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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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