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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 textis to deduce a remarkable formula for p(n): the convergent infinite series obtained by Hans Rademaer.Initially the reader is familiarised with the basic concepts of generating functions, Mobius transformations,Farey fractions and Ford circles, before these are called upon to aid in the explanation of the proof ofRademaer’s formula. ereaer a few auxillary results are provided to conclude the text, among themthe 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 ofpositive 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 leastone more partition — namely (n+ 1).
e theory of partitions owes mu to Leonhard Euler — he penned numerous important theorems onthe topic, whi would garner aention from many of the most prolific mathematicians of the last threecenturies (Andrews 1984, p. xv). e most illustrious result in this area, however, is due to Hardy &Ramanujan (1918). e pair obtained an asymptotic formula for p(n); a finite sum with error O(n− 1
4 ),and in doing so, they also created and used what would later come to be known as the Hardy-Lilewoodcircle method. Two decades later, H. Rademaer 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 thepartition function remains its most endearing aievement. e intuitive idea of the method can be seenin the case where a function is defined inside the unit disk but has singularities along the unit circle. econtour integral of this function is sought, and a circle with radius slightly less than 1 is used as the curveof 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 integrablefunction whi approximates the integrand is then required. Replacing the integrand of the major arcswith this function and integrating, and showing that the contribution of the minor arcs is of an order lessthan the major arcs, will hopefully yield an expression with a small enough error term to render it anasymptotic equivalent of the original integral. An introduction to the circle method is given in Chapter13 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 appli-cation to p(n), as the handiwork of three mathematicians make this application rather striking. e proofof the exact formula for p(n) follows the same course as that outlined in Chapter 5 of Apostol (1990), andas 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
1− xk=
∞∑k=0
p(k)xk
contain the sequence (p(n)). We seek a formula for p(n). With a lile bit of manipulation,
F (x)
xn+1=
∞∑k=−n−1
p(k + n+ 1)xk,
1
so that the coefficient of x−1 is p(n), and Cauy’s residue theorem gives us
p(n) =1
2πi
∫C
F (x)
xn+1dx.
We are thus interested in the behaviour of F (x)xn+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 complexroot of unity, F (x) can be approximated by a function whi can be integrated. us we oose a path Cnear the roots of unity on the unit circle. We split our path up so that we consider it as the sum of integralsrelated to ea root of unity, and hope that the difference between our approximation of F near ea rootand the actual value of F (x) is very small. It turns out that Rademaer’s method for integrating near theroots 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 itsterms. For this cause, generating functions (Chapters 5, 6 ofWagner (2009)) provide an elegant and intuitiveprocess. To a sequence of complex numbers a0, a1, a2, . . . , we associate the power series
A(x) =
∞∑n=0
anxn,
where the coefficient of xn is an. roughout this text, unless otherwise stated, x and z will be regardedas complex. If we consider the geometric progression 1, r, r2, r3, . . . , we obtain the generating function
R(x) =∞∑r=0
rnxn
=∞∑r=0
(rx)n =1
1− rx,
whi gives the important representation of 1, 1, 1, . . . :
∞∑n=0
xn =1
1− x.
In combinatorics, one is oen concerned with structures of different sizes, and specifically, counting thenumber of structures of a given size. Approaing generating functions from this perspective, we can viewxn 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) =
∞∑n=1
xn = x+ x2 + x3 + · · ·
= x(1 + x+ x2 + · · · )
= x
∞∑n=0
xn =x
1− x.
is view of generating functions lets us handle families of structures simply. IfA and B are two disjointfamilies, 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 tothe sum of the sizes of a and b. us the number of pairs of size n is
n∑k=0
akbn−k,
and the generating function for pairs of structures from A and B is
∞∑n=0
n∑k=0
akbn−kxn = A(x)B(x).
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, anddifferentiating them as su provides a simple way of understanding their generating function. A sequenceof length 1 would be any a ∈ A. Sequences of length 2 are all the pairs (a1, a2); of length 3 are all thetriples; and so forth. is implies that
Seq(A) = {ε} ∪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) =
∞∑n=0
A(x)n =1
1−A(x).
2.1 Integer compositions
Rather remarkably, these tools are already sufficient to solve a problem only slightly different from thefocus of this text. Namely, howmany partitions of an integern are there when these partitions are ordered?Ordered partitions are known as compositions, and we shall denote the number of compositions of apositive 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 asequence 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
I(x) =x
1− x,
so that the generating function for sequences of positive integers Seq(I) = C is
C(x) =1
1− x1−x
=1− x
1− 2x,
and with a small amount of manipulation we find that
C(x) =12
1− 2x+
12 − x
1− 2x
=1
2
(1
1− 2x+ 1
)=
1
2
(1 +
∞∑n=0
2nxn
)
=1
2+
∞∑n=0
2n−1xn.
e 12 only affects the constant term in the sum, so that c(0) = 1
2 + 2−1 = 1, giving us 1 composition of0. For any n ≥ 1, c(n) = 2n−1.
4
2.2 Integer partitions
We can expand on the idea of sequences of structures with powersets and multisets. e powerset of afamily A is the set of all subsets of A, and is represented by
PSet(A) =⊗a∈A
{a, ε},
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 + zn), and if there are an su structures, we find the generatingfunction of the powerset to be:
B1(x) =∞∏n=1
(1 + zn)an .
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) =⊗a∈A
Seq({a}),
and since the generating function of Seq({a}) is (1− zn)−1,
B2(x) =∞∏n=1
(1− zn)−an .
Now that we possess a foundation of descriptive generating functions, it should be clear that a partitionof an integer n is tenically just a multiset of size n over the family of integers I . Recalling that an = 1as there is only one integer of size n, this yields a generating function for p(n) quite easily, namely
P (x) =
∞∏n=1
(1− xn)−1 =
∞∑n=0
p(n)xn.
e problem of finding an exact formula for p(n) is hardly as simple as it was for c(n), but is far moreinteresting to investigate. In time we shall come upon this formula, but before we do we must establishan 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 andRamanujan 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 necessaryproperties of Mobius transformations and the modular group, as presented in Chapter 2 of Apostol (1990),will be helpful.
Firstly, considering a transformation of the form
f(z) =az + b
cz + d
where a, b, c, d are complex numbers, we can see that f is defined on the extended complex numbersystem for all values of z except z = −d
c and z = ∞. en f can be extended, according to the intuitionthat z
0 = ∞ for non-zero z, by leing
f
(−d
c
)= ∞ and f(∞) =
a
c.
If one then considers another w in the extended complex system, then
f(w)− f(z) =aw + b
cw + d− az + b
cz + d
=acwz + bcz + adw + bd
(cw + d)(cz + d)− acwz + bcw + adz + bd
(cw + d)(cz + d)
=(ad− bc)(w − z)
(cw + d)(cz + d),
so that if ad− bc = 0, then f(w)− f(z) = 0 and f is constant. We thus consider only transformationswhi satisfy ad− bc 6= 0, and call these functions on the extended complex systemMobius transforma-tions.
Furthermore, if ad− bc = k 6= 0, we can instead consider the analogous transformation
f(z) =
a√kz + b√
kc√kz + 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 ′, then the set of Mobiustransformations of the form
τ ′ =aτ + b
cτ + d
is called themodular group and is depicted by Γ. Ea transformation can also be wrien as a 2×2 integermatrix with determinant 1:
A =
(a bc d
),
with A and −A both representing the same transformation. en, if two transformations τ ′ and µ′ haverelevant matrices A and B, the matrix product BA represents the composition transformation µ′ ◦ τ ′.Hence, because the set of 2 × 2 matrices with integer coefficients and determinant 1 form a group withrespect to multiplication, the modular group is, in fact, a group.
6
4 Farey Fractions
It was Rademaer who noted that one can make use of the theory of Farey fractions to obtain a moreaccurate 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 ofFarey 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
F1 =
{0
1,1
1
}F2 =
{0
1,1
2,1
1
}F3 =
{0
1,1
3,1
2,2
3,1
1
}F4 =
{0
1,1
4,1
3,1
2,2
3,3
4,1
1
}and so forth. We can concisely define Fn as {h
k : 0 ≤ h ≤ k ≤ n; (h, k) = 1}. It can also be easily seenin the examples above that Fn ⊂ Fn+1.
If we consider two fractions in Fn, say ab < c
d , then an important new fraction called the mediant isdefined as
h
k=
a+ c
b+ d.
e mediant lies between these two fractions, as ab < c
d implies that bc− ad > 0, so
a+ c
b+ d− a
b=
bc− ad
b(b+ d)> 0
andc
d− a+ c
b+ d=
bc− ad
d(b+ d)> 0.
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 c
d are consecutivefor the case where bc− ad = 1: if p
q is any reduced fraction withab < p
q < cd then
q = b(cq − dp) + d(bp− aq),
and because cq− dp > 0 and bp−aq > 0must be integer values, we have cq− dp ≥ 1 and bp−aq ≥ 1,so the denominator of any fraction whi lies between a
b and cd must be at least b + d. Combining this
7
with the fact that max(b, d) ≤ n because ab and c
d are in Fn, we find that ab and c
d are consecutive in Fn
for any n satisfyingmax(b, d) ≤ n ≤ b+ d− 1.
is provides us with enough information to fully describe the way in whi the sequence of Farey frac-tions F1, F2, . . . is built. Consider the construction of Fn+1 from Fn. If a
b and cd are consecutive in Fn
and bc− ad = 1 then they will remain consecutive until n ≥ b+ d. Forming the mediant
h
k=
a+ c
b+ d,
we know that (h, k) = 1 because
bh− ak = 1 and ck − dh = 1,
and any twhi divides both h and k must divide 1 as well. us, hk is in Fk, and these two equations alsoimply that both the pairs a
b < hk and h
k < cd 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 consecutiveFarey fractions satisfy a relation of the form bc − ad = 1, as when passing from Fn to Fn+1, ea newfraction is the mediant of two consecutive fractions satisfying su a relation and the new consecutivepairs satisfy a similar relation.
8
5 Ford Circles
Originally, Rademaer used only Farey fractions to construct his exact formula for p(n). However, helater refined this approa by making use of Ford circles — geometric constructs whi relate elegantly tosets of Farey fractions.
e Ford circle of a reduced fraction hk is defined as the circle in the complex plane centered at h
k + i 12k2
with radius 12k2
, and is denoted byC(h, k). ese circles shall be of high importance when addressing theproof of Rademaer’s formula later in this text, and as su some valuable properties shall be developedhere.
hk
12k2
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 ab and c
d . esecircles have centers
a
b+ i
1
2b2and
c
d+ i
1
2d2
respectively, and ifD is the distance between these centers, then
D2 =(ab− c
d
)2+
(1
2b2− 1
2d2
)2
.
e square of the sum of their radii is
(r +R)2 =
(1
2b2+
1
2d2
)2
.
e difference between these two lengths is
D2 − (r +R)2 =
(ad− bc
bd
)2
+
(1
2b2− 1
2d2
)2
−(
1
2b2+
1
2d2
)2
=
(ad− bc
bd
)2
− 41
4b2d2
=(ad− bc)2 − 1
b2d2.
Because ab and
cd are different, ad−bc 6= 0, andD2−(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 onlyif, bc − ad = ±1. More importantly, this tells us that the Ford circles of consecutive Farey fractions aretangent to ea other.
9
h1k1
hk
α1 ba
Figure 5.2
Let us investigate this further, and determine these points of tangency. Let h1k1
< hk < h2
k2be three
consecutive Farey fractions. To find α1(h, k), the point of contact of C(h1, k1) and C(h, k) as in Figure5.2, consider the large right-angled triangle above the line joining the centers of the two circles, with thisline 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,
α1(h, k) =
(h
k− a
)+ i
(1
2k2+ b
). (5.1)
e large triangle has a hypotenuse of length 12k1
2 +1
2k2, vertical length 1
2k12 − 1
2k2and horizontal length
hk−
h1k1, whereas the only known side of the smaller triangle is the hypotenuse —with length 1
2k2. However,
because the two triangles are similar, we have
ahk − h1
k1
=1
2k2
12k1
2 + 12k2
=1
2k2· 2k1
2k2
k2 + k12
=k1
2
k2 + k12
and, also using the hypotenuse of the large triangle,
b1
2k12 − 1
2k2
=k1
2
k2 + k12 ,
so that, remembering that hk1 − h1k = 1,
a =k1
2
k2 + k12 · 1
kk1=
k1
k(k2 + k12)
and
b =k1
2
k2 + k12 · k
2 − k12
2k2k12 =
k2 − k12
2k2(k2 + k12).
10
Combining this with equation (5.1), we find that
α1(h, k) =
(h
k− k1
k(k2 + k12)
)+ i
(1
2k2+
k2 − k12
2k2(k2 + k12)
)=
h
k− k1
k(k2 + k12)
+ i2k2
2k2(k2 + k12)
=h
k− k1
k(k2 + k12)
+i
k2 + k12 .
In this investigation we have subtly assumed that k1 > k (see Figure 5.2). If the converse holds, we wouldhave
α1(h, k) =
(h
k− a
)+ i
(1
2k2− b
).
But in this case we would then divide b by 12k2
− 12k1
2 to compare the ratio with the larger triangle, and
this would yield the term k12 − k2 in place of k2 − k1
2 in the formula for b, and leave the formula for α1
unanged.
Following an analogous course, it is shown that
α2(h, k) =h
k+
k2
k(k2 + k22)
+i
k2 + k22 . (5.2)
is provides us with sufficient knowledge of the structure of the Ford circles formed by sets of Fareyfractions. Let us draw nearer to p(n), and with it, the proof of an exact formula.
11
6 Rademaer’s Series
So far we have a generating function for p in the infinite product
F (x) =∞∏
m=1
1
1− xm=
∞∑n=0
p(n)xn (6.1)
and a basic knowledge of the properties of Farey fractions and Ford circles. Our first order of business willbe 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 infiniteproduct, as our results will also apply to the series. An infinite product is said to converge when its limitexists and is not zero. us, for an infinite product to converge, it is necessary that its terms approa 1.
If |x| ≥ 1, |xm| cannot have limit 0, so lim 11−xm 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∑
xm 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)
xn+1=
∞∑k=0
p(k)xk
xn+1
for n ≥ 0. is gives us the Laurent series representation of F (x)xn+1 , as
∞∑k=0
p(k)xk
xn+1= p(0)x−n−1 + p(1)x−n + · · ·+ p(n)x−1 + p(n+ 1) + · · ·
So F (x)xn+1 has a pole of order n + 1 at x = 0 and residue p(n). is allows us to apply Cauy’s residue
theorem and find that
p(n) =1
2πi
∫C
F (x)
xn+1dx. (6.2)
6.2 e Rademaer path P (N)
ough we are yet to prove it, F (x) can be approximated by an elementary function whenever x liesnear a complex root of unity — a result whi follows from a functional equation satisfied by F . ereason for postponing this result is that the functional equation’s form is more understandable once a fewtransformations have been made. We shall encounter these transformations while elaborating on the pathof integration used by Rademaer.
So, assuming that we have reliable information about F ’s behaviour near the roots of unity, we wouldpreferC to lie as near as possible to ea root of unity. is is where Rademaer was inspired with a more
12
elegant approa than that of Hardy and Ramanujan. He originally divided the circular contour used by hisfellows 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 tointegrate nearer to the singularities and thus simplify the estimations made (Apostol 1990). is laerapproa has proved useful in other areas of modular function theory (Andrews 1984, p. 85), and shall beour elected course. e Ford circles of Section 5 reside in the upper half-planeH = {τ : Im(τ) > 0}, andwe thus need to relate our domain, the unit disk, to H. An elegant mapping between the open unit diskandH 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} inH.
01
i
0 1
x-plane V
Figure 6.1: e transformation from the x-plane to the τ -plane
is can be seen by taking a point τ = a+ bi inH 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. edenominator 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 ofradius e−2π with center 0.
Rademaer’s idea was to follow the curve in the τ -plane created by the upper arcs of the Ford circlesformed from Fn — the set of Farey fractions of order n. ese circles fit neatly into V , bounded by i on theimaginary axis. If h1
k1, hk and h2
k2are 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). Notethat for the fractions 0
1 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 toapproa infinity to account for infinitely many complex roots of unity.
13
1 0 1
1
0
1
x-plane
0.00 0.25 0.50 0.75 1.00
0.25
0.50
0.75
1.00τ-plane
Figure 6.2: e Rademaer 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
yieldsdx = x · 2πidτ,
so that (6.2) becomes
p(n) =2πi
2πi
∫P (N)
F (x)
xndτ
=
∫P (N)
F (e2πiτ )e−2πinτdτ.
If γ(h, k) denotes the upper arc of C(h, k), then we can split P (N) into separate arcs for ea Ford circleso that we may treat ea root of unity independently:∫
P (N)=
N∑k=1
∑0≤h<k(h,k)=1
∫γ(h,k)
=∑h,k
∫γ(h,k)
where∑
h,k denotes the double sum over k and h.
14
ere remains one transformation to be made before we can deduce an alternate form of F :
z = −ik2(τ − h
k
),
so that
τ =h
k+
iz
k2.
To see the effect on a Ford circle C(h, k), the transformation can be broken down into elementary opera-tions. e term τ − h
k shis C(h, k) a distance of hk to the le, centering the circle above the origin. e
factor k2 expands the radius to
k2 · 1
2k2=
1
2,
while multiplication by −i = e−πi/2 rotates the figure through −π2 radians. C(h, k) is thus transformed
from the τ -plane onto a circleK in the z-plane with center 12 and a radius of 1
2 .
K
z1(h, k)
z1(h, k)
0 12
Figure 6.3
e point of contact α1(h, k) with the preceding circle becomes
z1(h, k) = −ik2(α1(h, k)−
h
k
)= −ik2
(i
k2 + k12 − k1
k(k2 + k12)
)=
k2
k2 + k12 + i
kk1
k2 + k12 (6.3a)
and similarly the other point of contact α2(h, k) is now
z2(h, k) =k2
k2 + k22 − i
kk2
k2 + k22 . (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 curvewhi does not tou the imaginary axis in the z-plane. Noting that
dτ =i
k2dz,
15
the effect of this ange of variable on p(n) is:
p(n) =∑h,k
∫γ(h,k)
F (e2πiτ )e−2πinτdτ
=∑h,k
∫ z2(h,k)
z1(h,k)F
(exp(2πih
k− 2πz
k2
))i
k2e−2πinh/ke2πnz/k
2dz
=∑h,k
i
k2e−2πinh/k
∫ z2(h,k)
z1(h,k)e2πnz/k
2F
(exp(2πih
k− 2πz
k2
))dz.
e allenge of finding a usable form of F can now be aempted.
F (x) is an elliptic modular function, and in general this class of functions satisfy functional equationswhi describe them near the unit circle. We proceed by obtaining su an equation for F from theknown 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-planeH = {τ ∈C : Im(τ) > 0} as
η(τ) = eπiτ/12∞∏
m=1
(1− e2πimτ ). (6.4)
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 differentpoints, e.g.,
Γ(1 + x) = xΓ(x).
To Dedekind’s functional equation: if(a bc d
)∈ Γ, c > 0 and τ ∈ H, then
η
(aτ + b
cτ + d
)= ε(a, b, c, d){−i(cτ + d)}
12 η(τ), (6.5)
where
ε(a, b, c, d) = exp{πi
(a+ d
12c+ s(−d, c)
)}and
s(h, k) =
k−1∑r=1
r
k
(hr
k−⌊hr
k
⌋− 1
2
).
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 basicequations (6.1) and (6.4). ankfully, if x = e2πiτ ,
η(τ) = eπiτ/12∞∏
m=1
(1− e2πimτ )
=eπiτ/12
F (e2πiτ ).
Dedekind’s functional equation for η (6.5) can in fact be represented in terms of F . If we remember that(a bc d
)∈ Γ and τ ′ = aτ+b
cτ+d , with c > 0, then the relation between η and F implies
eπiτ′/12
F (e2πiτ ′)=
eπiτ/12
F (e2πiτ ){−i(cτ + d)}
12 exp
{πi
(a+ d
12c+ s(−d, c)
)},
so that
F (e2πiτ ) = F (e2πiτ′) exp
(πi(τ − τ ′)
12
){−i(cτ + d)}
12 exp
{πi
(a+ d
12c+ s(−d, c)
)}.
With a careful oice of(a bc d
), it can be shown (see Section A.1) that
F
(exp(2πih
k− 2πz
k2
))= ω(h, k)
(zk
) 12F
(exp(2πiH
k− 2π
z
))exp( π
12z− πz
12k2
)where
ω(h, k) = eπis(h,k), hH ≡ −1 (mod k), (h, k) = 1.
Note that s(h, k) ∈ Q so that ω(h, k) is a complex root of unity, with modulus 1.
Aer two transformations we have reaed the point where
x = e2πiτ = exp(2πih
k− 2πz
k2
).
Leing
x′ = e2πiτ′= exp
(2πiH
k− 2π
z
),
the functional equation can be rewrien as
F (x) = ω(h, k)(zk
) 12exp( π
12z− πz
12k2
)F (x′), (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 x′, note that x is near to the root of unity e2πih/k when |z|
17
is small. However, the smaller z becomes, the larger 2πz becomes, and the nearer x′ moves toward 0, the
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)(zk
) 12exp( π
12z− πz
12k2
),
whi is an elementary function. e error in approximating F in this way will be determined by justhow close to 1 the value F (x′) really is, i.e., by the difference (F (x′)− 1).
Returning to our equation for p(n),
p(n) =∑h,k
ik−2e−2πinh/k
∫ z2(h,k)
z1(h,k)e2πnz/k
2F
(exp(2πih
k− 2πz
k2
))dz,
we can apply Dedekind’s functional equation and obtain
p(n) =∑h,k
ik−5/2ω(h, k)e−2πinh/k
∫ z2(h,k)
z1(h,k)e2πnz/k
2z
12 exp
( π
12z− πz
12k2
)F (x′) dz.
We now perform the substitution F (x′) = 1+ (F (x′)− 1) to separate our elementary function from theerror made in our approximation, and find
p(n) =∑h,k
ik−5/2ω(h, k)e−2πinh/k
∫ z2(h,k)
z1(h,k)e2πnz/k
2Ψk(z){1 + (F (x′)− 1)} dz,
with Ψk(z) = z12 exp
(π12z − πz
12k2
).
If we define
I1(h, k) =
∫ z2(h,k)
z1(h,k)Ψk(z)e
2πnz/k2 dz (6.7a)
and
I2(h, k) =
∫ z2(h,k)
z1(h,k)Ψk(z)e
2πnz/k2{F
(exp(2πiH
k− 2π
z
))− 1
}dz (6.7b)
then we are finally le with the more visually appealing
p(n) =∑h,k
ik−5/2ω(h, k)e−2πinh/k(I1(h, k) + I2(h, k)), (6.8)
in whi I1(h, k) is the elementary integral with whi we wish to work and I2(h, k) is the error madein approximating our integral in this way.
Let us take a moment to evaluate our position. We began with an integral (6.2) obtained from Cauy’sresidue theorem and proceeded to find an integration curve P (N) whi we could split into a sum ofintegrals over the Farey fractions in Fn. We have now applied Dedekind’s functional equation and splitea integral into two integrals I1 and I2, ea along the arc from z1(h, k) to z2(h, k) on the circleK .
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 isour difference term. Luily, I2(h, k) is O(N−1/2), and thus becomes small for large values of N . To seethis, we show that the integrand is bounded and then integrate along the direct ord from z1(h, k) toz2(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∣∣∣∣Ψk(z)
{F
(exp(2πiH
k− 2π
z
))− 1
}e2πnz/k
2
∣∣∣∣=
∣∣∣∣∣z 12 exp
( π
12z− πz
12k2
)exp(2πnz
k2
){ ∞∑m=1
p(m) exp(2πimH
k− 2πm
z
)}∣∣∣∣∣= |z|
12 exp
(π Re(1/z)
12− π Re(z)
12k2
)exp(2πnRe(z)
k2
) ∞∑m=1
p(m)∣∣∣e2πiHm/k
∣∣∣ e−2πm Re(1/z).
Because z is a point on the ord from z1 to z2, 0 < Re(z) < 1. To continue, we also need informationabout Re(1/z). is is thus a good time to investigate the transformation 1/z for a point z = a + bi onor within the circleK (see Figure 6.3): (
a− 1
2
)2
+ b2 ≤ 1
4
soa2 − a+
1
4+ b2 ≤ 1
4, and a2 + b2 ≤ a.
We can write1
z=
a− bi
a2 + b2,
and then investigate ea part separately:
Re(1/z) =a
a2 + b2≥ a
a= 1
and
Im(1/z) = − b
a2 + b2.
In the case where z lies on the circleK , we have a2 + b2 = a, so that Re(1/z) = 1 and
Im(1/z) = − b
a= ±
√a− a2
a2= ±
√1
a− 1,
19
whi can take on any real value as 0 ≤ a ≤ 1.
Noting then that
exp(−π Re(z)
12k2
)≤ 1,
exp(2πnRe(z)
k2
)< e2πn,∣∣∣e2πiHm/k∣∣∣ = 1,
we can continue with our estimation, seeing that the modulus of the integrand is
< |z|12 exp
(π Re(1/z)
12
)e2πn
∞∑m=1
p(m)e−2πm Re(1/z)
= |z|12 e2πn
∞∑m=1
p(m)e−2π Re(1/z)(m−1/24)
≤ |z|12 e2πn
∞∑m=1
p(m)e−π(24m−1)/12
< |z|12 e2πn
∞∑m=1
p(24m− 1)e−π(24m−1)/12.
Taking y = e−π/12 and recognising that the infinite sum is F (y) and thus convergent (|y| < 1), we arele with
c |z|12 ,
where
c = e2πn∞∑
m=1
p(24m− 1)y24m−1,
so that c is independent of N and z, and thus remains unanged throughout the double summation inour current formula for p(n).
We have yet to consider the term |z|12 , and thereaer 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),
|z1(h, k)|2 =∣∣∣∣ k2
k2 + k12
∣∣∣∣2 + ∣∣∣∣ kk1
k2 + k12
∣∣∣∣2=
k4 + k2k12
(k2 + k12)
2
=k2
k2 + k12 .
20
A similar equation holds for z2(h, k), so that
|z1(h, k)| =k√
k2 + k12
|z2(h, k)| =k√
k2 + k22
(6.9)
Now, the Cauy-Swarz inequality implies that
(k · 1 + k1 · 1)2 ≤ (k2 + k12)(1 + 1)
so that √k2 + k1
2 ≥ k + k1√2
≥ N + 1√2
≥ N√2
because we have consecutive Farey fractions, so N ≤ k1 + k − 1. e same inequality holds when k1 isreplaced by k2, so that
|z1| <√2k
Nand |z2| <
√2k
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
c |z|12 ≤ c 2
14
(k
N
) 12
.
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 theord, so that
|I2(h, k)| < C
(k
N
) 32
21
where C is a constant. Hence, noting the roots of unity ω(h, k) and e−2πinh/k,∣∣∣∣∣∣∑h,k
ik−5/2ω(h, k)e−2πinh/kI2(h, k)
∣∣∣∣∣∣ ≤∑h,k
k−5/2 |I2(h, k)|
<
N∑k=1
∑0≤h<k(h,k)=1
C
kN3/2
=C
N3/2
N∑k=1
1
k
∑0≤h<k(h,k)=1
1
≤ C
N3/2
N∑k=1
1
=C√N
Equation (6.8) then becomes
p(n) =∑h,k
ik−5/2ω(h, k)e−2πinh/kI1(h, k) + O(N−1/2). (6.10)
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 allowN → ∞. But we are geing ahead of ourselves. Before we hastily get rid of our error term, we can showthat integrating I1(h, k) along the entire circleK instead of just a section of its circumference will provideus with an error also of O(N−1/2).
Recall equation (6.7a):
I1(h, k) =
∫ z2(h,k)
z1(h,k)Ψk(z)e
2πnz/k2dz.
IfK− denotes the curve along the circumference ofK in the negative direction then
I1(h, k) =
∫K−
−∫ z1(h,k)
z2(h,k)
=
∫K−
−∫ 0
z2(h,k)−∫ z1(h,k)
0.
We split the difference into two separate integrals (call them J1 and J2 respectively) because 1z does not
exist when z = 0. It remains to show that both J1 and J2 are O(N−1/2).
22
K
z1(h, k)
0 12
|z1|
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 arconK from 0 to z1 lies within the upper half of this circle, so that the length of the arc is less than half ofthe new circle’s circumference, i.e.,
<π |z1|2
< π
√2k
N.
Similarly, this same inequality holds for the length of the arc from z2 to 0. Clearly, any z on either of thesearcs 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, soRe(1/z) = 1. We consider the integrand:∣∣∣Ψk(z)e
2πnz/k2∣∣∣
=∣∣∣z 1
2 exp( π
12z− πz
12k2
)e2πnz/k
2∣∣∣
= |z|12 exp
(π Re(1/z)
12− π Re(z)
12k2
)e2πn Re(z)/k2
≤ |z|12 eπ/12e−πe2πn
< c1
(k
N
) 12
,
using the inequality for |z| obtained above. Combining this with our knowledge of the length of the arcs,we find that both
|J1| < C1
(k
N
) 32
and |J2| < C1
(k
N
) 32
,
where C1 is constant. Remembering that these together form the error when using the curve K− to
23
calculate I1(h, k), we find∣∣∣∣∣∣∑h,k
ik−5/2ω(h, k)e−2πinh/k(J1 + J2)
∣∣∣∣∣∣ ≤∑h,k
k−5/2 |J1 + J2|
<N∑k=1
∑0≤h<k(h,k)=1
2C1
kN−3/2
=2C1
N−3/2
N∑k=1
1
k
∑0≤h<k(h,k)=1
1
≤ 2C1√N
.
is, when applied to (6.10), gives us
p(n) =∑h,k
ik−5/2ω(h, k)e−2πinh/k
∫K−
Ψk(z)e2πnz/k2dz + O(N−1/2).
6.7 Letting go
At present we are considering only a finite number of Ford circles — namely those based on FN . LeingN → ∞, our equation becomes
p(n) =
∞∑k=1
∑0≤h<k(h,k)=1
ik−5/2ω(h, k)e−2πinh/k
∫K−
Ψk(z)e2πnz/k2dz
=
∞∑k=1
ik−5/2Ak(n)
∫K−
z12 exp
( π
12z− πz
12k2
)e2πnz/k
2dz,
=∞∑k=1
ik−5/2Ak(n)
∫K−
z12 exp
{π
12z− 2πz
k2
(n− 1
24
)}dz,
where we letAk(n) =
∑0≤h<k(h,k)=1
ω(h, k)e−2πinh/k
as it shall remain unanged for the remainder of the proof.
We then apply the ange of variable
z =π
12twith dz = − π
12t2dt,
noting that this is just a constant multiple of the transform 1z whi was investigated earlier, and thus
maps the circleK (and with it our path of integration) onto the line Re(z) = π12 .
24
en
p(n) =∞∑k=1
−ik−5/2Ak(n)
∫ π/12+∞i
π/12−∞i
( π
12t
) 12exp{t+
π2
6k2t
(n− 1
24
)}π
12t2dt
=
∞∑k=1
−ik−5/2Ak(n)( π
12
) 32
∫ π/12+∞i
π/12−∞it−5/2 exp
{t+
π2
6k2t
(n− 1
24
)}dt
= 2π( π
12
) 32
∞∑k=1
k−5/2Ak(n)1
2πi
∫ π/12+∞i
π/12−∞it−5/2 exp
{t+
π2
6k2t
(n− 1
24
)}dt.
We can evaluate the integral in terms of Bessel functions, but for the sake of continuity the tenicalities ofthe evaluation are performed in Section A.2, with a narrative sket provided here. Equation (B.1) foundin Section B.1 states that (
12z)ν
2πi
∫ π/12+∞i
π/12−∞it−ν−1 exp
(t+
z2
4t
)dt = Iν(z),
and if z and ν in this integral are taken as
z
2=
{π2
6k2
(n− 1
24
)} 12
and ν =3
2,
we find that our integral can now be expressed as
1
2πi
∫ π/12+∞i
π/12−∞it−5/2 exp
{t+
π2
6k2t
(n− 1
24
)}dt =
{π2
6k2
(n− 1
24
)}− 34
I 32
(π
k
√2
3
(n− 1
24
)).
Combining this equation with that of p(n) and simplifying gives us
p(n) =2π(n− 1
24
)− 34
(24)34
∞∑k=1
k−1Ak(n)I 32
(π
k
√2
3
(n− 1
24
)).
I 32(z) is a Bessel function of half odd order, and, specifically (B.2) gives
I 32(z) =
√2z
π
d
dz
(sinh zz
),
whi implies that
I 32
(π
k
√2
3
(n− 1
24
))=
634k
32
π2
(n− 1
24
) 34 d
dn
sinh{
πk
√23
(n− 1
24
)}√n− 1
24
.
Once again simplifying our equation for p(n), we arrive at our infinite series representation:
p(n) =1
π√2
∞∑k=1
k12Ak(n)
d
dn
sinh{
πk
√23
(n− 1
24
)}√n− 1
24
, (6.11)
25
where, as beforeAk(n) =
∑0≤h<k(h,k)=1
ω(h, k)e−2πinh/k.
We have now deduced the convergent series for p(n), our main objective for this text. A few auxillaryresults will yet be presented.
6.8 Asymptotics
When Hardy and Ramanujan published their results on p(n), they included the asymptotic formula
p(n) ∼ eπ√
2n3
4n√3
as n → ∞.
is formula can in fact be deduced from the first term of Rademaer’s infinite series. To prove this, weneed to show that the first term (denote the k-th term byRk(n)) is of this form and that the weight of therest of the terms combined is of an order less than R1(n). Recall from (6.11) that
Rk(n) = k12Ak(n)
d
dn
sinh{
πk
√23
(n− 1
24
)}√n− 1
24
,
so that for the first term, where k = 1:
A1(n) =∑
0≤h<k(h,k)=1
eπis(h,k)e−2πinh/k
= eπi·0e−2πn·0
= 1,
because s(h, k) is a sum from 1 to k−1 = 0, and is thus 0. e remaining part ofR1(n) is the derivative,
R1(n) =d
dn
sinh(π√
23
(n− 1
24
))√n− 1
24
=
1
4
(eπ√
23(n−
124) + e
−π√
23(n−
124))π
√2
3
(n− 1
24
)−1
− 1
4
(eπ√
23(n−
124) − e
−π√
23(n−
124))(
n− 1
24
)− 32
.
Also, we can make use of the binomial series and obtain:√n− 1
24=
√n
(1− 1
24n
) 12
=√n
∞∑m=0
( 12
m
)(−1)k
(1
24n
)m
=√n
(1− 1
2 · 24n− 1
8 · (24n)2− · · ·
)=
√n
(1 + O
(1
n
))=
√n+ O
(1√n
),
26
so that, by the Taylor expansion of e,
eπ√
23(n−
124) = e
π√
2n3
(1 + O
(1√n
)).
Following on from the above, and noting that the exponential terms with negative exponents diminishrapidly, we can further simplify our former result to:
R1(n) =π
4n
√2
3eπ√
2n3
(1 + O
(1√n
)).
We now have a representation of the first term whi seems strikingly familiar to the asymptotic formulawe wish to obtain. To further our knowledge of the weight of the subsequent terms, we must investigatethe derivative term more closely. Leing
t =π
k
√2
3
(n− 1
24
), so that
(n− 1
24
)=
(kt
π
)2 3
2,
we find that
d
dn
sinh t√n− 1
24
=π
2k
√2
3
cosh t(n− 1
24
) − 1
2
sinh t(n− 1
24
) 32
=π
2k
(2
3
) 32 ( π
kt
)2cosh t− 1
2
(2
3
) 32 ( π
kt
)3sinh t
=2
3√6
π3
k3cosh tt2
−√2
3√3
π3
k3sinh tt3
=2
3√6
π3
k3
(cosh tt2
− sinh tt3
). (6.12)
With f(t) =( cosh t
t2− sinh t
t3
), the Taylor series for cosh and sinh imply
f(t) =
(1
t2+
t2
2!t2+
t4
4!t2+ · · ·
)−(
t
t3+
t3
3!t3+
t5
5!t3+ · · ·
)=
(1
2!− 1
3!
)+ t2
(1
4!− 1
5!
)+ t4
(1
6!− 1
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 withless than k terms, it is clear that |Ak(n)| ≤ k. With this in mind (and C a constant), it follows that:∣∣∣∣∣∣
∑k≥2
k12Ak(n)
2
3√6
π3
k3f
(π
k
√2
3
(n− 1
24
))∣∣∣∣∣∣ ≤ C ·∑k≥2
k−32 f
(π
k
√2
3
(n− 1
24
))
≤ C · f
(π
2
√2
3
(n− 1
24
))∑k≥2
k−32 ,
27
because f(t) decreases as k increases. Furthermore, f(t) does not approa −∞ as k increases, and theinfinite sum is a p-series with p > 1, and thus converges. So∣∣∣∣∣∣
∑k≥2
Rk(n)
∣∣∣∣∣∣ = O
{f
(π
2
√2
3
(n− 1
24
))}= O(R2(n)).
It is clear from the above processes that ea term is dominated by an exponential function, and althoughthe exponent of e is similar for ea term, the constant part of the exponent in R2(n) (and subsequentterms) is smaller than that of R1(n), so that the first term dominates the sum. From (6.11) then,
p(n) ∼ 1
π√2R1(n)
=1
π√2
π
4n
√2
3eπ√
2n3
(1 + O
(1√n
))
∼ eπ√
2n3
4n√3
as n → ∞. (6.13)
e actual weight of 1π√2R1(n) (and the growth of p(n)), is demonstrated below:
n p(n) 1π√2R1(n)
1 1 12 2 23 3 34 5 55 7 710 42 4211 56 5720 627 626100 190569292 190568945500 2300165032574324047872 23001650325737465446401000 24061467864032624306850057158656 240614678640327233860418593095685000 1.6982e+74 1.6982e+7410000 3.61673e+106 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 ata far slower rate than the increase of the partition function. Because the contribution of the other termsbecomes less important as n becomes large, the ratio of 1
π√2R1(n) to p(n) tends to 1.
28
0 20 40 60 80 100n
0.980
0.985
0.990
0.995
1.000
1.005
1.010
1.015
Figure 6.5: e ratio of 1π√2R1(n) to p(n).
6.9 Distinct summands
Another interesting result due to Goh & Smutz (1995) is that the average number of distinct summandsin a partition of n is
D(n) =
√6n
π
(1 + O
(n− 1
6
)). (6.14)
Firstly, let us obtain a combinatoric formula for this quantity. Instead of considering ea partition indi-vidually, regard the set of all partitions of n. Summing the number of partitions containing a 1 with thenumber of partitions containing a 2, then a 3, and so forth, counts ea distinct summand in every parti-tion once, and so gives the total number of distinct summands over all partitions. Dividing this quantityby p(n) gives the average number of distinct summands in a typical partition. e number of partitionscontaining the integer r is given by p(n− r), as removing the r gives a partition of n− r, and converselyaaing 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
p(n− r) =eπ
√2(n−r)
3
4√3(n− r)
(1 + O
(1√n− r
)).
Making use of the binomial series to expand and noting that rn ≤ 1 gives
√n− r =
√n(1− r
n
) 12=
√n
(1− 1
2n− r2
8n2− r3
16n3− · · ·
)=
√n
(1− r
2n+ O
(r2
n2
)),
29
1
n− r=
1
n
(1− r
n
)−1=
1
n
(1 +
r
n+
r2
n2+ · · ·
)=
1
n
(1 + O
( rn
)),
so that
p(n− r) =exp{π√
23
√n− π√
6r√n+ O
(r2
n32
)}4√3n
(1 + O
( rn
))(1 + O
(1√n− r
))= p(n) · exp
{− πr√
6n
}exp{O(r2
n32
)}(1 + O
( rn
)+ O
(1√n− r
)+ O
(r
n√n− r
))= p(n) · exp
{− πr√
6n
}exp{O(r2
n32
)}(1 + O
( rn
)+ O
(1√n− r
)),
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:
r√n=
r2
r√n≥ r2
n32
because r < n. So also, if r > n23 , then r√
n> n
16 , 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 suan r, so their contribution tends to 0. We thus consider terms p(n − r) with r ≤ n
23 . e Taylor series
for exp(z) implies
exp(r2
n32
)= 1 +
r2
n32
+1
2!
(r2
n32
)2
+ · · · = 1 + O(r2
n32
),
as r2 < n32 . Applying this to our equation for p(n− r) yields
p(n− r) = p(n) · exp{− πr√
6n
}(1 + O
(r2
n32
)+ O
( rn
)+ O
(1√n− r
)).
Because r ≤ n23 , the fractions can be ordered
r2
n32
≤ n− 16 and
r
n≤ n− 1
3 ,
and for sufficiently sized n (n ≥ 8), n23 ≤ n
2 , so that
√n ≥
√n− r ≥
√n
2,
implying that 1√n−r
= O(n− 1
2
). is knowledge combined simplifies our equation to
p(n− r) = p(n) · exp{− πr√
6n
}(1 + O
(n− 1
6
)),
30
and, when substituted into (6.14):
D(n) =1
p(n)
bn2/3c∑r=1
p(n) · exp{− πr√
6n
}(1 + O
(n− 1
6
))
=(1 + O
(n− 1
6
)) bn2/3c∑r=1
(exp{− π√
6n
})r
.
e sum is recognisable as a finite geometric series, and can be simplified to
D(n) =1−
(exp{− π√
6n
(⌊n2/3
⌋+ 1)})
1− exp{− π√
6n
} (1 + O
(n− 1
6
)).
Once again, the exponential term in the numerator approaes 0 quily, as⌊n2/3
⌋>
√n. Expanding
the other exponential function yields
D(n) =1
1−(1 + π√
6n+ O
(π2
6n
)) (1 + O(n− 1
6
))=
1
π√6n
+ O(π2
6n
) (1 + O(n− 1
6
))
=
√6n
π
(1 + O
(π√6n
))−1 (1 + O
(n− 1
6
)).
Lastly, the binomial series can once again be applied, this time to the reciprocal term:
D(n) =
√6n
π
(1 + O
(π√6n
))(1 + O
(n− 1
6
))=
√6n
π
(1 + O
(n− 1
6
)),
and the result is obtained. With a similar but more extensive argument it can also be shown that theaverage number of summands (counting repeated summands) of a partition of n is
√6n
2πlogn+ O(
√n).
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(a bc d
)∈ Γ, c > 0 and τ ′ = aτ+b
cτ+d , then
F (e2πiτ ) = F (e2πiτ′) exp
(πi(τ − τ ′)
12
){−i(cτ + d)}
12 exp
{πi
(a+ d
12c+ s(−d, c)
)}.
Now if c = k and d = −h so that s(−d, c) = s(h, k) and if a = H , then b = −hH+1k because
(a bc d
)∈ Γ.
In our case, where τ = iz+hkk2
,
aτ + b =Hiz + hHk
k2− hHk + k
k2
=Hiz − k
k2,
cτ + d =iz + hk
k− h
=iz
k
so
τ ′ =Hiz − k
kiz=
iz−1k +H
k.
Seing this into our functional equation, we have
F
(exp(2πihk
k2− 2πz
k2
))= F
(exp(2πiH
k− 2πk
zk
)){−i
(iz
k
)} 12
× exp(πi
12
(iz − iz−1k2 + hk −Hk
k2
))exp{πi
(H − h
12k+ s(h, k)
)}= F
(exp(2πiH
k− 2π
z
))(zk
) 12
× exp(πi
12
(h−H
k+
H − h
k+
iz
k2− i
z
))exp (πis(h, k))
= F
(exp(2πiH
k− 2π
z
))(zk
) 12exp( π
12z− πz
12k2+ πis(h, k)
),
as required.
32
A.2 Evaluation of the Bessel Integral
We have the function
p(n) = 2π( π
12
) 32
∞∑k=1
k−5/2Ak(n)1
2πi
∫ π/12+∞i
π/12−∞it−5/2 exp
{t+
π2
6k2t
(n− 1
24
)}dt,
and wish to evaluate the integral. Equation (B.1) implies that(12z)ν
2πi
∫ π/12+∞i
π/12−∞it−ν−1 exp
(t+
z2
4t
)dt = Iν(z),
and if z and ν in this integral are taken as
z
2=
{π2
6k2
(n− 1
24
)} 12
and ν =3
2,
we have
1
2πi
∫ π/12+∞i
π/12−∞it−5/2 exp
{t+
π2
6k2t
(n− 1
24
)}dt =
{π2
6k2
(n− 1
24
)}− 34
I 32
(π
k
√2
3
(n− 1
24
)).
Hence,
p(n) = 2π( π
12
) 32
∞∑k=1
k−5/2Ak(n)π− 3
2
(n− 1
24
)− 34
6−34k−
32
I 32
(π
k
√2
3
(n− 1
24
))
=2π(n− 1
24
)− 34
2434
∞∑k=1
k−1Ak(n)I 32
(π
k
√2
3
(n− 1
24
)).
As obtained in (B.2), Bessel functions of half odd order can be reduced, in this case
I 32(z) =
√2z
π
d
dz
(sinh zz
).
is yields
I 32
(π
k
√2
3
(n− 1
24
))=
212
π12
π12
k12
214
314
(n− 1
24
) 14 d
dz
sinh{
πk
√23
(n− 1
24
)}πk
√23
(n− 1
24
) .
Because k is constant we have
dz =π
2k
(2
3
(n− 1
24
))− 12 2
3dn,
33
and thus
I 32
(π
k
√2
3
(n− 1
24
))=
212
k12
214
314
(n− 1
24
) 14 2k
π
212
312
(n− 1
24
) 12 3
2
k
π
212
312
d
dn
sinh{
πk
√23
(n− 1
24
)}√n− 1
24
=
634k
32
π2
(n− 1
24
) 34 d
dn
sinh{
πk
√23
(n− 1
24
)}√n− 1
24
.
Inserting this into our equation for p(n) we obtain
p(n) =2π(n− 1
24
)− 34
2434
634
(n− 1
24
) 34
π2
∞∑k=1
k−1k32Ak(n)
d
dn
sinh{
πk
√23
(n− 1
24
)}√n− 1
24
=
1
π√2
∞∑k=1
k12Ak(n)
d
dn
sinh{
πk
√23
(n− 1
24
)}√n− 1
24
.
34
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
Jν(z) =∞∑
m=0
(−1)m(z2
)ν+2m
m!Γ(ν +m+ 1).
e modified Bessel functions of the first kind are consequently defined as
Iν(z) = e−12νπiJν(iz)
= i−νJν(iz)
=
∞∑m=0
(z2
)ν+2m
m!Γ(ν +m+ 1).
ese functions can be described in a myriad of forms, and for our purposes a specific integral represen-tation is required. Making use of the Gamma integral
1
Γ(z)=
1
2πi
∫ c+∞i
c−∞it−zetdt,
we find that
Iν(z) =
(z2
)ν2πi
∞∑m=0
∫ c+∞i
c−∞i
(z2
)2mm!
t−ν−m−1etdt.
Replacing the sum of integrals by an integral of sums and continuing:
Iν(z) =
(z2
)ν2πi
∫ c+∞i
c−∞i
∞∑m=0
((z2
)2t−1)m
m!t−ν−1etdt
=
(z2
)ν2πi
∫ c+∞i
c−∞iexp{(z
2
)2 1t
}t−ν−1etdt.
is yields the necessary integral equation, in whi c > 0 and Re(ν) > 0:
Iν(z) =
(12z)ν
2πi
∫ c+∞i
c−∞it−ν−1 exp
(t+
z2
4t
)dt. (B.1)
Bessel functions of half-odd order, i.e., those with ν = (n + 12), where n ∈ Z, can be expressed in finite
terms by means of algebraic and trigonometrical functions of z. In particular,
I 32(z) =
√2z
π
d
dz
(sinh zz
). (B.2)
35
e proof of this result is rather tenical. We rework the right and le hand sides respectively to a mutual‘middle-point’.√
2z
π
d
dz
(sinh zz
)=
√2z
π
(cosh zz
− sinh zz2
)=
√2z
π
{(1
2!− 1
3!
)z +
(1
4!− 1
5!
)z3 +
(1
6!− 1
7!
)z5 + · · ·
}=
√2z
π
∞∑m=0
{1
(2m+ 2)!− 1
(2m+ 3)!
}z2m+1
=
√2z
π
∞∑m=0
{2m+ 2
(2m+ 3)!
}z2m+1
=
√2z
π
∞∑m=0
z2m+1
(2m)!(4m2 + 8m+ 3).
Before we alter the le hand side, we require a few extra properties of the Gamma function, namely that
Γ(z + 1) = zΓ(z) and Γ(n+1
2) =
(2n− 1)!!
2n√π,
and also a tri involving (double) factorials:
2mm!(2m− 1)!! = (2m)!!(2m− 1)!! = (2m)!.
Now, we proceed as follows:
I 32(z) =
(z2
) 32
∞∑m=0
(z2
)2mm!Γ(m+ 5
2)
=(z2
) 32
∞∑m=0
(z2
)2mm!Γ(m+ 1
2)(m+ 32)(m+ 1
2)
=1√π
(z2
) 32
∞∑m=0
(z2
)2m2m
m!(2m− 1)!!(m2 + 2m+ 34)
=1
4
√2z
π
∞∑m=0
z2m+1
2mm!(2m− 1)!!(m2 + 2m+ 34)
=
√2z
π
∞∑m=0
z2m+1
(2m)!(4m2 + 8m+ 3),
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 Rademaer path P (N)
1 0 1
1
0
1
x-plane
0.00 0.25 0.50 0.75 1.00
0.25
0.50
0.75
1.00τ-plane
1 0 1
1
0
1
x-plane
0.00 0.25 0.50 0.75 1.00
0.25
0.50
0.75
1.00τ-plane
Figure C.1: P(5) and P(20).
37
References
Andrews, G. E. (1984),e eory of Partitions, Cambridge University Press.
Apostol, T. M. (1990),Modular Functions and Dirilet Series in Number eory, 2nd edn, Springer-Verlag.
Goh, W. M. Y. & Smutz, 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 UniversityPress.
Rademaer, H. (1974), Collected Papers of Hans Rademaer, Vol. 2, MIT Press.
Vaughan, R. C. (1997),e Hardy-Lilewood 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
