Post on 27-Jun-2020
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
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
�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.�
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
THE PATH FROM THE LAST LETTER TO THE LOSTNOTEBOOK
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
�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.�
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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)+ · · · .
. . . . . .
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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.
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
THE MOCK THETA CONJECTURES
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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)+ · · ·
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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)
}.
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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.
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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)
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
FROM W. N. BAILEY TO THE FIFTH ORDER MOCK THETAFUNCTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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)
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
A BAILEY PAIR IS A PAIR OF SEQUENCES (αn, βn) RELATEDBY
βn =n∑
r=0
αr
(q; q)n−r (aq; q)n+r
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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
.
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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)
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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?
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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;
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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)
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
FRANK GARVAN'S THESIS AND THE ATKIN SWINNERTONDYER PROOFS OF THE DYSON CONJECTURES
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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.
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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:
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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!
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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).
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
EVEN MORE SURPRISING, GARVAN REVEALS THAT F (q)CONTAINS WITHIN IT, THE CONJECTURED CRANK,FAMOUSLY PREDICTED BY DYSON.
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
HADAMARD PRODUCTS FOR ENTIRE FUNCTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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),
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
�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.�
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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 LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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,
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
THE PITFALLS OF GUESSING HOW RAMANUJAN THOUGHT
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
G (a, λ; b; q)
G (aq, λq; b; q)
=1
1+aq + λq
1+bq + λq2
1+aq2 + λq3
1+bq2 + λq4
1+. . .
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
∞∏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
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
∞∑n=0
(−1)nq(n2+n)/2
=1
1+q
1+q2 − q
1+q3
1+q4 − q2
. . .
a = 0
b = −1λ = 1
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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 LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S MISCHIEVOUS GHOST
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
JUST BEFORE HIS TALK AT THE 100TH BIRTHDAY OFRAMANUJAN (UNIVERSITY OF ILLINOIS), BILL GOSPERSHOWED ME TWO BEAUTIFUL IDENTITES HE HAD JUSTDISCOVERED.
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
∞∑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
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
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:
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
�HOW CAN WE LOVE THIS MAN WHEN HE REACHES OUTFROM THE GRAVE TO SNATCH AWAY OUR BEST RESULTS?�
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
IS THIS A MISTAKE?
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
1− x + x3 − x6 + · · ·
=1
1+x + x2
1+x3 + x4
1+x5 + x6
1+. . .
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
YES, BUT
1
1+x + x2
1+x3 + x4
1+x5 + x6
1+. . .
=1− x + x3 − x6 + x8 − x9 − x10 · · ·
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS
Thank You!Alles Gute zum Geburtstag!Happy Birthday, Peter!
RAMANUJAN'S LOST NOTEBOOKIN FIVE VOLUMESSOME REFLECTIONS