RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESS SOME REFLECTIONS

Transcription

1 RAMANUJAN'S LOST NOTEBOOK IN FIVE VOLUMES SOME REFLECTIONS PAULE60 COMBINATORICS, SPECIAL FUNCTIONS AND COMPUTER ALGEBRA May 17, 2018 This talk is dedicated to my good friend Peter Paule

2 I. BACKGROUND

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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.

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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 ) +.

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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.

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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 + q 2 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 ) +

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19 φ 0( q) = +1 n=0 n=0 (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 ) n=0 { (1 q 5n+5 ) q + (1 q 1 ) n=0 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= n=0 ( 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 FROM W. N. BAILEY TO THE FIFTH ORDER MOCK THETA FUNCTIONS

24 THE BAILEY TRANSFORM IF AND THEN n β n = α r u n r v n+r, r=0 γ n = δ r u r n v r+n r=n α n γ n = β n δ n n=0 n=0

25 THE MOST FRUITFUL INSTANCE δ r = (p 1; q) r (p 2 ; q) r (q N ; q) r q r (p 1 p 2 q N /a; q) r γ r = ( aq p 1 ; q) N ( aq p 2 ; q) N (aq; q) N ( aq ( aq ) n p 1 p 2 ; q) N p 1 p 2 ( 1)n (p 1 ; q) n (p 2 ; q) n (q N ; q) n q n(2n n+1)/2 ( aq p 1 ; q) n ( aq p 2 ; q) n (aq N=1 ; q) n u n = 1 (q; q) n v n = 1 (aq; q) n THIS IS BAILEY'S LEMMA. HERE (A; q) n = (1 A)(1 Aq) (1 Aq n 1 )

26 A BAILEY PAIR IS A PAIR OF SEQUENCES (α n, β n ) RELATED BY n α r β n = (q; q) n r (aq; q) n+r r=0

27 BAILEY'S LEMMA IF (α n, β n ) IS A BAILEY PAIR, SO IS (α n, β n) WHERE α n = (p 1; q) n (p 2 ; q) n ( aq p 1 p 2 ) n α n ( aq p 1 ; q) n ( aq p 2 ; q) n β n = n j=0 (p 1 ; q) j (p 2 ; q) j ( aq p 1 p 2 ) j ( aq p 1 p 2 ; q) n j β j (q; q) n j ( aq p 1 ; q) j ( aq. p 2 ; q) j

28 THIS ITERATIVE FORM OF BAILEY'S LEMMA HAS TURNED OUT TO HAVE EXTENSIVE APPLICATIONS: E.G. q n2 k +n2 k 1 + +n2 1 n k n k 1 n 1 0 n=1 n 0,±(k+1)( mod 2k+3) (k = 1 is the rst Rogers-Ramanujan) (q) nk n k 1 (q) n2 n 1 (q) n1 1 1 q n.

29 TO TREAT THE FIFTH ORDER MOCK THETA FUNCTIONS WE NEED ONLY BAILEY'S LEMMA WITH p 1, p 2. THUS 1 β n q n2 = αn q n2 (q; q) WITH WHAT IS α n? n=0 β = 1 ( q; q) n.

30 α 0 =1 α 1 =q 2 3q α 2 =q 7 2q 6 q 5 + 2q 4 + 2q 3 α 3 =q 15 2q 14 q q 11 2q 8 2q 6 α 4 =q 26 2q 25 + q q 21 2q 18 2q q q 10 =q 2 6(1 2q 1 + 2q 4 2q 9 + 2q 16 ) q 22 (1 2q 1 + 2q 4 2q 9 ) AND THE REST IS HISTORY;

31 α n =q n(3n+1)/2 n ( 1) j q j2 j= n q n(3n 1)/2 n 1 j= n+1 ( 1) j q j2, AND q n2 f 0 (q) := ( q; q) n=0 n 1 = ( 1) j q n(5n+1)/2 j2 (1 q 4n+2 ) (q; q) j= n j

32 FRANK GARVAN'S THESIS AND THE ATKIN SWINNERTON DYER PROOFS OF THE DYSON CONJECTURES

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

34 DYSON CONJECTURED THAT FOR 0 a 4, r a (5n + 4) = 1 p(5n + 4). 5 SIMILARLY, DYSON CONJECTURED THAT THE RANK mod 7 SPLIT THE PARTITIONS OF 7n + 5 INTO 7 EQUAL SETS OF EQUAL SIZE. THE SAME IDEA DIDNT WORK FOR p(11n + 6) mod 11. THIS LED TO DYSON'S FAMOUS CONCLUDING PARAGRAPH:

35 I hold in fact: That there exists an arithmetical coecient similar to, but more recondite than, the rank of a partition; I shall call this hypothetical coecient the crank of the partition, and denote by M(m, q, n) the number of partitions of n whose crank is congruent to m modulo q: that M(m, q, n) = M(q m, q, n); that M(0, 11, 11n + 6) = M(1, 11, 11n + 6) = M(2, 11, 11n + 6) = M(3, 11, 11n + 6) = M(4, 11, 11n + 6);. Whether these quesses are warranted by the evidence, I leave to the reader to decide. Whatever the nal verdict of posterity may be, I believe the crank is unique among arithmetical functions in having been named before it was discovered. May it be preserved from the ignominious fate of the planet Vulcan!

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37 FRANK GARVAN'S PH.D THESIS IS DEVOTED TO AN EXPANDED STUDY OF THIS PAGE AND PROOF OF THE IDENTITIES. HE NOTES THAT THE ATKIN/SWINNERTON-DYER PROOF OF DYSON'S CONJECTURE IS ESSENTIALLY EQUIVALENT TO THE IDENTITY FOR f (q).

38 EVEN MORE SURPRISING, GARVAN REVEALS THAT F (q) CONTAINS WITHIN IT, THE CONJECTURED CRANK, FAMOUSLY PREDICTED BY DYSON.

39 HADAMARD PRODUCTS FOR ENTIRE FUNCTIONS

40 IN THE LOST NOTEBOOK, WE FIND: a n q ( n2 = 1 + (q; q) n where n=0 y 1 = y 2 = 0, y 3 = n=1 1 (1 q)ψ 2 (q), aq 2n 1 1 q n y 1 q 2n y 2 q 3n y 3 q + q 3 (1 q)(1 q 2 )(1 q 3 )ψ 2 (q) y 4 = y 1 y 3, ψ(q) = q n(n+1)/2 = (q2 ; q 2 ) (q; q 2 ) and where n=0 n=0 ), (2n+1)q 2n+1 1 q 2n+1 (1 q) 3 ψ 6 (q), (A; q) n = (1 A)(1 Aq)(1 Aq 2 ) (1 Aq n 1 ),

41 The most important property of a polynomial is that it can be expressed uniquely as a product of linear factors of the form ( Az p 1 z ) ( 1 z ) ) (1 zzn, z 1 z 2 Where A is a constant, p is a positive integer or zero, and z 1, z 2,..., z n the points, other than the origin, at which the polynomial vanishes, multiple zeros being repeated in the set according to their order. Conversely, if the zeros are given, the polynomial is determined apart from an arbitrary constant multiplier. Now, a polynomial is an integral function [i.e. entire function] of a very simple type, its singularity at innity being a pole. We naturally ask whether it is possible to exhibit in a similar manner the way in which any integral function depends on its zeros.

42 Theorem (Hadarmard's factorization theorem (Weak case)) Suppose f (z) is an entire function with simple zeros at z 1, z 2, z 3,..., f (0) = 1, and n=1 z n 1 <, then f (z) = n=1 (1 zzn ).

43 RAMANUJAN'S ASSERTION FOLLOWS FROM A STUDY OF THE ROGERS-SZEGO POLYNOMIALS, where [ ] n = j q K n (a) = n j=0 [ ] n q j2 a j, j q 0 if j 0 or j > n 1 if j = 0 or n otherwise (1 q n )(1 q n 1 ) (1 q n j+1 ) (1 q j )(1 q j 1 ) (1 q)

44 ASSUMING 0 < q < 1 (q REAL), K n (a) FORM A FAMILY OF ORTHOGONAL POLYNOMIALS WITH SIMPLE ZERO'S ON THE NEGATIVE REAL AXIS. A CAREFUL STUDY OF WHAT HAPPENS WHEN n YIELDS RAMANUJAN'S IDENTITY,

45 THE PITFALLS OF GUESSING HOW RAMANUJAN THOUGHT

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48 = G(a, λ; b; q) G(aq, λq; b; q) aq + λq bq + λq aq2 + λq bq2 + λq

49 a = b = 0, λ = 1 is the Rogers-Ramanujan Continued Fraction. ( 1) n q 3n2 +2n (1 + q 2n+1 ) n=0 = q 2 q 1 + q4 q q6 q q q 2 a = q 1 b = 1 λ = 1

50 (1 q 8n+1 )(1 q 8n+7 ) (1 q 8n+3 )(1 q 8n+5 ) n=0 = q 2 + q q q6 + q q q 2 a = q 1 b = 0 λ = 1

51 ( 1) n q (n2 +n)/2 n=0 = q q 2 q q q4 q 2... a = 0 b = 1 λ = 1

52 AFTER CONFIDENTLY DEDUCING ALL OF RAMANUJAN'S LIST FROM THE G(a, λ) CONT'D FRACTION, K. G. RAMANATHAN POINTED OUT THAT IN 1936 A. SELBERG HAD DONE THE ENTIRE LIST IN A VERY RAMANUJAN-LIKE MANNER.

53 RAMANUJAN'S MISCHIEVOUS GHOST

54 JUST BEFORE HIS TALK AT THE 100TH BIRTHDAY OF RAMANUJAN (UNIVERSITY OF ILLINOIS), BILL GOSPER SHOWED ME TWO BEAUTIFUL IDENTITES HE HAD JUST DISCOVERED.

55 ( aq; q 2 ) n ( q/a; q 2 ) n q 2n2 (q 2 ; q 2 ) n=0 2n 1 = a n q 3n2 (q 2 ; q 2 ) n= AND (aq; q 2 ) n (q/a; q 2 ) n q n ( q; q) 2n+1 n=0 ( a = ( 1) n n+1 + a n ) q n2 +n a + 1 n=0

56 I WAS COMPELLED TO POINT OUT TO HIM THAT BOTH WERE IN RAMANUJAN'S LOST NOTEBOOK. DURING GOSPER'S TALK HE DESCRIBED HIS DISCOVERY AND ACKNOWLEDGED RAMANUJAN'S PRIORITY WITH THE REMARK:

57 HOW CAN WE LOVE THIS MAN WHEN HE REACHES OUT FROM THE GRAVE TO SNATCH AWAY OUR BEST RESULTS?

58 IS THIS A MISTAKE?

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60 1 x + x 3 x 6 + = x + x x 3 + x x 5 + x

61 YES, BUT x + x x 3 + x x 5 + x =1 x + x 3 x 6 + x 8 x 9 x 10

62 Thank You! Alles Gute zum Geburtstag! Happy Birthday, Peter!