RAMANUJAN'S LOST NOTEBOOK IN FIVE VOLUMES SOME REFLECTIONS · RAMANUJAN'S LOST NOTEBOOK IN FIVE VOLUMES SOME REFLECTIONS APULE60 COMBINATORICS, SPECIAL FUNCTIONS AND COMPUTER ALGEBRA

RAMANUJAN'S LOST NOTEBOOK

IN FIVE VOLUMES

SOME REFLECTIONS

PAULE60COMBINATORICS, SPECIAL FUNCTIONS AND COMPUTER

ALGEBRAMay 17, 2018

This talk is dedicated to my good friendPeter Paule

RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS

I. BACKGROUND



�His memory, and his powers of calculation, were very unusual, butthey could not be reasonably be called �abnormal�. If he had tomultiply two very large numbers, he multiplied them in the ordinaryway; he would do it with unusual rapidity and accuracy, but notmore rapidly than any mathematician who is naturally quick andhas the habit of computation. There is a table of partitions at theend of our paper. . . . This was, for the most part, calculatedindependently by Ramanujan and Major MacMahon; and MajorMacMahon was, in general the slightly quicker and more accurateof the two.�



THE PATH FROM THE LAST LETTER TO THE LOSTNOTEBOOK


�He returned from England only to die, as the saying goes. He livedfor less than a year. Throughout this period, I lived with himwithout break. He was only skin and bones. He often complainedof severe pain. In spite of it he was always busy doing hisMathematics. That, evidently helped him to forget the pain. I usedto gather the sheets of papers which he �lled up. I would also givethe slate whenever he asked for it. He was uniformly kind to me. Inhis conversation he was full of wit and humour. Even while mortallyill, he used to crack jokes. One day he con�ded in me that hemight not live beyond thirty-�ve and asked me to meet the eventwith courage and fortitude. He was well looked after by his friends.He often used to repeat his gratitude to all those who had helpedhim in his life.�


UNIVERSITY OF MADRAS,12th January 1920.

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 (studiedpartially by Prof. Rogers in his interesting paper) they enter intomathematics as beautifully as the ordinary θ-functions. I amsending you with this letter some examples . . . .Mock θ-functions

φ(q) = 1+q

1+ q2+

q4

(1+ q2)(1+ q4)+ · · · ,

ψ(q) =q

1− q+

q4

(1− q)(1− q3)+

q9

(1− q)(1− q3)(1− q5)+ · · · .

. . . . . .




When the Royal Society asked me to write G. N. Watson'sobituary memoir I wrote to his widow to ask if I could examine hispapers. She kindly invited me to lunch and afterwards his son tookme upstairs to see them. They covered the �oor of a fair sizedroom to a depth of about a foot, all jumbled together, and were tobe incinerated in a few days. One could make lucky dips and, asWatson never threw anything away, the result might be a sheet ofmathematics but more probably a receipted bill or a draft of hisincome tax for 1923. By extraordinary stroke of luck one of my dipsbrought up the Ramanujan material which Hardy must have passedon to him when he proposed to edit the earlier notebooks.






THE MOCK THETA CONJECTURES


f0(q) = 1+q

1+ q+

q4

(1+ q)(1+ q2)+ · · · ,

φ0(q) = 1+ q(1+ q) + q4(1+ q)(1+ q3) + q9(1+ q)(1+ q3)(1+ q5) + · · · ,

ψ0(q) = q + q3(1+ q) + q6(1+ q)(1+ q2) + q10(1+ q)(1+ q2)(1+ q3) + · · · ,

F0(q) = 1+q2

1− q+

q8

(1− q)(1− q3)+ · · · ,

χ0(q) = 1+q

1− q2+

q2

(1− q3)(1− q4)+

q3

(1− q4)(1− q5)(1− q6)+ · · ·

= 1+q

1− q+

q3

(1− q2)(1− q3)+

q5

(1− q3)(1− q4)(1− q5)+ · · ·



φ0(−q) =∞∏n=0

(1− q5n+5)(1+ q5n+2)(1+ q5n+3)

(1− q10n+2)(1− q10n+8)

+1−∞∑n=0

q5n2

(1− q)(1− q6) · · · (1− q5n+1)(1− q4)(1− q9) · · · (1− q5n−1).

φ0(−q) =∞∏n=0

(1− q5n+5)(1+ q5n+2)(1+ q5n+3)

(1− q10n+2)(1− q10n+8)

+1−∞∏n=0

(1− q5n+5)−1{

1

1− q+ (1− q−1)

∞∑n=1

(−1)nqn(15n+5)/2(1+ q5n)

(1− q5n+1)(1− q5n−1)

}.


EQUIVALENTLY

∞∏n=1

(1− q10n)∞∑n=0

qn2

(1+ q) · · · (1+ qn)

=∞∑

n=−∞

(−1)nq2n

1− q5n+1

− 2∞∑

n=−∞

(−1)nq15n2+15n+2

1− q10n+2


MOCK THETA CONJECTURES

ra(n) = # OF PTNS OF n WITH RANK ≡ a( mod 5)

rank = (largest part) − (# of parts)

The rank of 5+ 4+ 2+ 1+ 1+ 1 is 5− 6 = −1.


1ST MOCK CONJECTURE(Proved by Hickerson)

r1(5n)− r0(5n)

EQUALS THE # OF PTNS OF n IN WHICH THE LARGESTPART IS ODD AND EACH PART IS = 1

2(LARGEST PART)


FROM W. N. BAILEY TO THE FIFTH ORDER MOCK THETAFUNCTIONS


THE BAILEY TRANSFORM

IF

βn =n∑

r=0

αrun−rvn+r ,

AND

γn =∞∑r=n

δrur−nvr+n

THEN∞∑n=0

αnγn =∞∑n=0

βnδn


THE MOST FRUITFUL INSTANCE

δr =(p1; q)r (p2; q)r (q

−N ; q)rqr

(p1p2q−N/a; q)r

γr =(aqp1 ; q)N(

aqp2; q)N

(aq; q)N(aq

p1p2; q)N

(aq

p1p2)n

× (−1)n(p1; q)n(p2; q)n(q−N ; q)nqn(2N−n+1)/2

(aqp1 ; q)n(aqp2; q)n(aqN=1; q)n

un =1

(q; q)nvn =

1

(aq; q)n

THIS IS BAILEY'S LEMMA.HERE (A; q)n = (1− A)(1− Aq) · · · (1− Aqn−1)


A BAILEY PAIR IS A PAIR OF SEQUENCES (αn, βn) RELATEDBY

βn =n∑

r=0

αr

(q; q)n−r (aq; q)n+r


BAILEY'S LEMMA

IF (αn, βn) IS A BAILEY PAIR, SO IS (α′n, β′n) WHERE

α′n =(p1; q)n(p2; q)n(

aqp1p2

)nαn

(aqp1 ; q)n(aqp2; q)n

β′n =n∑

j=0

(p1; q)j(p2; q)j(aq

p1p2)j( aq

p1p2; q)n−jβj

(q; q)n−j(aqp1; q)j(

aqp2; q)j

.


THIS ITERATIVE FORM OF BAILEY'S LEMMA HAS TURNEDOUT TO HAVE EXTENSIVE APPLICATIONS:E.G.

∑nk=nk−1=···=n1=0

qn2

k+n2k−1+···+n21

(q)nk−nk−1 · · · (q)n2−n1(q)n1∞∏n=1

n 6≡0,±(k+1)( mod 2k+3)

1

1− qn.

(k = 1 is the �rst Rogers-Ramanujan)


TO TREAT THE FIFTH ORDER MOCK THETA FUNCTIONSWE NEED ONLY BAILEY'S LEMMA WITH p1, p2 →∞.THUS

∞∑n=0

βnqn2 =

1

(q; q)∞

∑αnq

n2

WITH

β =1

(−q; q)n.

WHAT IS αn?


α0 =1

α1 =q2 − 3q

α2 =q7 − 2q6 − q5 + 2q4 + 2q3

α3 =q15 − 2q14 − q12 + 4q11 − 2q8 − 2q6

α4 =q26 − 2q25 + q22 + 2q21 − 2q18 − 2q17 + 2q13 + 2q10

=q26(1− 2q−1 + 2q−4 − 2q−9 + 2q−16)

− q22(1− 2q−1 + 2q−4 − 2q−9)

AND THE REST IS HISTORY;


αn =qn(3n+1)/2n∑

j=−n(−1)jq−j2

− qn(3n−1)/2n−1∑

j=−n+1

(−1)jq−j2 ,

AND

f0(q) :=∞∑n=0

qn2

(−q; q)n

=1

(q; q)∞

∞∑j=−∞

∑n≥|j |

(−1)jqn(5n+1)/2−j2(1− q4n+2)


FRANK GARVAN'S THESIS AND THE ATKIN SWINNERTONDYER PROOFS OF THE DYSON CONJECTURES


ra(n) = # OF PTNS OF n WITH RANK≡ a( mod 5)

rank = (largest part)− (# of parts)

The rank of 5+ 4+ 2+ 1+ 1+ 1 is 5− 6 = −1.


DYSON CONJECTURED THAT FOR 0 ≤ a ≤ 4,

ra(5n + 4) =1

5p(5n + 4).

SIMILARLY, DYSON CONJECTURED THAT THE RANKmod 7 SPLIT THE PARTITIONS OF 7n + 5 INTO 7 EQUALSETS OF EQUAL SIZE.THE SAME IDEA DIDNT WORK FOR p(11n+ 6) mod 11. THISLED TO DYSON'S FAMOUS CONCLUDING PARAGRAPH:


I hold in fact:That there exists an arithmetical coe�cient similar to, but morerecondite than, the rank of a partition; I shall call this hypotheticalcoe�cient the �crank� of the partition, and denote by M(m, q, n)the number of partitions of n whose crank is congruent to mmodulo 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 thereader to decide. Whatever the �nal verdict of posterity may be, Ibelieve the �crank� is unique among arithmetical functions in havingbeen named before it was discovered. May it be preserved from theignominious fate of the planet Vulcan!



FRANK GARVAN'S PH.D THESIS IS DEVOTED TO ANEXPANDED STUDY OF THIS PAGE AND PROOF OF THEIDENTITIES. HE NOTES THAT THEATKIN/SWINNERTON-DYER PROOF OF DYSON'SCONJECTURE IS ESSENTIALLY EQUIVALENT TO THEIDENTITY FOR f (q).


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


HADAMARD PRODUCTS FOR ENTIRE FUNCTIONS


IN THE LOST NOTEBOOK, WE FIND:

∞∑n=0

anqn2

(q; q)n=∞∏n=1

(1+

aq2n−1

1− qny1 − q2ny2 − q3ny3 − · · ·

),

where

y1 =1

(1− q)ψ2(q),

y2 = 0,

y3 =q + q3

(1− q)(1− q2)(1− q3)ψ2(q)−∑∞

n=0

(2n+1)q2n+1

1−q2n+1

(1− q)3ψ6(q),

y4 = y1y3,

ψ(q) =∞∑n=0

qn(n+1)/2 =(q2; q2)∞(q; q2)∞

and where

(A; q)n = (1− A)(1− Aq)(1− Aq2) · · · (1− Aqn−1),


�The most important property of a polynomial is that it can beexpressed uniquely as a product of linear factors of the form

Azp(1− z

z1

)(1− z

z2

)· · ·(1− z

zn

),

Where A is a constant, p is a positive integer or zero, andz1, z2, . . . , zn the points, other than the origin, at which thepolynomial vanishes, multiple zeros being repeated in the setaccording to their order. Conversely, if the zeros are given, thepolynomial is determined apart from an arbitrary constantmultiplier.Now , a polynomial is an integral function [i.e. entire function] of avery simple type, its singularity at in�nity being a pole. Wenaturally ask whether it is possible to exhibit in a similar mannerthe way in which any integral function depends on its zeros.�


Theorem (Hadarmard's factorization theorem (Weak case))

Suppose f (z) is an entire function with simple zeros at

z1, z2, z3, . . . , f (0) = 1, and∑∞

n=1|zn|−1 <∞, then

f (z) =∞∏n=1

(1− z

zn

).


RAMANUJAN'S ASSERTION FOLLOWS FROM A STUDY OFTHE ROGERS-SZEGO POLYNOMIALS,

Kn(a) =n∑

j=0

[n

j

]q

qj2

aj ,

where

[n

j

]q

=

0 if j ≤ 0 or j > n

1 if j = 0 or n(1−qn)(1−qn−1)···(1−qn−j+1)

(1−qj )(1−qj−1)···(1−q) otherwise


ASSUMING 0 < q < 1 (q REAL), Kn(a) FORM A FAMILY OFORTHOGONAL POLYNOMIALS WITH SIMPLE ZERO'S ONTHE NEGATIVE REAL AXIS. A CAREFUL STUDY OF WHATHAPPENS WHEN n→∞ YIELDS RAMANUJAN'S IDENTITY,


THE PITFALLS OF GUESSING HOW RAMANUJAN THOUGHT




G (a, λ; b; q)

G (aq, λq; b; q)

=1

1+aq + λq

1+bq + λq2

1+aq2 + λq3

1+bq2 + λq4

1+. . .


a = b = 0, λ = 1 is the Rogers-Ramanujan Continued Fraction.

∞∑n=0

(−1)nq3n2+2n(1+ q2n+1)

=1

1+q2 − q

1+q4 − q2

1+q6 − q3

1+. . .

q → q2

a = −q−1

b = −1λ = 1


∞∏n=0

(1− q8n+1)(1− q8n+7)

(1− q8n+3)(1− q8n+5)

=1

1+q2 + q

1+q4

1+q6 + q3

1+. . .

q → q2

a = q−1

b = 0

λ = 1


∞∑n=0

(−1)nq(n2+n)/2

=1

1+q

1+q2 − q

1+q3

1+q4 − q2

. . .

a = 0

b = −1λ = 1


AFTER CONFIDENTLY DEDUCING ALL OF RAMANUJAN'SLIST FROM THE G (a, λ) CONT'D FRACTION, K. G.RAMANATHAN POINTED OUT THAT IN 1936 A. SELBERGHAD DONE THE ENTIRE LIST IN A VERY RAMANUJAN-LIKEMANNER.


RAMANUJAN'S MISCHIEVOUS GHOST


JUST BEFORE HIS TALK AT THE 100TH BIRTHDAY OFRAMANUJAN (UNIVERSITY OF ILLINOIS), BILL GOSPERSHOWED ME TWO BEAUTIFUL IDENTITES HE HAD JUSTDISCOVERED.


∞∑n=0

(−aq; q2)n(−q/a; q2)nq2n2

(q2; q2)2n

=1

(q2; q2)∞

∞∑n=−∞

anq3n2

AND

∞∑n=0

(aq; q2)n(q/a; q2)nq

n

(−q; q)2n+1

=∞∑n=0

(−1)n(an+1 + a−n

a+ 1

)qn

2+n


I WAS COMPELLED TO POINT OUT TO HIM THAT BOTHWERE IN RAMANUJAN'S LOST NOTEBOOK.DURING GOSPER'S TALK HE DESCRIBED HIS DISCOVERYAND ACKNOWLEDGED RAMANUJAN'S PRIORITY WITH THEREMARK:


�HOW CAN WE LOVE THIS MAN WHEN HE REACHES OUTFROM THE GRAVE TO SNATCH AWAY OUR BEST RESULTS?�


IS THIS A MISTAKE?



1− x + x3 − x6 + · · ·

=1

1+x + x2

1+x3 + x4

1+x5 + x6

1+. . .


YES, BUT

1

1+x + x2

1+x3 + x4

1+x5 + x6

1+. . .

=1− x + x3 − x6 + x8 − x9 − x10 · · ·


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


Documents

RAMANUJAN'S LOST NOTEBOOK IN FIVE VOLUMES SOME REFLECTIONS · RAMANUJAN'S LOST NOTEBOOK IN FIVE VOLUMES SOME REFLECTIONS APULE60 COMBINATORICS, SPECIAL FUNCTIONS AND COMPUTER ALGEBRA