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3 Jacobi's triple product identity
3 Jacobi's triple product identity
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80
Chapter 8. Durfee squares
Theorem 11 (Jacobi's triple product identity)
00
L:
n(l00
znq~ =
for
lql
qn)(l
+ zqn)
(8.5)
n=l
n=-oo
< 1, z =I= 0.
In order to prove Jacobi's triple product identity, we define
n(l
00
J(z) =
+zqn)(l +z-!qn-l).
n=l
We may expand 1 (z) in a Laurent series around
z = 0, so
00
l(z)=
L
(8.6)
An(q)zn.
n=-oo
Furthermore,
n(l +
00
J(zq) =
zqn+l)(l
n=l
+ z-!qn-2)
n(l
00
= (1 +z-lq-')
n=2
+z-!qn-!)
n=l
n(l + n(l +
00
= z-!q-!
n(l
00
+zqn)
00
zqn)
n=l
z-!qn-!)
n=l
Comparing coefficients of zn on both sides, we see that
Iteration of this recursion reveals that for all n,
An(q) = q
!!!!!±.!1
2
Ao(q).
(8.7)
But by Eq. (8.4) and the comments preceding it,
n--,
00
Ao(q)=
1
n=l 1 - qn
(8.8)
81
8.4 The Rogers-Ramanujan identities
Combining (8.8), (8.7), and (8.6), we deduce
no
00
L
00
= l(z) = Ao(q)
+zqn)(l +z-lqn-1)
n=l
znq"
n=-oo
n
1
00
n!!l!!±!l
-zq2
1
n
'
n=l - q n=-oo
L
00
=
which is clearly equivalent to Jacobi's triple product identity.
EXERCISES
106. By replacing q by q 3 and z by -q- 1 , deduce Euler's pentagonal number
theorem from Jacobi's triple product. (Difficulty rating: 1)
107. Prove the identity
r:;::_
00 (
-1tqn 2 =
fl::: g~::~. (Difficulty rating: 2)
1
2
108 ' Prove the identity "'oo
(l-q 2"l
(Difficulty rating·• 2)
L...n=O qn
109. Prove Jacobi's identity
-l)nqn(n+l)f 2(2n + 1) =
(1 - qn)3.
(Difficulty rating: 3)
00
r::::o (
n:::l
8.4 The Rogers-Ramanujan identities
There is no really easy proof of the Rogers-Ramanujan identities. However, our
study of Gaussian polynomials and Jacobi's triple product puts us in a position
to prove some polynomial identities of David Bressoud (1981), which in turn
will yield the Rogers-Ramanujan identities.
Following Chapman (2002), we shall take things step by step. First we
consider five sequences of polynomials:
(8.9)
(8.10)
(8.11)
(8.12)
Chapter 8. Durfee squares
82
and
r"(q)
= jf;oo(-l)jqj(5j-3)/2 [~n++2~
J.
(8.13)
In the following pages, we shall prove using mathematical induction that
(8.14)
and
Once this has been achieved, it will be a straightforward matter to show that
00
f=t
1 +"
·2
ql
. = lim S11 (q) = lim an(q)
(1- q)(1- q 2 ) ..• (1- qJ)
n---+00
n---+00
1
00
=
!]
(1 _ q5n+i )(1 _ q5n+4)
and
+"
f=t
.
qj2+j
00
1
(1- q)(1- q 2) ... (1- qJ)
= lim f 11 (q) = lim
n---+00
1
00
=
!]
T11 (q)
n--->00
(1 _ q5n+2)(1 _ q5n+3) ·
Indeed it follows immediately from (7. 7) of Chapter 7, that
•
qj2
00
hms 11 (q)=1+L
j=i
n---+00
·
2
(1- q)(1- q ) ... (1- qJ)
and
•
n---+00
+"
f=t
=
1
..
qj2+j
00
hm t11 (q)
(1- q)(1- q2) ... (1- ql)
(8.15)
Thus we are treating polynomials that, in the limit as n -+ oo, converge to the
left-hand sides of the Rogers-Ramanujan identities.
Now we are going to prove two recurrences for s11 (q) and t11 (q). First,
=Sn-!(q)+qnLqjz+j
j'?.O
= Sn-!(q)
+ q 11 fn-!(q).
[n~
1]
}
(8.16)
83
8.4 The Rogers-Ramanujan identities
Second,
It is now a simple problem in mathematical induction to show that (8.13) and
(8.14) plus the initial values s 0(q) = t0(q) = 1 uniquely define sn(q) and tn(q)
for all n.
We now want to show that an(q) and a;(q) are identical. This is because
a;(q) =
f: (-
1)jqj(5j+l)/2 ([
j=-oo
= an(q) + qn+l
(f:c
-1)j qj(5j+5)/2 [
j=O
+
2n .] +qn+i+2j [
2n
.])
n + 2]
n + 1 + 2]
2n
.]
n + 1 + 2]
~ ( _ 1)j qj(5j+5)/2 [
.~
J=-00
=an(q)+qn+i
2n
.
n+1+2J
J)
(~(2n
.]
1 )jqj(5j+5)/2[ n -1-2]
~
j=O
+
~(-1)-j-Iq(-j-1)(5(-j))/2 [
~
j=O
2n
.
n -1-2]
J)
(8.17)
So in the following we will use either (8.11) or (8.12) to represent an(q). In
particular,
an(q)- an-J(q)
f: (f: (-
.
J=-oo
=
1)jqj(5j+I)/2 [
2n .] _ ~ (- 1)jqj(5j+I)/2 [2n- ~]
n + 2]
.~
n + 2]
J=-oo
1)jqj(5j+IJ/2 [
2n ~ 1 ] qn- 2 j
n + 2] - 1
j=-oo
= qn
(by (7.2))
(8.18)
84
Chapter 8. Durfee squares
Finally,
in(q)- qnan(q)
=
=
J_qn+2j [ n -2n2]. ])
~
( _ 1)j qj(5j-3)/2 ([
~
( -1)i qj(5j-3)/2 [
~
(- 1)jqj(5j-3)/2 ([2n- ~] +qn+1-2j [ 2n -1 .])
n - 2}
n + 1 - 2]
.~
;=-oo
.~
J=-00
.~
;=-00
= Tn-J(q) _ qn
~
.~
J=-00
2n
+1
.
n + 1 - 2}
2n
.
n+1-2j
J
(- 1)jqj(5j-3)/2 [
(by (7.2))
2n- 1 .
n -1 +2J
J
(8.19)
But as we remarked earlier, the recurrences (8.15) and (8.16) plus so(q) =
t0(q) = 1 uniquely determine all the sn(q) and tn(q). Now we have just shown
in (8.18) and (8.19) that an(q) and Tn(q) satisfy precisely the same recurrences.
Furthermore, ao(q) = r 0 (q) = 1. Consequently for each n ~ 0,
Sn(q)
= an(q)
and
Therefore, by (8.9),
00
1+
L:
ql
j=l (1- q)(l- q2) ... (1 - qi)
·2
= lim Sn(q) = lim an(q)
n----+-oo
=
~
~
j=-oo
n~oo
(-1)jqj(5j+l)/2
1
floo
(1- q m)
m=l
(by (7.8))
n::=l (1 - q5m)(1 - q5m-2)(1 - q5m-3)
n::=l (1 - qm)
(by Theorem 11 with q replaced by q 5 and z by -q- 2)
00
1
=!] (1 _ q5m-4)(l _ q5m-1)'
8.5 Successive Durfee squares
85
proving the first Rogers-Ramanujan identity. By (8.10),
qi2+j
00
1+
I:
j=l (1-q)(l-q
2
·
)···(l-ql)
= lim tn(q) = lim
n-+oo
=
=
=
n--HXJ
~ ( -l)jqj(5j-3)/2
1
.~
floo
J=-oo
m=l (1- q m)
n:=l(l- q5m)(l- q5m-l)(l- q5m-4)
(by (7.8))
n:=l (1 - qm)
(by Theorem 11 with q replaced by q 5 and z by -q- 1)
00
1
D
(1 _ q5m-3)(1 _ q5m-2)'
proving the second Rogers-Ramanujan identity.
8.5 Successive Durfee squares
We shall here present a nice generalization of the first Rogers-Ramanujan
identity. Let us reconsider the polynomials sn(q) defined in equation (8.10):
Sn(q) = t q l
j=O
[~].
}
Now we already know from Chapter 7 that [;] is the generating function
for partitions with at most j parts, each at most n - j. Noting that } 2 = j +
j + j + · · · + j, we see immediately that sn (q) is the generating function for
partitions into parts each ~ n, wherein each part is ;;:; the number of parts (our
j above).
j
n-j
,--,-----,-------,+-------+-
j
•••••••
•• • • • •
•••
••••
For example,
+ ql + q2 + l + q4 + q2+2 + q3+2 + q4+2 + q3+3
+ q4+3 + q4+4 + q3+3+3 + q4+3+3 + q4+4+3 + q4+4+4 + q4+4+4+4.
s4(q) = 1
If we examine such partitions with regard to their Durfee squares, we see
that these partitions have no parts below their Durfee square.
86
Chapter 8. Durfee squares
With this beginning, it is possible to extend the concept of Durfee square
to that of successive Dwfee squares. The idea here is to find a succession of
Durfee squares by examining the portion of the partition below a given square.
For example, the partition
7+5+4+4+4+3+2+2+1+1
has five successive Durfee squares:
•••••••
•••••
••••
••••
••••
•••
Once one considers the idea of successive Durfee squares, it is fairly easy
to determine the generating function for partitions with at most k successive
Durfee squares. Indeed, we already know the answer when k = 1, namely,
Sn(q)
= tqj2
j=O
[~].
J
But for general k, we need only consider k possible squares jf :::: /f ~ jf ~
· · · ~ jf while noting how the interstices of the Ferrers board may be filled.
This is most easily understood in the following schematic of the Ferrers board:
8.5 Successive Durfee squares
87
Hence, the generating function for partitions with at most k successive Durfee
squares, sk. 11 (q) is
where we have used the notation (q; q)m = (1 - q)(1 - q 2 ) · · · (1 - qm). If we
let n -+ oo, we find the complete generating function for all partitions with at
most k successive Durfee squares
It is possible to prove (Andrews, 1979) that, in fact,
00
=
n
1
(8.20)
n =I
n "f; 0, ±(k +I)
(mod 2k + 3)
Thus we may deduce that the number of partitions of n, each with at most k
successive Durfee squares, equals the number of partitions of n into parts not
congruent to 0, ±(k + 1) (mod 2k + 3).
When k = 1, this reduces precisely to the first Rogers-Ramanujan identity.
Chapter 9
Euler refined
In Chapters 1 and 2, we examined in detail Euler's simple and elegant theorem:
The number of partitions of an integer n into distinct parts equals the number
of partitions of n into odd parts.
As we have seen, this theorem foreshadows a number of further results,
the Rogers-Ramanujan identities being the most celebrated. In this chapter,
we shall delve deeper into Euler's theorem (2.1), presenting two combinatorial
variations and concluding with a consideration of a very recent refinement.
Highlights of this chapter
• Sylvester refined Euler's theorem by considering the number of odd
parts, and the number of sequences of consecutive distinct parts,
respectively.
• Fine's refinement of Euler's theorem instead considers the largest of the
odd parts and Dyson's rank ofthe partition into distinct parts.
• Bousquet-Melou and Eriksson's refinement considers odd parts
of bounded size and so-called "lecture hall partitions" into distinct
parts.
9.1 Sylvester's refinement of Euler
In the late nineteenth century, the colorful mathematician J. J. Sylvester (1884)
first discovered that there is more to Euler's theorem than meets the eye. The
theorem we now consider appears in a gigantic paper, entitled A Constructive
Theory of Partitions in Three Acts, an Interact, and an Exodion.
Theorem 12 (Sylvester's refinement) The number of partitions of n using
exactly k odd parts (each of which may be repeated) equals the number of
88
9.1 Sylvester's refinement of Euler
89
partitions of n into k separate sequences of consecutive integers. (A sequence
might have only one term.)
For example, when n
are
= 15 and k = 3, the eleven partitions in the first class
11 + 3 + 1, 9 + 5 + 1, 9 + 3 + 1 + 1 + 1, 7 + 5 + 3,
7 + 5 + 1 + 1 + 1, 7 + 3 + 1 + 1 + 1 + 1 + 1, 7 + 3 + 3 + 1 + 1,
5 + 5 + 3 + 1 + 1, 5 + 3 + 3 + 3 + 1, 5 + 3 + 3 + 1 + 1 + 1 + 1,
5 + 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1,
and the eleven partitions in the second class are
11+3+1, 10+4+1, 9+5+1, 9+4+2,
8 + 6 + 1, 8 + 5 + 2, 8 + 4 + 2 + 1, 7 + 5 + 3, 7 + 5 + 2 + 1,
7 + 4 + 3 + 1, 6 + 5 + 3 + 1.
We now prove this result by producing a variation on the Ferrers graph for
partitions into odd parts. Instead of lining up rows of dots or boxes to represent
parts by left justification, we justify them centrally as follows:
m
m
• • • • • • • • • • •
Partitions with odd parts can clearly be represented with centrally justified
graphs. For example, the partition 13 + 13 + 13 + 11 + 9 + 3 + 3 + 1 is thus
represented:
... ... ... ... ... ...
.. ..
.
•
.... .
...... ...
.• .
•
•
Now we take this representation and associate the dots in a new way:
and the new partition is 14 + 12 + 11 + 8 + 7 + 6 + 4 + 3 + 1.
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