A very large proportion of the most interesting arithmetical functions–of the functions, for example, which occur in the theory of partitions, the theory of the divisors of numbers, or the theory of the representation of numbers by sums of squares–occur as the coefficients in the expansions of elliptic modular functions in powers of the variable $q=e^{\pi i\tau}$. All of these functions have a restricted region of existence, the unit circle $|q|=1$ being a ``natural boundary'' or line of essential singularities. The most important of them, such as the functions¹

are regular inside the unit circle; and many, such as the functions (1.1) and (1.2), have the additional property of having no zeros inside the circle, so that their reciprocals are also regular.

In a series of recent papers² we have applied a new method to the study of these arithmetical functions. Our aim has been to express them as series which exhibit explicitly their order of magnitude, and the genesis of their irregular variations as $n$ increases. We find, for example, for $p(n)$, the number of unrestricted partitions of $n$, and for $r_s(n)$, the number of representations of $n$ as the sum of an even number $s$ of squares, the series

where $\ds\lambda_n=\sqrt{\left ({n-\frac{1}{24}}\right)}$ and $\ds C=\pi\sqrt{\left ({\frac{2}{3}}\right )}$, and and our methods enable us to write down similar formulæ for a very large variety of other arithmetical functions.

The study of series such as (1.5) and (1.6) raises a number of interesting problems, some of which appear to be exceedingly difficult. The first purpose for which they are intended is that of obtaining approximations to the functions with which they are associated. Sometimes they give also an exact representation of the functions, and sometimes they do not. Thus the sum of the series (1.6) is equal to $r_s(n)$if $s$ is 4, 6, or 8, but not in any other case. The series (1.5) enables us, by stopping after an appropriate number of terms, to find approximations to $p(n)$ of quite startling accuracy; thus six terms of the series give $p(200)=3972999029388$, a number of 13 figures, with an error of 0.004. But we have never been able to prove that the sum of the series is $p(n)$ exactly, nor even that it is convergent.

There is one class of series, of the same general character as (1.5) or (1.6), which lends itself to comparatively simple treatment. These series arise when the generating modular function $f(q)$ of $\phi (\tau )$ satisfies an equation $$\phi(\tau )=(a+b\tau )^n\phi\left ({\ds\frac{c+d\tau}{a+b\tau}}\right),$$ where $n$ is a positive integer, and behaves, inside the unit circle, like a rational function; that is to say, possesses no singularities but poles. The simplest examples of such functions are the reciprocals of the functions (1.3) and (1.4). The coefficients in their expansions are integral, but possess otherwise no particular arithmetical interest. The results, however, are very remarkable from the point of view of approximation; and it is in any case, well worthwhile, in view of the many arithmetical applications of this type of series, to study in detail any example in which the results can be obtained by comparatively simple analysis.

We begin by proving a general theorem (Theorem 1) concerning the expression of a modular function with poles as a series of partial fractions. This series is (as appears in Theorem 2) a ``Poincaré's series'': what our theorem asserts is, in effect, that the sum of a certain Poincaré's series is the only function which satisfies certain conditions. It would, no doubt, be possible to obtain this result as a corollary from propositions in the general theory of automorphic functions; but we thought it best to give an independent proof, which is interesting in itself and demands no knowledge of this theory.

2.

The proof requires certain geometrical preliminaries.

The half-plane ${\bf I}(\tau)>0$, which corresponds to the inside of the unit circle in the plane of $q$, is divided up, by the substitutions of the modular group, into a series of triangles whose sides are arcs of circles and whose angles are $\pi /3, \pi /3$, and 0³. One of these, which is called the fundamental polygon ($P$)⁴, has its vertices at the points $\rho , \rho^2$, and $i\infty$, where $\rho=e^{\pi i/3}$, and its sides are parts of the unit circle $|\tau |=1$ and the lines ${\bf R}(\tau )=\pm \frac{1}{2}$.

Suppose that $F_m$ is the ``Farey's series'' of order $m$, that is to say the aggregate of the rational fractions between 0 and 1, whose denominators are not greater than $m$, arranged in order of magnitude⁵, and that $h'/k'$ and $h/k$, where $0\lt h'/k'\lt h/k\lt 1$, are two adjacent terms in the series. We shall consider what regions in the $\tau$-plane correspond to $P$ in the $T$-plane, when

Both of these substitutions belong to the modular group, since $hk'-h'k=1$. The points $i\infty , 1/2, -1/2$, in the $T$-plane correspond to $h/k, (h+2h')/(k+2k'), (h-2h')/(k-2k')$ in the $\tau$-plane. Thus the lines ${\bf R}(T)=\frac{1}{2}, {\bf R}(T)=-\frac{1}{2}$ correspond to semicircles described on the segments $$\left ({\frac{h}{k}, \frac{h+2h'}{k+2k'}}\right), \left({\frac{h}{k}, \frac{h-2h'}{k-2k'}}\right)$$ respectively as diameters. Similarly the upper half of the unit circle corresponds to a semicircle on the segment $$\left ({\frac{h+h'}{k+k'}, \frac{h-h'}{k-k'}}\right).$$

The polygon $P$ corresponds to the region bounded by these three semicircles. In particular, the right hand edge of $P$ corresponds to a circular arc stretching from $h/k$ (where it cuts the real axis at right angles) to the point $$ \frac{h'k'+hk+\frac{1}{2}(hk'+h'k)+\frac{1}{2}i\sqrt{3}} {k^2+kk'+k'^2} \tag{3.3} $$ corresponding to $\tau=\rho$.

Similarly we find that the substitution (3.2) correlates to $P$ a triangle bounded by semicircles on the segments $$\left ({\frac{h'}{k'}, \frac{h'-2h}{k'-2k}}\right ) , \left ({\frac{h'}{k'}, \frac{h'+2h}{k'+2k}}\right ), \left ({\frac{h'-h}{k'-k}, \frac{h'+h}{k'+k}}\right ).$$ In particular, the left hand edge of $P$ corresponds to a circular arc from $h'/k'$ to the point (3.3). These two arcs, meeting at the point (3.3), form a continuous path $\omega$, connecting $h/k$ and $h'/k'$, every point of which corresponds, in virtue of one or other of the substitutions (3.1) and (3.2), to a point on one of the rectilinear boundaries of $P$⁶.

Performing a similar construction for every pair of adjacent fractions of $F_m$, we obtain a continuous path from $\tau=0$ to $\tau=1$. This path, and its reflexion in the imaginary axis, give a continuous path from $\tau=-1$ to $\tau=1$, which we shall denote by $\Omega_m$. To $\Omega_m$ corresponds a path in the $q$-plane, which we call $H_m$; $H_m$ is a closed path, formed entirely by arcs of circles which cut the unit circle at right angles.

Since $$\frac{h'}{k'}\lt \frac{h'+2h}{k'+2k}, \frac{h+2h'}{k+2k'}\lt\frac{h}{k},$$ the path $\omega$ from $h'/k'$ to $h/k$ is always passing from left to right, and its length is less than twice that of the semicircle on $(h'/k',h/k)$, i.e., than $\pi /kk'$. The total length of $\Omega_m$ is less than $2\pi$; and, since $$\left |{\frac{dq}{d\tau}}\right |=\left |{\pi ie^{\pi i\tau}}\right |\leq \pi ,$$ the length of $H_m$ is less than $2\pi^2$. Finally, we observe that the maximum distance of $\Omega_m$ from the real axis is less than half the maximum distance between two adjacent terms of $F_m$, and so less than $1/2m$⁷. Hence $\Omega_m$ tends uniformly to the real axis, and $H_m$ to the unit circle, when $m\longrightarrow \infty$.

4. The function $\phi (\tau)$ can have but a finite number of poles in $P$; we suppose, for simplicity, that none of them lie on the boundary. There is then a constant $K$ such that $|f(q)|\lt K$ on the boundary of $P$.

We now consider the integral

where $|q|\lt 1$ and the contour of integration is $H_m$⁸. By Cauchy's Theorem, the integral is equal to $$f(q)-\Sigma R,$$ where $R$ is a residue of $f(x)/(q-x)$ at a pole of $f(x)$ inside $H_m$⁹. To prove our theorem, then, we have merely to shew that the integral (4.1) tends to zero when $m\longrightarrow \infty$.

Let $\omega_1'$ and $\omega_1$ be the left- and right-hand parts of $\omega$, and $\zeta_1' , \zeta_1$ and $\zeta$ the corresponding arcs of $H_m$. The length of $\omega_1$ is, as we have seen, less than $\frac{1}{2}\pi/kk'$, and that of $\zeta_1$ than $\frac{1}{2}\pi^2/kk'$. Further, we have, on $\zeta_1$, $$|f(x)|=|\phi (\tau )|= |h-k\tau |^n|\phi (T)| \lt K\left\{{k\left ({\frac{h}{k}-\frac{h'}{k'}}\right )}\right\}^n=\frac{K}{k'^n}.$$ Thus the contribution of $\zeta_1$ to the integral is numerically less than $C/(kk'^{n+1})$, where $C$ is independent of $m$; and the whole integral (4.1) is numerically less than

where the summation extends to all pairs of adjacent terms of $F_m$.

When $\nu$ is fixed and $m>\nu $, the number of terms of $F_m$ whose denominators are less than $\nu$ is a function of $\nu$ only, say $N(\nu )$. If $h/k$ is one of these, and $h'/k'$ is adjacent to it, $k+k'>m$¹⁰, and so $k'>m-\nu$. Thus the terms of (4.2) in which either $k$ or $k'$ is less than $\nu$ contribute less than $8CN(\nu )/(m-\nu )$. The remaining terms contribute less than $$\frac{4C}{\nu^n}\sum\frac{1}{kk'}=\frac{4C}{\nu^n}.$$ Hence the sum (4.2) is less than $$\frac{8CN(\nu )}{m-\nu}+\frac{4C}{\nu^n},$$ and it is plain that, by choice of first $\nu$ and then $m$, this may be made as small as we please. Thus (4.1) tends to zero and the theorem is proved. It should be observed that $\Sigma R$ must, for the present at any rate, be interpreted as meaning the limit of the sum of terms corresponding to poles inside $H_m$; we have not established the absolute convergence of the series.

We supposed that no pole of $\phi (\tau )$ lies on the boundary of $P$. This restriction, however, is in no way essential; if it is not satisfied, we have only to select our ``fundamental polygon'' somewhat differently. The theorem is consequently true independently of any such restriction.

So far we have supposed $|q|\lt 1$. It is plain that, if $|q|>1$, the same reasoning proves that

5. Suppose in particular that $\phi (\tau )$ has one pole only, and that a simple pole at $\tau =\alpha$, with residue $A$. The complete system of poles is then given by

If $a$ and $b$ are fixed, and $(c,d)$ is one pair of solutions of $ad-bc=1$, the complete system of solutions is $(c+ma,d+mb)$, where $m$ is an integer. To each pair $(a,b)$ correspond an infinity of poles in the plane of $\tau$; but these poles correspond to two different poles only in the plane of $q$, viz, the positive and negative signs corresponding to even and odd values of $m$ respectively. It is to be observed, moreover, that different pairs $(a,b)$ may give rise to the same pole ${\bf q}$. The residue of $\phi (\tau )$ for $\tau={\bf a}$ is , in virtue of the functional equation (2.2), $$\frac{A}{(a+b\alpha )^{n+2}};$$ and the residue of $f(q)$ for $q={\bf q}$ is $$\frac{A}{(a+b\alpha )^{n+2}}\left ({\frac{dq}{d\tau}}\right )_{\tau ={\rm a}}= \frac{\pi iA{\bf q}}{(a+b\alpha )^{n+2}}.$$

Thus the sum of the terms of our series which correspond to the poles (5.2) is $$\frac{\pi iA}{(a+b\alpha )^{n+2}}\left ({\frac{\bf q}{q-{\bf q}}-\frac{\bf q}{q+{\bf q}}}\right ) = \frac{2\pi iA}{(a+b\alpha )^{n+2}}\frac{\bf q^2}{q^2-{\bf q}^2}.$$

We thus obtain:

If $\phi (\tau )$ has several poles in $P$, $f(q)$, of course, will be the sum of a number of series such as (5.3). Incidentally, we may observe that it now appears that the series in question are absolutely convergent.

6. As an example, we select the function

say, where $x=q^2$. It is evident that $p_n$ is always an integer; the values of the first 13 coefficients are $$ \begin{array}{l} p_0 =1,\\ p_1 =504,\\ p_2 =270648,\\ p_3 =144912096,\\ p_4 =77599626552,\\ p_5 =41553943041744,\\ p_6 =22251789971649504,\\ p_7 =11915647845248387520,\\ p_8 =6380729991419236488504,\\ p_9 =3416827666558895485479576,\\ p_{10} =1829682703808504464920468048,\\ p_{11} =979779820147442370107345764512,\\ p_{12} =524663917940510191509934144603104; \end{array} $$ so that $p_{12}$ is a number of 33 figures.

By means of the formulæ¹¹ $$g_3=\frac{8}{27}(e_1-e_3)^2(1+k^2)(1-\frac{1}{2}k^2)(1-2k^2),$$ $$e_1-e_3= \left ({\frac{\pi}{2\omega_1}}\right )^2\{\vartheta_3 (0)\}^4, \frac{2K}{\pi}= \{\vartheta_3 (0)\}^2,$$ we find that $$\frac{1}{f(q)}=\left ({\frac{2K}{\pi}}\right )^6(1+k^2)(1-\frac{1}{2}k^2)(1-2k^2).$$

The value of $n$ is 6. The poles of $f(q)$ correspond to the value of $\tau$ which make $K=k^2$ equal to $-1,2$ or $\frac{1}{2}$. It is easily verified¹² that these values are given by the general formula $$\tau =\frac{c+di}{a+bi}\quad (ad-bc=1),$$ so that

The value of $\alpha$ is $i$¹³. In order to determine $A$ we observe that $$-504\frac{d}{dq}\left ({\frac{1^5q^2}{1-q^2}+\frac{2^5q^4}{1-q^4}+\cdots}\right )=-\frac{1008}{q}\left\{\frac{1^6q^2}{(1-q^2)^2}+ \frac{2^6q^4}{(1-q^4)^2}+\cdots\right\}.$$

The series in curly brackets is the function called by Ramanujan¹⁴ $\Phi_{1,6}$ and¹⁵ $$1008\Phi_{1,6}=Q^2-PR,$$ where $$P=\frac{12\eta_1\omega_1}{\pi^2}, \quad Q=12g_2\left ({\frac{\omega_1}{\pi}}\right )^4, \quad R=216g_3\left ({\frac{\omega_1}{\pi}}\right )^6.$$ Here $R=0$, so that $$1008\Phi_{1,6}=Q^2=1+480\Phi_{0,7}\href{#p37_en16}{^{16}} = 1+480\left ({\frac{1^7q^2}{1-q^2}+\frac{2^7q^4}{1-q^4}+\cdots}\right ).$$ Hence we find that $$A=i/\pi C, \quad 2\pi iA=-2/C,$$ where

Another expression for $C$ is where We have still to consider more closely the values of $a$ and $b$, over which the summation is effected. Let us fix $k$, and suppose that $(a,b)$ is a pair of positive solutions of the equation $a^2+b^2=k$. This pair gives rise to a system of eight solutions, viz., $$(\pm a,\pm b), (\pm b, \pm a).$$ But it is obvious that, if we change the signs of both $a$ and $b$, we do not affect the aggregate of values of ${\bf a}$. Thus we need only consider the four pairs $$(a,b), (a,-b) ,(b,a), (b,-a).$$ If $a$ or $b$ is zero, or if $a=b$, these four pairs reduce to two.

It is easily verified that, if $(a,b)$ leads to the pair of poles $$q=\pm{\bf q}=\pm\exp\left ({\frac{ac+bd}{a^2+b^2}\pi i-\frac{\pi}{a^2+b^2}}\right ),$$ then $(a,-b)$ and $(b,a)$ each lead to $q=\pm\bf{\bar{q}}$, where $\bf{\bar {q}}$ is the conjugate of ${\bf q}$. Thus, in general $(a,b)$ and the solutions derived from it lead to four distinct poles, viz., $\pm{\bf q}$ and $\pm\bf{\bar{q}}$. These four reduce to two in two cases, when ${\bf q}$ is real, so that ${\bf q}=\bf{\bar{q}}$, and when ${\bf q}$ is purely imaginary, so that ${\bf q}=-\bf{\bar{q}}$. It is easy to see that the first case can occur only when $k=1$, and the second when $k=2$¹⁷.

If $k=1$ we take $a=1, b=0, c=0, d=1$; and ${\bf q}={\bf \bar{q}}=e^{-\pi}$. If $k=2$ we take $a=1, b=1, c=0, d=1$; and ${\bf q}=-{\bf \bar{q}}=ie^{-\pi/2}$. The corresponding terms in our series are $$\frac{1}{1-q^2e^{2\pi}}, \frac{1}{2^4(1+qe^{\pi})}.$$

If $k>2$, and is a sum of two coprime squares $a^2$ and $b^2$, it gives rise to terms $$\frac{1}{(a+bi)^8}\frac{1}{1-(q/{\bf q})^2}+\frac{1}{(a-bi)^8}\frac{1}{1-(q/{\bf{\bar{q}})^2}}.$$ There is, of course, a similar pair of terms corresponding to every other distinct representation of $k$ as a sum of coprime squares. Thus finally we obtain the following result:

7. It follows that

say. Here $\lambda$ is the sum of two coprime squares, so that $$\lambda=2^{a_2}5^{a_5}13^{a_{13}}17^{a_{17}}\cdots ,$$ where $a_2$ is 0 or 1 and 5, 13, 17, $\ldots$ are the primes of the form $4k+1$; and the first few values of $c_{\lambda}(n)$ are $$c_1(n)=1, c_2(n)=(-1)^n, c_5(n)=2\cos \left (\mbox{$\frac{4}{5}$}n\pi +8\arctan 2\right ),$$ $$c_{10}(n)=2\cos\left (\mbox{$\frac{3}{5}$}n\pi - 8\arctan 2\right), c_{13}(n)=2\cos\left (\mbox{$\frac{10}{13}$}n\pi +8\arctan 5\right).$$ The approximations to the coefficients given by the formula (7.1) are exceedingly remarkable. Dividing by $\frac{1}{2}C$, and taking $n=0,1,2,3,6,$ and 12, we find the following results: $$\begin{array}{r} (0)\phantom{+}0.944\\ +0.059\\ -0.003\\ \hline p_0=1.000 \end{array}\qquad \begin{array}{r} (1)\phantom{+}505.361\\ -1.365\\ +0.004\\ \hline p_1=504.000 \end{array}\qquad \begin{array}{r} (2)\phantom{+}270616.406\\ +31.585\\ +0.009\\ \hline p_2=270648.000 \end{array} $$ $$\begin{array}{r} (3)\phantom{+}144912827.002\\ -730.900\\ -0.101\\ -0.001\\ \hline p_3=144912096.000 \end{array} \begin{array}{r} (6)\phantom{+}22251789962592450.237\\ +9057051.688 \\ +2.081 \\ -0.006 \\ \hline p_6=22251789971649504.000 \end{array} $$ $$\begin{array}{r} (12)\phantom{++}524663917940510190119197271938395.329 \\ +1390736872662028.140 \\ +2680.418 \\ +0.130 \\ -0.014\\ -0.003\\ \hline p_{12}=524663917940510191509934144603104.000 \end{array}$$

An alternative expression for $C$ is $$C=96^2e^{-8\pi /3}\{(1-e^{-4\pi})(1-e^{-8\pi})\cdots \}^{16},$$ by means of which $C$ may be calculated with great accuracy¹⁸. To five places we have $2/C=0.94373$, which is very nearly equal to $352/373=0.94370$.

It is easy to see directly that $p_n$ lies between the coefficients of $x^n$ in the expansions of $$\frac{1}{(1-535x)(1+31x)}, \quad \frac{1-7.5x}{(1-535.5x)(1+24x)},$$ and so that $$\frac{(535)^{n+1}-(-31)^{n+1}}{566}\leq p_n \leq \frac{352(535.5)^{n}+21(-24)^{n}}{373}.$$

The function $$ \Omega (x)=\sum_{(\lambda )}\frac{c_{\lambda}(x)}{\lambda^4}e^{2x\pi/\lambda} $$ has very remarkable properties. It is an integral function of $x$, whose maximum modulus is less than a constant multiple of $e^{2\pi |x|}$. It is equal to $p_n$, an integer, when $x=n$, a positive integer; and to zero when $x=-n$. But we must reserve the discussion of these peculiarities for some other occasion.