IDENTITY
PROFESSOR D.M.JACKSON
1.
The Jacobi Triple Product Identity
These are notes on the
Jacobi Triple Product Identity
and its use in proving
the
Euler Pentagonal Number Theorem
and the mod 5 and 7
congruences for the
partition number.
Ihave included in a few more of the details than I included in
the lectures.
Let
P
be the set of all partitions and let
D
n
be the set of all partitions of
n
into
distinct parts only. Let
T
k
denote the partition (
k,k
−
1
,...,
1)
.
Lemma 1.1.
[Sylvester’s Decomposition]
P×{
T
k
}
f
→
[
j
≥
0
D
k
+
j
×
(
D
j
∪D
j
−
1
)
where
D
0
∪D
−
1
=
D
0
.
Proof.
Append the reverse (1
,
2
,...,k
)o
f
T
k
to the top of the Ferrers diagram for
π
∈
P
, and consider the staircase that continues the pro±le of the Ferrers diagram
for
π.
The length of the staircase is
k
+
j.
The staircase partitions the diagram into
a partition
α
obtained by summing the columns of
?
’s below the staircase, and a
partition
β
obtained by summing the
?
’s in rows above the staircase. The number
of rows in
β
is
j
or
j
−
1
.
The partitions
α
and
β
necessarily have distinct parts,
induced by the staircase. The construction is clearly reversible.
Theorem 1.2.
[Jacobi Triple Product Identity]
Y
m
≥
1
(
1
−
q
2
m
)(
1+
yq
2
m
−
1
)(
1+
y
−
1
q
2
m
−
1
)
=
∞
X
k
=
−∞
y
k
q
k
2
.
Date
: October 5, 2002.
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R
OFESSOR D.M.JACKSON
Proof.
From Lemma 1.1, by counting partitions with respect to the sum of their
parts, marked by
q
,wehave
q
(
k
+1
2
)
Y
m
≥
1
(1
−
q
m
)
−
1
=
X
j
≥
0
£
s
k
+
j
¤
Y
a
≥
1
(1 +
sq
a
)
·
£
t
j
¤
Y
b
≥
1
(
1+
tq
b
)
+
£
t
j
−
1
¤
Y
b
≥
1
(
1+
tq
b
)
=
X
j
≥
0
£
s
k
+
j
t
j
¤
(1 +
t
)
Y
m
≥
1
(1 +
sq
m
)(1+
tq
m
)
=
X
j
≥
0
£
s
k
+
j
t
j
¤
Y
m
≥
1
(1 +
sq
m
)
(
1+
tq
m
−
1
)
.
We now change variables from
s
and
t
to
s
and
u
through
st
=
u.
Then
q
(
k
+1
2
)
Y
m
≥
1
(1
−
q
m
)
−
1
=
£
s
k
¤
X
j
≥
0
£
u
j
¤
Y
m
≥
1
(1 +
sq
m
)
(
1+
us
−
1
q
m
−
1
)
so
q
(
k
+1
2
)
Y
m
≥
1
(1
−
q
m
)
−
1
=
£
s
k
¤
Y
m
≥
1
(1 +
sq
m
)
(
1+
s
−
1
q
m
−
1
)
.
(1.1)
We next sum over
k
from
−∞
to +
∞
by making use of the following symmetry in
k.
Replacing
s
by
s
−
1
,
we have
q
(
k
+1
2
)
Y
m
≥
1
(1
−
q
m
)
−
1
=
£
s
−
k
¤
Y
m
≥
1
(
1+
s
−
1
q
m
)(
1+
sq
m
−
1
)
.
Now replace
s
by
qS,
noting that
£
s
−
k
¤
=
q
k
£
S
−
k
¤
.
Then
q
(
k
+1
2
)
Y
m
≥
1
(1
−
q
m
)
−
1
=
q
k
£
S
−
k
¤
Y
m
≥
1
(
1+
S
−
1
q
m
−
1
)
(1 +
Sq
m
)
so, replacing
S
by
s,
q
(
−
k
+1
2
)
Y
m
≥
1
(1
−
q
m
)
−
1
=
£
s
−
k
¤
Y
m
≥
1
(1 +
sq
m
)
(
1+
s
−
1
q
m
−
1
)
since
(
k
+1
2
)
−
k
=
(
−
k
+1
2
)
.
Thus (1
.
1) holds with
k
replaced by
−
k.
Thus summing
(1
.
1) over
k
from
−∞
to +
∞
we have
∞
X
k
=
−∞
s
k
q
(
k
+1
2
)
Y
m
≥
1
(1
−
q
m
)
−
1
=
∞
X
k
=
−∞
s
k
£
s
k
¤
Y
m
≥
1
(1 +
sq
m
)
(
1+
s
−
1
q
m
−
1
)
=
Y
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 Spring '05
 R.Metzger
 Number Theory, jacobi triple product, Jacobi Triple Product Identity, Pentagonal number theorem, triple product identity

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