Rademacher’s Formula for the Partition Function

The Dedekind eta function is introduced by Dedekind in 1877 and is defined in the upper half plane $\displaystyle\mathbb{H}=\left\{\tau\,|\,\text{Im}\,\tau>0\right\}$ by the equation

$\displaystyle\eta(\tau)=e^{\pi i\tau/12}\prod_{n=1}^{\infty}\left(1-e^{2\pi in\tau}\right).$

(2.1)

The generating function $F(x)$ (1.1) for the sequence $\{p(n)\}$ is related to $\eta(\tau)$ by

$\displaystyle\eta(\tau)=e^{\pi i\tau/12}F(e^{2\pi i\tau}).$

(2.2)

The eta function $\eta(\tau)$ is closely related to the theory of modular forms [Apo90]. Its 24th power, $\eta(\tau)^{24}$, is a modular form of weight 12.

To prove Rademacher’s formula, one needs a transformation formula for the Dedekind eta function under a fractional linear transformation

$$\tau\mapsto\frac{a\tau+b}{c\tau+d}$$

defined by an element $\displaystyle\begin{bmatrix}a&b\\ c&d\end{bmatrix}$ of the modular group $\Gamma=\text{PSL}\,(2,\mathbb{Z})$.

Before stating the transformation formula, let us define the Dedekind sum.

A proof of this theorem is given in [Apo90] using a more general formula proved by Iseki [Ise57]. In [KT23], we use a simpler approach to derive that formula.

Let $\nu$ be a complex number. The modified Bessel function $I_{\nu}(x)$ is defined as

$\displaystyle I_{\nu}(x)=\left(\frac{x}{2}\right)^{\nu}\sum_{j=0}^{\infty}\frac{1}{j!\,\Gamma(\nu+j+1)}\left(\frac{x}{2}\right)^{2j}.$

(2.5)

$I_{\nu}(x)$ has a contour integral representation given by

$$I_{\nu}(x)=\left(\frac{x}{2}\right)^{\nu}\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}t^{-\nu-1}\exp\left(t+\frac{x^{2}}{4t}\right)dt,$$

where $c$ is a positive number. To prove this, we need the Hankel’s formula.

Let $K$ be the contour which is a loop around the negative real axis. As shown in Figure 1, it consists of three parts $K_{1}$, $K_{2}$ and $K_{3}$, where $K_{1}$ and $K_{3}$ are the lower and upper edges of a cut in the $z$-plane along the negative real-axis, and $K_{2}$ is a positively oriented circle of radius $c<2\pi$ about the origin.

Using this lemma, we can prove the contour integral formula for the modified Bessel function $I_{\nu}(x)$.

For our applications, we are interested in the case where $\nu=3/2$.

Let $n$ be a positive integer. The set of Farey fractions of order $n$, denoted by $F_{n}$, is the set of all reduced fractions in the closed interval $[0,1]$ whose denominator is not larger than $n$, listed in increasing order of magnitude.

Obviously, for $n\geq 2$, $F_{n-1}$ is a subset of $F_{n}$. Moreover,

$$|F_{n}|-|F_{n-1}|=\varphi(n),$$

which is the number of positive integers less than or equal to $n$ that are relatively prime to $n$.

To be more precise, one adds in $\varphi(n)$ fractions of the form $\displaystyle\frac{h}{n}$ with $1\leq h\leq n$ and $(h,n)=1$ into the set $F_{n-1}$ to obtain $F_{n}$. To determine the precise places to insert these fractions, one need the following lemmas.

From this, we can deduce the following theorem.

Now we can give the precise algorithm to construct $F_{n}$ from $F_{n-1}$. First notice that for the two terms

$$\frac{a}{b}=\frac{0}{1}\quad\text{and}\quad\frac{c}{d}=\frac{1}{1}$$

in $F_{1}$, $a=0$, $b=c=d=1$ and thus

$$bc-ad=1.$$

Now to construct $F_{2}$, we insert the term

$$\frac{h}{k}=\frac{a+c}{b+d}=\frac{1}{2}$$

in between them. It is easy to check that if $bc-ad=1$, $h=a+c$ and $k=b+d$, then

$$bh-ak=1\quad\text{and}\quad ck-dh=1.$$

In general, assume that $F_{n-1}$ has been constructed and for any two consecutive terms $a/b$ and $c/d$ in $F_{n-1}$, $bc-ad=1$. To construct $F_{n}$, we need to insert the $\varphi(n)$ fractions $h/n$ with $1\leq h\leq n$ and $(h,n)=1$ into $F_{n-1}$. To do this, we look for consecutive pairs of reduced fractions $a/b$ and $c/d$ in $F_{n-1}$ for which $a/b<c/d$ and $b+d=n$. Let $h=a+c$. Since $0\leq a\leq b$ and $0\leq c\leq d$, we have

$$0\leq h=a+c\leq b+d.$$

Lemma 2.7 implies that

$$\frac{a}{b}<\frac{h}{n}<\frac{c}{d}.$$

By induction hypothesis, $bc-ad=1$. As we have shown, this implies that

$$bh-an=1\quad\text{and}\quad cn-dh=1.$$

It remains to show that we can find exactly $\varphi(n)$ pairs of consecutive terms $a/b$ and $c/d$ in $F_{n-1}$ such that $b+d=n$.

For $n\geq 2$, there are exactly $\varphi(n)$ positive integers $b$ with $1\leq b\leq n$ and $(b,n)=1$. In fact, we must also have $b\leq n-1$. For each of these integers $b$, $d=n-b$ is also relatively prime to $n$ and $1\leq d\leq n-1$. Since $b+d=n$, $b$ and $d$ must also be relatively prime. There exists $s_{0}$ and $t_{0}$ such that

$$bs_{0}-dt_{0}=1.$$

Any pairs of integers $(s,t)$ with $bs-dt=1$ can be written as

$\displaystyle s=s_{0}+du,\quad t=t_{0}+bu$

for some integer $u$. Let $u_{0}$ be the smallest one so that $c=s_{0}+du_{0}\geq 1$. Then $c\leq d$. If $a=t_{0}+bu_{0}$, then $bc-ad=1$. Since

$$ad=bc-1<bc\leq bd,$$

we find that $a<b$. Since $b\geq 1,c\geq 1$, we have $ad\geq 0$ and thus $a\geq 0$. In other words, we have shown that for each of the $\varphi(n)$ positive integers $b$ so that $1\leq b\leq n$ and $(b,n)=1$, there are unique integers $a,c$ and $d$ such that

$$0\leq\frac{a}{b}<\frac{c}{d}\leq 1$$

and $bc-ad=1$. This shows that we can find exactly $\varphi(n)$ consecutive pairs in $F_{n-1}$ for us to insert the fractions $h/n$ with $1\leq h\leq n$ and $(h,n)=1$.

Given a reduced fraction $h/k$, the Ford circle $C(h,k)$ (Figure 2) is the circle in the complex plane with center at

$$c_{h,k}=\frac{h}{k}+\frac{i}{2k^{2}}$$

and radius

$$r_{h,k}=\frac{1}{2k^{2}}.$$

Obviously, each Ford circle $C(h,k)$ touches the $x$-axis at the point $(h/k,0)$.

Now we study the points of tangency of two Ford circles that are tangent to each other.

For each positive integer $N$, we construct a path $P(N)$ joining the points $i$ and $1+i$ in the complex plane as follows. Let

$$0,\frac{h_{1}}{k_{1}},\ldots,\frac{h_{m_{N}-1}}{k_{m_{N}-1}},1$$

be the Farey fractions in $F_{N}$ listed in increasing order. Let

$$\frac{h_{0}}{k_{0}}=0\quad\text{and}\quad\frac{h_{m_{N}}}{k_{m_{N}}}=1.$$

For each $1\leq j\leq m_{N}-1$, let $A_{j}$ be the point of tangency between $C(h_{j-1},k_{j-1})$ and $C(h_{j},k_{j})$; and let $B_{j}$ be the point of tangency between $C(h_{j},k_{j})$ and $C(h_{j+1},k_{j+1})$. Then $B_{j}=A_{j+1}$ for $1\leq j\leq m_{N}-2$. $A_{j}$ and $B_{j}$ divide the circle $C(h_{j},k_{j})$ into two arcs, the upper arc and the lower arc. Let $U_{j}$ be the upper arc. Finally, let $U_{0}$ be the arc of the circle $C(h_{0},k_{0})$ that joins the point $i$ to $A_{1}$ that is in the right half plane $\text{Re}\,z\geq 0$, and let $U_{m_{N}}$ be the arc of the circle $C(h_{m_{N}},k_{m_{N}})$ that joins $B_{m_{N}-1}$ to the point $1+i$ that is in the left half-plane $\text{Re}\;z\leq 1$. The path $P(N)$ is defined as

$\displaystyle P(N)=\bigcup_{j=0}^{m_{N}}U_{j},$

(2.9)

following the orientation from the point $i$ to the point $1+i$.