theta functions: the problem of the...

Sao Paulo State University - UNESP

Department of Mathematics

Theta Functions: The Problem of the Representation of Numbers

as Sums of Squares

Junior Paper

Julio C. Andrade

Advisor: Stavros Christodoulou

Rio Claro - Sao Paulo

10 May 2005

Julio C. Andrade

Theta Functions: The Problem of the Representation of

Numbers as Sums of Squares

Junior Paper

Rio Claro - Sao Paulo

10 May 2005

Andrade, Julio Cesar

Theta Functions: The Problem of the Representation of Numbers

as Sums of Squares / Julio Cesar Andrade - 10 May 2005

55.pp

1. Analysis 2. Number Theory 3. Algebraic Number Theory.

This paper is dedicated to the memory of my

father Antonio de Andrade.

To my mother Inez Bueno, for all kinds of sup-

port.

For Carolina Esteves for your affection, love

and support.

To my friends, for the friendship.

Abstract

In this paper we use techniques developed by Jacobi, Ramanujan (such as the Ramanujan’s

sums) and modular functions to prove some theorems on the representation of numbers

as sums of squares. We start with a theorem of Jacobi, which allows to obtain formulas

for r2(n) e r4(n) (formula for the representation of a number as a sum of two squares and

four squares respectively) as corollaries of an analytic theory. We then use some theorems

of Ramanujan to deduce an expression for ϑ4(x). We then studied formulas to determine

r2s(n) when 2s exceeds 8 and so we need to use some sums developed by Ramanujan. The

use of modular functions and the modular group is required to prove that r8(s) = ρ8(s)

and a large part of this paper is concerned to proof this fact. The rest of the paper is

concerned to study the issue of representation of numbers as sums of 24 squares and ends

with a brief discussion of the Ramanujan’s function τ(n) and its relation to the problem

of representation of numbers as sums of squares.

Keywords: modular functions, modular group, Ramanujan’s Function τ(n), Ramanu-

jan’s Sums, sum of squares, theta function.

Acknowledgements

Many thanks to Prof. Stavros Christodoulou for advising me on this project. Thanks also

to Professor Freeman Dyson of the Institute for Advanced Study - Princeton and Bruce

C. Berndt of the University of Illinois at Urbana-Champaign for having suggested this

topic during my stay in Princeton. I am also grateful to Professor Bruce C. Berndt, M.

Ram Murty and Stavros Christodoulous for answering my questions on various aspects of

this paper. I am also grateful to Professor Romulo Campos Lins of the Sao Paulo State

University for having introduced and presented me to the richness and beauty of Number

Theory.

“And all its clear relations,

Its divisions and precisions,

Every street lamp that I pass

Beats like a fatalistic drum,”.

T.S. Elliot.

(Rhapsody on a Windy Night.)

Contents

1 A Historical Overview of the Problem. 7

1.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Some Results of Jacobi and Ramanujan. 9

2.1 Ramanujan’s Theorems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Other Values for k = 2s. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Some Comments on Ramanujan and the Following Work. . . . . . . . . . . 19

3 Ramanujan’s Sums. 21

3.1 Ramanujan’s Sum cq(n). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2 The series∑

q−scq(n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.3 The series∑

εqq−scq(n). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4 The Singular Series in the problem of 2s squares. 30

4.1 Summation of the singular series when 2s ≡ (mod 8) . . . . . . . . . . . . 32

5 Modular Functions. 34

5.1 Modular Group and Fundamental Region. . . . . . . . . . . . . . . . . . . 34

5.2 Functions associated with the sub–group Γ3 . . . . . . . . . . . . . . . . . 37

5.3 Proof of r8(n) = ρ8(n). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.3.1 Proof of (A). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5.3.2 Proof of (B). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6 The case of 24 squares. 44

7 A Brief Discussion of the Ramanujan’s Function τ(n). 47

7.1 The Order of τ(n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

8 Conclusion. 50

Bibliography 51

7

1 A Historical Overview of the Problem.

1.1 Introduction.

It is known that Diphantus wrote the number 13 as the sum of two square, 13 = 22 + 32.

Also Diophantus, treated the division of unity into two parts such that, if a given number

a is added to each part, the sums are (rational) squares. The problem is equivalent to the

representation of 2a + 1 as a sum of two squares.

Mohammed Ben Alhocain, an Arab of the tenth century, gave a table of num-

bers equal to a sum of two squares.

Leonardo Pisano in his Liber Quadratorum of 1225, proved that

(a2 + b2)(c2 + d2) = (ac± bd)2 + (ad∓ bc)2

where (a2− b2, 2ab, a2 + b2) and (c2− d2, 2cd, c2 + d2) are right triangles and we obtain by

multiplication two triangles with the above hypotenuse.

A. Girard had already made a determination of the numbers expressible as a

sum of two integral squares: every square, every prime 4n + 1, a product formed of such

numbers, and the double of one of the foregoing.

This brief historical memories show us the nature of the problem. The problem

of the representation of an integer n as the sum of a given number of integral squares is

one of the most celebrated in the number theory. The modern history of this problem

begins effectively with A. Girard and Fermat.

The problem that we are interested in discussing in this paper can be stated

as follows:

Problem 1. How we can represent an integer n as the sum of k integral squares?

Almost every arithmetician of note since Fermat has contributed to the solu-

tion of the problem, and it has its puzzles for us still.

1.1 Introduction. 8

Result 1. (Girard, Fermat): A prime p of form 4m + 1 is the sum of two squares.

This result is know as Fundamental Theorem of Right Triangles, for more

details and a historical treatment of this problem see [Dickson]. Therefore this initial

discussion lead us to a theoretical approach of the problem. It should be noted that there

is a full account of the history of the classical theorems concerning representation by two

or four squares in [5], vol.2 chs. VI and VIII.

9

2 Some Results of Jacobi and Ramanujan.

Definition 2.0.1. We denote the number of representations of n by k squares, i.e., the

number of integral solutions of

n = x21 + x2

2 + · · ·+ x2k,

por rk(n).

Remark 2.0.1. We must to pay attention to the sign and order of x1, x2, . . . , xk.

Example 2.0.1. :

i) 1 = (±1)2 + 02 = 02 + (±1)2

and therefore r2(1) = 4.

ii)5 = (±2)2 + (±1)2 = (±1)2 + (±2)2

and therefore r2(5) = 8.

And in this way we are interested in determining rk(n). Thus a new problem

arises naturally.

Problem 2. How can we determine rk(n) in terms of simpler arithmetical functions of

n? For example, we can express rk(n) in terms of the number or the sum of divisors of

n?

It is a good deal easier if k is an even number 2s, and I shall suppose this

throughout this paper. The great mathematician Jacobi solved the problem for 2s = 2, 4, 6

and 8 in his work known as Fundamenta Nova. Thus he proved that:

Theorem 2.0.1 (Jacobi).

(i)

r2(n) = 4∑

d odd, d|n(−1)

12(d−1) = 4d1(n)− d3(n); (2.1)

(ii)

r4(n) = 8∑

d|nd = 8σ(n), (2.2)

or


(iii)

r4(n) = 24∑

d odd, d|nd = 24σ0(n), (2.3)

The formulas for r6(n) and r8(n) is a little more complicated, but of the same

general character, and will occur later.

Remark 2.0.2. In the theorem above the sums extend over all divisor d of n, or all odd

divisor; d1(n) and d3(n) are the numbers of the divisors of n of the forms 4m + 1 and

4m + 3 respectively; σ(n) is the sum of the divisors of n, and σ0(n) the sum of its odd

divisors.

Proof.

(i) We write n = 2αN = 2αµν = 2α∏

pr∏

qs in the form

n = (1 + i)(1− i)α∏(a + bi)(a− bi)r

∏qs,

where a and b are positives and a 6= b and p = a2 + b2. This expression of p is unique

except for the order of a and b. The factors

1± i, a± bi, q

are primes of k(i). If n = A2 + B2 = (A + Bi)(A−Bi), entao

A + Bi = it(1 + i)α1(i− 1)α2∏(a + bi)r1(a− bi)r2∏

qs1 ,

A−Bi = i−t(1− i)α1(i + 1)α2∏(a− bi)r1(a + bi)r2∏

qs2 ,

where

t = 0, 1, 2, or 3, α1 + α2 = α, r1 + r2 = r, s1 + s2 = s.

Plainly s1 = s2, so that every s is even, and ν is a square. Unless this is so there is no

representation. We suppose then that

ν =∏

qs =∏

q2s1

is a square. There is no choice in the division of the factors q between A+Bi and A−Bi.

There are

4(α + 1)∏

(r + 1)


choices in the division of the other factors. But

1− i

1 + i= −i

is a unity, so that a change in α1 and α2 produces no variation in A and B beyond that

produced variation of t. We are thus left with

4∏

(r + 1) = 4d(µ)

possibly effective choices, i.e., choices which may produce variation in A and B.

The trivial variation in a representation n = A2 + B2 correspond (i) to multiplication of

A + Bi by a unity and (ii) to exchange of A + Bi with its conjugate. Thus

1(A + Bi) = A + Bi, i(A + Bi) = −B + Ai,

i2(A + Bi) = −A−Bi, i3(A + Bi) = B − Ai,

and A − Bi, −B − Ai, −A + Bi, B + Ai are the conjugates of these four numbers.

Any change in t varies the representation. Any change in the r1 and r2 also varies the

representation, and in a manner not accounted for by any change in t; for

it(1 + i)α1(1− i)α2

∏(a + bi)r1(a− bi)r2 = iθit

′(1 + i)α

′1(1− i)α

′2

∏(a + bi)r

′1(a− bi)r

′2

is impossible, unless r1 = r′1 and r2 = r

′2. There are therefore 4d(µ) different sets of values

of A and B, or of representations of n; and this proves (2.1).

(ii) and (iii) We will see later.

Jacobi found his formulae from the theory of the elliptic theta-functions. Then

is natural to consider the theta functions.

Definition 2.0.2. The theta functions is by definition

ϑ(x) = 1 + 2x + 2x4 + · · · =∞∑−∞

xn2

, (2.4)

As immediate consequence of the definition we can calculate ϑ2(x) and ϑ4(x)

and then we get as result:

ϑ2(x) = 1 + 4

(x

1− x− x3

1− x3+

x5

1− x5− x7

1− x7+ · · ·

)(2.5)


and

ϑ4(x) = 1 + 8

(x

1− x+

2x2

1 + x2+

3x3

1− x3+

4x4

1 + x4+ · · ·

), (2.6)

Proposition 2.0.1. Let r2(n) = 4∑

d odd, d|n(−1)12(d−1), then r2(n) is the result of the

equating coefficient with ϑ2(x). (ignoring the 1).

Proof. Consider ϑ2(x). Thus the right-hand side of (2.5) is ignoring the 1

ϑ2(x)− 1 = 4∑

d odd

(−1)12(d−1) xd

1− xd=

= 4∑

d odd

(−1)12(d−1)

∞∑i=1

xid =

= 4∑

d odd

∞∑i=1

(−1)12(d−1)xid

i.e.,

4

(x

1− x− x3

1− x3+

x5

1− x5− x7

1− x7+ · · ·

)= 4

∑

d odd, d|n(−1)

12(d−1) xd

1− xd

And in this way,

4

(x

1− x− x3

1− x3+

x5

1− x5− x7

1− x7+ · · ·

)= 4

∑

d odd, d|n

∞∑i=1

(−1)12(d−1)xid

Then we can conclude,(

x

1− x− x3

1− x3+

x5

1− x5− x7

1− x7+ · · ·

)=

∑

d odd, d|n(−1)

12(d−1) xd

1− xd

and thus the coefficient of xn is,

4∑

d odd, d|n(−1)

12(d−1)

In this way the formulae for r2(n) and r4(n) appear as corollaries of an ana-

lytical theory.

What we did here was get r2(n) as a corollary of analytical theory, i.e., r2(n) is the result

of equating coefficients with ϑ2(x). A natural question in this point is:

2.1 Ramanujan’s Theorems. 13

Problem 3. Is it possible to reverse the procedure above, to prove the arithmetical for-

mulae directly and deduce the analytical identities?

It is easy to deduce (2.1), the Theorem 1 (i) did this. But to deduce (2.2) and

(2.3) present greater difficulties. For this reason it is interesting to have an elementary

deduction of (2.6) from (2.5). And so we would have the following relationship:

Gaussian Complex Integers // (2.1)1

// (2.5)2

// (2.6)

Remark 2.0.3.

(1) - equaling coefficients.

(2) - elementary deduction?

2.1 Ramanujan’s Theorems.

Ramanujan found an elementary deduction of (2.6) from (2.5). I repeat it here because,

though it has very little bearing on the substance of my paper, it is a very characteristic

specimen of Ramanujan’s work.

Theorem 2.1.1 (Ramanujan). Consider by hypothesis ϑ2(x). Then is possible to obtain

as thesis ϑ4(x).

ϑ4(x) = 1 + 8

(x

1− x+

2x2

1 + x2+

3x3

1− x3+

4x4

1 + x4+ · · ·

)

Proof. We can write conveniently that,

ur =xr

1− xr,

and thus,xr

(1− xr)2= ur(1 + ur)

And we will need two preliminary lemmas to proof the theorem.

Lemma 2.1.1. ∞∑m=1

um(1 + um) =∞∑

n=1

nun.


Proof.

∞∑m=1

xm

(1− xm)2=

∞∑m=1

∞∑n=1

nxmn =∞∑

n=1

n

∞∑m=1

xmn =∞∑

n=1

nxn

1− xn=

∞∑n=1

nun.

Lemma 2.1.2. ∞∑m=1

(−1)m−1u2m(1 + u2m) =∞∑

n=1

(2n− 1)u4n−2.

Proof.

∞∑m=1

(−1)m−1x2m

(1− x2m)2=

∞∑m=1

(−1)m−1

∞∑r=1

rx2mr =

=∞∑

r=1

r

∞∑m=1

(−1)m−1x2mr =∞∑

r=1

rx2r

1 + x2r=

=∞∑

r=1

(rx2r

1− x2r− 2rx4r

1− x4r

)=

∞∑n=1

(2n− 1)x4n−2

1− x4n−2=

=∞∑

n=1

(2n− 1)u4n−2.

To prove the theorem we just need to prove that

(14+u1−u3+u5−· · · )2 = 1

16+ 1

2

∑′ mum. (2.7)

where the dash indicates the omission of multiples of 4.

But Ramanujan made a more general proof and the proof of the theorem follow as simple

corollary as we will see.

Theorem 2.1.2 (Ramanujan). If θ is real and is not an even multiple of π, and if

S = S(x, θ) =1

4cot

1

2θ + u1 sin θ + u2 sin 2θ + · · · ,

T1 = T1(x, θ) =

(1

4cot

1

2θ

)2

+ u1(1 + u1) cos θ + u2(1 + u2) cos 2θ + · · · ,

T2 = T2(x, θ) =1

2

u1(1− cos θ) + 2u2(1− cos 2θ) + 3u3(1− cos 3θ) + · · ·

.

then

S2 = T1 + T2


Proof. We have that

S2 =

1

4cot

1

2θ +

∞∑n=1

un sin nθ

2

=

(1

4cot

1

2θ

)+

1

2

∞∑n=1

un cot1

2θ sin nθ +

+∞∑

m=1

∞∑n=1

umun sin mθ sin nθ =

(1

4cot

1

2θ

)2

+ S1 + S2

We use now the follows identities,

1

2cot

1

2θ sin nθ =

1

2+ cos θ + cos 2θ + · · ·+ cos(n− 1)θ +

1

2cos nθ

and

2 sin mθ sin nθ = cos(m− n)θ − cos(m + n)θ,

and this leads us to the following equations,

S1 =∞∑

n=1

un1/2 + cos θ + cos 2θ + · · ·+ cos(n− 1)θ + 1/2 cos nθ.

S2 =1

2

∞∑m=1

∞∑n=1

umuncos(m− n)θ − cos(m + n)θ,

we get S1 + S2 as follows:

S1 + S2 =∞∑

k=0

Ck cos kθ

and thus

S2 =

(1

4cot

1

2θ

)2

+ C0 +∞∑

k=1

Ck cos kθ,

therefore S1 and S2 are cosines series that are multiples of θ.1

1To justify the argument above, we need to prove that

∞∑n=1

|un|(1/2 + | cos θ|+ · · ·+ 1/2| cos nθ|) and∞∑

m=1

∞∑n=1

|um||un|(| cos(m + n)θ|+ | cos(m− n)θ|)

are convergent. But this is an immediate consequence of the absolute convergence of the

∞∑n=1

nun,

∞∑m=1

∞∑n=1

umun.


(i) The contribution of S1 to C0 is 12

∑um, and that of S2 is 1

2

∑u2

m. Hence

C0 =1

2

∞∑1

um(1 + um) =1

2

∞∑1

xm

(1− xm)2=

1

2

∞∑1

∞∑1

nxmn =

=1

2

∞∑1

nxn

1− xn=

1

2

∞∑1

nun.

(ii) If k > 0, then the contribution of S1 to Ck is

1

2uk +

∞∑

k+1

um =1

2uk +

∞∑

l=1

uk+l.

That of S2 is,

1

2

∑

m−n=k

umun +1

2

∑

n−m=k

umun − 1

2

∑

m+n=k

umun =∞∑

l=1

ulkk+l − 1

2

k−1∑

l=1

uluk−l.

where m ≥ 1 and n ≥ 1 in each of the sums. Hence

Ck =1

2uk +

∞∑

l=1

uk+l +∞∑

l=1

uluk+l − 1

2

k−1∑

l=1

uluk−l.

It is easily verified that

uk+l(1 + ul) = uk(ul − uk+l), uluk−l = uk(1 + ul + uk−l),

so that

Ck = uk

1

2+

∞∑

l=1

(ul − uk+l)− 1

2

k−1∑

l=1

(1 + ul + uk−l)

=

= uk

1

2+ u1 + u2 + · · ·+ uk − 1

2(k − 1)− (u1 + u2 + · · ·+ uk−1)

=

= uk

(1 + uk − 1

2k

).

Hence finally

S2 =

(1

4cot

1

2θ

)2

+1

2

∞∑n=1

nun +∞∑

k=1

uk

(1 + uk

1

2k

)cos kθ =

=

(1

4cot

1

2θ

)2

+∞∑

m=1

um(1 + um) cos mθ +1

2

∞∑m=1

mum(1− cos mθ) =

= T1 + T2.

2.2 Other Values for k = 2s. 17

Remark 2.1.1. The identity above is equivalent to

ζ(u)− η1u

ω1

2

− ℘(u) =

(2π

ω1

)2− 1

24+

∞∑1

q2m

(1− q2m)2cos

mπu

ω1

,

in the ordinary notation of elliptic functions.

Corollary 2.1.1. We can proof that,

(14

+ u1 − u3 + u5 − u7 + · · · )2 = 116

+ 12(u1 + 2u2 + 3u3 + 5u5 + 6u6 + 7u7 + 9u9 + · · · ).

Proof. Let θ = 12π in the Theorem 2.1.2. Then we have,

T1 =1

16−

∞∑m=1

(−1)m−1u2m(1 + u2m),

T2 =1

2

∞∑m=1

(2m− 1)u2m−1 + 2∞∑

m=1

(2m− 1)u4m−2.

Now, by the Lemma 2.1.2,

T1 =1

16−

∞∑m=1

(2m− 1)u4m−2,

and in this way we have

T1 + T2 =1

16+

1

2(u1 + 2u2 + 3u3 + 5u5 + · · · ).

And then this proves the Theorem 2.1.1.

2.2 Other Values for k = 2s.

It is known that Jacobi solved the problem for k = 2, 4, 6 and 8. The next values for

k = 2s = 10 and k = 12 were obtained by Liouville and Eisenstein. But in these cases

r2s(n) is not usually expressible as a simple divisor function of n. In modern notation of

Glaisher we have,

r10(n) =4

5E4(n) + 16E

′4(n) + 8χ4(n),

where

E4(n) =∑d odd

d|n

(−1)12(d−1)d4,

2.2 Other Values for k = 2s. 18

E′4(n) =

∑

d′

oddd|n

(−1)12(d′−1)d4

(d′being n/d, the divisor of n “conjugate” to d), and

χ4(n) =1

4

∑

a2+b2=n

(a + bi)4

(a sum extended over the Gaussian complex divisor of n), and

r12(n) =

r12(2n) = −8ξ5(2n)

r12(2n + 1) = 8∆5(2n + 1) + 2Ω(2n + 1),

where,

∆r(n) denotes the sum of the r-th powers of the uneven divisors of n.

ξr(n) denotes the sum of the r-th powers of the even divisors of n whose conjugates are

even and of the uneven divisors of n whose conjugates are uneven diminished by the sum

of the r-th powers of the even divisors of n whose conjugates are uneven and of the uneven

divisors of n whose conjugates are even.

Thus, if d is any divisor of n, and dd′= n, then

ξr(n) =∑

(−1)(d+d′)dr.

The function Ω(n) can be defined as coefficients of expansions of elliptic functions.

k2k′2ρ6 = 16∑∞

1 Ω(n)qn, where ρ = 2Kπ

.

But is possible to define Ω(n) through of arithmetic by representation of n as

sum of 4 squares in the following way.

Ω(n) =1

8

∑(r)[a4]− 2[a2b2].

If a2 + b2 + c2 + d2 is any representation of n as sum of four squares, where∑(r) refers

to all the representations of n as a sum of 4 squares and the [ ] denotes the sums of all

types are taken:

Example 2.2.1. [a4] = a4 + b4 + c4 + d4.

In the case of r6(n) is more complicated, and was found by Jacobi. Now I will

present the formulae for r6(n) in the modern notation of Glaisher

r6(n) = 44E ′2(n)− E2(n),

2.3 Some Comments on Ramanujan and the Following Work. 19

where,

E′2(n) denotes the sum of the 2-nd powers of the divisors of n whose conjugates are of the

form 4k + 1 diminished by the sum of the 2-nd powers of those whose conjugates are of

the form 4k + 3.

E2(n) denotes the sum of the 2-nd powers of the divisors of n which are of the form 4k+1

diminished by the sum of the 2-nd powers of the divisors of n which are of the form 4k+3.

As promised then we have the formula for r6(n).

There are similar formulae for k = 2s = 12, 14, ..., r2s(n) being in each case

the sum of a divisor-function and one or more supplementary functions. For a more

detailed account about this formulae see [8]. The complexity of these supplementary

functions increases with s, and it is only the simplest of them and the majority can only

be recognised as coefficients in the expansions of modular functions.

We shall find, however, that there is always a divisor-function which dominates r2s(n). In

all cases

r2s(n) = δ2s(n) + e2s(n),

where δ2s(n) is a divisor function and e2s(n) is much smaller than δ2s(n) for large n, so

that

r2s(n) ∼ δ2s(n)

when n tends to infinity.

2.3 Some Comments on Ramanujan and the Follow-

ing Work.

In this first part I propose to work with two special cases, 2s = 8 and 2s = 24. The analysis

is more simpler than usual when 2s is a multiple of 8. In the first case I will obtain the

classic formula of Jacobi and the second will have the characteristics of the new theorems

of Ramanujan. I think that any demonstration of this never have been published before,

except for [9] to which this work is based. Here I will not use the methods of Jacobi

and Ramanujan, but the latest methods of functions obtained by Mordell and Hardy,

which take us far more clearly the fundament for the theory. I though I must begin

with some comments on the contributions of Ramanujan in the area, which are basically

two substantial papers published in Transactions of the Cambridge Philosophical Society

2.3 Some Comments on Ramanujan and the Following Work. 20

Papers Numbers 18 and 21. It is always difficult to say how Ramanujan is liable for other

writers, and the difficulty reaches its maximum when he developed work before he go to

England that was where he have had contact with the theory of elliptic functions. There

were no books on elliptic functions which he could have found in India and even worse

should not exist shadows of those books with arithmetic applications of the theory. I, as

Hardy and others, believe that Ramanujan rediscovered the formulas of Jacobi, which was

fully in its capacity. But at the time that Ramanujan published these two papers he had

read more and knew everything about Jacobi and the most recent work. In particular he

had read the papers from Glaisher, and dealt with their content as well known. And the

reader who knows Glaisher and the papers from Ramanujan will estimate the originality

of both, and shows the better in Ramanujan. They contain many memorable theorems,

which were confirmed later by Mordell, the general level of analysis is astonishing high.

In particular the second paper contains all the formal theory of “sums of Ramanujan”,

which is crucial to my purpose here. The interesting thing is that I begin my discussion

with the formulas of Ramanujan even when I’m developing a theory about this entirely

in different ways. For some observations, speculations and the importance of the great

mathematician Ramanujan see [3].

21

3 Ramanujan’s Sums.

3.1 Ramanujan’s Sum cq(n).

Definition 3.1.1. The Ramanujan’s sum is

cq(n) =∑

p(q)

e−2npπi/q,

where the notation indicates a sum over values of p less than and prime to q1.

cq(n) may also be written as

Proposition 3.1.1.

cq(n) =∑

p(q)

cos2npπ

q

(a form which shows that it is real), or again cq(n) may be written as

Proposition 3.1.2.

cq(n) =∑

ρnq

where ρq is a primitive q–th root of 1.

Definition 3.1.2 (Mobius Function). The Mobius function is defined by

(i) µ(1) = 1,

(ii) µ(n) = (−1)ν if n = p1p2 . . . pν is the product of ν different prime factos, and

(iii) µ(n) = 0 if n has a repeated factor.

Now will we see some properties of µ(n).

Proposition 3.1.3. The chief properties of µ(n) are

(a)∑

d|qµ(d) = 0 para todo q > 1.

1Or any complete system of residues prime to q.

3.1 Ramanujan’s Sum cq(n). 22

(b) And the identities

g(q) =∑

d|qf(d) (3.1)

f(q) =∑

d|qµ

(q

d

)g(d) (3.2)

are equivalent to one another. This is usually described as the “Mobius inversion

formula”.

Theorem 3.1.1 (Ramanujan).

cq(n) =∑d|q,d|n

µ(q

d

)d, i.e., (3.3)

cq(n) is a sum extended over all common divisors of q and n

Proof. To prove this Ramanujan observes that

ηq(n) =

q−1∑

h=0

e−2nhπi/q

is q if q | n and 0 otherwise. But is plain that

ηq(n) =∑

d|qcd(n),

and therefore, by the Mobius formula,

cq(n) =∑

d|qµ

(q

d

)ηd(n),

which is (3.3).

There is another proof which is a little longer but depends on principles which

will be useful later.

Definition 3.1.3. We say that f(q) is multiplicative if,

f(qq′) = f(q)f(q

′) (3.4)

whenever (q, q′) = 1. In particular this involves f(1) = 1.

It is plain that, if we want to prove that

f(q) = F (q) (3.5)

3.1 Ramanujan’s Sum cq(n). 23

and know both f(q) and F (q) to be multiplicative, then it is enough to prove the result

when q is a power of a prime.

Now, if (q, q′) = 1, we have

cq(n)cq′ (n) =

∑

p(q)

e−2npπi/q∑

p′ (q′ )

e−2np′πi/q

′

=∑p(q),

p′(q′)

e−2nPπi/qq′.

where

P = pq′+ p

′q;

and P runs over the range P (qq′) when p and p

′run over p(q) and p

′(q′). Hence

∑p(q),

p′(q′)

e−2nPπi/qq′=

∑

P (qq′ )

e−2nPπi/qq′= cqq

′ (n),

and Ramanujan’s sum is multiplicative.

Now I will present another proof of the Theorem 3.1.1

Proof. (Theorem 3.1.1)

Again, if we denote the right–hand side of (3.3) by Cq(n), we have

Cq(n)Cq′ (n) =∑

µ(q

d

)µ

(q′

d′

)dd

′=

∑µ

(qq

′

dd′

)dd

′. (3.6)

Here d | q, d | n, d′ | q′ , d

′ | n, and these relations are equivalent to

dd′ | qq′ , dd

′ | n

(since q and q′are coprime). Hence the right-hand side of (3.6) is Cqq′ (n), and Cq(n) also

is multiplicative.

We have only to prove that

c$k(n) = C$k(n)

when $ is prime. The values of p in p($k) are

p = $k−1z + p1,

where z = 0, 1, 2, . . . , $ − 1 and p1 runs through p1($k−1). Hence

c$k(n) =∑

p1($k−1)

e−2np1πi/$k$−1∑z=0

e−2nzπi/$,

3.2 The series∑

q−scq(n) 24

and the inner sum is $ when $ | n and zero otherwise. It follows that

c$k(n) = $∑

p1($k−1)

e−2n1p1πi/$k−1

= $c$k−1(n1) (3.7)

if $ | n and n = $n1; and that c$k(n) = 0 otherwise.

Now

c$(n) =$−1∑

1

e−2npπi/$,

so that

c$(n) = −1 ($ - n), c$(n) = $ − 1 ($ - n); (3.8)

and we can now calculate c$k(n) by (3.7) for every k. We find that

c$2(n) = 0 ($ - n), −$ ($ | n,$2 - n), $($ − 1) ($2 | n), (3.9)

and generally

c$k(n) = 0 ($k−1 - n), −$k−1 ($k−1 | n,$k - n), $k−1($−1) ($k | n); (3.10)

and we can verify at once that these are also the values of C$k(n). Thus cq(n) = Cq(n)

whenever q = $k, and therefore for all q.

3.2 The series∑

q−scq(n)

Ramanujan summed a large number of series of the form∑

aqcq(n). The simplest is

U(n) =∞∑

q=1

cq(n)

qs. (3.11)

There are a number of interesting ways of summing this series. It is absolutely convergent

when s > 1, since |cq(n)| ≤ σ(n) for every q.

(i) The shortest way is due to Estermann. We can write cq(n) in the form

cq(n) =∑lm=qm|n

µ(l)m,

so thatcq(n)

qs=

∑lm=q,m|n

µ(l)l−sm1−s.

3.2 The series∑

q−scq(n) 25

When we sum with respect to q we remove the restriction on l, which now assumes

all positive integral values; and so

U(n) =∑

l,m|nµ(l)l−sm1−s (3.12)

=∑

m|nm1−s

∞∑

l=1

µ(l)

ls=

σ1−s(n)

ζ(s)=

n1−sσ1−s(n)

ζ(s),

where σν(n) is the sum of the ν–th powers of the divisors of n.

(ii) Ramanujan argued as follows. We suppose that F (x, y) is any function of the two

variables x and y, and that

D(n) =∑

d|nF

(d,

n

d

);

and define ην(n) as in the section §3.1, so that ην(n) is ν or 0 according as ν is or

is not a divisor of n. Then

D(n) =t∑

ν=1

ην(n)

νF

(ν,

n

ν

)

for any value of t not less than n; and so

D(n) =t∑

ν=1

1

νF

(ν,

n

ν

) ∑

d|νcd(n).

Now cj(n) occurs in this series when j | ν or ν = jµ, in which case µ ≤ tj; and

therefore

D(n) = c1(n)t∑1

1

µF

(µ,

n

µ

)+ (3.13)

+ c2(n)

12t∑

1

1

2µF

(2µ,

n

2µ

)+ c3(n)

13t∑

1

1

3µF

(3µ,

n

3µ

)+ · · ·

Suppose in particular that F (x, y) = x1−s. Then

D(n) =∑

d|nd1−s = σ1−s(n)

and

σ1−s(n) =c1(n)

1s

t∑1

µ−s +c2(n)

2s

12t∑

1

µ−s +c3(n)

3s

13t∑

1

µ−s + · · ·

Finally, if s > 1, and we make t →∞, we obtain (3.12)

3.2 The series∑

q−scq(n) 26

(iii) A third proof proceeds on lines like those of the second proof of the Theorem 3.1.1.

We start from the identity

∞∑q=1

f(q)

qs=

∏$

1 +

f($)

$s+

f($2)

$2s+ · · ·

=

∏$

χ$,

where f(q) is any multiplicative function and $ runs through the primes. The identity

reduces to “Euler’s product” when f(q) = 1.

In this case

χ$ = 1 +c$(n)

$s+

c$2(n)

$2s+ · · · ,

and we can quote the formulae (3.10). Suppose that $a os the highest power of $ which

divides n. Then

χ$ = 1−$−s

if a = 0, and generally

χ$ = 1 +$ − 1

$s+

$($ − 1)

$2s+ · · ·+ $a−1($ − 1)

$as− $a

$(a+1)s

= (1−$−s)1−$(a+1)(1−s)

1−$1−s.

Hence∑

q

cq(n)

qs=

∏$

(1−$−s)∏

$|n

1−$(a+1)(1−s)

1−$1−s=

σ1−s(n)

ζ(s),

by ordinary formula for a sum of powers of divisors of n.

In particular, when s = 2, we obtain the formula

σ(n) =1

6π2n

1 +

(−1)n

22+

2 cos 23nπ

32+

2 cos 12nπ

42+

2(cos 25nπ + cos 4

5nπ)

52+ · · ·

for the sum of the divisors n. The formula shows in a very striking way the oscillations

of σ(n) about its “average” 16π2n.

We have supposed that s > 1, when our series are absolutely convergent. We

can write (3.12) as∑

q

cq(n)

qs=

∑

d|n

d

ds

∑m

µ(m)

ms. (3.14)

The first factor on the right is a finite, and therefore absolutely convergent, Dirichlet’s

series, and the second a Dirichlet’s series convergent for s ≥ 1; and therefore, by a familiar

theorem on the multiplication of Dirichlet’s series, the series on the left is convergent, and

(3.14) is true, for s ≥ 1. Putting s = 1, we obtain

c1(n) +c2(n)

2+

c3(n)

3+ · · · = 0

3.3 The series∑

εqq−scq(n). 27

(a theorem of the same depht as the prime number theorem).

Ramanujan made a number of similar summations, among which

c1(n) log 1 +c2(n)

2log 2 +

c3(n)

3log 3 + · · · = −d(n)

and

π

c1(n)− c3(n)

3+

c5(n)

5− · · ·

= r2(n)

are two of the most striking.

3.3 The series∑

εqq−scq(n).

The series which is important for our present purpose is not (3.11) but

V (n) =∞∑

q=1

εqcq(n)

qs, (3.15)

where

εq = 1 (q ≡ 1, 3), 0 (q ≡ 2), 2s (q ≡ 0),

the congruence being to modulus 4. This series may be summed by any of the methods

of the section §3.2; I select the first as the shortest. I define σ∗ν(n) by

σ∗ν(n) = σν(n) (n odd),

σ∗ν(n) = σeν(n)− σo

ν(n) (n even),

σeν(n) and σo

ν(n) being the sums of the ν–th powers of the even and odd divisors of n; and

I prove that

Theorem 3.3.1.

V (n) =n1−s

(1− 21−s)ζ(s)σ∗s−1.

2 (3.16)

Proof. We have

V (n) =∑

q=1,3,...

q−scq(n) + 2s∑

q=4,8,...

q−scq(n) = V1(n) + V2(n),

2We have not now two alternative forms like those in (3.12), since

n1−sσ∗s−1(n) 6= σ∗1−s(n)

when n is even.

3.3 The series∑

εqq−scq(n). 28

say. Here, first,

V1(n) =∑

q=1,3,...

q−s∑lm=q,m|n

µ(l)m

=∑

lm=1,3,...,m|n

µ(l)l−sm1−s = σo1−s(n)

∑

l=1,3,...

µ(l)

ls

= σo1−s(n)

∏$>2

(1− 1

$s

)=

σo1−s(n)

(1− 2−s)ζ(s).

Next,

V2(n) = 2s∑

q=4,8,...

q−s∑lm=q,m|n

µ(l)m = 2s∑

lm=4,8,...

µ(l)l−sm1−s.

If 4 | l, then µ(l) = 0. Hence we may suppose that either (i) l is odd and 4 | m or (ii)

l = 2l1, where l1 is odd, and 2 | m. The terms of type (i) give

2s∑

l=1,3,...

µ(l)

ls

∑4|m,m|n

m1−s = − 2sσee1−s(n)

(1− 2−s)ζ(s),

where the double index indicates a sum over doubly even divisors; and the terms of type

(ii) give

−∑

l1=1,3,...

µ(l1)

ls1

∑2|m,m|n

m1−s = − σe1−s(n)

(1− 2−s)ζ(s).

Collecting our results, we find that

(1− 2−s)ζ(s)V (n) = σo1−s(n)− σe

1−s(n) + 2sσee1−s(n);

and we have only to verify that

σo1−s(n)− σe

1−s(n) + 2sσee1−s(n) = n1−sσ∗s−1(n). (3.17)

This is obvious if n is odd. If n = 2N , where N is odd, and δ runs through the divisors

of N , then the left–hand side of (3.17) is

∑δ1−s −

∑(2δ)1−s = (1− 21−s)

∑δ1−s = (1− 21−s)N1−s

∑δs−1

= (2s−1 − 1)n1−s∑

δs−1 = n1−s∑

(2δ)s−1 −∑

δs−1

= n1−sσ∗s−1(n).

Finally, if n = 2αN , where α > 1, then it is

∑δ1−s−

21−s + 41−s + · · ·+ 2α(1−s) ∑

δ1−s +2s41−s + 81−s + · · ·+ 2α(1−s)

∑δ1−s

3.3 The series∑

εqq−scq(n). 29

=

1 + 21−s + 41−s + · · ·+ 2(α−1)(1−s)−2α(1−s) ∑

δ1−s

= N1−s

1 + 21−s + 41−s + · · ·+ 2(α−1)(1−s)−2α(1−s) ∑

δs−1

= n1−s2α(s−1) + 2(α− 1)(s− 1) + · · ·+ 2s−1 − 1

∑δs−1

= n1−sσe

s−1(n)− σos−1(n)

= n1−sσ∗s−1(n).

This completes the proof of (3.16).

30

4 The Singular Series in the problem of 2s

squares.

I must now introduce ideas which are not to be found (at any rate explicitly) in Ra-

manujan’s work.1 They are the ideas which Littlewood and Hardy started in the work on

Waring’s problem [11].

It is easy to find asymptotic formulae for the behaviour of

f(s) = ϑ2s(x) = (1 + 2x + 2x4 + · · · )2s,

where x tends radially to a “rational point” e2pπi/q on the unit circle. We may suppose

that q = 1, p = 0 or that q > 1, 0 < p < q and (p, q) = 1. If

x = re2pπi/q

and r → 1, then

ϑ(x) = 1 + 2∞∑1

rn2

e2n2pπi/q

= 1 + 2

q∑j=1

∞∑

l=0

r(lq+j)2e2(lq+j)2pπi/q

= 1 + 2

q∑j=1

e2j2pπi/q

∞∑

l=0

r(lq+j)2 .

If r = e−δ, so that δ → 0, then

∞∑

l=0

r(lq+j)2 =∞∑

l=0

e−δ(lq+j)2 ∼∫ ∞

0

e−δ(xq+j)2dx ∼∫ ∞

0

e−δx2q2

dx =1

2q

√(π

δ

)=

√π

2q

(log

1

r

)− 12

;

and hence we have the following theorem

Theorem 4.0.2.

ϑ(x) ∼√

π

qSp,q

(log

1

r

)− 12

, (4.1)

where

Sp,q =

q∑j=1

e2j2pπi/q (4.2)

is one of “Gauss’s sums”.

1Except in the work from Hardy and Ramanujan[10]

4 The Singular Series in the problem of 2s squares. 31

It Sp,q = 0, as happens when q ≡ 2(mod 4), then (4.1) is to be interpreted as

ϑ(x) = o

(log

1

r

)− 12

.

Corollary 4.0.1. It follows that

f(x) ∼ π−s

(Sp,q

q

)2s (log

1

r

)−s

. (4.3)

We now form an auxiliary function which mimics the behaviour of f(x) when

x approaches the unit circle in this way. It is know that

Theorem 4.0.3. If

Fs =∞∑1

ns−1xn,

then

Fs(x)− Γ(s)

(log

1

x

)−s

is regular at x = 1.

Corollary 4.0.2. If

fp,q(x) =πs

Γ(s)

(Sp,q

q

)2s

Fs(xe−2pπiq/),

then

fp,q(x) ∼ πs

(Sp,q

q

)2s (log

1

r

)−s

∼ f(x)

when x tends to e2pπi/q.

Thus fp,q(x) “mimics” f(x) near this point; and, if we write

Θ2s(x) = 1 +∑p,q

fp,q(x),

then we may expect that Θ2s(x) will mimic f(x) near all rational points e2pπi/q. To say

this is to say that Θ2s(x) mimics f(x) very comprehensively, so comprehensively that

there should be a very close relation between the coefficients of the two functions. If this

be so, then the way will be open at any rate to an approximate determination of r2s(n).

Now

Θ2s(x) = 1 +πs

Γ(s)

∑p,q

(Sp,q

q

)2s ∞∑n=1

ns−1e−2npπi/qxn

= 1 +∞∑

n=1

ρ2s(n)xn,

4.1 Summation of the singular series when 2s ≡ (mod 8) 32

where

ρ2s(n) =πs

Γ(s)ns−1

∑p,q

(Sp,q

q

)2s

e−2npπi/q.

We can write this as

ρ2s(n) =πs

Γ(s)ns−1

∞∑q=1

Aq(n), (4.4)

where A1(n) = 1 and

Aq(n) = q−2s∑

p(q)

S2sp,qe

−2npπi/q (4.5)

when q > 1.

We are entitled to expect a pretty close relation between r2s(n) and ρ2s(n). It

is only an expectation, since our analysis has been entirely “heuristic”; but it is plainly

one worth pursuing.

Definition 4.0.1. We call (4.4) the singular series. This construction is typical of that

of the “singular series” in the general Waring problem.[11]

4.1 Summation of the singular series when 2s ≡ (mod 8)

The singular series can be summed for all s. The analysis is simplest when 2s ≡ 0(mod 8),

as I shall suppose.

The Gaussian sum Sp,q (or simply Sq, if we omit the explicit reference to p)

can be calculated, for all, p, q, from the formulae

Sp,qq′ = Spq

′,qSpq,q

′ ,

S1 = 1, S2 = 0, S22µ = 2µ(1 + ip), S22µ+1 = 2µ+1e14pπi,

S$ =( p

$

)i

14($−1)2

√$, S$2µ = $µ, S$2µ+1 = $µS$.

Here q and q′

are coprime, and $ is an odd prime. The formulae for the 8–th powers

become much simpler, and we find that

S2sp,q = εqq

s

when 2s ≡ 0(mod 8), εq being the symbol of the section §3.3. It follows that

Aq = εqq−s

∑

p(q)

e−2npπi/q = εqq−scq(n),

4.1 Summation of the singular series when 2s ≡ (mod 8) 33

and that, in the notation of the section §3.3,

ρ2s(n) =πsns−1

Γ(s)V (n). (4.6)

Thus the singular series is (apart from the outside factor) Ramanujan’s series (3.15), and

so we have the following theorem

Theorem 4.1.1.

ρ2s(n) =πs

Γ(s)(1− 2−s)ζ(s)σ∗s−1(n). (4.7)

In particular, if 2s = 18, then

(1− 2−s)ζ(s) =π4

96,

ρ8(n) = 16σ∗3(n). (4.8)

If 2s = 24, then

(1− 2−12)ζ(12) = (1− 2−12)211π12 B6

12!, B6 =

691

2730,

and

ρ24(n) =16

691σ∗11(n). (4.9)

34

5 Modular Functions.

We have strong reasons for expecting to find a fairly close resemblance between r2s(n)

and ρ2s(n), and further analysis does more than would be required to justify our hopes,

the correspondence being extremely close. Indeed when 2s ≤ 8 the two functions are

identical; in particular

r8(n) = ρ8(n).

The proof of this depends on arguments of a quite different character, first applied to this

problem by Mordell [18].

5.1 Modular Group and Fundamental Region.

We need the elements of the theory of the modular group and the functions associated

with it. The modular group Γ has two forms. In the homogeneous form it is defined as:

Definition 5.1.1. The modular group Γ in the homogeneous for is defined as the group

of substitutions

ω′1 = aω1 + bω2, ω

′2 = cω1 + dω2, (5.1)

where a, b, c, d are integers and

ad− bc = 1. (5.2)

Definition 5.1.2. In the non–homogeneous form we write

τ =ω2

ω1

, (5.3)

and the group is defined by the substitutions

τ′=

c + dτ

a + bτ.

We shall use any of the symbols

S,

a b

c d

,

(c + dτ

a + bτ

)

5.1 Modular Group and Fundamental Region. 35

to denote the substitutions (5.1) or (5.3). Γ is generated by repeated application of the

two substitution 1 0

1 1

,

0 −1

1 0

or

τ′= τ + 1, τ

′= −1

τ. (5.4)

In what follows we shall suppose that

=(τ) > 0, (5.5)

so that

|x| = |eπiτ | < 1.1 (5.6)

If τ = u + iv, τ′= u

′+ iv

′, then

v′=

(ad− bc)v

(a + bu)2 + b2v2> 0,

so that one point in the upper half–plane is transformed into another, and only such

points are relevant.

Definition 5.1.3. We call the region D defined by

−1

2< v <

1

2, u2 + v2 > 1

the fundamental region of Γ.

Each substitution of Γ transforms D into a curvilinear triangle, whose sides are

circles and whose angles are(

13π, 1

3π, 0

). These triangles cover up the half–plane without

overlapping.

The fundamental function associated with the modular group is Klein’s “ab-

solute invariant” J(τ), which is defined as follows.

Definition 5.1.4. We write

g2 = g2(ω1, ω2) =1

12

(π

ω1

)4 1 + 240

(13x2

1− x2+

23x4

1− x4+ · · ·

),

g3 = g3(ω1, ω2) =1

216

(π

ω1

)6 1− 504

(15x2

1− x2+

25x4

1− x4+ · · ·

)

1I use x instead of the usual q.

5.1 Modular Group and Fundamental Region. 36

Figure 5.1: Fundamental Region of Γ

(these being the ordinary invariants of the Weierstrassian theory),

∆ = ∆(ω1, ω2) = g32 − 27g2

3 =

(π

ω1

)12

x2(1− x2)(1− x4) . . .

24,

and

J(τ) =g32

∆.

Then J(τ) is a function of τ only, invariant for the substitutions of Γ. It

assumes every value just once in D (when proper conventions have been laid down about

the boundary of D), and represents D conformally on the whole complex plane (regarded

as bounded by the points 0, 1, and ∞).

Definition 5.1.5. We call a function which, like J(τ), is invariant for Γ a modular in-

variant : the phrase “modular function” is used more vaguely.

The function

J = J(τ)

plays, for modular invariants, the part played by z in the ordinary theory of one–valued

functions f(z). Thus a modular variant, with properly restricted singularities, is a one–

valued function of J . In particular, if a modular invariant is regular and bounded in D,

then it is a one–valued function of J bounded in the whole plane of J , and accordingly it

is constant.

5.2 Functions associated with the sub–group Γ3 37

5.2 Functions associated with the sub–group Γ3

We shall be concerned now with functions which are not modular invariants in the full

sense just defined, but which are invariant, or “all but” invariant, for the substitutions of

a certain sub–group of Γ.

It easily verified that the substitutions of Γ which satisfy the congruence con-

ditions2

a b

c d

≡

1 0

0 1

or

0 1

1 0

(mod 2)

Figure 5.2: Fundamental Region of D3

form a group, a sub–group of Γ which we call Γ3. Γ3 is generated by

2Either a and d are odd and b and c even, or conversely.

5.3 Proof of r8(n) = ρ8(n). 38

τ′= τ + 2, τ

′= −1

τ. (5.7)

It has a “fundamental region” D3 defined by

−1 < u < 1, u2 + v2 > 1;

and the substitutions of Γ3 transform D3 into a system of triangles, all of whose angles

are 0, which just fill up the half–plane.3

There is a principal invariant J3(τ) of Γ3, which is related to Γ3 as J(τ) is to

Γ (but whose expression we shall not require). A one–valued function, invariant for Γ3,

is a one–valued function of J3 and a three–valued function of J . Finally we have,

Theorem 5.2.1. A function invariant for Γ3, and regular and bounded in D3, is constant.

5.3 Proof of r8(n) = ρ8(n).

We will proof in this section that,

Theorem 5.3.1.

r8(n) = ρ8(n) = 16σ∗3(n).

Proof. The functions

ϑ8 = ϑ8(x) = ϑ8(0, τ) = (1 + 2x + 2x4 + · · · )8

and

Θ8 = Θ8(x) = 1 +∞∑1

ρ8(n)xn,

where x = eπiτ , are one–valued functions of τ . If we can prove that

(A) ϑ−8Θ8 is invariant for Γ3,

(B) ϑ−8Θ8 is regular and bounded in D3,

3D3 may be described roughly as formed by fitting together three regions congruent to D (i.e. trans-

forms of D for substitutions of Γ). More strictly, it is formed by D and one–half of each of the four

regions

D(τ + 1), D(1− τ), D

(−1

τ+ 1

), D

(−1

τ− 1

)

(the transforms of D by τ′= τ + 1, etc.). In Fig.2, D3 is bounded by the thicker lines.

5.3 Proof of r8(n) = ρ8(n). 39

then using the general theorem (5.2.1) will follow that ϑ−8Θ8 is a constant, which is

plainly 1 and from this will follow our theorem.

Then, we will prove (A) and (B) now.

5.3.1 Proof of (A).

It is sufficient to prove that ϑ−8Θ8 is invariant for the two substitutions

S1 (τ + 2) , S2

(−1

τ

)

which generate Γ3. We can prove, quite generally, that ϑ−2sΘ2s is invariant.4

In the first place

ϑ2s(0, τ + 2) = ϑ2s(0, τ), (5.8)

ϑ2s

(0,−1

τ

)= τ sϑ2s(0, τ), (5.9)

by the familiar formulae for the linear transformation of the ϑ–function.5It remains to

determine the behaviour of Θ2s. It is obvious that Θ2s is invariant for S1, since x is

invariant for S1 (and so that ϑ−2sΘ2s is invariant). This will also appear incidentally from

the analysis which follows.

4We are supposing 2 ≡ 0(mod 8), but the proof is very much the same in other cases.5We shall require the full table for the functions

ϑ2(0, τ) = 2x14 + 2x

94 + 2x

254 + · · · ,

ϑ3(0, τ) = 1 + 2x + 2x4 + 2x9 + · · · ,

ϑ4(0, τ) = 1− 2x + 2x4 − 2x9 + · · · ,

which is

ϑ2(0, τ + 1) =√

iϑ2(0, τ), ϑ3(0, τ + 1) = ϑ4(0, τ), ϑ4(0, τ + 1) = ϑ3(0, τ),

ϑ2

(0,−1

τ

)=√

τ√iϑ4(0, τ),

ϑ3

(0,−1

τ

)=√

τ√iϑ3(0, τ),

ϑ4

(0,−1

τ

)=√

τ√iϑ2(0, τ).

Here√

i = e14 πi and

√τ has its real and imaginary parts positive. The notation is Tannery and Molk’s,

and ϑ3(0, τ) = ϑ(0, τ).

5.3 Proof of r8(n) = ρ8(n). 40

The function Fs(x) of chapter §4 is elementary. In fact, if x = e−y, so that

y = −πiτ ,

Fs(x) =∑

ns−1xn =∑

ns−1e−ny =

(d

dy

)s−2 ∑ne−ny

=

(d

dy

)s−2e−y

(1− e−y)2=

(d

dy

)s−21

4cosech2 1

2y

=

(d

dy

)s−2 ∞∑−∞

1

(y + 2nπi)2= Γ(s)

∞∑−∞

1

(y + 2nπi)s=

Γ(s)

πs

∞∑−∞

1

(2n− τ)s.

Also

xe−2pπi/q = eπiτ−(2p/q),

and so

Fs(xe−2pπi/q) =Γ(s)

πs

∞∑−∞

1

2n− τ + (2p/q)s,

fp,q(x) =πs

Γ(s)

(Sp,q

q

)2s

Fs(xe−2pπi/q)

=πs

Γ(s)

εq

qsFs(xe−2pπi/q) = εq

∞∑−∞

1

2(nq + p)− qτs.

Hence

Θ2s(x) = 1 +∑p,q

fp,q(x) = 1 +∑p,q,n

εq

2(nq + p)− qτs, (5.10)

where the range of summation is defined by

q = 1, 2, 3, . . . ; 0 < p < q, (p, q) = 1; −∞ < n < ∞

(except that p = 0 when q = 1).

We can write (5.10) in the form

Θ2s = 1 +∑p,q

εq

(2p− qτ)s, (5.11)

where q = 1, 2, . . . and p runs through all values (positive or negative) prime to q; and

this is

Θ2s = 1 +∑

q=1,3,...

1

(2p− qτ)s+

∑q=4,8,...

2s

(2p− qτ)s(5.12)

= 1 +∑

q=1,3,...

1

(2p− qτ)s+

∑q=2,4,8,...

1

(p− qτ)s

= 1 +∑p,q

1

(p− qτ)s,

5.3 Proof of r8(n) = ρ8(n). 41

where now q = 1, 2, . . . and p runs through all values prime to and of opposite parity to

q.

We wish to remove the restriction that p should be prime to q. We can do this

by multiplying both sides of (5.12) by

η(s) = (1− 2−s)ζ(s) = 1 + 3−s + 5−s + · · ·

We thus obtain6

η(s)Θ2s = η(s) +∑p,q

1

(p− qτ)s, (5.13)

where q = 1, 2, 3, . . ., and p runs through all values of opposite parity to q. We may also

write this in either of the forms

η(s)Θ2s = −η(s) +∑p,q

1

(p− qτ)s(q = 0, 1, 2, . . . ; p + q ≡ 1), (5.14)

η(s)Θ2s =1

2

∑ 1

(p− qτ)s(p + q ≡ 1). (5.15)

Here the congruences are to modulus 2, and q, in (5.15), runs through all integral values.

If we write

η(s)Θ2s = χ(τ),

then it follows at once from any of (5.12)–(5.15) that

χ(τ + 2) = χ(τ).

This, as I pointed out, obvious from the beginning (but a useful check on our analysis).

We use (5.15) in investigating the effect of the substitution S2. It gives

χ

(−1

τ

)=

1

2τ s

∑p+q≡1

1

(pτ + q)s

=1

2τ s

∑p+q≡1

1

(p− qτ)s= τ sχ(τ),

on replacing p, q by −q, p.

Thus χ(τ) is affected by S1 and S2 in just the same way as ϑ2s; and ϑ−2sΘ2s

is invariant for S1 and S2 and therefore for Γ3.

We have thus proved (A). More generally, we have proved the invariance of

ϑ−2sΘ2s whenever 2s ≡ 0(mod 8).

6A pair (p, q) of opposite parity is one of the forms (P,Q), (3P, 3Q, . . .), where P and Q are coprime

and of opposite parity

5.3 Proof of r8(n) = ρ8(n). 42

5.3.2 Proof of (B).

The function ϑ8 and Θ8 are defined by power series in x = eπiτ convergent in the circle

|x| < 1 or the half–plane v > 0. Also

ϑ(x) =∞∏1

(1− x2n)(1 + x2n−1)2

has no zeros in the circle. Hence ϑ−8Θ8 is regular for |x| < 1 or v > 0. It is bounded in

any closed part of D3 which excludes the two points in which D3 abuts on the real axis;

and therefore it is bounded in D3 if it is bounded in the neighbourhood of the two points

τ = ±1. It is plain that we need consider only the point τ = 1.

We write

τ = 1− 1

T, T =

1

1− τ, (5.16)

and suppose that τ → 1 inside D3, so that 0 < u < 1 and u2 + v2 > 1. If T = U + iV

then

U =1− u

(1− u)2 + v2, V =

v

(1− u)2 + v2,

so that 0 < U < 1 and V →∞. Hence T →∞ inside D3, and

X = eπiT → 0.

Now

ϑ8(x) = ϑ83(0, τ) = ϑ8

3

(0, 1− 1

T

)= T 4ϑ8

2(0, T )

= T 4(2X14 + 2X

94 + · · · )8,

so that

ϑ8 ∼ 256T 4X2 (5.17)

when T →∞ and X → 0. We require a similar formula for Θ8.

We have from (5.15)

π4

96Θ8 = χ(τ) = χ

(1− 1

T

)(5.18)

=1

2T 4

∑ 1

q + (p− q)T4(p + q ≡ 1)

=1

2T 4

∑ 1

(Q + PT )4,

where now Q runs through all integers and P through all odd integers. If

|P |T = ζ, ξ = eπiζ = X |P |,

5.3 Proof of r8(n) = ρ8(n). 43

then

∑Q

1

(Q + PT )4=

∑Q

1

(Q + ζ)4=

1

6

(d

dζ

)2 ∑ 1

(Q + ζ)2

=1

6

(d

dζ

)2

π2cosec2πζ = −1

6

(d

dζ

)24π2e2πiζ

(1− e2πiζ)2

= −2

3π2

(d

dζ

)2

(e2πiζ + 2e4πiζ + 3e6πiζ + · · · )

=8

3π4(e2πiζ + 23e4πiζ + 33e6πiζ + · · · )

=8

3π4(X2|P | + 23X4|P | + 33X6|P | + · · · ).

We have to sum this with respect to P , but we need only the lowest powers of X, and X2

occurs only for P = ±1. Hence (5.18) gives

π4

96Θ8 ∼ 1

2· 2 · 8

3π4T 4X2,

or

Θ8 ∼ 256T 4X2. (5.19)

Finally (5.17) and (5.19) show that ϑ−8Θ8 is bounded when τ → 1, and therefore bounded

in D3.

It follows that ϑ−8Θ8 is a constant, which must be 1, and that r8(n) = ρ8(n).

44

6 The case of 24 squares.

We proved in the section §5.3.1 that ϑ−2sΘ2s is invariant for Γ3 whenever 2s ≡ 0(mod 8),

and this is true in particular when 2s = 24. If ϑ−24Θ24 were bounded in D3, it would

follow that ϑ24 = Θ24 and r24(n) = ρ24(n); but this is untrue. The correct formula, which

was found first by Ramanujan, and which I shall proceed to prove, is:


ϑ24(x) = Θ24(x)− 33152

691g(−x)− 65536

691g(x2), (6.1)

where

g(x) = x(1− x)(1− x2)(1− x3) . . .24, (6.2)

so that

g(x2) = x2(1− x2)(1− x4)(1− x6) . . .24 = h24(τ),

in Tannery and Molk’s notation. This last function is, apart from a factor of homogeneity,

the discriminant ∆(ω1, ω2).

Proof. We require the formulae for the linear transformation of h(τ). These are

h(τ + 1) = e112

πih(τ), (6.3)

h

(−1

τ

)= e−

14πi√τh(τ). (6.4)

We shall also use one other formula, viz.

h2

(τ + 1

2

)= e

112

πih(τ)ϑ3(0, τ). (6.5)

This belongs to the theory of quadratic transformation, but is a simple corollary of the

product formulae for h(τ) and ϑ3(0, τ). The three functions

ϑ−24Θ24, ϑ−24g(−x), ϑ−24g(x2) (6.6)

are all invariant for Γ3. We have already proved the first invariant. The invariance of the

third follow from the formulae

g(x2) = h24(τ), h24(τ + 2) = h24(τ), h24

(−1

τ

)= τ 12h24(τ).


Finally

g(−x) = h24

(τ + 1

2

)= −h12(τ)ϑ12

3 (0, τ),

by (6.5), and the invariance of the second of the functions (6.6) follows from the formulae

(5.9) and (6.4).

Hence

ϑ−24Θ∗24 = ϑ−24Θ24 + αg(−x) + βg(x2)

is invariant, for any α and β. We shall prove that it is possible to choose α and β so that

ϑ−24Θ∗24 is bounded in D3. For this, we use the substitution (5.16) of §5.3.2, and examine

the behaviour of all the functions concerned when T →∞.

(i) First, by (5.17),

ϑ24

(0, 1− 1

T

)∼ 224T 12X6. (6.7)

(ii) Secondly

g(−x) = −h12(τ)ϑ12(0, τ) = −h12

(1− 1

T

)ϑ12

3

(0, 1− 1

T

)

= h12

(− 1

T

)ϑ12

4

(0,− 1

T

)=

(T

i

)12

h12(T )ϑ122 (0, T )

= T 12X(1−X2)(1−X4) . . .24(2X14 + 2X

19 + · · · )12.

Hence

g(−x) = 212T 12X4 + O(X6). (6.8)

(iii) Thirdly

g(x2) = h24(τ) = h24

(1− 1

T

)= h24

(− 1

T

)

= T 12h24(T ) = T 12X2(1−X2)(1−X4) . . .24,

g(x2) = T 12X2 − 24X4 + O(X6). (6.9)

(iv) Finally we have to determine the behaviour of Θ24, which can be done by calculations

loke those §5.3.2. We have now

η(12)Θ24 =1

2T 12

∑Q

∑P

1

(Q + PT )12,


∑Q

1

(Q + PT )12=

∑Q

1

(Q + ζ)12=

1

11!

(d

dζ

)10 ∑Q

1

(Q + ζ)2

=(2π)12

11!(e2πiζ + 211e4πiζ + · · · )

=(2π)12

11!(X2|P | + 211X4|P | + · · · ).

It is again only the terms with P = ±1 that matter, since P = ±3 yields O(X6). Hence

we obtain

Θ24 =(2π)12

11!η(12)T 12X2 + 211X4 + O(X6) (6.10)

=214

691T 12X2 + 211X4 + O(X6),

on inserting the value η(12).

We have to choose α and β so that

214

691X2 + 211X4 + O(X6)+ α212X4 + O(X6)+ βX2 − 24X4 + O(X6) = O(X6).

It will then follow from (6.7), (6.8), (6.9), and (6.10) that ϑ−24Θ∗24 is bounded. Equating

coefficients, we find that

α = −33152

691, β = −65536

691,

and (6.1) follows and the theorem is proved.

47

7 A Brief Discussion of the Ramanujan’s

Function τ (n).

Definition 7.0.1. We define τ(n)1 as the coefficient of xn in

g(x) = x(1− x)(1− x2) . . .24 =∞∑1

τ(n)xn.

Then, after §4.1,

Θ24 = 1 +∞∑1

ρ24(n)xn = 1 +16

691

∞∑1

σ∗11(n)xn.

Also

g(−x) =∞∑1

(−1)nτ(n)xn,

and

g(x2) =∞∑1

τ

(1

2n

)xn,

if we agree that τ(y) means 0 when y is not an integer. Hence finally we obtain Ramanu-

jan’s formula


r24(n) =16

691σ∗11(n) + e24(n), (7.1)

where

e24(n) =128

691

(−1)n−1259τ(n)− 512τ

(1

2n

). (7.2)

7.1 The Order of τ (n)

The function τ(n) has been defined only as a coefficient, and it is natural to ask whether

there is any reasonably simple “arithmetical” definition; but none has yet been found. For

more details on the properties of the functionτ(n) see [9–12–13–19–24]. I must however

1I retain Ramanujan’s notation. The collision of the τ in τ(n) and the τ in eπiτ is a little unfortunate

but is not likely to cause confusion.

7.1 The Order of τ(n) 48

prove something about the order of τ(n), since I have to justify my assertion, in §2.2, that

r24(n) is “dominated” by ρ24(n).

It follows from a formula of Jacobi which I have quoted several times already

that

Proposition 7.1.1 (Jacobi).

∑τ(n)xn = x(1− x)(1− x2) . . .24 = x(1− 3x + 5x3 − 7x6 + · · · )8,

the exponents in the series being the triangular numbers.

Now (1− 3x + · · · )8 is majorised by ∞∑

n=0

(2n + 1)x12n(n+1)

8

,

which is of order (1− x)−8 when x → 1.2 Hence

|τ(n)|xn <∑

|τ(n)|xn < A(1− x)−8,

where A is a constant, for all n and x. Taking x = 1−n−1, when xn is about e−1, we find

that

Theorem 7.1.1.

τ(n) = O(n8). (7.3)

On other hand

16

691σ∗11(n) = ρ24(n) =

π12n11

11!

∞∑q=1

εqcq(n)

q12;

and the series is greater than3

1− 3

312− 212 · 4

412− 5

512− 7

712− 212 · 8

812− · · · > 1

2.

Hence σ∗11(n) is greater than a constant multiple of n11, and

r24(n) =16

691σ∗11(n)

1 + O

(1

n3

)

is dominated heavily by its leading term.

The order of τ(n) is really a good deal smaller than is shown by (7.3). Ra-

manujan showed, by a more sophisticated method, that

2That of(∑

nx12 n2

)8

or of(∫ ∞

0

te−12 yt2dt

)8

, where e−y = x. This is that of y−s or (1− x)−8 .3Using the crude inequality |cq(n)| ≤ n.

7.1 The Order of τ(n) 49


τ(n) = O(n7). (7.4)

Hardy showed later, by a function–theoretic method, that

Theorem 7.1.3 (Hardy).

τ(n) = O(n6). (7.5)

It is known a better result, due to Rankin [24] about the order of τ(n). It is

very plausible at this point to suppose that (as Ramanujan conjectured)

Conjecture 1 (Ramanujan Hypothesis).

τ(n) = O(n112

+ε)

for every positive ε.

The Ramanujan hypothesis was generalized by H. Petersson [20] for Fourier

coefficients of cusp forms which are eigenvalues of Hecke operators. The Ramanujan–

Petersson conjecture was proved in 1974 by P. Deligne [4]. In this very important paper,

Deligne proved the Riemann Hypothesis for varieties over finite fields, from which the

Ramanujan–Peterson conjecture follows. For excellent surveys on problems on the size of

τ(n), see papers by Murty [19] and Rankin [24].

50

8 Conclusion.

I conclude by repeating that the results which we have proved are typical of those in the

general problem of 2s squares. We can always express e2s(n) as the sum of a number,

fixed by Ramanujan and Mordell, of terms defined as modular coefficients; and each of

the coefficients gives rise to a series of problems resembling those arising from τ(n), and

thus would be desirable to understand more deeply the Ramanujan’s Function τ(n). In

some cases it is possible to define them fairly simply in arithmetical terms. In all cases

the number of representations is dominated by the divisor function ρ2s(n).

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theta functions: the problem of the...

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