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THE JACOBI TRIPLE PRODUCT
RAGIB ZAMAN
1. Summary
The Jacobi Triple Product is a beautiful identity that arises natu-
rally during the study of Elliptic functions, Lie Algebras and the theory
of integer partitions. After an elementary proof of the Triple Product,
we give an application to the problem of counting lattice points on the
surface of N dimensional spheres. We then develop Euler’s Pentagonal
Number theorem which has an interesting combinatoric interpretation
in terms of partitions, and leads us to an elegant recurrence relation
involving the partition function. Throughout this paper we will con-
sider every expression formally. For an analytic treatment, we refer the
reader to Hardy and Wright [1].
2. The Jacobi Triple Product
The statement of the Jacobi Triple Product is the following:
Theorem 2.1.∞∏n=1
(1− x2n
) (1 + x2n−1z2
)(1 +
x2n−1
z2
)=
∞∑m=−∞
xm2
z2m
Definition 2.2. The proof presented here is well known and is a slight
alteration of the one present at [2]. We first produce some preliminary
results.
We define the generating function F(z) by the following:
F(z) ≡∞∏n=1
(1 + x2n−1z2
)(1 +
x2n−1
z2
)
Date: Semester 1, 2011.1
2 RAGIB ZAMAN
Lemma 2.3.
F(z) = xz2 F(xz)
Proof. By replacing z with xz in the above definition, we have
F(xz) =∞∏n=1
(1 + x2n+1z2
) ∞∏n=1
(1 +
x2n−3
z2
)We now multiply the first factor by (1+xz2)/(1+xz2), and extract the
first term from the second factor outside of the product. Then after
re-indexing to start at n = 1 again we have
F(xz) =1
1 + xz2·∞∏n=1
(1 + x2n−1z2
)·(
1 +1
xz2
) ∞∏n=1
(1 +
x2n−1
z2
)
=
(1 +
1
xz2
)(1
1 + xz2
)F(z)
=
(1 + xz2
xz2
)(1
1 + xz2
)F(z)
=F(z)
xz2
which yields the result. �
Definition 2.4. Define
G(z) ≡ F(z)∞∏n=1
(1− x2n
)Lemma 2.5.
G(z) = xz2 G(xz)
Proof. Replacing z by xz gives
G(xz) = F(xz)∞∏n=1
(1− x2n
)By (2.3) this becomes
G(xz) =F(z)
xz2
∞∏n=1
(1− x2n
)=
G(z)
xz2
which proves our claim. �
THE JACOBI TRIPLE PRODUCT 3
Lemma 2.6.
F(z) = F
(1
z
)and
G(z) = G
(1
z
)Proof. The results follow immediately after setting x = 1/z2 into (2.3)
and (2.5). �
Lemma 2.7. There exists a set of coefficients {am : m ∈ Z} such that
G(z) =∞∑
m=−∞
amz2m
and am = a−m for all m ∈ Z.
Proof. Note that we assume that we may write
G(z) =∞∑
m=−∞
cmzm
This type of expansion is called a Laurent Expansion. We justify this
assumption by noting that, in practice, a great variety of functions
have a Laurent Expansion.
Now, by substituting −z into (2.2) we clearly have F(−z) = F(z).
Using this result with (2.4) yields G(−z) = G(z). Thus, when we
expand G(z) into a Laurent series, only even powers of z occur like
such:
G(z) =∞∑
m=−∞
amz2m
By this result and (2.6) we have∞∑
m=−∞
amz2m =
∞∑m=−∞
amz−2m
=∞∑
m=−∞
a−mz2m
where the last equality was produced by reversing the index of sum-
mation. Equating coefficients of z2m yields the result. �
4 RAGIB ZAMAN
Lemma 2.8.
am = am−1x2m−1
for all m ∈ Z.
Proof. By (2.5) and (2.7) we have∞∑
m=−∞
amz2m = xz2
∞∑m=−∞
am(xz)2m =∞∑
m=−∞
amx2m+1z2m+2
If we now re-index the last sum with m = n− 1 we have∞∑
m=−∞
amx2m+1z2m+2 =
∞∑n=−∞
an−1x2(n−1)+1z2(n−1)+2
=∞∑
m=−∞
am−1x2m−1z2m
where the last equality follows by simplifying the exponents and chang-
ing the index of summation from n to m. From this we have∞∑
m=−∞
amz2m =
∞∑m=−∞
am−1x2m−1z2m
and the result follows by equating coefficients of z2m.
�
Lemma 2.9.
am = a0xm2
Proof. Clearly the case m = 0 is true. Suppose that ak = a0xk2 for
some k ∈ N. Then by (2.8) we have
ak+1 = akx2k+1 = a0x
k2 · x2k+1 = a0x(k+1)2
so the claim is true for all m ∈ N by induction. Using (2.7) produces
the result for m ∈ Z. �
THE JACOBI TRIPLE PRODUCT 5
Lemma 2.10.
a0 = 1
Proof. Letting z = 1 into (2.7) and using (2.9) gives
G(1) = a0
∞∑m=−∞
xm2
.
On the other hand, letting z = 1 into (2.2) gives
F(1) =∞∏n=1
(1 + x2n−1
) (1 + x2n−1
)=∞∏n=1
(1 + x2n−1
)2Then substituting z = 1 into (2.4) with this result gives
G(1) = F(1)∞∏n=1
(1− x2n
)=∞∏n=1
(1 + x2n−1
)2 ∞∏n=1
(1− x2n
)=∞∏n=1
(1 + x2n−1
)2 (1− x2n
)Comparing these two equations for G(1) shows that the constant
term a0 is 1, completing the proof. �
We are now in a position to prove the Jacobi Triple Product (2.1).
Proof. By (2.10), (2.9) and (2.7), we have
G(z) =∞∑
m=−∞
xm2
z2m
By inserting (2.2) into (2.4) we have
G(z) = F(z)∞∏n=1
(1− x2n
)=∞∏n=1
(1 + x2n−1z2
)(1 +
x2n−1
z2
) ∞∏n=1
(1− x2n
)=∞∏n=1
(1 + x2n−1z2
)(1 +
x2n−1
z2
)(1− x2n
)Equating these forms the result. �
6 RAGIB ZAMAN
3. Application to counting lattice points on spheres
The problem of counting the number of lattice points inside the
boundary of a circle of radius r centered at the origin is called ”Gauss’s
Circle Problem” as it was Gauss who made the first substantial progress
on it with an asymptotic approximation. More generally, the problem
of finding a closed formula for the number of lattice points inside N
dimensional spheres is currently unresolved. In this section, by an
application of the Jacobi Triple Product, we will arrive at a method
(but not a closed formula) for counting the number of lattice points
inside a N dimensional sphere.
Definition 3.1. We call a ∈ RN a lattice point if all of its components
are integer valued. Define the N-sphere of radius r to be the set
SN ={
(x1, x2, · · · , xN) ∈ RN : x21 + x22 + · · ·+ x2N = r2}
Then the set of lattice points on the surface of the N -sphere with radius
r is
L(N, r) ≡{
(x1, x2, · · · , xN) ∈ ZN : x21 + x22 + · · ·+ x2N = r2}
and the set of lattice points inside the N -sphere of radius r is
L(N, r) ≡{
(x1, x2, · · · , xN) ∈ ZN : x21 + x22 + · · ·+ x2N ≤ r2}.
Lemma 3.2. The number of lattice points on the surface of the N-
sphere is given by
|L(N, r)| = [xr2
]
( ∞∑m=−∞
xm2
)N
where [xk] f(x) denotes the coefficient of xk in the series expansion of
f(x).
Proof. Consider( ∞∑m=−∞
xm2
)( ∞∑m=−∞
xm2
)· · ·( ∞∑
m=−∞
xm2
)︸ ︷︷ ︸
N factors
When expanding this product, we must pick a single term from each
sum. If we choose xz21 from the first factor, xz
22 from the second, and
so on, then the final term produced by this is
xz21+z22+···z2N
THE JACOBI TRIPLE PRODUCT 7
Thus, the coefficient of xr2
is the number of ways we could have
picked the terms such that z21 + z22 + · · · z2N = r2, which is precisely
when (z1, z2, · · · , zN) is a lattice point of the N -sphere. This proves
the claim. �
Lemma 3.3.∣∣L(N, r)∣∣ = [xr
2
]∞∏n=1
(1− x2n
)N (1 + x2n−1
)2NProof. Letting z = 1 into the Jacobi Triple Product (2.1) gives
∞∏n=1
(1− x2n
) (1 + x2n−1
)2=
∞∑m=−∞
xm2
Combining this with (3.2) yields the result. �
Theorem 3.4. The number of lattice points bounded by the N-sphere
of radius r centered at the origin is given by∣∣L(N, r)∣∣ =
r2∑k=0
([xk]
∞∏n=1
(1− x2n
)N (1 + x2n−1
)2N)Proof. By definition, every component of a lattice point is an integer, so
the sum of the squares of its components is also an integer. Therefore
lattice points will lie precisely on the surfaces of spheres whose radius
squared is an integer. This gives us
L(N, r) ={
(x1, x2, · · · , xN) ∈ ZN : x21 + x22 + · · ·+ x2N ≤ r2}
=⋃
k∈[0,r2]
{(x1, x2, · · · , xN) ∈ ZN : x21 + x22 + · · ·+ x2N = k
}=
r2⋃k=0
L(N,√k).
Thus, by the sum principle of disjoint sets we have∣∣L(N, r)∣∣ =
r2∑k=0
∣∣L(N,√k)∣∣
=r2∑k=0
([xk]
∞∏n=1
(1− x2n
)N (1 + x2n−1
)2N)where the last equality followed from (3.3).
�
8 RAGIB ZAMAN
Example 3.5. The number of lattice points on a 3-dimensional sphere of
radius 3 centered at the origin can be found by taking the first 5 terms
of the above product. Note that terms further along in the product
produce terms that do not effect the coefficient of x9. It is simple (but
tedious) to compute
5∏n=1
(1− x2n
)3 (1 + x2n−1
)6= 1 + 6x+ 12x2 + 8x3 + 6x4 + 24x5
+ 24x6 + 12x8 + 30x9 +O(x10)
where O(xk)
indicates a sum of terms with degree greater than or
equal to k.
Since the coefficient of x9 is 30, by (3.3)∣∣L(3, 3)∣∣ = 30
In other words, there are 30 lattice points on the surface of the 3-
dimensional sphere of radius 3 centered at the origin. Similarly, there
are 6 lattice points on the surface of the 3-sphere of radius 2.
Example 3.6. By (3.4) and the above expansion, the number of lattice
points inside the 3-sphere of radius 3 is given by∣∣L(3, 3)∣∣ =
9∑k=0
([xk]
∞∏n=1
(1− x2n
)3 (1 + x2n−1
)6)= 1 + 6 + 12 + 8 + 6 + 24 + 24 + 12 + 30
= 123
THE JACOBI TRIPLE PRODUCT 9
4. Euler’s Pentagonal Number Theorem
Euler’s Pentagonal Number Theorem is a special case of the Jacobi
Triple Product and is named as such due to the appearance of the
”Pentagonal Numbers” in the exponent of the summand. These num-
bers have an appealing geometric interpretation many will be familiar
with, but as it is not required here we will investigate that no further.
The statement of the theorem is presented below.
Theorem 4.1.
∞∏n=1
(1− zn) =∞∑
m=−∞
(−1)mz(3m−1)m/2
Proof. We proceed by substituting x = y3/2 and z2 = −√y into the
following:∞∏n=1
(1− x2n
) (1 + x2n−1z2
)(1 +
x2n−1
z2
)
=∞∏n=1
(1 + y3(2n−1)/2(−√y)
)(1 +
y3(2n−1)/2
−√y
)(1− y3n
)=∞∏n=1
(1− y(6n−3+1)/2
) (1− y(6n−3−1)/2
) (1− y3n
)=∞∏n=1
(1− y3n−1
) (1− y3n−2
) (1− y3n
)=∞∏n=1
(1− yn)
By the same substitution,
∞∑m=−∞
xm2
z2m =∞∑
m=−∞
y3m2/2 (−√y)m
=∞∑
m=−∞
(−1)my(3m2+m)/2
=∞∑
m=−∞
(−1)my(3m−1)m/2
where the last equality is obtained by reversing the index of summation
and factoring the exponent. By the Jacobi Triple Product (2.1) these
two are equal, which concludes the proof. �
10 RAGIB ZAMAN
5. Application to the theory of integer partitions
Definition 5.1. Let P(n) denote the number of partitions of an integer
n. For convenience, also set the conventions that P(0) = 1 and P(n) =
0 for negative integers n.
Example 5.2. Some values of the partition function are P(−1) = 0,P(0) =
P(1) = 1,P(2) = 2,P(3) = 3,P(4) = 5,P(5) = 7, P(100) = 190, 569, 292
and P(200) = 3, 972, 999, 029, 388.
Definition 5.3. Define a distinct partition to be a partition all of
whose parts are distinct. Define a soft partition to be a distinct parti-
tion which has an even number of parts and a hard partition to be a
distinct parition which has an odd number of parts. Denote the num-
ber of distinct, soft and hard paritions of an integer n by Pd(n),Ps(n)
and Ph(n) respectively.
Example 5.4. 1 + 3 + 4 + 7 is a soft partition of 15. 2 + 9 + 12 is a hard
partition of 23. 4 + 7 + 7 is a partition of 18 that is not distinct, hence
it is neither hard nor soft.
Theorem 5.5.
Ps(n)− Ph(n) =
{(−1)m for n = (3m−1)m
2for some m ∈ Z
0 otherwise
Proof. For a fixed integer n, it is well known (see [1]) that we have the
following equality of formal expressions∞∏k=1
(1 + zk
)=
∞∑m=0
Pd(n)zn
The idea behind this is that the coefficient of zn in
(1 + z)(1 + z2)(1 + z3) · · · (1 + zn) · · ·
is determined by the number of possible ways to pick 1 or zk from
each of the factors so that the product of the terms chosen is zn. This
corresponds to choosing to include or exclude k as a part of a partition
of n.
In a similar manner, consider the coefficient of zn in∞∏k=1
(1− zk
)= (1− z)(1− z2)(1− z3) · · · (1− zn) · · ·
THE JACOBI TRIPLE PRODUCT 11
Since we can pick zk at most once, the partition is necessarily dis-
tinct. For every term we pick we also get a factor of (−1) from it. Thus
if we pick an even number of terms (corresponding to a soft partition)
then +1 is contributed to the coefficient of zn. Similarly, if we choose
an odd number of terms, which corresponds to to a hard partition, then
−1 is contributed to the coefficient of zn. Therefore, the coefficient of
zn in∞∏k=1
(1− zk
)is Ps(n)−Ph(n). Now by equating coefficients of zn in (4.1) we get the
result. �
Theorem 5.6.
P(n) =∞∑
m=−∞
(−1)m−1P(n− qm)
= P(n− 1) + P(n− 2)− P(n− 5)− P(n− 7) + · · ·
where qm = (3m− 1)m/2.
Proof. Using (4.1) we find
(∞∏n=1
(1− zn)
)(∞∑
m=0
P(m)zm
)=
(∞∑
m=−∞
(−1)mz(3m−1)m/2
)(∞∑
m=0
P(m)zm
)On the other hand, we have the well known identity
∞∑m=0
P(m)zm =
( ∞∏n=1
(1− zn
))−1which can be found in [1]. Combining these gives(
∞∑m=−∞
(−1)mz(3m−1)m/2
)(∞∑
m=0
P(m)zm
)= 1
and comparing coefficients of powers of z establishes the result.
�
Remark 5.7. This sum at first sight may appear to be an infinite sum,
but note that eventually all terms are 0 as P(n) = 0 for negative
integers n.
12 RAGIB ZAMAN
Remark 5.8. This recurrence relation gives us a relatively convenient
way of calculating partition numbers for small values of n. However,
the values of P(n) grow quite quickly. In fact, it can be shown (see [1])
that
P(n) ∼ 1
4√
3nexp
(π√
2n/3)
as n→∞
where we write f(n) ∼ g(n) as n→∞ if limn→∞
f(n)/ g(n) = 1. This fol-
lowing method relies on repeatedly breaking down the term into smaller
pieces, each with small values. Thus, with this recurrence relation the
calculation of P(n) becomes difficult if n is much larger than 15.
Example 5.9. As an example we evaluate P(8). Applying (5.6) we have
P(8) = P(7) + P(6)− P(3)− P (1)
From (5.2) we have P(1) = 1,P(3) = 3. We can find P(6) by
applying (5.6) again
P(6) = P(5) + P(4)− P(1)
= 7 + 5− 1
= 11
Doing the same for P(7)
P(7) = P(6) + P(5)− P(2)− P(1)
= 11 + 7− 2− 1
= 15
Thus,P(8) = 15 + 11− 3− 1
= 22
Example 5.10. If we take n = 200 then by (5.8)
P(200) ≈ 1
800√
3exp
(π√
400/3)
≈ 4.10025× 1012
This differs to the precise value given in (5.2) by less than 4%.
THE JACOBI TRIPLE PRODUCT 13
6. Exercises
Exercise 1. - Compute the following product to 5 decimal places.∞∏n=1
3n − 1
3n
Exercise 2. - Express the the following sum in terms of the product in
the previous exercise and hence estimate it to 5 decimal places.
∞∑n=1
1
k(1− 3k)
Exercise 3. - Compute P(13).
Exercise 4. - Compute Ph(7526)− Ps(7526).
Exercise 5. - Give a rough estimate for P(1000).
Exercise 6. - Find the number of lattice points inside a 4-sphere of
radius 4. You may use the following in your calculation:8∏
n=1
(1− x2n
)4 (1 + x2n−1
)8= 1 + 8x+ 24x2 + 32x3 + 24x4 + 48x5
+ 96x6 + 64x7 + 24x8 + 104x9 + 144x10
+ 96x11 + 96x12 + 112x13 + 192x14 + 192x15
+ 24x16 +O(x10)
Exercise 7. - Show that for all k ∈ N
[xk]∞∏n=1
(1− x2n
)3 (1 + x2n−1
)6 ≥ 0
and find the minimal k where equality holds.
Exercise 8. - Find the minimal N ∈ N such that
[xk]∞∏n=1
(1− x2n
)N (1 + x2n−1
)2N ≥ 1
for all k ∈ N. Hint: A theorem of Lagrange is useful here.
14 RAGIB ZAMAN
References
[1] G.H. Hardy and E.M. Wright, An Introduction to the Theory of Numbers, Ox-
ford Science Publications
[2] E.W. Weisstein, Jacobi Triple Product From MathWorld–A Wolfram Web Re-
source. http://mathworld.wolfram.com/JacobiTripleProduct.html
E-mail address: [email protected]
