a 2 q 2 /(r1 2r 2 2); q2 ) (a 2 q 2 /r1 2; q2 ) (a 2 q 2 /r2 2; q2 ) (r1 2 ; q2 ) n (r2 2; ( q2 ) n ( aq/b; q) 2n a 2 q 2 (q 2 ; q 2 ) n (a 2 q 2 /b 2 ; q 2 ) n ( aq; q) 2n r1 2r 2 2 ) n br 2 1 r 2 2

62 lim N 10 φ 9 ( a, q 2 a, q 2 a, p 1, p 1 q, p 2, p 2 q, f, q 2N, q 2N+1 ; q 2, a3 q 4N+3 ) (aq; q) (aq/(p 1 p 2 ); q) (aq/p 1 ; q) (aq/p 2 ; q) a, a, aq 2 p 1, aq p 1, aq2 p 2, aq p 2, aq2 f, aq 2N+2, aq 2N+1 ( aq (p 1 ; q) n (p 2 ; q) n (aq/f ; q 2 ) n (q; q) n (aq; q 2 ) n (aq/f ; q) n p 1 p 2 p 2 1 p2 2 f ) n.

63 ( a, q a, q a, b1, c 1, b 2, c 2, b 3, c 3, q N ; q, 10φ 9 a, a, aq, aq, aq, aq, aq, aq, aq b 1 c 1 b 2 c 2 b 3 c N+1 3 a 3 q N+3 b 1 b 2 b 3 c 1 c 2 c 3 = (aq; q) N(aq/(b 3 c 3 ); q) N (aq/b 3 ; q) N (aq/c 3 ; q) N (aq/(b 1 c 1 ); q) m1 (aq/(b 2 c 2 ); q) m2 (b 2 ; q) m1 (c 2 ; q) m1 (q; q) m1 (q; q) m2 (aq/b 1 ; q) m1 (aq/c 1 ; q) m1 m 1,m 2 =0 (b 3 ; q) m1 +m 2 (c 3 ; q) m1 +m 2 (q N ; q) m1 +m 2 (aq) m 1q m 1+m 2 (aq/b 2 ; q) m1 +m 2 (aq/c 2 ; q) m1 +m 2 (b 3 c 3 q N /a; q) m1 +m 2 (b 3 c 3 q N /a; q) m1 +m 2 (b 2 c 2 ) m 1 )

64 SURELY THERE MUST BE AN EASIER PROOF

65 TRINKETS

66 q n (1 + q 2 )(1 + q 4 ) (1 + q 2n ) =1 + q + q 2 + q 5 q 7 q 12 q 15 q 22 + q 26 + q 35 + q 40 + q 51 = ( 1) n q n(6n+1) (1 + q 4n+1 ) + ( 1) n q (2n+1)(3n+2) (1 + q 4n+3 )

67 q n (1 + q)(1 + q 2 )(1 + q 3 ) (1 + q 2n ) =(1 q 11 + q 13 ) + q(1 q 5 + q 19 ) = q 12n2 +n (1 q 22n+11 ) +q q 12n2 +7n (1 q 10n+5 )

68 q n (1 + q)(1 + q 3 ) (1 + q 2n+1 ) = ( 1) n q 6n2 +4n q n (1 + q)(1 + q 2 )(1 + q 3 ) (1 + q 2n+1 ) = 1 q 12n2 n (1 q 2n ) + q n=1 q 12n2 7n (1 q 14n ) n=1

69 q n (q; q 2 ) n ( q; q) 2n =1 + q q 2 q 5 + q 7 + q 12 = ( 1)q n(3n+1)/2 (1 + q 2n+1 )

70 Thank You! Alles Gute zum Geburtstag! Happy Birthday, Christian!