e Integer Partition Function
Kevin Durant
October 2010
Abstract
e function p(n) counts the number of integer partitions of a positive integer n. e purpose of this text
is to deduce a remarkable formula for p(n): the convergent infinite series obtained by Hans Rademaer.
Initially the reader is familiarised with the basic concepts of generating functions, Möbius transformations,
Farey fractions and Ford circles, before these are called upon to aid in the explanation of the proof of
Rademaer’s formula. ereaer a few auxillary results are provided to conclude the text, among them
the asymptotic formula first presented by Hardy and Ramanujan.
1 Introduction
First things first: the partition function p(n) counts the ways in whi n can be represented as sums of
positive integers; su a representation is known as a partition. ese partitions are unordered, and p(0)
is defined as 1. As an example, the partitions of 5 are
{(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)},
so that p(5) = 7. e partition function is strictly positive, as p(n) ≥ n. It is also an increasing function,
as every partition of n can be made into a partition of n + 1 by adding 1 to the set, and n + 1 has at least
one more partition — namely (n + 1).
e theory of partitions owes mu to Leonhard Euler — he penned numerous important theorems on
the topic, whi would garner aention from many of the most prolific mathematicians of the last three
centuries (Andrews 1984, p. xv). e most illustrious result in this area, however, is due to Hardy &
1
Ramanujan (1918). e pair obtained an asymptotic formula for p(n); a finite sum with error O(n− 4 ),
and in doing so, they also created and used what would later come to be known as the Hardy-Lilewood
circle method. Two decades later, H. Rademaer refined the approa of Hardy and Ramanujan (1974,
pp. 100-103) and was rewarded with an exact formula for p(n), whi shall be the focus of this text.
e circle method has become an important tool in additive number theory, but its application to the
partition function remains its most endearing aievement. e intuitive idea of the method can be seen
in the case where a function is defined inside the unit disk but has singularities along the unit circle. e
contour integral of this function is sought, and a circle with radius slightly less than 1 is used as the curve
of integration. is circle is dissected into arcs related to ea singularity; two sets of integrals are formed:
the ‘major’ and ‘minor’ arcs, whose singularities are more and less ‘important’ respectively. An integrable
function whi approximates the integrand is then required. Replacing the integrand of the major arcs
with this function and integrating, and showing that the contribution of the minor arcs is of an order less
than the major arcs, will hopefully yield an expression with a small enough error term to render it an
asymptotic equivalent of the original integral. An introduction to the circle method is given in Chapter
13 of Miller & Takloo-Bighash (2006), and a popular reference for its applications is Vaughan (1997).
e motivation behind the circle method will hopefully become clear during the investigation of its application to p(n), as the handiwork of three mathematicians make this application rather striking. e proof
of the exact formula for p(n) follows the same course as that outlined in Chapter 5 of Apostol (1990), and
as far as possible the same or similar function labels have been used.
1.1 Proof outline
e coefficients of the function
∞
∏
F (x) =
k=1
∞
∑
1
=
p(k)xk
1 − xk
k=0
contain the sequence (p(n)). We seek a formula for p(n). With a lile bit of manipulation,
F (x)
=
xn+1
∞
∑
p(k + n + 1)xk ,
k=−n−1
1
so that the coefficient of x−1 is p(n), and Cauy’s residue theorem gives us
∫
1
F (x)
p(n) =
dx.
2πi C xn+1
(x)
We are thus interested in the behaviour of xFn+1
within the unit disk, as this is where it is defined, and we
must integrate it. e cornerstone of the proof is the fact that when F (x) is calculated near a complex
root of unity, F (x) can be approximated by a function whi can be integrated. us we oose a path C
near the roots of unity on the unit circle. We split our path up so that we consider it as the sum of integrals
related to ea root of unity, and hope that the difference between our approximation of F near ea root
and the actual value of F (x) is very small. It turns out that Rademaer’s method for integrating near the
roots of unity is so effective that this difference vanishes as we consider the infinite sum of these integrals.
2
2 Generating Functions
To begin, we must obtain an infinite series whi contains information about the function p within its
terms. For this cause, generating functions (Chapters 5, 6 of Wagner (2009)) provide an elegant and intuitive
process. To a sequence of complex numbers a0 , a1 , a2 , . . . , we associate the power series
A(x) =
∞
∑
an xn ,
n=0
where the coefficient of
is an . roughout this text, unless otherwise stated, x and z will be regarded
as complex. If we consider the geometric progression 1, r, r2 , r3 , . . . , we obtain the generating function
xn
R(x) =
∞
∑
r n xn
r=0
=
∞
∑
(rx)n =
r=0
1
,
1 − rx
whi gives the important representation of 1, 1, 1, . . . :
∞
∑
xn =
n=0
1
.
1−x
In combinatorics, one is oen concerned with structures of different sizes, and specifically, counting the
number of structures of a given size. Approaing generating functions from this perspective, we can view
xn as a structure of size n and the coefficient an as the number of su structures present.
For an example, consider the family of positive integers I = {1, 2, 3, . . . }. If we say integer k has size k,
then there is exactly one integer of ea size, providing us with the generating function
I(x) =
∞
∑
xn = x + x2 + x3 + · · ·
n=1
= x(1 + x + x2 + · · · )
∞
∑
x
=x
xn =
.
1−x
n=0
is view of generating functions lets us handle families of structures simply. If A and B are two disjoint
families, then it is not difficult to see that the generating function of A ∪ B will be A(x) + B(x).
Pairs of structures (a, b), with a ∈ A and b ∈ B, can also be considered. A pair (a, b) has size equal to
the sum of the sizes of a and b. us the number of pairs of size n is
n
∑
ak bn−k ,
k=0
and the generating function for pairs of structures from A and B is
∞ ∑
n
∑
ak bn−k xn = A(x)B(x).
n=0 k=0
3
A similar result holds for other tuples (triples, quadruples, etc).
Sequences of structures from a family A build upon this principle. A sequence can be of any length, and
differentiating them as su provides a simple way of understanding their generating function. A sequence
of length 1 would be any a ∈ A. Sequences of length 2 are all the pairs (a1 , a2 ); of length 3 are all the
triples; and so forth. is implies that
Seq(A) = {} ∪ A ∪ (A × A) ∪ (A × A × A) ∪ · · ·
is the set of all sequences, and thus that the generating function S(x) for sequences of structures on A is
S(x) =
∞
∑
A(x)n =
n=0
1
.
1 − A(x)
2.1 Integer compositions
Rather remarkably, these tools are already sufficient to solve a problem only slightly different from the
focus of this text. Namely, how many partitions of an integer n are there when these partitions are ordered?
Ordered partitions are known as compositions, and we shall denote the number of compositions of a
positive integer as c(n). en
c(1) = 1;
{(1)}
c(2) = 2;
{(2), (1, 1)}
c(3) = 4;
{(3), (2, 1), (1, 2), (1, 1, 1)}
c(4) = 8;
{(4), (3, 1), (1, 3), (2, 2), (2, 1, 1), (1, 2, 1), (1, 1, 2), (1, 1, 1, 1)}
Compositions can easily be deconstructed in terms of generating functions, as a composition of n is a
sequence of any length with size n. us the set of compositions of n is the set of sequences of size n.
Recall that the generating function for the positive integers I is
x
,
I(x) =
1−x
so that the generating function for sequences of positive integers Seq(I) = C is
1
1−x
C(x) =
,
x =
1 − 1−x
1 − 2x
and with a small amount of manipulation we find that
C(x) =
1
2
+
−x
1 − 2x
)
1
2
1−
( 2x
1
1
+1
=
2 1 − 2x
(
)
∞
∑
1
n n
=
1+
2 x
2
n=0
=
1
+
2
∞
∑
2n−1 xn .
n=0
1
2
e only affects the constant term in the sum, so that c(0) =
0. For any n ≥ 1, c(n) = 2n−1 .
4
1
2
+ 2−1 = 1, giving us 1 composition of
2.2 Integer partitions
We can expand on the idea of sequences of structures with powersets and multisets. e powerset of a
family A is the set of all subsets of A, and is represented by
⊗
PSet(A) =
{a, ε},
a∈A
where the set {a, ε} shows that structure a can either be present or absent in the subset. If a is of size n,
then {a, ε} has generating function (1 + z n ), and if there are an su structures, we find the generating
function of the powerset to be:
∞
∏
B1 (x) =
(1 + z n )an .
n=1
A multiset is similar to a powerset except that ea structure can be included multiple times — essentially,
a sequence (whi may be of length 0) of ea structure occurs. Hence,
⊗
MSet(A) =
Seq({a}),
a∈A
and since the generating function of Seq({a}) is (1 − z n )−1 ,
B2 (x) =
∞
∏
(1 − z n )−an .
n=1
Now that we possess a foundation of descriptive generating functions, it should be clear that a partition
of an integer n is tenically just a multiset of size n over the family of integers I. Recalling that an = 1
as there is only one integer of size n, this yields a generating function for p(n) quite easily, namely
P (x) =
∞
∏
n −1
(1 − x )
n=1
=
∞
∑
p(n)xn .
n=0
e problem of finding an exact formula for p(n) is hardly as simple as it was for c(n), but is far more
interesting to investigate. In time we shall come upon this formula, but before we do we must establish
an elementary knowledge of a few related topics.
5
3 e Modular Group
e theory of partitions is linked to that of elliptic modular functions, and the approa of Hardy and
Ramanujan in obtaining their asymptotic formula for p(n) makes use of the Dedekind eta function η(τ ),
an important modular function in number theory. As su, a very simple overview of the necessary
properties of Möbius transformations and the modular group, as presented in Chapter 2 of Apostol (1990),
will be helpful.
Firstly, considering a transformation of the form
az + b
cz + d
where a, b, c, d are complex numbers, we can see that f is defined on the extended complex number
system for all values of z except z = − dc and z = ∞. en f can be extended, according to the intuition
that z0 = ∞ for non-zero z, by leing
)
(
a
−d
= ∞ and f (∞) = .
f
c
c
f (z) =
If one then considers another w in the extended complex system, then
aw + b az + b
−
cw + d cz + d
acwz + bcz + adw + bd acwz + bcw + adz + bd
−
=
(cw + d)(cz + d)
(cw + d)(cz + d)
(ad − bc)(w − z)
=
,
(cw + d)(cz + d)
f (w) − f (z) =
so that if ad − bc = 0, then f (w) − f (z) = 0 and f is constant. We thus consider only transformations
whi satisfy ad − bc 6= 0, and call these functions on the extended complex system Möbius transformations.
Furthermore, if ad − bc = k 6= 0, we can instead consider the analogous transformation
f (z) =
√a z
k
c
√ z
k
+
+
√b
k
d
√
k
=
az + b
cz + d
whi satisfies ad − bc = 1. We can thus replace the requirement that ad − bc 6= 0 by that of ad − bc = 1.
If we restrict our coefficients a, b, c, d to the integers, and denote f (τ ) as τ 0 , then the set of Möbius
transformations of the form
aτ + b
τ0 =
cτ + d
is called the modular group and is depicted by Γ. Ea transformation can also be wrien as a 2×2 integer
matrix with determinant 1:
(
)
a b
A=
,
c d
with A and −A both representing the same transformation. en, if two transformations τ 0 and µ0 have
relevant matrices A and B, the matrix product BA represents the composition transformation µ0 ◦ τ 0 .
Hence, because the set of 2 × 2 matrices with integer coefficients and determinant 1 form a group with
respect to multiplication, the modular group is, in fact, a group.
6
4 Farey Fractions
It was Rademaer who noted that one can make use of the theory of Farey fractions to obtain a more
accurate estimate of p(n), and as su we shall require a few simple properties of these fractions.
A reduced fraction is one in whi the numerator and denominator have no common factors. e set of
Farey fractions of order n is the set of reduced fractions lying in [0, 1] whose denominator is at most n,
and shall be denoted by Fn . So we have
{
}
0 1
F1 =
,
1 1
{
}
0 1 1
F2 =
, ,
1 2 1
{
}
0 1 1 2 1
F3 =
, , , ,
1 3 2 3 1
{
}
0 1 1 1 2 3 1
F4 =
, , , , , ,
1 4 3 2 3 4 1
and so forth. We can concisely define Fn as { hk : 0 ≤ h ≤ k ≤ n; (h, k) = 1}. It can also be easily seen
in the examples above that Fn ⊂ Fn+1 .
If we consider two fractions in Fn , say
defined as
a
b
<
c
d,
then an important new fraction called the mediant is
h
a+c
=
.
k
b+d
e mediant lies between these two fractions, as ab < dc implies that bc − ad > 0, so
bc − ad
a+c a
− =
>0
b+d b
b(b + d)
and
c a+c
bc − ad
−
=
> 0.
d b+d
d(b + d)
Also, in the case that bc − ad = 1, then
k = (bc − ad)k = b(ck) − d(ak) + (d(bh) − b(dh))
= b(ck − dh) + d(bh − ak),
(4.1)
so that ck − dh = bh − ak = 1. is means that if the relation bc − ad = 1 holds for two Farey fractions,
it holds between ea fraction and the mediant as well.
Equation (4.1) also gives some information revealing when two Farey fractions ab and
for the case where bc − ad = 1: if pq is any reduced fraction with ab < pq < dc then
c
d
are consecutive
q = b(cq − dp) + d(bp − aq),
and because cq − dp > 0 and bp − aq > 0 must be integer values, we have cq − dp ≥ 1 and bp − aq ≥ 1,
so the denominator of any fraction whi lies between ab and dc must be at least b + d. Combining this
7
with the fact that max(b, d) ≤ n because ab and dc are in Fn , we find that
for any n satisfying
max(b, d) ≤ n ≤ b + d − 1.
a
b
and
c
d
are consecutive in Fn
is provides us with enough information to fully describe the way in whi the sequence of Farey fractions F1 , F2 , . . . is built. Consider the construction of Fn+1 from Fn . If ab and dc are consecutive in Fn
and bc − ad = 1 then they will remain consecutive until n ≥ b + d. Forming the mediant
h
a+c
=
,
k
b+d
we know that (h, k) = 1 because
bh − ak = 1 and ck − dh = 1,
and any t whi divides both h and k must divide 1 as well. us, hk is in Fk , and these two equations also
imply that both the pairs ab < hk and hk < dc are consecutive in Fk , as max(b, k) = k = max(k, d). Noting
that the consecutive fractions in F1 satisfy 1 · 1 − 0 · 1 = 1, we have inductively shown that consecutive
Farey fractions satisfy a relation of the form bc − ad = 1, as when passing from Fn to Fn+1 , ea new
fraction is the mediant of two consecutive fractions satisfying su a relation and the new consecutive
pairs satisfy a similar relation.
8
5 Ford Circles
Originally, Rademaer used only Farey fractions to construct his exact formula for p(n). However, he
later refined this approa by making use of Ford circles — geometric constructs whi relate elegantly to
sets of Farey fractions.
e Ford circle of a reduced fraction hk is defined as the circle in the complex plane centered at hk + i 2k12
with radius 2k12 , and is denoted by C(h, k). ese circles shall be of high importance when addressing the
proof of Rademaer’s formula later in this text, and as su some valuable properties shall be developed
here.
1
2k2
h
k
Figure 5.1: e Ford circle C(h, k)
Firstly, consider C(a, b) and C(c, d), the Ford circles of two different reduced fractions
circles have centers
1
1
a
c
+ i 2 and
+i 2
b
2b
d
2d
respectively, and if D is the distance between these centers, then
2
D =
(a
c )2
−
+
b d
(
1
1
− 2
2
2b
2d
a
b
and dc . ese
)2
.
e square of the sum of their radii is
(
2
(r + R) =
1
1
+ 2
2
2b
2d
)2
.
e difference between these two lengths is
(
D − (r + R) =
2
2
(
ad − bc
bd
)2
(
+
1
1
− 2
2
2b
2d
)
ad − bc 2
1
=
−4 2 2
bd
4b d
2
(ad − bc) − 1
.
=
b2 d2
)2
(
−
1
1
+ 2
2
2b
2d
)2
Because ab and dc are different, ad − bc 6= 0, and D2 − (r + R)2 ≥ 0. Equality holds when (ad − bc)2 = 1.
Geometrically, this means that two different Ford circles never intersect, and that they tou if, and only
if, bc − ad = ±1. More importantly, this tells us that the Ford circles of consecutive Farey fractions are
tangent to ea other.
9
α1
a
h1
k1
b
h
k
Figure 5.2
Let us investigate this further, and determine these points of tangency. Let hk11 < hk < hk22 be three
consecutive Farey fractions. To find α1 (h, k), the point of contact of C(h1 , k1 ) and C(h, k) as in Figure
5.2, consider the large right-angled triangle above the line joining the centers of the two circles, with this
line as hypotenuse. is triangle is similar to the triangle whi has as hypotenuse the radius of C(h, k),
and horizontal and vertical lengths a and b respectively. en,
)
(
(
)
1
h
−a +i
α1 (h, k) =
+b .
(5.1)
k
2k 2
e large triangle has a hypotenuse of length
1
2k1 2
+ 2k12 , vertical length
1
2k1 2
− 2k12 and horizontal length
1
h h1
k − k1 , whereas the only known side of the smaller triangle is the hypotenuse — with length 2k2 .
because the two triangles are similar, we have
h
k
a
=
− hk11
1
2k2
1
2k1 2
+
1
2k2
=
1
2k1 2 k 2
·
2k 2 k 2 + k1 2
=
k1 2
k 2 + k1 2
and, also using the hypotenuse of the large triangle,
1
2k1 2
b
−
1
2k2
=
k1 2
,
k 2 + k1 2
so that, remembering that hk1 − h1 k = 1,
a=
and
b=
1
k1
k1 2
·
=
2
k 2 + k1 kk1
k(k 2 + k1 2 )
k1 2
k 2 − k1 2
k 2 − k1 2
.
·
=
k 2 + k1 2 2k 2 k1 2
2k 2 (k 2 + k1 2 )
10
However,
Combining this with equation (5.1), we find that
(
)
(
)
h
k1
1
k 2 − k1 2
α1 (h, k) =
−
+i
+
k k(k 2 + k1 2 )
2k 2 2k 2 (k 2 + k1 2 )
2k 2
h
k1
+
i
= −
k k(k 2 + k1 2 )
2k 2 (k 2 + k1 2 )
h
k1
i
= −
+
.
2
k k(k 2 + k1 ) k 2 + k1 2
In this investigation we have subtly assumed that k1 > k (see Figure 5.2). If the converse holds, we would
have
)
(
)
(
1
h
−a +i
−b .
α1 (h, k) =
k
2k 2
But in this case we would then divide b by
1
2k2
this would yield the term k1 2 − k 2 in place of
unanged.
−
1
to compare the ratio with the larger triangle, and
2k1 2
2
k 2 − k1 in the formula for b, and leave the formula for α1
Following an analogous course, it is shown that
α2 (h, k) =
k2
h
i
+
+ 2
.
2
2
k k(k + k2 ) k + k2 2
(5.2)
is provides us with sufficient knowledge of the structure of the Ford circles formed by sets of Farey
fractions. Let us draw nearer to p(n), and with it, the proof of an exact formula.
11
6 Rademaer’s Series
So far we have a generating function for p in the infinite product
F (x) =
∞
∏
m=1
∞
∑
1
=
p(n)xn
1 − xm
(6.1)
n=0
and a basic knowledge of the properties of Farey fractions and Ford circles. Our first order of business will
be to investigate further the function F , and ultimately its behaviour near the complex roots of unity.
6.1 e function F
Let us consider F (x) as in (6.1): to see where the function converges we need only consider the infinite
product, as our results will also apply to the series. An infinite product is said to converge when its limit
exists and is not zero. us, for an infinite product to converge, it is necessary that its terms approa 1.
1
If |x| ≥ 1, |xm | cannot have limit 0, so lim 1−x
m 6= 1, and the infinite product diverges. However, consider
∏
the case where |x| < 1. If we can show that the reciprocal product (1 − xm ) converges to l, it implies
that F (x) has limit 1l , and thus that F (x) converges.
e reciprocal product does converge throughout
∏
the
unit
disk
as
an
infinite
product
of
the
form
(1
−
xm ) converges if and only if the geometric series
∑ m
x converges.
So we shall restrict our focus to within the unit disk. If we divide F (x) through by xn+1 we obtain
∞
F (x) ∑ p(k)xk
=
xn+1
xn+1
k=0
for n ≥ 0. is gives us the Laurent series representation of
∞
∑
p(k)xk
k=0
xn+1
F (x)
,
xn+1
as
= p(0)x−n−1 + p(1)x−n + · · · + p(n)x−1 + p(n + 1) + · · ·
(x)
So xFn+1
has a pole of order n + 1 at x = 0 and residue p(n). is allows us to apply Cauy’s residue
theorem and find that
∫
1
F (x)
p(n) =
dx.
(6.2)
2πi C xn+1
6.2 e Rademaer path P (N )
ough we are yet to prove it, F (x) can be approximated by an elementary function whenever x lies
near a complex root of unity — a result whi follows from a functional equation satisfied by F . e
reason for postponing this result is that the functional equation’s form is more understandable once a few
transformations have been made. We shall encounter these transformations while elaborating on the path
of integration used by Rademaer.
So, assuming that we have reliable information about F ’s behaviour near the roots of unity, we would
prefer C to lie as near as possible to ea root of unity. is is where Rademaer was inspired with a more
12
elegant approa than that of Hardy and Ramanujan. He originally divided the circular contour used by his
fellows into arcs whose endpoints were defined by Farey fractions, as seen in Chapters 5 and 8 of Andrews
(1984) and Hardy (1978) respectively, but later refined this even further by making use of Ford circles to
integrate nearer to the singularities and thus simplify the estimations made (Apostol 1990). is laer
approa has proved useful in other areas of modular function theory (Andrews 1984, p. 85), and shall be
our elected course. e Ford circles of Section 5 reside in the upper half-plane H = {τ : Im(τ ) > 0}, and
we thus need to relate our domain, the unit disk, to H. An elegant mapping between the open unit disk
and H is found in the first transformation:
x = e2πiτ ,
whi maps the open unit disk onto the vertical strip V = {τ : 0 ≤ Re(τ ) < 1, Im(τ ) > 0} in H.
V
x-plane
i
1
0
0
1
Figure 6.1: e transformation from the x-plane to the τ -plane
is can be seen by taking a point τ = a + bi in H from whi we have
x = e2πi(a+bi)
= e2aπi−2bπ
=
e2aπi
.
e2bπ
Because e2aπi traverses the unit circle as a varies from 0 to 1, the integer part of a can be ignored. e
denominator is a real constant > 1, as b > 0, so x lies within the punctured disk.
If we then let b = 1, so τ varies from i to 1 + i along a horizontal line, we see that x traverses a circle of
radius e−2π with center 0.
Rademaer’s idea was to follow the curve in the τ -plane created by the upper arcs of the Ford circles
formed from Fn — the set of Farey fractions of order n. ese circles fit neatly into V, bounded by i on the
imaginary axis. If hk11 , hk and hk22 are consecutive fractions in Fn then there is an upper arc (whi does not
tou the real axis) from C(h, k)’s intersection with C(h1 , k1 ) to its intersection with C(h2 , k2 ). Note
that for the fractions 10 and 11 we consider only the parts of the circles contained within the vertical strip
V. e path P (N ), whi begins at i and ends at 1 + i, is the union of these arcs.
From Figure 6.2 it can be seen that P (N ) arcs away from every k-th complex root of unity, where k < N .
roughout the proof, n will remain fixed, as we seek to determine p(n), while N will be allowed to
approa infinity to account for infinitely many complex roots of unity.
13
x-plane
τ-plane
1.00
1
0.75
0
0.50
0.25
1
1
1
0
0.00
0.25
0.50
0.75
1.00
Figure 6.2: e Rademaer path P (N = 10). See Appendix C.1 for paths of different orders.
To make use of this path of integration, we note that the ange of variable
x = e2πiτ
yields
dx = x · 2πidτ,
so that (6.2) becomes
2πi
p(n) =
2πi
∫
=
∫
P (N )
F (x)
dτ
xn
F (e2πiτ )e−2πinτ dτ.
P (N )
If γ(h, k) denotes the upper arc of C(h, k), then we can split P (N ) into separate arcs for ea Ford circle
so that we may treat ea root of unity independently:
∫
=
P (N )
N
∑
∑ ∫
k=1 0≤h<k γ(h,k)
(h,k)=1
=
∑∫
h,k
where
∑
h,k
denotes the double sum over k and h.
14
γ(h,k)
ere remains one transformation to be made before we can deduce an alternate form of F :
(
)
h
2
z = −ik τ −
,
k
so that
h
iz
+ 2.
k k
To see the effect on a Ford circle C(h, k), the transformation can be broken down into elementary operations. e term τ − hk shis C(h, k) a distance of hk to the le, centering the circle above the origin. e
factor k 2 expands the radius to
1
1
k2 · 2 = ,
2k
2
−πi/2
while multiplication by −i = e
rotates the figure through − π2 radians. C(h, k) is thus transformed
from the τ -plane onto a circle K in the z-plane with center 12 and a radius of 12 .
τ=
z1 (h, k)
0
1
2
z1 (h, k)
K
Figure 6.3
e point of contact α1 (h, k) with the preceding circle becomes
(
)
h
2
z1 (h, k) = −ik α1 (h, k) −
k
)
(
k1
i
2
= −ik
−
k 2 + k1 2 k(k 2 + k1 2 )
k2
kk1
= 2
2 +i 2
k + k1
k + k1 2
(6.3a)
and similarly the other point of contact α2 (h, k) is now
z2 (h, k) =
k2
kk2
.
2 −i 2
2
k + k2
k + k2 2
(6.3b)
Because γ(h, k) does not tou the real axis in the τ -plane, the arc from z1 (h, k) to z2 (h, k) is the curve
whi does not tou the imaginary axis in the z-plane. Noting that
dτ =
i
dz,
k2
15
the effect of this ange of variable on p(n) is:
∑∫
F (e2πiτ )e−2πinτ dτ
p(n) =
h,k
γ(h,k)
(
(
))
2πih 2πz
i −2πinh/k 2πnz/k2
=
F exp
− 2
e
e
dz
k
k
k2
h,k z1 (h,k)
(
(
))
∫ z2 (h,k)
∑ i
2πih 2πz
−2πinh/k
2πnz/k2
e
F exp
=
e
dz.
− 2
k2
k
k
z1 (h,k)
∑∫
z2 (h,k)
h,k
e allenge of finding a usable form of F can now be aempted.
F (x) is an elliptic modular function, and in general this class of functions satisfy functional equations
whi describe them near the unit circle. We proceed by obtaining su an equation for F from the
known functional equation for η(τ ).
6.3 Dedekind’s functional equation
e eta function was introduced by Dedekind in 1877, and is defined on the upper half-plane H = {τ ∈
C : Im(τ ) > 0} as
∞
∏
πiτ /12
η(τ ) = e
(1 − e2πimτ ).
(6.4)
m=1
Dedekind’s eta function satisfies an important functional equation whi we shall require (Apostol 1990,
p. 52). Briefly, a functional equation is a aracterisation of a function whi relates its values at different
points, e.g.,
Γ(1 + x) = xΓ(x).
(a b)
To Dedekind’s functional equation: if c d ∈ Γ, c > 0 and τ ∈ H, then
)
(
1
aτ + b
= ε(a, b, c, d){−i(cτ + d)} 2 η(τ ),
η
(6.5)
cτ + d
where
{ (
)}
a+d
ε(a, b, c, d) = exp πi
+ s(−d, c)
12c
and
s(h, k) =
(
⌊ ⌋
)
k−1
∑
r hr
hr
1
−
−
.
k k
k
2
r=1
16
6.4 Application to F
One could hope that F and η are in some way linked when one considers the similarities in their basic
equations (6.1) and (6.4). ankfully, if x = e2πiτ ,
η(τ ) = eπiτ /12
=
∞
∏
(1 − e2πimτ )
m=1
πiτ
/12
e
F (e2πiτ )
.
functional equation for η (6.5) can in fact be represented in terms of F . If we remember that
(Dedekind’s
)
a b ∈ Γ and τ 0 = aτ +b , with c > 0, then the relation between η and F implies
c d
cτ +d
{ (
)}
0
1
eπiτ /12
eπiτ /12
a+d
2 exp
=
{−i(cτ
+
d)}
πi
+
s(−d,
c)
,
F (e2πiτ )
12c
F (e2πiτ 0 )
so that
F (e
2πiτ
) = F (e
2πiτ 0
With a careful oice of
(
) exp
(a b)
c d
πi(τ − τ 0 )
12
)
{ (
)}
a+d
{−i(cτ + d)} exp πi
+ s(−d, c)
.
12c
1
2
, it can be shown (see Section A.1) that
(
(
(
))
))
( π
(z )1 (
2πih 2πz
2πiH
2π
πz )
2
F exp
− 2
F exp
−
exp
−
= ω(h, k)
k
k
k
k
z
12z 12k 2
where
hH ≡ −1 (mod k),
ω(h, k) = eπis(h,k) ,
(h, k) = 1.
Note that s(h, k) ∈ Q so that ω(h, k) is a complex root of unity, with modulus 1.
Aer two transformations we have reaed the point where
)
(
2πih 2πz
2πiτ
− 2 .
x=e
= exp
k
k
Leing
0
x0 = e2πiτ = exp
(
2πiH
2π
−
k
z
)
,
the functional equation can be rewrien as
F (x) = ω(h, k)
(z )1
2
k
exp
( π
πz )
−
F (x0 ),
12z 12k 2
(6.6)
providing us, at length, with an alternate formula for F (x).
is equation does indeed provide us with valuable information about F ’s behaviour near the unit circle.
In light of the above representations of x and x0 , note that x is near to the root of unity e2πih/k when |z|
17
0
is small. However, the smaller z becomes, the larger 2π
z becomes, and the nearer x moves toward 0, the
0
origin. So when x is near to a root of unity, F (x ) ≈ F (0) = 1 and F (x) can be approximated by
ω(h, k)
(z )1
2
exp
k
( π
πz )
−
,
12z 12k 2
whi is an elementary function. e error in approximating F in this way will be determined by just
how close to 1 the value F (x0 ) really is, i.e., by the difference (F (x0 ) − 1).
Returning to our equation for p(n),
p(n) =
∑
ik
−2 −2πinh/k
∫
z2 (h,k)
e
e
2πnz/k2
z1 (h,k)
h,k
))
(
(
2πih 2πz
dz,
F exp
− 2
k
k
we can apply Dedekind’s functional equation and obtain
p(n) =
∑
ik
−5/2
ω(h, k)e
−2πinh/k
∫
z2 (h,k)
2
1
e2πnz/k z 2 exp
z1 (h,k)
h,k
( π
πz )
−
F (x0 ) dz.
12z 12k 2
We now perform the substitution F (x0 ) = 1 + (F (x0 ) − 1) to separate our elementary function from the
error made in our approximation, and find
p(n) =
∑
ik
−5/2
ω(h, k)e
−2πinh/k
1
z2 (h,k)
e2πnz/k Ψk (z){1 + (F (x0 ) − 1)} dz,
2
z1 (h,k)
h,k
with Ψk (z) = z 2 exp
∫
(
π
12z
−
πz
12k2
)
.
If we define
∫
z2 (h,k)
I1 (h, k) =
2
Ψk (z)e2πnz/k dz
(6.7a)
z1 (h,k)
and
∫
z2 (h,k)
I2 (h, k) =
Ψk (z)e
z1 (h,k)
2πnz/k2
{ (
(
))
}
2πiH
2π
F exp
−
− 1 dz
k
z
then we are finally le with the more visually appealing
∑
p(n) =
ik −5/2 ω(h, k)e−2πinh/k (I1 (h, k) + I2 (h, k)),
(6.7b)
(6.8)
h,k
in whi I1 (h, k) is the elementary integral with whi we wish to work and I2 (h, k) is the error made
in approximating our integral in this way.
Let us take a moment to evaluate our position. We began with an integral (6.2) obtained from Cauy’s
residue theorem and proceeded to find an integration curve P (N ) whi we could split into a sum of
integrals over the Farey fractions in Fn . We have now applied Dedekind’s functional equation and split
ea integral into two integrals I1 and I2 , ea along the arc from z1 (h, k) to z2 (h, k) on the circle K.
18
6.5 e contribution of I2 (h, k)
We now possess an almost reasonable-looking formula for p(n) — except for the term I2 (h, k), whi is
our difference term. Luily, I2 (h, k) is O(N −1/2 ), and thus becomes small for large values of N . To see
this, we show that the integrand is bounded and then integrate along the direct ord from z1 (h, k) to
z2 (h, k) instead of the arc. Before we investigate the integrand, note that
|ez | = eRe(z) eIm(z)i
= eRe(z)
as eIm(z)i is a point on the unit circle with modulus 1.
Now, considering the integrand of I2 (h, k) and making use of the infinite sum (with first term 1) in (6.1),
we have
))
}
{ (
(
2πiH
2π
2
− 1 e2πnz/k
Ψk (z) F exp
−
k
z
) {∑
)}
(
(
∞
( π
1
πz )
2πm
2πnz
2πimH
= z 2 exp
−
−
exp
p(m) exp
12z 12k 2
k2
k
z
m=1
(
)
(
)∑
∞
1
π Re(1/z) π Re(z)
2πn Re(z)
−
exp
p(m) e2πiHm/k e−2πm Re(1/z) .
= |z| 2 exp
12
12k 2
k2
m=1
Because z is a point on the ord from z1 to z2 , 0 < Re(z) < 1. To continue, we also need information
about Re(1/z). is is thus a good time to investigate the transformation 1/z for a point z = a + bi on
or within the circle K (see Figure 6.3):
(
)
1 2
1
a−
+ b2 ≤
2
4
so
a2 − a +
1
1
+ b2 ≤ ,
4
4
We can write
and a2 + b2 ≤ a.
1
a − bi
= 2
,
z
a + b2
and then investigate ea part separately:
Re(1/z) =
a2
a
a
≥ =1
2
+b
a
and
Im(1/z) = −
a2
b
.
+ b2
In the case where z lies on the circle K, we have a2 + b2 = a, so that Re(1/z) = 1 and
√
√
b
a − a2
1
Im(1/z) = − = ±
=±
− 1,
2
a
a
a
19
whi can take on any real value as 0 ≤ a ≤ 1.
Noting then that
)
(
π Re(z)
≤ 1,
exp −
12k 2
(
)
2πn Re(z)
exp
< e2πn ,
k2
e2πiHm/k = 1,
we can continue with our estimation, seeing that the modulus of the integrand is
1
(
< |z| 2 exp
1
2
= |z| e2πn
1
2
≤ |z| e2πn
1
2
< |z| e2πn
π Re(1/z)
12
∞
∑
m=1
∞
∑
m=1
∞
∑
)
e2πn
∞
∑
p(m)e−2πm Re(1/z)
m=1
p(m)e−2π Re(1/z)(m−1/24)
p(m)e−π(24m−1)/12
p(24m − 1)e−π(24m−1)/12 .
m=1
Taking y = e−π/12 and recognising that the infinite sum is F (y) and thus convergent (|y| < 1), we are
le with
1
c |z| 2 ,
where
c = e2πn
∞
∑
p(24m − 1)y 24m−1 ,
m=1
so that c is independent of N and z, and thus remains unanged throughout the double summation in
our current formula for p(n).
1
We have yet to consider the term |z| 2 , and thereaer to investigate the length of the ord between
z1 (h, k) and z2 (h, k). We will then be able to combine our results and obtain an estimate for |I2 (h, k)|.
Firstly, let us investigate |z| along the ord.
Making use of (6.3),
2
kk1
k2
+ 2
|z1 (h, k)| = 2
2
k + k1
k + k1 2
k 4 + k 2 k1 2
=
2
(k 2 + k1 2 )
k2
= 2
.
k + k1 2
2
20
2
A similar equation holds for z2 (h, k), so that
|z1 (h, k)| = √
|z2 (h, k)| = √
k
k2
+ k1 2
k
(6.9)
k 2 + k2 2
Now, the Cauy-Swarz inequality implies that
(k · 1 + k1 · 1)2 ≤ (k 2 + k1 2 )(1 + 1)
so that
√
k 2 + k1 2 ≥
k + k1
N +1
N
√
≥ √
≥√
2
2
2
because we have consecutive Farey fractions, so N ≤ k1 + k − 1. e same inequality holds when k1 is
replaced by k2 , so that
√
√
2k
2k
and |z2 | <
.
|z1 | <
N
N
Because z is on the ord,
|z| = |αz1 + (1 − α)z2 |
≤ α |z1 | + (1 − α) |z2 |
≤ max(|z1 | , |z2 |)
√
2k
.
<
N
is implies that
1
2
c |z| ≤ c 2
1
4
(
k
N
)1
2
.
√
Also, by way of the triangle inequality, the length of the ord is at most |z1 | + |z2 | = 2 2k/N . We can
thus bound the integral by multiplying the length of the ord by an upper bound for the integrand on the
ord, so that
( )3
k 2
|I2 (h, k)| < C
N
21
where C is a constant. Hence, noting the roots of unity ω(h, k) and e−2πinh/k ,
∑
ik −5/2 ω(h, k)e−2πinh/k I2 (h, k) ≤
∑
k −5/2 |I2 (h, k)|
h,k
h,k
<
N
∑
∑
k=1 0≤h<k
(h,k)=1
C
kN 3/2
N
C ∑1 ∑
= 3/2
1
k
N
0≤h<k
k=1
(h,k)=1
≤
N
C ∑
1
N 3/2 k=1
C
=√
N
Equation (6.8) then becomes
p(n) =
∑
ik −5/2 ω(h, k)e−2πinh/k I1 (h, k) + O(N −1/2 ).
(6.10)
h,k
6.6 e contribution of I1 (h, k)
We now know that I2 (h, k) is of an encouragingly small order, and will even vanish when we allow
N → ∞. But we are geing ahead of ourselves. Before we hastily get rid of our error term, we can show
that integrating I1 (h, k) along the entire circle K instead of just a section of its circumference will provide
us with an error also of O(N −1/2 ).
Recall equation (6.7a):
∫
z2 (h,k)
2
Ψk (z)e2πnz/k dz.
I1 (h, k) =
z1 (h,k)
If K − denotes the curve along the circumference of K in the negative direction then
∫
I1 (h, k) =
K−
∫
−
∫
=
K−
z1 (h,k)
z2 (h,k)
0
∫
−
z2 (h,k)
∫
−
z1 (h,k)
.
0
We split the difference into two separate integrals (call them J1 and J2 respectively) because
exist when z = 0. It remains to show that both J1 and J2 are O(N −1/2 ).
22
1
z
does not
z1 (h, k)
K
|z1 |
1
2
0
Figure 6.4
If one considers the straight line from 0 to z1 to be the diameter of a new circle (Figure 6.4), then the arc
on K from 0 to z1 lies within the upper half of this circle, so that the length of the arc is less than half of
the new circle’s circumference, i.e.,
π |z1 |
2
√
2k
<π
.
N
<
Similarly, this same inequality holds for the length of the arc from z2 to 0. Clearly, any√
z on either of these
arcs lies within or on the circle about the origin with radius max(|z1 | , |z2 |), so |z| < 2k/N .
We already know that when applied to the circle K, 1/z maps z onto the vertical line through 1, so
Re(1/z) = 1. We consider the integrand:
2
Ψk (z)e2πnz/k
( π
1
πz ) 2πnz/k2
−
= z 2 exp
e
2
(12z 12k
)
1
π Re(1/z) π Re(z) 2πn Re(z)/k2
−
= |z| 2 exp
e
12
12k 2
1
≤ |z| 2 eπ/12 e−π e2πn
( )1
k 2
< c1
,
N
using the inequality for |z| obtained above. Combining this with our knowledge of the length of the arcs,
we find that both
( )3
( )3
k 2
k 2
|J1 | < C1
and |J2 | < C1
,
N
N
where C1 is constant. Remembering that these together form the error when using the curve K − to
23
calculate I1 (h, k), we find
∑
ik −5/2 ω(h, k)e−2πinh/k (J1 + J2 ) ≤
h,k
∑
k −5/2 |J1 + J2 |
h,k
<
N
∑
∑
k=1 0≤h<k
(h,k)=1
=
2C1
kN −3/2
N
2C1 ∑ 1 ∑
1
N −3/2 k=1 k 0≤h<k
(h,k)=1
2C1
≤√ .
N
is, when applied to (6.10), gives us
p(n) =
∑
ik −5/2 ω(h, k)e−2πinh/k
∫
K−
h,k
Ψk (z)e2πnz/k dz + O(N −1/2 ).
2
6.7 Letting go
At present we are considering only a finite number of Ford circles — namely those based on FN . Leing
N → ∞, our equation becomes
∫
∞
∑
∑
2
ik −5/2 ω(h, k)e−2πinh/k
p(n) =
Ψk (z)e2πnz/k dz
k=1 0≤h<k
(h,k)=1
∞
∑
−5/2
K−
∫
( π
πz ) 2πnz/k2
−
e
dz,
12z 12k 2
K−
k=1
{
)}
(
∫
∞
∑
1
π
2πz
1
−5/2
=
ik
Ak (n)
z 2 exp
− 2 n−
dz,
12z
k
24
K−
=
ik
Ak (n)
1
z 2 exp
k=1
where we let
Ak (n) =
∑
ω(h, k)e−2πinh/k
0≤h<k
(h,k)=1
as it shall remain unanged for the remainder of the proof.
We then apply the ange of variable
z=
π
12t
with dz = −
π
dt,
12t2
noting that this is just a constant multiple of the transform z1 whi was investigated earlier, and thus
π
maps the circle K (and with it our path of integration) onto the line Re(z) = 12
.
24
en
(
{
)}
1
π2
π
π ) 12
p(n) =
exp t + 2 n −
dt
−ik
Ak (n)
12t
6k t
24
12t2
π/12−∞i
k=1
{
(
)}
∞
( π ) 3 ∫ π/12+∞i
∑
π2
1
2
=
−ik −5/2 Ak (n)
t−5/2 exp t + 2 n −
dt
12
6k t
24
π/12−∞i
k=1
{
(
)}
∫ π/12+∞i
∞
( π )3 ∑
π2
1
1
2
−5/2
−5/2
t
exp t + 2 n −
= 2π
k
Ak (n)
dt.
12
2πi π/12−∞i
6k t
24
∞
∑
∫
−5/2
π/12+∞i (
k=1
We can evaluate the integral in terms of Bessel functions, but for the sake of continuity the tenicalities of
the evaluation are performed in Section A.2, with a narrative sket provided here. Equation (B.1) found
in Section B.1 states that
( 1 )ν ∫ π/12+∞i
(
)
z2
−ν−1
2z
t
exp t +
dt = Iν (z),
2πi π/12−∞i
4t
and if z and ν in this integral are taken as
z
=
2
{
π2
6k 2
)} 21
(
1
n−
24
3
and ν = ,
2
we find that our integral can now be expressed as
1
2πi
∫
π/12+∞i
π/12−∞i
( √ (
{
(
)}
{ 2 (
)}− 34
))
2
π
1
π
1
π
2
1
t−5/2 exp t + 2 n −
dt =
n−
I3
n−
.
2
6k t
24
6k 2
24
k 3
24
Combining this equation with that of p(n) and simplifying gives us
( √ (
) 3 ∞
(
))
1 −4 ∑
2π n − 24
π 2
1
−1
k Ak (n)I 3
n−
.
p(n) =
3
2
k 3
24
(24) 4
k=1
I 3 (z) is a Bessel function of half odd order, and, specifically (B.2) gives
2
√
I 3 (z) =
2
2z d
π dz
(
sinh z
z
)
,
whi implies that

{ √ (
( √ (
π
2
))
)3
3 3 (
sinh
k
3 n−
π 2
1
64 k2
1 4 d 
√
n−
=
n
−
I3

2
k 3
24
π2
24
dn
n− 1
1
24
)} 

.
24
Once again simplifying our equation for p(n), we arrive at our infinite series representation:

{ √ (
)} 
2
1
π
∞
d  sinh k 3 n − 24 
1 ∑ 1
√
k 2 Ak (n)
p(n) = √
,

dn
π 2 k=1
n− 1
24
25
(6.11)
where, as before
Ak (n) =
∑
ω(h, k)e−2πinh/k .
0≤h<k
(h,k)=1
We have now deduced the convergent series for p(n), our main objective for this text. A few auxillary
results will yet be presented.
6.8 Asymptotics
When Hardy and Ramanujan published their results on p(n), they included the asymptotic formula
√
π
2n
3
e
√
p(n) ∼
as n → ∞.
4n 3
is formula can in fact be deduced from the first term of Rademaer’s infinite series. To prove this, we
need to show that the first term (denote the k-th term by Rk (n)) is of this form and that the weight of the
rest of the terms combined is of an order less than R1 (n). Recall from (6.11) that

{ √ (
)} 
π
2
1
sinh
n
−
1
k
3
24
d 

√
Rk (n) = k 2 Ak (n)

,
dn
n− 1
24
so that for the first term, where k = 1:
A1 (n) =
∑
eπis(h,k) e−2πinh/k
0≤h<k
(h,k)=1
= eπi·0 e−2πn·0
= 1,
because s(h, k) is a sum from 1 to k − 1 = 0, and is thus 0. e remaining part of R1 (n) is the derivative,

( √ (
)) 
1
2
d  sinh π 3 n − 24 
√
R1 (n) =


dn
n− 1
24
( √
) √ (
)
√
1
1
1 −1
1 π 23 (n− 24
2
−π 23 (n− 24
)
)
=
e
+e
π
n−
4
3
24
( √
)− 3
)(
√
2
1
1
1 π 23 (n− 24
1
) − e−π 23 (n− 24
)
−
e
n−
.
4
24
Also, we can make use of the binomial series and obtain:
√
(
)1
(
)
∞ (1)
2
√
√ ∑
1
1
1 m
k
2
n−
= n 1−
= n
(−1)
24
24n
m
24n
m=0
(
)
√
1
1
= n 1−
−
− ···
2 · 24n 8 · (24n)2
(
( ))
(
)
√
√
1
1
= n 1+O
= n+O √
,
n
n
26
so that, by the Taylor expansion of e,
√
e
π
2
3
1
(n− 24
) = eπ
√
2n
3
(
(
))
1
.
1+O √
n
Following on from the above, and noting that the exponential terms with negative exponents diminish
rapidly, we can further simplify our former result to:
√
(
(
))
√
π
2 π 2n
1
3
R1 (n) =
e
1+O √
.
4n 3
n
We now have a representation of the first term whi seems strikingly familiar to the asymptotic formula
we wish to obtain. To further our knowledge of the weight of the subsequent terms, we must investigate
the derivative term more closely. Leing
√ (
)
(
) ( )2
π 2
1
1
kt
3
t=
n−
, so that
n−
=
,
k 3
24
24
π
2
we find that


d  sinh t 
π
√
=
dn
2k
n− 1
24
With f (t) =
( cosh t
t2
−
sinh t
t3
√
2 cosh t
1 sinh t
)− (
(
1
3 n − 24
2 n − 1 ) 32
24
( )3 ( )
( )3
π 2 2 π 2
1 2 2 ( π )3
=
cosh t −
sinh t
2k 3
kt
2 3
kt
√
2 π 3 cosh t
2 π 3 sinh t
= √ 3 2 − √ 3 3
3 6k t
3 3k t
(
)
3
2 π
cosh t sinh t
= √ 3
− 3
.
t2
t
3 6k
(6.12)
)
, the Taylor series for cosh and sinh imply
(
) (
)
1
t2
t4
t
t3
t5
f (t) = 2 + 2 + 2 + · · · − 3 + 3 + 3 + · · ·
t
2!t
4!t
t
3!t
5!t
)
(
)
(
)
(
1
1
1
1
1
1
−
+ t2
−
+ t4
−
+ ··· ,
=
2! 3!
4! 5!
6! 7!
so that f is increasing and limt→0 f (t) 6= ±∞.
We can now apply this knowledge to the remaining terms. Because Ak (n) is a sum of roots of unity with
less than k terms, it is clear that |Ak (n)| ≤ k. With this in mind (and C a constant), it follows that:
( √ (
( √ (
))
))
∑ 1
∑ 3
2 π3
π
2
π 2
1
1
k 2 Ak (n) √ 3 f
≤C·
k− 2 f
n−
n−
k
k 3
24
k 3
24
3
6
k≥2
k≥2
√
(
(
)) ∑
3
1
π 2
n−
k− 2 ,
≤C ·f
2 3
24
k≥2
27
because f (t) decreases as k increases. Furthermore, f (t) does not approa −∞ as k increases, and the
infinite sum is a p-series with p > 1, and thus converges. So
∑
k≥2
{ ( √ (
))}
π 2
1
Rk (n) = O f
n−
2 3
24
= O(R2 (n)).
It is clear from the above processes that ea term is dominated by an exponential function, and although
the exponent of e is similar for ea term, the constant part of the exponent in R2 (n) (and subsequent
terms) is smaller than that of R1 (n), so that the first term dominates the sum. From (6.11) then,
1
p(n) ∼ √ R1 (n)
π 2
√
(
(
))
√
1
1 π
2 π 2n
3
1+O √
= √
e
n
π 2 4n 3
√
π
2n
3
e
√
∼
4n 3
e actual weight of
n
1
2
3
4
5
10
11
20
100
500
1000
5000
10000
1
√
R (n)
π 2 1
as n → ∞.
(6.13)
(and the growth of p(n)), is demonstrated below:
p(n)
1
2
3
5
7
42
56
627
190569292
2300165032574324047872
24061467864032624306850057158656
1.6982e+74
3.61673e+106
1
√
R (n)
π 2 1
1
2
3
5
7
42
57
626
190568945
2300165032573746544640
24061467864032723386041859309568
1.6982e+74
3.61673e+106
Table 1: Comparison of p(n) and its first term approximation (rounded to the nearest integer).
e difference between the first term approximation and p(n) does grow infinitely large with n, but at
a far slower rate than the increase of the partition function. Because the contribution of the other terms
1
becomes less important as n becomes large, the ratio of π√
R (n) to p(n) tends to 1.
2 1
28
1.015
1.010
1.005
1.000
0.995
0.990
0.985
0.9800
20
40
60
n
Figure 6.5: e ratio of
1
√
R (n)
π 2 1
80
100
to p(n).
6.9 Distinct summands
Another interesting result due to Goh & Smutz (1995) is that the average number of distinct summands
in a partition of n is
√ (
( 1 ))
6n
.
(6.14)
D(n) =
1 + O n− 6
π
Firstly, let us obtain a combinatoric formula for this quantity. Instead of considering ea partition individually, regard the set of all partitions of n. Summing the number of partitions containing a 1 with the
number of partitions containing a 2, then a 3, and so forth, counts ea distinct summand in every partition once, and so gives the total number of distinct summands over all partitions. Dividing this quantity
by p(n) gives the average number of distinct summands in a typical partition. e number of partitions
containing the integer r is given by p(n − r), as removing the r gives a partition of n − r, and conversely
aaing r to a partition of n − r results in a partition of n. Hence,
D(n) =
p(n − 1) + p(n − 2) + · · · + p(0)
.
p(n)
To understand the numerator, we recall equation (6.13), and see that
√
π
2(n−r)
3
e
p(n − r) = √
4 3(n − r)
(
(
))
1
1+O √
.
n−r
Making use of the binomial series to expand and noting that nr ≤ 1 gives
(
)
(
( 2 ))
√
√ (
√
r ) 12 √
1
r2
r3
r
r
n−r = n 1−
− 2−
− ··· = n 1 −
+O
,
= n 1−
3
n
2n 8n
16n
2n
n2
29
1
1(
r )−1
1
=
1−
=
n−r
n
n
n
(
)
( r ))
r
r2
1(
1 + + 2 + ··· =
1+O
,
n n
n
n
so that
{ √ √
( 2 )}
(
))
(
( r )) (
exp π 23 n − √π6 √rn + O r 3
1
2
n
√
p(n − r) =
1+O
1+O √
n
n−r
4 3n
(
)
(
))
{
}
{ ( 2 )} (
(
)
r
1
r
πr
r
√
1+O
+O √
+O
= p(n) · exp − √
exp O
3
n
n−r
n n−r
6n
2
n
{
}
{ ( 2 )} (
(
))
(r)
πr
r
1
= p(n) · exp − √
exp O
1+O
+O √
,
3
n
n−r
6n
n2
√
Now, large values of r (greater than n) will contract the value of p(n − r), as the effect of the first
exponential function outweighs the second:
r2
r2
r
√ = √ ≥ 3
n
r n
n2
2
1
because r < n. So also, if r > n 3 , then √rn > n 6 , and p(n − r) → 0 exponentially as n → ∞. is
shall prove to be an opportune value at whi to separate the r’s. ere are less than n terms with su
2
an r, so their contribution tends to 0. We thus consider terms p(n − r) with r ≤ n 3 . e Taylor series
for exp(z) implies
(
exp
r2
3
n2
)
r2
1
=1+ 3 +
2!
n2
(
r2
)2
(
+ ··· = 1 + O
3
n2
r2
3
)
,
n2
3
as r2 < n 2 . Applying this to our equation for p(n − r) yields
))
{
}(
( 2)
(
(r)
πr
r
1
√
√
+O
.
p(n − r) = p(n) · exp −
1+O
+O
3
n
n−r
6n
n2
2
Because r ≤ n 3 , the fractions can be ordered
r2
n
3
2
≤ n− 6
1
and
1
r
≤ n− 3 ,
n
2
and for sufficiently sized n (n ≥ 8), n 3 ≤ n2 , so that
√
√
n≥ n−r ≥
implying that
√1
n−r
√
n
,
2
( 1)
= O n− 2 . is knowledge combined simplifies our equation to
}
{
( 1 ))
πr (
1 + O n− 6
p(n − r) = p(n) · exp − √
,
6n
30
and, when substituted into (6.14):
bn2/3 c
{
}
( 1 ))
1 ∑
πr (
p(n) · exp − √
D(n) =
1 + O n− 6
p(n)
6n
r=1
{
})r
(
( 1 )) bn∑ c (
π
= 1 + O n− 6
exp − √
.
6n
r=1
2/3
e sum is recognisable as a finite geometric series, and can be simplified to
(
{
(⌊
⌋
)})
( 1 ))
(
1 − exp − √π6n n2/3 + 1
{
}
D(n) =
1 + O n− 6
.
1 − exp − √π6n
⌊
⌋
√
Once again, the exponential term in the numerator approaes 0 quily, as n2/3 > n. Expanding
the other exponential function yields
D(n) =
(
1− 1+
1
(
√π
6n
(
( 1 ))
)) 1 + O n− 6
2
π
+ O 6n
(
( 1 ))
1
( ) 1 + O n− 6
=
2
√π + O π
6n
6n
√ (
(
))−1 (
( 1 ))
π
6n
=
1+O √
1 + O n− 6
.
π
6n
Lastly, the binomial series can once again be applied, this time to the reciprocal term:
√ (
(
)) (
( 1 ))
π
6n
1+O √
D(n) =
1 + O n− 6
π
6n
√ (
))
(
1
6n
,
=
1 + O n− 6
π
and the result is obtained. With a similar but more extensive argument it can also be shown that the
average number of summands (counting repeated summands) of a partition of n is
√
√
6n
log n + O( n).
2π
31
A Appendix of Calculations
A.1 Dedekind’s functional equation in terms of F
is deduction is outlined in Section 6.4.
e relationship between F and η implies that if
F (e
2πiτ
) = F (e
2πiτ 0
(
) exp
πi(τ − τ 0 )
12
)
(a b)
c d
∈ Γ, c > 0 and τ 0 =
aτ +b
cτ +d ,
then
{ (
)}
a+d
{−i(cτ + d)} exp πi
+ s(−d, c)
.
12c
1
2
Now if c = k and d = −h so that s(−d, c) = s(h, k) and if a = H, then b = − hH+1
because
k
iz+hk
In our case, where τ = k2 ,
(a b)
c d
∈ Γ.
Hiz + hHk hHk + k
−
k2
k2
Hiz − k
=
,
k2
iz + hk
cτ + d =
−h
k
iz
=
k
aτ + b =
so
Hiz − k
iz −1 k + H
=
.
kiz
k
Seing this into our functional equation, we have
τ0 =
(
(
))
(
(
)) { ( )} 1
2
2πihk 2πz
2πiH
2πk
iz
F exp
−
=
F
exp
−
−i
k2
k2
k
zk
k
))
{ (
)}
( (
−1
2
H −h
πi iz − iz k + hk − Hk
exp
πi
+
s(h,
k)
× exp
12
k2
12k
(
(
)) ( ) 1
2πiH
2π
z 2
= F exp
−
k
z
k
( (
))
πi h − H
H −h
iz
i
+
+ 2−
exp (πis(h, k))
× exp
12
k
k
k
z
(
(
)) ( ) 1
( π
)
2πiH
2π
z 2
πz
= F exp
exp
−
−
+
πis(h,
k)
,
k
z
k
12z 12k 2
as required.
32
A.2 Evaluation of the Bessel Integral
We have the function
{
(
)}
∫ π/12+∞i
∞
( π )3 ∑
1
π2
1
2
−5/2
−5/2
p(n) = 2π
k
Ak (n)
t
exp t + 2 n −
dt,
12
2πi π/12−∞i
6k t
24
k=1
and wish to evaluate the integral. Equation (B.1) implies that
( 1 )ν ∫ π/12+∞i
)
(
z2
−ν−1
2z
dt = Iν (z),
t
exp t +
2πi π/12−∞i
4t
and if z and ν in this integral are taken as
z
=
2
we have
1
2πi
∫
π/12+∞i
π/12−∞i
{
π2
6k 2
(
)} 21
1
n−
24
3
and ν = ,
2
( √ (
{
(
)}
{ 2 (
)}− 34
))
2
π
1
π
π
2
1
1
t−5/2 exp t + 2 n −
dt =
I3
n−
.
n−
2
6k t
24
6k 2
24
k 3
24
Hence,
(
) 3 ( √ (
))
∞
− 32
1 −4
( π )3 ∑
π
n
−
π
2
1
2
24
k −5/2 Ak (n)
p(n) = 2π
n−
I3
− 34 − 23
2
12
k 3
24
6
k
k=1
√
(
)
3
)
(
(
)
∞
1 −4 ∑
2π n − 24
π 2
1
−1
k Ak (n)I 3
n−
=
.
3
2
k 3
24
24 4
k=1
As obtained in (B.2), Bessel functions of half odd order can be reduced, in this case
√
(
)
2z d sinh z
I 3 (z) =
.
2
π dz
z
is yields

{ √ (
)} 
( √ (
π
1
2
))
)1
1
1
1 (
sinh
n
−
4
k
3
24
π 2
1
22 π 2 24
1
d 

√ (
I3
n−
= 1 1 1 n−

.
)
2
k 3
24
24
dz
π
2
π 2 k2 34
n− 1
k
Because k is constant we have
π
dz =
2k
( (
))− 1
2 2
2
1
n−
dn,
3
24
3
33
3
24
and thus

{ √ (
( √ (
π
2
))
)1
)1
1
1 (
1 (
1
22 24
π 2
1
1 4 2k 2 2
1 2 3 k 2 2 d  sinh k 3 n −
√
I3
n−
= 1 1 n−
n
−

2
k 3
24
24
π 3 12
24
2 π 3 21 dn
1
k2 34
n − 24

{ √ (
)} 
2
π
1
)3
3 3 (
sinh
n
−
k
3
24
64 k2
1 4 d 

√
=
n−

.
2
π
24
dn
1
n − 24
Inserting this into our equation for p(n) we obtain
(
p(n) =
) 3
1 −4
24
2π n −
24
3
4
6
3
4
(
n−
π2
1
24
)3
4

∞
∑
1
1
d  sinh
= √
k 2 Ak (n)

dn
π 2 k=1

∞
∑
k=1
3
d  sinh
k −1 k 2 Ak (n)

dn
{ √ (
π
2
k
√
3
n−
n−
1
24
34
1
24
)} 

.
{ √ (
2
π
k
√
3
n−
n−
1
24
1
24
)} 


1
24
)} 


B Appendix of Required Results
B.1 Bessel functions
Bessel functions of the first kind, whi are solutions to the Bessel differential equation, are defined as
( )ν+2m
∞
∑
(−1)m z2
Jν (z) =
.
m!Γ(ν + m + 1)
m=0
e modified Bessel functions of the first kind are consequently defined as
Iν (z) = e− 2 νπi Jν (iz)
1
= i−ν Jν (iz)
( z )ν+2m
∞
∑
2
=
.
m!Γ(ν + m + 1)
m=0
ese functions can be described in a myriad of forms, and for our purposes a specific integral representation is required. Making use of the Gamma integral
1
1
=
Γ(z)
2πi
we find that
( z )ν
Iν (z) =
2
2πi
∞ ∫
∑
∫
c+∞i
t−z et dt,
c−∞i
c+∞i
( z )2m
2
m=0 c−∞i
m!
t−ν−m−1 et dt.
Replacing the sum of integrals by an integral of sums and continuing:
(
)
( z )ν ∫ c+∞i ∞ ( z )2 −1 m
t
∑
2
Iν (z) = 2
t−ν−1 et dt
2πi c−∞i
m!
m=0
( z )ν ∫ c+∞i
{( ) }
z 2 1 −ν−1 t
2
=
exp
t
e dt.
2πi c−∞i
2 t
is yields the necessary integral equation, in whi c > 0 and Re(ν) > 0:
( 1 )ν ∫ c+∞i
(
)
z2
−ν−1
2z
Iν (z) =
t
exp t +
dt.
2πi c−∞i
4t
(B.1)
Bessel functions of half-odd order, i.e., those with ν = (n + 21 ), where n ∈ Z, can be expressed in finite
terms by means of algebraic and trigonometrical functions of z. In particular,
√
(
)
2z d sinh z
I 3 (z) =
.
(B.2)
2
π dz
z
35
e proof of this result is rather tenical. We rework the right and le hand sides respectively to a mutual
‘middle-point’.
√
(
) √ (
)
2z d sinh z
2z cosh z sinh z
=
−
π dz
z
π
z
z2
√ {(
)
(
)
(
)
}
2z
1
1
1
1
1
1
3
5
=
−
z+
−
z +
−
z + ···
π
2! 3!
4! 5!
6! 7!
√
}
∞ {
1
2z ∑
1
=
−
z 2m+1
π
(2m + 2)! (2m + 3)!
m=0
√
}
∞ {
2z ∑
2m + 2
=
z 2m+1
π
(2m + 3)!
m=0
√
∞
2z ∑
z 2m+1
=
.
π
(2m)!(4m2 + 8m + 3)
m=0
Before we alter the le hand side, we require a few extra properties of the Gamma function, namely that
Γ(z + 1) = zΓ(z)
1
(2n − 1)!! √
and Γ(n + ) =
π,
2
2n
and also a tri involving (double) factorials:
2m m!(2m − 1)!! = (2m)!!(2m − 1)!! = (2m)!.
Now, we proceed as follows:
I 3 (z) =
2
∞
(z )3 ∑
2
( z )2m
2
m!Γ(m + 52 )
m=0
( z )2m
∞
(z )3 ∑
2
2
=
1
2
m!Γ(m
+
)(m
+ 32 )(m + 12 )
2
m=0
( z )2m m
∞
2
1 ( z ) 32 ∑
2
=√
π 2 m=0 m!(2m − 1)!!(m2 + 2m + 34 )
√
∞
1 2z ∑
z 2m+1
=
4 π
2m m!(2m − 1)!!(m2 + 2m + 43 )
m=0
√
∞
z 2m+1
2z ∑
,
=
π
(2m)!(4m2 + 8m + 3)
2
m=0
and obtain the desired result, equation (B.2).
For a thorough resource on Bessel functions the reader is referred to Watson (1966), specifically pages 15,
52-54, 77 and 181.
36
C Supplementary Figures
C.1 e Rademaer path P (N )
x-plane
τ-plane
1.00
1
0.75
0
0.50
0.25
1
1
0
1
x-plane
0.00
0.25
0.75
1.00
0.75
1.00
τ-plane
1.00
1
0.50
0.75
0
0.50
0.25
1
1
0
1
0.00
Figure C.1: P(5) and P(20).
37
0.25
0.50
References
Andrews, G. E. (1984), e eory of Partitions, Cambridge University Press.
Apostol, T. M. (1990), Modular Functions and Dirilet Series in Number eory, 2nd edn, Springer-Verlag.
Goh, W. M. Y. & Smutz, E. (1995), ‘e number of distinct part sizes in a random integer partition’, J.
Combin. eory Ser. A 69(1), 149–158.
Hardy, G. H. (1978), Ramanujan: Twelve Lectures on Subjects Suggested by his Life and Work, 3rd edn,
American Mathematical Society.
Hardy, G. H. & Ramanujan, S. (1918), ‘Asymptotic formulae in combinatory analysis’, Proc. London Math.
Soc. 17, 75–115.
Miller, S. J. & Takloo-Bighash, R. (2006), An Invitation to Modern Number eory, Princeton University
Press.
Rademaer, H. (1974), Collected Papers of Hans Rademaer, Vol. 2, MIT Press.
Vaughan, R. C. (1997), e Hardy-Lilewood Method, 2nd edn, Cambridge University Press.
Wagner, S. (2009), Mathematics 344: Combinatorics, Stellenbos University.
Watson, G. N. (1966), A Treatise on the eory of Bessel Functions, 2nd edn, Cambridge University Press.
38