Download THE IRRATIONALITY OF ζ(3) AND AP´ERY NUMBERS

THE IRRATIONALITY OF ζ(3)
AND APÉRY NUMBERS
A Thesis Submitted
in Partial Fulfillment of the Requirements
for the Degree of
Master Of Science
by
Sneha Chaubey
to the
School of Mathematical Sciences
National Institute of Science Education and Research
Bhubaneswar
April-2012
To
Maa and Dadi
i
DECLARATION
I hereby declare that I am the sole author of this thesis in partial fulfillment
of the requirements for an undergraduate degree from National Institute of Science
Education and Research (NISER). I authorize NISER to lend this thesis to other
institutions or individuals for the purpose of scholarly research.
Sneha Chaubey
Date:
The thesis work reported in the thesis entitled “The Irrationality of ζ(3) and
Apéry Numbers” was carried out under my supervision, in the School of Mathematical Sciences at NISER, Bhubaneswar, India.
Dr. Brundaban Sahu
Date:
ii
ACKNOWLEDGMENT
This dissertation would not have been possible without the guidance and the help
of several individuals who in one way or another contributed and extended their
valuable assistance in the preparation and completion of this study, to only some of
whom it is possible to give particular mention here.
Foremost, I would like to express my deepest gratitude to my supervisor Dr.
Brundaban Sahu for the continuous support he provided me during the entire period
of my dissertation, for his patience, motivation, enthusiasm, and immense knowledge.
His guidance helped me throughout my study and writing of this thesis. I could not
have imagined having a better advisor and mentor for my Master’s Dissertation.
My sincere thanks also goes to Prof. P. C. Das, Prof. Muruganandam and all the
faculty in the School of Mathematical Sciences for their encouragement, insightful
comments and guiding me for the past several years and helping me develop a strong
background in algebra, analysis and number theory.
I would like to thank my classmate Ashish Kumar Pandey who as a good friend
was always willing to help and give his best suggestions. Many thanks to all my
friends for their encouragement and constant support.
Last but by no means the least I thank my parents and my sister and brother for
being there for me always and bestowing their best wishes on me forever.
iii
Contents
1 Introduction
1
2 Algebraic and Transcendental numbers
2.1 Algebraic numbers . . . . . . . . . . .
2.1.1 Properties of Algebraic numbers
2.2 Transcendence of e and π . . . . . . .
2.2.1 Transcendence of e . . . . . . .
2.2.2 Transcendence of π . . . . . . .
2.3 Rational Approximations . . . . . . . .
.
.
.
.
.
.
3
3
4
5
6
8
10
3 Apéry’s Proof of Irrationality of ζ(3)
3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Irrationality of ζ(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
12
17
4 Theory of Modular forms
4.1 SL(2, Z) and its congruence subgroups . . . . . . .
4.2 Modular functions and modular forms . . . . . . .
4.3 q-expansion of a modular form . . . . . . . . . . . .
4.3.1 Estimates for the coefficients in q-expansion
4.3.2 The Eisenstein series of weight 2 . . . . . . .
4.4 L-functions and Modular forms . . . . . . . . . . .
4.4.1 Mellin Transform . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
27
27
28
31
32
33
35
35
5 Irrationality Proofs using Modular Forms
5.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 The group Γ1 (6) and the irrationality of ζ(3) . . . . . . . . . . . . . .
5.3 Irrationality of L-functions using modular forms . . . . . . . . . . . .
37
37
40
44
6 Congruence Properties of Apéry numbers
6.1 Congruences of Apéry numbers . . . . . . . . . . . . . . . . . . . . .
6.2 Congruences of Apéry Numbers relating modular forms . . . . . . . .
47
48
57
7 Conclusion
62
References
65
iv
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Chapter 1
Introduction
The Riemann zeta function, ζ(s) =
P∞
1
n=1 ns
is one of the most important objects
in modern number theory and the role played by their special values at integral
arguments is the theme of many interesting mathematical questions. The prototypical
result on special values of Riemann zeta function is the theorem that
−(2πis)s Bs
ζ(s) =
2s!
for s > 0 and s is even and Bs are the Bernoulli numbers. This was proved by Euler
in 1735. Using this identity and the transcendence of π proved by Lindemann, the
value of Riemann zeta function at even integers was proved to be transcendental. The
motivation behind this thesis is to investigate the arithmetic nature of the value of
Riemann zeta function at odd integers.
The most affirmative result on the arithmetic nature of ζ(s) for odd s is attributed
to Apéry who showed ζ(3) is irrational. In this thesis we have studied about the
irrationality of ζ(3) through two perspectives. First, Apéry’s proof of irrationality
of ζ(3) and then Beuker’s proof of irrationality using modular forms. Though both
seem to be entirely different at a glance, they are indeed related. We have seen
that their connection is justified using some special modular forms of the congruence
group Γ1 (6). The sequences an and bn used to prove ζ(3) irrational are called Apéry
numbers. These numbers satisfy congruence relation modulo prime powers. The
congruence properties satisfied by the Apéry numbers are important in a way as these
relations are similar to the Atkin Swinnerton-Dyer congruence relations for certain
congruence subgroups [T2]. Apéry’s proof is also significant as their relation with
1
1 Introduction
modular forms can be explained with the help of some differential equations satisfied
by the generating function of the Apéry numbers. This is more vividly explained at
the end of Chapter 6.
Chapter 2 of the thesis gives the basic results on algebraicity and transcendence
and we give Hermite’s and Lindemann’s proofs of transcendence of e and π respectively. The chapter ends with a discussion on few results in diophantine approximation
theory, most prominent being Dirichlet’s theorem.
Chapter 3 contains the proof of irrationality of ζ(3) from Van der Poorten’s informal report on Apéry’s proof of the irrationality of ζ(3) written in 1978.
Chapter 4 introduces modular forms. It also discusses the Dedekind eta function
and the Eisenstein series. It gives the associated Dirichlet series for a modular from,
its L-series. The chapter ends with a discussion on Mellin Transforms and the Hecke’s
lemma by Weil and Razar.
Chapter 5 discusses Beuker’s proof of irrationality of ζ(3) using modular forms.
This proof is taken from F. Beukers paper on irrationality proofs using modular
forms. The last section of the chapter proves the irrationality of certain combinations
of L-functions using Hecke’s Lemma and the results obtained in previous two sections.
The sequences obtained in the course of Apéry’s proof of irrationality of ζ(3)
satisfy some beautiful congruence properties. Chapter 6 gives proofs of two of these
properties.
The thesis ends by discussing some of the observations made from the proof of
Beukers of irrationality of ζ(3) and the relation of irrationality proof of ζ(3) with the
theory of differential equations.
2
Chapter 2
Algebraic and Transcendental numbers
2.1
Algebraic numbers
Definition 2.1.1. (Algebraic numbers) A complex number α is called an algebraic
number if there exists p(x) ∈ Z[x], p 6= 0 with p(α) = 0.
Definition 2.1.2. (Transcendental numbers) A complex number α is called transcendental if it is not algebraic.
Remark 2.1.1. The property of a number being rational, irrational, algebraic or transcendental is called its arithmetic property. Many interesting properties of numbers
depend on whether or not a number is algebraic or transcendental. Transcendental
numbers are not scarce, in fact Cantor showed that the set of algebraic numbers form
a countable set, so transcendental numbers exist and are a measure 1 set in [0, 1] and
hence essentially all numbers are transcendental.
Example 2.1.1. (Algebraic numbers) All rational numbers, square root of any rap
tional number, the cube root of any rational number, numbers of the form r q where
p √
√
r, p, q ∈ Q, i = −1, 3 2 − 5.
Example 2.1.2. (Transcendental numbers)
√ √2
e, π, e , 2 , ζ(3)
π
Remark 2.1.2. The transcendence of the number
1 1
1
γ = limn→∞ 1 + + + ... + − loge (n)
2 3
n
3
2 Algebraic and Transcendental numbers
(called Euler’s constant) is a well known open question in transcendental number
theory. It is not even known if γ is rational or not.
2.1.1
Properties of Algebraic numbers
Theorem 2.1.1. The set of algebraic numbers forms a countable set.
Theorem 2.1.2. (Cantor) The set of all real numbers is uncountable.
The proof of the above theorem is a consequence of the following result.
Lemma 2.1.3. Let S be the set of all binary sequences, i.e. sequences {yi }i∈Z with
yi ∈ {0, 1}. Then S is uncountable.
The proof of Cantor’s theorem follows from this.
Proof. (Proof of Cantor’s Theorem) Consider all numbers in the interval [0, 1] whose
decimal expansion consists entirely of 0’s and 1’s. Then there is a bijection between
this subset of real numbers and the set S defined in the previous lemma. But, from
the Lemma 2.2.3, S is uncountable. Consequently R has an uncountable subset and
thus R is uncountable.
As a consequence of Cantor’s theorem we have,
Corollary 2.1.4. The set of transcendental numbers is uncountable.
Proof. As R = A ∪ T where A is the set of algebraic numbers in R and T is the set
of transcendental numbers in R. As A is the subset of all algebraic numbers, A is
countable and since R is uncountable from above hence T is uncountable. Now set of
all transcendental numbers in R forms a subset of set of all transcendental numbers,
thus the set of transcendental numbers is uncountable.
4
2 Algebraic and Transcendental numbers
Lemma 2.1.5. If α and β are algebraic numbers, then α±β and αβ are also algebraic
numbers. Also, if α, β are algebraic integers, then α ± β and αβ are also algebraic
integers.
Lemma 2.1.6. If α is an algebraic number with minimal polynomial g(x) ∈ Z with
b as the leading coefficient of g(x), then bα is an algebraic integer.
Lemma 2.1.7. If α is an algebraic integer and α is rational, then α is an integer.
2.2
Transcendence of e and π
We know there are more transcendental numbers than algebraic; we finally show the
numbers e and π to be transcendental. Before that we will see a brief history on the
development of proofs of transcendence of e and π. Liouville proved the number
∞
X
10−n!
n=1
to be transcendental, and this was one of the first numbers proven to be transcendental. Lambert (1761), Legendre (1794), Hermite (1873) and others proved π is
irrational. Lindemann (1882) proved π transcendental. What about e? Euler(1737)
proved that e and e2 are irrational. Liouville(1844) proved e is not an algebraic number of degree 2, and Hermite(1873) proved e is transcendental. Liouville (1851) gave
a construction for an infinite (in fact uncountable) family of transcendental numbers.
In this section we will give Lindemann’a and Hermite’s proof of the transcendence of
π and e respectively. To prove the transcendence of both e and π we shall make use
of the function,
Z
I(t) =
t
et−u f (u)du
0
where t is a complex number and f (x) is a polynomial with complex coefficients which
will be specified as per our need. Integration by parts of the expression on the right
5
2 Algebraic and Transcendental numbers
gives:
I(t) =
−f (u)et−u |t0
Z
t
f 0 (u)et−u du
+
0
−t
0
0
−t
Z
t
= −f (t) + f (0)e (−f (t) + f (0)e ) +
f 00 (u)et−u du
0
df
= et
X
j=0
df
f j (0) −
X
f j (t)
j=0
where df is the degree of f also note that if f (x) −
Pdf
f¯(x) = j=0
| aj | xj . Then,
Z
| I(t) |≤|
Pdf
j=0
aj xj , we set
t
| et−u f (u) | du |≤| t | max | et−u || max(| f (u) |) ≤| t | e|t| f¯(| t |).
0
Hence we get an upper bound for I(t).
2.2.1
Transcendence of e
Theorem 2.2.1. The number e is transcendental
Proof. Assume e is algebraic. There is a minimal polynomial
g(x) =
r
X
bk xk ∈ Z[x],
k=0
with
g(e) =
r
X
bk ek = 0.
k=0
Let p be a large prime greater than max {r, |b0 |} to be chosen suitably and define
f (x) = x
p−1
r
Y
(x − j)p .
j=1
6
2 Algebraic and Transcendental numbers
The degree of f is n = (r + 1)p − 1. Consider
J = b0 I(0) + b1 I(1) + ... + br I(r) =
r
X
bk I(k)
k=0
=
=
r
X
k=0
n
X
bk
ek
f j (0)
n
X
f j (0) −
!
f j (k)
j=0
r
X
j=0
r
n
XX
k=0
k=0 j=0
bk ek −
j=0
= −
n
X
r X
n
X
bk f j (k)
bk f j (k)
k=0 j=0
= −b0
n
X
f j (0) −
j=0
p−1
r
X
X
bk f j (k) −
j=0
k=1
n
X
!
bk f j (k) .
j=p
Here we have used the fact that the degree of f is (r + 1)p − 1. With a simple
computation we get,

if j ≤ p − 2
 0
p
p
j
(p − 1)!(−1) ...(−n) if j = p − 1
f (0) =

≡ 0 (mod p!)
if j ≥ p.
For 1 ≤ k ≤ r,
j
f (k) =
if j ≤ p − 1
(mod p!) if j ≥ p.
0
≡0
Now since p >| b0 | the first summand is completely divisible by (p − 1)! and the
second summand is divisible by p! hence by (p − 1)! thus we find that J is an integer
which is divisible by (p − 1)! but not p and
| J |≥ (p − 1)!.
Since f (x) = xp−1
Qr
j=1
(x − j)p we get
f¯(k) ≤ k p−1
r
Y
(k + j)p ≤ (2r)n < (2r)2rp
j=1
for 0 ≤ k ≤ r. From this we deduce that,
| J |≤
r
X
j=0
| bj || I(j) |≤
r
X
| bj | jej f¯(j) ≤ r(r + 1)Cer (2r)2rp < K p ,
j=0
7
2 Algebraic and Transcendental numbers
where C = max1≤k≤r {| bk |}, and K = r(r + 1)Cer (2r)2r , which is a constant not
depending on p. But | J | is bounded below by (p − 1)!, thus
(p − 1)! ≤| J |≤ K p ,
but p has been chosen arbitrarily large and this inequality bounds p hence we get a
contradiction. Thus, e is transcendental.
2.2.2
Transcendence of π
Theorem 2.2.2. The number π is transcendental.
Proof. Observe if π is algebraic, then iπ is also algebraic. Therefore it suffices to
show that θ = iπ is transcendental. Assume θ is algebraic. Let r be the degree of
the minimal polynomial g(x) for θ. For some r ∈ N let θ = θ1 , θ2 , ..., θr denote the
algebraic conjugates of θ. Let b be the leading coefficient of g(x) (so bθj is an algebraic
integer). Since eθ = eiπ = −1, called the Euler’s identity for e,
so (1 + eθ1 )(1 + eθ2 )...(1 + eθr ) = 0. Rewriting this expression into an expression
containing 2r number of terms we get,
r
Y
(1 + eθj ) =
X
eρ .
P
ρ= rj=1 bj θj
bj ∈{0,1}
j=1
Let θ1 , ..., θn denote the non-zero expressions of the form eρ so that
2r − n + eθ1 + eθ2 + ... + eθn = 0.
Let
np p−1
f (x) = b x
n
Y
(x − θi )p
i=1
where p is an arbitrarily chosen large prime. Consider θ1 , θ2 , ..., θ2r first n of which
are non zero. Use
r
2
n
Y
Y
2r −n
(x − θj ) = x
(x − θj )
j=1
j=1
8
2 Algebraic and Transcendental numbers
is symmetric in θ1 , ..., θn thus by the fundamental theorem of elementary symmetric
functions and with the use of Lemmas 2.1.5 and 2.1.6 we can check f (x) ∈ Z. Define
J=
n
X
I(θk )
k=1
where
I(t) = e
t
m
X
j
f (0) −
j=0
m
X
f j (t)
j=0
where m = (n + 1)p − 1 is the degree of f (x). Substituting the value of I(θk ) in J we
get
J =
n
X
m
X
eθ(k)
f j (0) −
j=0
k=1
m
n
XX
j
= −
f (θk ) +
m
X
m
X
f j (0) −
j=0
k=1
f j (θk )
j
f (0)
j=0
= −(2r − n)
Pn
!
j=0
j=0 k=1
Observe that
m
X
n
X
eθk
k=1
m X
n
X
f j (θk ).
j=0 k=1
f j (θk ) is a symmetric polynomial in bθ1 , ..., bθn with integer coef-
ficients and thus a symmetric polynomial with integer coefficients in the 2r numbers,
bθi . By the fundamental theorem of symmetric functions, this sum is a rational number and so is an integer. Since, f j (θk ) = 0 for j < p. The double sum in the expression
for J above is an integer divisible by p! for j ≥ p. Also,
f p−1 (0) = bnp (−1)np (p − 1)!(θ1 θ2 ...θn )p .
From the fundamental theorem of elementary symmetric functions and Lemmas 2.1.5
and 2.1.6, we find f p−1 (0) is an integer divisible by (p − 1)!, if p is sufficiently large,
then f p−1 (0) is not divisible by p. If p > 2r − n, then | J |≥ (p − 1)! On the other
hand using the upper bound for | I(t) |, we get
| J |≤
n
X
| θk | e|θk | f¯(θk ) ≤ C p
k=1
9
2 Algebraic and Transcendental numbers
where C is a constant independent of p. Hence we obtain a contradiction since p was
arbitrary. Thus, π is transcendental.
2.3
Rational Approximations
We discuss the problem of approximating a given number ξ (usually irrational), by
a rational fraction r =
p
q
(we suppose throughout that 0 < ξ < 1 and that
p
q
is
irreducible, i.e. it is expressed in its lowest terms) i.e.,
given ξ and > 0, there exists r =
p
q
such that
p
|r − ξ| = − ξ ≤ .
q
Dirichlet (1842) proved that for any real irrational number, there are infinitely many
rational numbers approximating it very closely.
Theorem 2.3.1. (Dirichlet’s Theorem) Given ξ be a real number and η, a positive
integer. If ξ is irrational, then there exists an infinite set of integers p, q where
p
0 < q ≤ n, such that ξ − q ≤ q12 .
Remark 2.3.1. If ξ is a rational and is equal to
a
b
where a and b are integers. If r 6= ξ,
then
p a bp − aq ≥ 1.
|r − ξ| = − = q
b
bq bq
From the theorem we get,
ξ −
p 1
< 2
q
q
hence
p a
1
1
≤ − < 2 .
bq
q
b
q
From this we get, q < b and hence there are only finite number of solutions to the
inequality,
ξ −
p 1
< 2.
q
q
10
2 Algebraic and Transcendental numbers
Proof. (Dirichlet’s Theorem) Denote [x] as the integral part of a real number x and
{x} as the fractional part of x. Let Q be a fixed integer greater than 1. Consider
h
i
i i+1
the Q + 1 numbers, 0, {ξ}, {2ξ}, {3ξ}, ..., {Qξ} and the Q intervals, Q , Q where
i ∈ {0, 1, ..., Q − 1}. Hence there must be an interval, say [ Qj , j+1
] containing two of
Q
these Q + 1 numbers, say aξ and bξ with 0 ≤ a < b ≤ Q. Hence,
|bξ − aξ| ≤
1
.
Q
Write aξ = α + {aξ} and bξ = β + {bξ} where α, β ∈ Z are the integral parts of aξ
and bξ respectively. Then,
{bξ} − {aξ} = α − β + (b − a)ξ.
Put q = b − a so, 0 ≤ q ≤ Q. Let p = β − α then,
1
|qξ − p| ≤
⇒ ξ −
Q
This is true for any Q and as q ≤ Q,
p 1
≤
.
q
pQ
1
1
≤ 2 , we get an infinite number of solutions
qQ
q
to the inequality,
ξ −
p 1
< 2
q
q
thus proving the theorem.
¿From the Dirichlet’s theorem we get a criterion for proving a number irrational.
Corollary 2.3.2. A real number is irrational if there exists a δ > 0 and infinite pairs
p, q ∈ Z and (p, q) = 1 such that
ξ −
p 1
< 1+δ .
q
q
11
Chapter 3
Apéry’s Proof of Irrationality of ζ(3)
In this chapter we will discuss Apéry’s proof [J6] of irrationality of the Riemann zeta
function at odd integer 3. We begin with giving some preliminary results required for
our proof.
3.1
Preliminaries
We begin with stating an identity involving the lowest common multiple of first n
natural numbers
Lemma 3.1.1.
dn =
Y
pd log p e
log n
p≤n
p is prime
where dn = lcm {1, 2, ..., n}.
n
Proof. Suppose p is a prime number and x is a real number. Then, n = px ⇒ x = log
log p
l
m
log n
so, dxe = log p is the highest integer power of p for which pdxe < n. Now suppose
{A1 , A2 , ..., An } is a set of positive integers where
Ai = p1 yi1 p2 yi2 ...pα yiα for i = 1, ..., n and pi is prime and
(yi will be 0 for many i). As lcm {A1 , A2 , ..., An } = p1 m1 p2 m2 ...pα mα where
mj = max{yi , i = 1, ..., n}. In the special case where Ai = i,
log n
.
mj =
log pj
12
3 Apéry’s Proof of Irrationality of ζ(3)
Lemma 3.1.2. Given n,
pd log p e =
log n
Y
p≤n
p is prime
Y
n.
p≤n
p is prime
Proof. To prove the lemma, it suffices to show
Y
log n
p log p =
Y
n.
p≤n
p≤n
log n
log n
Let p log p = y ⇒
log p = log y ⇒ log n = log y ⇒ n = y. Thus,
log p
log n
p log p = n.
The next lemma expresses ζ(3) in terms of binomial coefficients. Using this expression a double sequence converging to ζ(3) is obtained.
Lemma 3.1.3.
∞
∞
X
5 X (−1)n−1
1
.
=
ζ(3) =:
n3
2 n=1 n3 2n
n
n−1
Proof. We will prove the lemma step by step.
Step 1:
K
X
k=1
where
a0k s
a1 a2 ....ak−1
1
a1 a2 ...aK
= −
(x + a1 )(x + a2 )...(x + ak )
x x(x + a1 )...(x + aK )
are integers. Put
A0 =
1
a1 a2 ...aK
and AK =
.
x
x(x + a1 )...(x + aK )
Now,
a1 a2 ....ak−1
a1 a2 ....ak−1
a1 a2 ....ak
=
−
(x + a1 )(x + a2 )...(x + ak )
x(x + a1 )(x + a2 )...(x + ak−1 ) (x + a1 )(x + a2 )...(x + ak )
= Ak−1 − Ak
13
3 Apéry’s Proof of Irrationality of ζ(3)
Hence,
K
X
k=1
K
X
a1 a2 ....ak−1
=
Ak−1 − Ak = A0 − AK .
(x + a1 )(x + a2 )...(x + ak )
k=1
Step 2:
Putting x = n2 ; ak = −k 2 and letting 1 ≤ K ≤ n − 1 we have
K
X
−12 . − 22 ... − (k − 1)2
k=1
⇒
n−1
X
k=1
(n2 − 1)...(n2 − k 2 )
=
K
X
(−1)k−1 12 22 ....(k − 1)2
k=1
(n2 − 12 )...(n2 − k 2 )
.
(−1)k−1 (k − 1)!2
1
(−1)n−1 (n − 1)!2 n2n
=
−
(n2 − 12 )...(n2 − k 2 )
n2
n2 (2n)!2
1
(−1)n−1 2n!n!
=
−
n2
n2 (2n)!
1
(−1)n−1 2
.
=
−
n2
n2 2n
n
Write
n,k =
1 (k!)2 (n − k)!
.
2 k 3 (n + k)!
Since
(−1)k−1 (k − 1)!2
(−1) n(n,k − n−1,k ) = 2
,
(n − 12 )...(n2 − k 2 )
k
upon expanding we get,
n−1
N X
X
n=1 k=1
(−1)k (n,k −n−1,k ) =
N
X
(−1)k (N,k −k,k ) =
k=1
N
X
k=1
2k 3
(−1)k
N +k
k
N
+
N
k
1 X (−1)k−1
.
2 k=1 k 3 2k
k
From above, the right hand side of the equation
=
N
N
X
X
1
(−1)n−1
.
−
2
3
3 2n
n
n
n
n=1
k=1
Hence
N
N
N
X
X
1
(−1)n−1 X
(−1)k
−
2
=
3
3 2n
3 N +k
n
n
2k
n
k
n=1
k=1
k=1
14
N
+
N
k
1 X (−1)k−1
.
2 k=1 k 3 2k
k
3 Apéry’s Proof of Irrationality of ζ(3)
Take N −→ ∞, the first term vanishes and we get
∞
∞
X
1
5 X (−1)k−1
.
=
n3
2 k=1 k 3 2k
k
k=1
Thus, define the double sequence,
cn,k
n
k
X
X
1
(−1)m−1
n+m .
:=
+
3
3 n
m
2m
m
m
m=1
m=1
One may remark that cn,k → ζ(3) as n → ∞, uniformly in k i.e., the sequence
cn,n → ζ(3) but the following lemma will show that our conclusion will be false and
we cannot apply the irrationality criterion.
Lemma 3.1.4.
n+k
2dn cn,k
is an integer
k
3
where dn = lcm(1, 2, ..., n).
Proof. It suffices to show,
n+k
2cn,k
k
is a rational number whose denominator expressed in lowest terms divides dn 3 . Furthermore to prove this it suffices to show that the number of times any given prime
p divides the denominator of 2cn,k n+k
is less then the number of times that prime
k
divides dn 3 . Let ordp (x) mean the highest exponent of p that divides x. Observe,
n+k
n+k
k−m
k .
n+m =
k
m
m
Now,
n
n(n + 1)....(n − m − 1)
n!
ordp
= ordp
≤ ordp
m
m!
m!
= ordp (n!) − ordp (m!) ≤ ordp (dn ) − ordp (m)
log n
=
− ordp (m).
log p
15
3 Apéry’s Proof of Irrationality of ζ(3)
Next, we check that the denominator of each term in the expression for 2cn,k
divides dn 3 . Now,
n
k
n+k
n+k X 1
n + k X (−1)m−1
n+m .
2cn,k
=2
+2
n
m3
k
k
k
2m3 m
m
m=1
m=1
The denominator of
n
n+k X 1
2
k
m3
m=1
divides dn 3 . Hence we will check for the denominator of the second term,
n+k X
(−1)m−1
2
k 3 n n+m
k
2m m m
m=1
whose denominator is
2m3
n
n+m
m
m
n+k
k
=
n
k
m m
n+k
k−m
2m3
.
Hence,
ordp
m3
!
n
n+m
m
m
n+k
k
= ordp
m3
n
k
m m
n+k
k−m
!
log n
− ordp (m)
≤ 3ordp (m) +
log p
log k
n+k
+
− ordp (m) − ordp
log p
k−m
log n
log k
≤
+
+ ordp (m).
log p
log p
As m ≤ k ≤ n,
ordp (m) ≤ ordp (dm ) ≤ ordp (dk ) ≤ ordp (dn )
and therefore
log k
log n
log k
log n
+
+ ordp (m) ≤
+
+ ordp (dm ).
log p
log p
log p
log p
log m
log n
But ordp (dm ) =
≤3
= ordp (dn 3 ).
log p
log p
16
n+k
k
3 Apéry’s Proof of Irrationality of ζ(3)
Corollary 3.1.5. The Denominator of an divides dn 3 .
Proof. From the definition of an ,
an =
2
n 2 X
n
n+k
k=0
k
k
cn,k
where
cn,k
n
k
X
X
1
(−1)m−1
n+m
=
+
n
m3 m=1 2m3 m
m
m=1
and from the above lemma, denominator of cn,k divides dn 3 and thus denominator of
an divides dn 3 .
3.2
Irrationality of ζ(3)
Lemma 3.2.1. The equation,
n3 un + (n − 1)3 un−2 = (34n3 − 51n2 + 27n − 5)un−1 , n ≥ 2,
is equivalent to
(n + 1)3 un+1 − (34n3 + 51n2 + 27n + 5)un + n3 un−1 = 0, n ≥ 1.
Proof. The recurrence relation,
(n + 1)3 un+1 − (34n3 + 51n2 + 27n + 5)un + n3 un−1 = 0
can be obtained by replacing n by n − 1 in
n3 un + (n − 1)3 un−2 = (34n3 − 51n2 + 27n − 5)un−1 .
And hence we get
(n + 1)3 un+1 − (34n3 + 51n2 + 27n + 5)un + n3 un−1 = 0, n ≥ 1.
17
3 Apéry’s Proof of Irrationality of ζ(3)
Lemma 3.2.2. an =
Pn
k=0
n 2 n+k 2
cn,k
k
k
and bn =
Pn
k=0
n 2 n+k 2
k
k
satisfy the re-
currence relation,
n3 un + (n − 1)3 un−2 = (34n3 − 51n2 + 27n − 5)un−1 ,
n ≥ 2.
Proof. In order to show that the sequence an satisfies the recurrence relation, we
define a double sequence,
Bn,k
2 2
n
n+k
= 4(2n + 1)[k(2k + 1) − (2n + 1) ]
.
k
k
2
Step 1:
We observe
2 2
2 2
n+1
n+1+k
n
n+k
3
2
= (n + 1)
− (34n + 51n + 27n + 5)
k
k
k
k
2 2
n−1
n−1+k
+n3
.
k
k
3
Bn,k − Bn,k−1
Step 2:
Using the definition of cn,k =
Pn
m=1
cn,k − cn−1,k
Pk
1
(−1)m−1
n+m then
+
m=1
n
m3
2m3 m
m
(−1)k k!2 (n − k − 1)!
.
=
n2 (n + k)!
This can be obtained as follows:
cn,k − cn−1,k
n
k
n−1
k
X
X
X
X
1
(−1)m−1
1
(−1)m−1
n+m −
n−1+m
=
+
+
n
m3 m=1 2m3 m
m3 m=1 2m3 n−1
m
m
m
m=1
m=1
k
X (−1)m (m − 1)!2 (n − m − 1)!
1
=
+
.
n3 m=1
(n + m)!
The fraction
(−1)m (m − 1)!2 (n − m − 1)!
can be split into two parts,
(n + m)!
(−1)m (m − 1)!2 (n − m − 1)!
(−1)m (m − 1)!2 (n − m − 1)!(m2 + n2 − m2 )
=
(n + m)!
n2 (n + m)!
m
(−1) m!2 (n − m − 1)! − (−1)m+1 (m − 1)!2 (n − m)!(n + m)
=
n2 (n + m)!
(−1)m (m!)2 (n − m − 1)! (−1)m+1 (m − 1)!2 (n − m)!
=
−
.
n2 (n + m)!
n2 (n + m − 1)!
18
3 Apéry’s Proof of Irrationality of ζ(3)
Thus,
k
k
X
1
(−1)m (m!)2 (n − m − 1)! X (−1)m−1 (m − 1)!2 (n − m)!
=
−
+
n3 m=1
n2 (n + m)!
n2 (n + m − 1)!
m=1
!
1
(−1)k (k!)2 (n − k − 1)!
(n − 1)!
=
+
.
−
n3
n2 n!
n2 (n + k)!
cn,k − cn−1,k
Step 3:
If bn,k =
n 2 n+k 2
,
k
k
define Bn,k = 4(2n − 1)(k(2k + 1) − (2n + 1)2 )bn,k then
(Bn,k −Bn,k−1 )cn,k +(n + 1)3 bn−1,k (cn−1,k −cn,k )−n3 bn−1,k (cn,k −cn−1,k ) = An,k −An,k−1
where
An,k
5(2n + 1)(−1)k−1 k n n + k
.
= Bn,k cn,k +
n(n + 1)
k
k
Step 4:
Using the previous definition of an and cn,k we will show an satisfies:
n3 un + (n − 1)3 un−2 = 34n3 − 51n2 + 27n − 5)un−1 ,
n ≥ 2.
From Lemma 3.2.1, the expression
n3 un + (n − 1)3 un−2 = 34n3 − 51n2 + 27n − 5)un−1 ,
n ≥ 2,
is equivalent to
(n + 1)3 un+1 − (34n3 + 51n2 + 27n + 5)un + n3 un−1 = 0,
n ≥ 1.
Substituting an for un in the above identity,
(n + 1)
3
n
X
bn+1,k cn+1,k − P (n)
k=1
where bn,k =
(n + 1)3
n
X
k=1
bn,k cn,k + n
3
n+k 2
k
n
X
bn−1,k cn−1,k = 0,
n ≥ 1,
k=1
k=1
n 2
k
n
X
and P (n) = 34n3 + 51n2 + 27n + 5. Since for r > n,
bn+1,k cn+1,k − P (n)
n
X
bn,k cn,k + n3
k=1
n
X
k=1
19
bn−1,k cn−1,k = 0,
n
r
= 0,
n ≥ 1,
3 Apéry’s Proof of Irrationality of ζ(3)
can be written as
n+1
X
(n + 1)3 bn+1,k cn+1,k − P (n)bn,k cn,k + n3 bn−1,k cn−1,k = 0,
n ≥ 1. (1)
k=0
We will now prove this expression. From Step 1,
Bn,k − Bn,k−1 = (n + 1)3 bn+1,k − P (n)bn,k + n3 bn−1,k
where Bn,k = 4(2n − 1)[k(2k + 1) − (2n + 1)2 ]bn,k and therefore
−P (n)bn,k cn,k = (Bn,k − Bn,k−1 )cn,k − (n + 1)3 bn+1,k cn,k − n3 bn−1,k cn,k .
Substituting in (1)
n+1
X
(n + 1)3 bn+1,k cn+1,k + (Bn,k − Bn,k−1 )cn,k − (n + 1)3 bn+1,k cn,k − n3 bn−1,k cn,k + n3 bn−1,k cn−1,k
k=0
=
n+1
X
(n + 1)3 bn+1,k (cn+1,k − cn,k ) + (Bn,k − Bn,k−1 )cn,k + n3 bn−1,k (cn−1,k − cn,k ) .
k=0
From the previous lemma this becomes
n+1
X
(An,k − An,k−1 )
k=0
where
An,k
5(2n + 1)9 − 1)k−1 k
= Bn,k cn,k +
n(n + 1)
n
k
n+k
k
.
Now
n+1
X
An,k − An,k−1 = An,n+1 − An,−1 .
k=0
But
(as Bn,n+1
2n+1
n
5(2n + 1)(−1)n (n + 1) n+1
n+1
An,n+1 = Bn,n+1 cn,n+1 +
=0
n(n + 1)
n
). Similarly An,−1 = 0. Thus, an satisfies the recurrence
has a factor n+1
n3 un + (n − 1)3 un−2 = 34n3 − 51n2 + 27n − 5)un−1 ,
20
n ≥ 2.
3 Apéry’s Proof of Irrationality of ζ(3)
Step 5:
Finally we will show
bn =
2
n 2 X
n
n+k
k=0
k
k
=
n
X
bn,k
k=0
satisfies the recurrence relation,
n3 un + (n − 1)3 un−2 = 34n3 − 51n2 + 27n − 5)un−1 ,
n ≥ 2.
From earlier,
Bn,k
2 2
n
n+k
= 4(2n+1)(k(2k+1)−(2n + 1) )
= 4(2n+1)(k(2k+1)−(2n + 1)2 )bn,k .
k
k
2
From Step (1),
Bn,k − Bn,k−1 = (n + 1)3 bn+1,k − P (n)bn,k − n3 bn−1,k := Gn,k .
So,
Bn,k = Gn,k + Bn,k−1 = Gn,k + Gn,k−1 + Bn,k−2 = Gn,k + Gn,k−1 + Gn,k−2 + ....
Continuing in this manner and using the fact that Bn,r = 0 for r < 0, we get
Bn,k =
k
X
Gn,i .
i=0
Thus,
Bn,n+1 =
n+1
X
Gn,i
i=0
=
n+1
X
(n + 1)3 bn+1,i − P (n)bn,i − n3 bn−1,i
i=0
= (n + 1)3
= (n + 1)3
n+1
X
i=0
n+1
X
bn+1,i − P (n)
bn+1,i − P (n)
i=0
n+1
X
i=0
n
X
i=0
21
bn,i − n3
bn,i − n3
n+1
X
i=0
n−1
X
i=0
bn−1,i
bn−1,i .
3 Apéry’s Proof of Irrationality of ζ(3)
Now since bn,n+1 = 0, bn−1,n+1 = 0, bn−1,n = 0,
Bn,n+1 = (n + 1)3 bn+1 − P (n)bn − n3 bn−1 .
But Bn,r = 0 for r > n. Thus,
(n + 1)3 bn+1 − P (n)bn − n3 bn−1 = 0.
And hence bn satisfies the recurrence relation.
Lemma 3.2.3. Given that an and bn satisfy the recurrence relation
n3 un − (n − 1)3 un−2 = P (n)un−1 ,
n≥2
where P (n) = 34n3 − 51n2 + 27n − 5 then
an bn−1 − an−1 bn =
6
n3
for an and bn as defined.
Proof. From the above lemma since an and bn satisfy the recurrence relation, we
substitute an and bn in the recurrence relation to get,
n3 an + (n − 1)3 an−2 = P (n)an−1 ,
n3 bn + (n − 1)3 bn−2 = P (n)bn−1 .
Multiplying the first equation by bn−1 and the second equation by an−1 and subtracting gives,
n3 (an bn−1 − bn an−1 ) = −(n − 1)3 (an−2 bn−1 − bn−2 an−1 ) .
22
3 Apéry’s Proof of Irrationality of ζ(3)
(n − 1)3
(an−1 bn−2 − bn−1 an−2 )
n3
(n − 1)3 (n − 2)3
(an−2 bn−3 − an−3 bn−2 )
=
n3 (n − 1)3
.
= ..
an bn−1 − bn an−1 =
(n − (n − 1))3
=
an−(n−1) bn−n − an−n bn−(n−1)
3
n
1
(a1 b0 − a0 b1 )
=
n3
1
=
.
n3
Now from definition a1 = 6, b0 = 1, a0 = 0, b1 = 5 and so, a1 b0 − a0 b1 = 6 hence we
get
an bn−1 − an−1 bn =
6
.
n3
Lemma 3.2.4. If
cn,k
n
k
X
X
1
(−1)m−1
n+m
=
+
n
m3 m=1 2m3 m
m
m=1
k≤n
then cn,k → ζ(3) as n → ∞ uniformly in k.
Proof.
∞
k
k
∞
X
X
(−1)m−1 X 1 X (−1)m−1 1
n+m ≤ n+m .
−
|cn,k − ζ(3)| = −
+
n
n
m3 m=1 2m3 m
m3 m=1 2m3 m
m
m
m=n+1
m=n+1
Since
P∞
m=1
1
is convergent, it follows that given > 0, ∀ n > N1 ,
m3
∞
X
1 < .
m3 m=n+1
Next we see that
k
X (−1)m−1 k
n
1
1
1
n+m ≤
≤
=
≤
<
,
∀
n
>
.
n
3
2n(n
+
1)
2n(n
+
1)
2(n
+
1)
2n
2m
m
m
m=1
23
3 Apéry’s Proof of Irrationality of ζ(3)
Let N2 =
1
,
2
N = max(N1 , N2 ) then for n > N,
| cn,k − ζ(3) |≤ 2 + = 3
since this convergence is independent of k so the convergence is uniform.
Corollary 3.2.5.
an
→ ζ(3) as n → ∞.
bn
Proof. Using the fact that
an =
n
X
bn,k cn,k ,
k=1
bn =
n
X
bn,k
k=1
and the result that if xn,k , yn,k ∈ R and yn,k → L as n → ∞ (uniformly in n) then
Pn
k=1 xn,k yn,k
P
→ L as n → ∞.
n
k=1 xn,k
We get
an
→ ζ(3) as n → ∞.
bn
Lemma 3.2.6. If an bn−1 − an−1 bn =
6
, then
n3
∞
X
an
6
ζ(3) −
=
.
3
bn
k
b
k bk−1
k=n+1
Proof. We have
an an−1
6
−
= 3
.
bn
bn−1
n bn bn−1
Let ζ(3) −
an
bn
= xn , then
xn − xn+1 =
an+1 an
6
−
=
.
bn+1
bn
(n + 1)3 bn+1 bn
24
3 Apéry’s Proof of Irrationality of ζ(3)
Hence
ζ(3) −
an
6
=
+ xn+1 ,
bn
(n + 1)3 bn+1 bn
and proceeding in this manner we get,
ζ(3) −
6
6
6
an
=
+
+ ··· +
+ xm+n
3
3
3
bn
(n + 1) bn+1 bn (n + 2) bn+2 bn+1
(n + m) bn+m bn+m−1
m
X
6
=
+ xn+m .
3b b
k
k k−1
k=n+1
Now, ζ(3) −
an
bn
→ 0 as n → ∞ and thus,
∞
X
an
6
| ζ(3) −
|=
.
3
bn
k
b
k bk−1
k=n+1
Theorem 3.2.7. ζ(3) is irrational
Proof. From the above estimation,
∞
X
an
6
| ζ(3) −
|=
3
bn
k bk bk−1
k=n+1
we have
ζ(3) −
an
= O(bn −2 ).
bn
The recurrence relation satisfied by the b0n s can be rewritten as
bn − (34 − 51n−1 + 27n−2 − 5n−3 )bn−1 + (1 − 3n−1 + 3n−2 − n−3 )bn−2 = 0.
From this relation we get the generating polynomial for bn which is x2 − 34x +1 and it
√
√ 4
√ 4
has zeros 17 ± 12 2 = (1 ± 2) . From this we get bn = O(αn ), where α = (1 + 2)
Now, we have already seen that a0n s are rationals with denominator dividing dn 3 where
pn
dn = lcm(1, 2, .., n) we define two new sequences pn and qn such that ζ(3) →
as
qn
n → ∞. Define
pn = 2dn 3 an , qn = 2dn 3 bn .
25
3 Apéry’s Proof of Irrationality of ζ(3)
Clearly pn , qn ∈ Z and
qn = O(αn e3n ), ζ(3) −
pn
an
= ζ(3) −
= O(bn −2 ) = O(α−2n ).
qn
bn
Choosing
δ=
log α − 3
= 0.080.. > 0,
log α + 3
we get
log α =
3(1 + δ)
⇒ α−1+δ = e−3(1+δ)
(1 − δ)
⇒ α−2 α1+δ = e−3(1+δ)
⇒ α−2 = α−1−δ e−3(1+δ)
⇒ α−2n = (αn e3 n)
−(1+δ)
.
Thus, O(α−2n ) = O(qn −(1+δ) ). Hence we have
ζ(3) −
pn
= O(qn −(1+δ) ).
qn
This implies ζ(3) is irrational.
26
Chapter 4
Theory of Modular forms
In this chapter we discuss the theory on modular forms, in particular Eisenstein series,
Dadekind eta function and L- functions associated to modular forms.
4.1
SL(2, Z) and its congruence subgroups
Let
SL(2, Z) =
a b
|a, b, c, d ∈ Z, ad − bc = 1 .
c d
We give an action of SL2 (Z) on the extended complex plane C̃ = C ∪ {∞} by
az+b
if z ∈ C
cz+d
γz =
a
if z = ∞
c
a b
where γ =
. The upper half plane H = {z ∈ C| Im z > 0} is stable under
c d
the given action of SL2 (Z). Let S and T be two elements of SL(2, Z) given by
0 −1
1 1
S=
and T =
. Note that SL(2, Z) is generated by S and T . For
1 0
0 1
any z ∈ H we have, Sz = −1
, T z = z + 1.
z
A fundamental domain D is a subset of H satisfying the properties:
1. ∀ z ∈ H, ∃ g ∈ G s.t. gz ∈ D,
2. If z1 , z2 ∈ int(D), then @ g ∈ G s.t. gz1 = z2 .
Theorem 4.1.1. The region D = {z ∈ H : |z| ≥ 1, − 12 ≤ Re(z) ≤
1
}
2
fundamental domain for the action of modular group SL(2, Z) on H.
Definition 4.1.1. The principal congruence subgroup of SL(2, Z) of level N is
a b
a b
1 0
Γ(N ) =
∈ SL(2, Z) :
≡
(mod N ) .
c d
c d
0 1
27
is a
4 Theory of Modular forms
Definition 4.1.2. A subgroup Γ of SL(2, Z) is called congruence subgroup if Γ ⊃
Γ(N ) for some N ∈ N, then Γ is called congruence subgroup of level N .
Besides the principal congruence subgroups, the important congruence subgroups are:
a b
a b
∗ ∗
Γ0 (N ) =
∈ SL(2, Z) :
≡
c d
c d
0 ∗
(mod N )
(where “ * ” means “unspecified” )and
Γ1 (N ) =
a b
a b
1 ∗
∈ SL(2, Z) :
≡
c d
c d
0 1
(mod N ) ,
satisfying
Γ(N ) ⊂ Γ1 (N ) ⊂ Γ0 (N ) ⊂ SL(2, Z).
4.2
Modular functions and modular forms
Let f (z) be a meromorphic function on the upper half plane H and let k be an integer.
Suppose that f (z) satisfies the relation
k
f (γz) = (cz + d) f (z) ∀ γ
a b
=
∈ SL(2, Z).
c d
1
0 1
and S =
above condition
1
−1 0
1
In particular, for the elements γ = T =
0
gives,
−1
f (z + 1) = f (z); f
= (−z)k f (z).
z
Hence f is a 1-periodic function. f (z) is called a modular function of weight k for
SL(2, Z) if there are finitely many non zero coefficients an with n < 0 in the fourier
series expansion of f ,
f (z) =
X
an q n ,
where q = e2πiz .
n∈Z
28
4 Theory of Modular forms
Definition 4.2.1. For k ∈ Z, a function f is a modular form of weight k for SL(2, Z)
if f is holomorphic on H and holomorphic at infinity (an = 0 ∀ n < 0) and satisfies
f (γz) = (cz + d)k f (z) ∀ γ ∈ SL( 2, Z), z ∈ H.
The space of all modular forms of weight k is denoted by Mk (SL(2, Z)).
Definition 4.2.2. A cusp form is a modular form with the constant term a0 = 0 in
its fourier series expansion. The space of all cusp forms of weight k are denoted by
Sk (SL(2, Z)).
−1 0
Remark 4.2.1. As the element
∈ SL(2, Z), f (z) = (−1)k f (z). Hence if k
0 −1
is odd, there are no nonzero modular functions of weight k for SL(2, Z).
Definition 4.2.3. Let γ ∈ SL2 (Z), then slash operator of weight k on functions on
H is defined as
(f |k γ)(z) := (cz + d)−k f (γz).
Definition 4.2.4. Let Γ be a congruence subgroup of SL2 (Z) of level N and let k be
an integer. A function f : H → C is a modular form of weight k with respect to Γ if
1. f is holomorphic on H,
2. f |k α = f ∀ f ∈ Γ,
3. f |k α is holomorphic at ∞ for all α ∈ SL2 (Z).
If in addition, we have a0 = 0 in the Fourier expansion of f |k α for all α ∈ SL2 (Z),
then we call f a cusp form of weight k with respect to Γ.
Example 4.2.1. Now we will explicitly construct a class of modular forms of all even
weights k > 2, the Eisenstein series of weight k.
29
4 Theory of Modular forms
Definition 4.2.5. Let V be a real vector space. A subgroup Γ of V is a lattice if there
exists a R−basis {e1 , ....en } of V , which is a Z−basis for Γ, i.e. Γ = Ze1 ⊕ ... ⊕ Zen .
Lemma 4.2.1. Let Γ be a lattice in C. Then the series
P
1
γ∈Γ,γ6=0 |γ|σ
is convergent for
σ > 2.
Definition 4.2.6. The Eisenstein series of index k is defined by
X
Gk (z) =
m,n∈Z
(m,n)6=(0,0)
1
,
(mz + n)2k
z ∈ H.
Proposition 4.2.2. Gk is a modular form of weight k. Also Gk (∞) = 2ζ(k).
Proof. First we see that
Gk (z + 1) = Gk (z)
−1
= z 2k Gk (z)
Gk
z
Next we show Gk is holomorphic in H. For this it is enough to show that the series
converges uniformly in all compact subsets of H. Let w be a complex number. For
z ∈ D since | z |> 1 and Re(z) >
−1
,
2
| mz + n |2 = (mz + n)(mz̄ + n)
= m2 | z |2 +2mnRe(z) + n2
≥ m2 − mn + n2
= | mω + n |2
1
1
≤
2
| mz + n |
| mw + n |2
X
X
1
⇒
≤
2k
|mz + n|
m,n∈Z
m,n∈Z
⇒
(m,n)6=(0,0)
1
|mω + n|2k
.
(m,n)6=(0,0)
But the latter series runs over all the lattice points of the lattice generated by ω
and 1, hence by Lemma 4.2.1 it is convergent for 2k > 2 i.e. k > 1 and hence the
30
4 Theory of Modular forms
Eisenstein series Gk . Since for any other point z ∈ H there exists g ∈ G such that
gz ∈ D. From the relation Gk (gz) = z 2k Gk (z), we show Gk converges for all z ∈ H
and hence it is holomorphic in H.
lim Gk (z) =
z→i∞
=
X
lim
z→i∞
m,n∈Z
(m,n)6=(0,0)
X
m,n∈Z
(m,n)6=(0,0)
1
(mz + n)2k
1
.
z→i∞ (mz + n)2k
lim
For m 6= 0 it vanishes but for m = 0 it contributes
P
1
n∈Z n2k
= 2ζ(2k). Thus
Gk (∞) = 2ζ(2k). Hence Gk (z) is holomorphic in H including infinity and so it is a
modular form.
4.3
q-expansion of a modular form
We recall, the q-expansion of a modular form f is
f (z) =
∞
X
an q n where q = ei2πz .
n=0
We compute the q-expansion coefficients for Gk .
Proposition 4.3.1. For even integer k > 2, z ∈ H. Then,
∞
2k X
Gk (z) = 2ζ(k) 1 −
σk−1 (n)q n
Bk n=1
where, σk (n) =
P
d|n
!
dk and the Bernoulli numbers are defined by the expansion,
∞
X xk
x
=
Bk .
ex − 1 k=0
k!
We normalize the value of Gk at infinity by defining Ek i.e.,
Ek (z) =
∞
X
Gk (z)
= 1 + ck
σk−1 (n)q n
2ζ(k)
n=1
31
4 Theory of Modular forms
where
ck =
(2iπ)k 1
−2k
=
.
(k − 1)! ζ(k)
Bk
We shall be mainly interested in the standard Eisenstein series,
E4 (z) = 1 + 240
∞
X
σ3 (n)q n .
n=1
4.3.1
Estimates for the coefficients in q-expansion
In this section, we give an estimate on the growth of the coefficients an in q-expansion
of modular form f .
Proposition 4.3.2. If f = Gk , the an = O(n2k−1 ).
Proof. From the q-expansion of Gk in Proposition 4.3.1, for some constant A,
an = A(−1)k σ2k−1 (n)
⇒| an |= Aσ2k−1 (n) ≥ An2k−1
∞
X d2k−1
X 1
X
| an |
1
⇒ 2k−1 = A
=A
≤A
=B<∞
2k−1
2k−1
2k−1
n
n
d
d
d=1
d|n
d|n
where B = Aζ(2k − 1), we have
An2k−1 ≤| an |≤ Bn2k−1
Thus, an = O(n2k−1 ).
The next theorem is on estimates of coefficients of cusp forms which is due to
Hecke.
Theorem 4.3.3. If f is a cusp form of weight 2k, then
an = O(nk ).
32
4 Theory of Modular forms
Proof. Since for a cusp form a0 = 0,
f (z) = q
∞
X
!
an q n−1 .
n=1
Hence, if q → 0, |f (z)| = O(q) = O(e−2πy ), where y = Im(z). Define a function
φ(z) = |f (z)|y k . Note that φ is invariant under G and continuous in the fundamental
domain D. Also as y → ∞, φ(z) → 0 ⇒ φ(z) is bounded in H, i.e. ∃ M s.t.
φ(z) ≤ M
∀ z ∈ H,
⇒| f (z) |≤ M y −k ∀ z ∈ H.
Using the formula to compute Fourier coefficients
Z 1
an =
f (z)e−i2πnz dx.
0
⇒ |an | ≤ M y −k e2πny ∀ y > 0.
Putting y =
1
n
we get,
|an | ≤ M nk e2π ⇒ an = O(nk ).
4.3.2
The Eisenstein series of weight 2
Define the Eisenstein series G2 ,
G2 (z) =
XX
c∈Z d∈Z0 c
1
(cz + d)2
where Z0 c = Z − {0} if c = 0 and Z0 c = Z otherwise. The normalized Eisenstein series
is,
E2 (z) = 1 − 24
∞
X
σ1 (n)q n = 1 − 24q − 72q 2 − ....
n=1
E2 satisfies all the properties of a modular form except the transformation law
Ek (−1/z) = z 2 Ek (z). In fact the Eisenstein series of weight 2 satisfies the following transformation law:
33
4 Theory of Modular forms
a b
Proposition 4.3.4. For z ∈ H and
∈ SL2 (Z), we have
c d
az + b
G2
= (cz + d)2 G2 (z) − πic(cz + d).
cz + d
Definition 4.3.1. The Dedekind eta function η : H −→ C is defined by an infinite
product
1
η(τ ) = q 24
∞
Y
(1 − q n ) for q = e2πiz .
n=1
The transformation law for the Dedekind eta function is
Theorem 4.3.5. Let
√
denote the branch of the square root having non negative real
part. Then
η
−1
z
r
=
z
η(z).
i
Proof. Compute the derivative of the logarithm of the eta function.
∞
X
dq d
d
πi
log(η(τ )) =
+ 2πi
dτ
12
1 − qd
d=1
=
∞
∞
X
X
πi
+ 2πi
d
q dm
12
d=1 m=1
∞ X
m
X
πi
+ 2πi
dq dm
=
12
m=1 d=1


∞
X
X
πi

=
+ 2πi
d q n
12
n=1
0<d|n
=
πi
E2 (τ ).
12
Thus,
d
πi
log(η(τ )) = τ −2 E2
dτ
12
−1
τ
and
√
1
πi
πi
d
log( −iτ η(τ )) =
+ E2 (τ ) =
dτ
2τ
12
12
34
12
E2 (τ ) +
.
2πiτ
4 Theory of Modular forms
Now
12
E2 (τ ) +
.
2πiτ
πi −2
−1
πi
τ E2 ( ) =
12
τ
12
Hence the desired relation holds up to a multiplicative constant. Setting τ = i shows
that the constant is 1.
Remark 4.3.1. η(τ )24 is a cusp form of weight 12 of SL(2, Z) and also of Γ1 (6) and
so η(τ )24 ∈ S12 (Γ1 (6)).
4.4
L-functions and Modular forms
Each modular form f ∈ Mk (Γ1 (N )) has an associated Dirichlet series, its L-function.
Definition 4.4.1. The L-function associated to a modular form, f (z) =
P∞
n=0
an q n
for a complex variable s ∈ C is
L(s, f ) =
∞
X
an n−s .
n=1
Theorem 4.4.1. If f ∈ Mk (Γ1 (N )) is a cusp form then L(s, f ) converges absolutely
for all s with Re(s) >
k
2
+ 1. If f is not a cusp form then L(s, f ) converges absolutely
for all s with Re(s) > k.
L(s, f ) satisfies the functional equation,
k
(2π)−s Γ(s)L(s, f ) = (−1) 2 (2π)−(k−s) Γ(k − s)L(k − s, f ).
4.4.1
Mellin Transform
If f (τ ) =
P∞
n=1
an q n ∈ Sk (Γ1 (N ). As T ∈ Γ1 (N ), we have a q-expansion rather than
qN , we set
Z
g(s) :=
i∞
f (z)z s−1 dz.
0
35
4 Theory of Modular forms
This integral is known as the Mellin transform of the function f . It converges for
P
an
n
c
Re s > c + 1 since for f (z) = ∞
n=1 an q , if |an | = O(n ) then | nc | < ∞ as n → ∞.
We conclude this section by stating the Hecke’s Lemma for Dirichlet L-functions.
Theorem 4.4.2 (J2). (Hecke’s Lemma by Weil’s and Razar) Let F (τ ) =
P
qn
G(τ ) = ∞
n=1 bn λ ,
P∞
n=0
n
an qλ ,
∞
a0 z k−1
2πi −(k−1) X an qn
f (τ ) =
+(
)
eλ,
k−1
(k − 1)!
λ
n
n=1
−(k−1) X
∞
2πi
bn q n
b0 τ k−1
+
eλ.
g(τ ) =
k−1
(k − 1)!
λ
n
n=1
P∞ an Let L(F, s) = n=1 ns be the associated L− function for F . Here k is a positive
), then
integer and γ ∈ C such that F (z) = γ(τ )−k G( −1
τ
f (τ ) − γτ
k−2
g
−1
τ
=
−(k−1−j)
k−2
X
L(F, k − 1 − j) 2πi
j!
j=0
36
λ
τ j.
Chapter 5
Irrationality Proofs using Modular
Forms
In this chapter we discuss Beuker’s proof of irrationality of ζ(3) using modular forms.
P
n
Let t(q) = ∞
n=0 tn q be a power series convergent for all | q |< 1. Let w(q) be another
analytic function on |q| < 1. Our aim is to study w as a function of t. In order to
make it a single-valued function, assume t0 = 0, t1 6= 0. Let q(t) be the inverse of
t(q) in a neighbourhood of 0 and q(0) = 0. Let w(q(t)) be defined for the value of w
around t = 0. We will be interested in determining the radius of convergence of the
P
n
power series w(q(t)) = ∞
n=0 wn t .
5.1
Preliminaries
Definition 5.1.1. Let t(q) =
P∞
n=0 tn q
n
. x is a branching value of t, if either x is
not in the image of t, or if t0 (y) = 0 for some y with t(y) = x.
Let us assume, that t has a discrete set of branching values, say t1 , t2 , ... where
ti 6= 0 ∀ i and |t1 | < |t2 | < ... and so the radius of convergence is |t1 |. We shall
now extend w(q(t) by analytic continuation to get a bigger radius of convergence. If
we continue w(q(t)) analytically to the disc |t| < |t2 | with exception of the possible
isolated singularity t1 and it remains bounded around t1 we can conclude that the
radius of convergence is at least |t2 |. Thus, we can continue doing so until we have a
large radius of convergence. The point of having a large radius of convergence is used
in the following theorem.
37
5 Irrationality Proofs using Modular Forms
Theorem 5.1.1. Let f0 (t), f1 (t), ..., fk (t) be power series in t. Suppose that for any
n ∈ N, i = 0, 1, ..., k the n-th coefficient in the Taylor series of fi is rational and has
denominator dividing dn dn r where r, d are certain fixed positive integers and dn is the
lowest common multiple of 1, 2, ..., n. Suppose there exist real numbers θ1 , θ2 , ..., θk
such that f0 (t) + θ1 f1 (t) + θ2 f2 (t) + ... + θk fk (t) has radius of convergence ρ and
infinitely many non zero Taylor coefficients. If ρ > der , then at least one of θ1 , ..., θk
is irrational.
Proof. As ρ > der , there exists an > 0 such that ρ − > der(1+) . Let fi (t) =
P∞
pin
n
, then qin |dn dn r . Since the radius of
n=0 ain t . From the hypothesis, if ain =
qin
convergence of fo (t) + θ1 f1 (t) + ... + θk fk (t) is ρ, we have for sufficiently large n,
|a0n + a1n θ1 + ... + akn θk | < (ρ − )−n .
Suppose θ1 , ..., θk are all rational and have common denominator D. Then,
An = Ddn dn r |a0n + a1n θ1 + ... + akn θk |
is an integer and
|An | < Ddn dn r (ρ − )−n
for sufficiently large n. But for such n dn < e(1+)n . Hence we get,
| An |< Ddn e(1+)nr (ρ − )−n .
Since,
der(1+)r (ρ − )−1 < 1
implies that An = 0 for sufficiently large n, which is a contradiction that An 6= 0 for
infinitely many n. Thus, at least one of θ1 , ..., θk is irrational.
The values for which irrationality results are obtained are integral points of Dirichlet series associated to modular forms.
38
5 Irrationality Proofs using Modular Forms
Proposition 5.1.2. Let F (τ ) =
P∞
n=1
an q n be a Fourier series convergent for | q |< 1
such that for some k, N ∈ N,
F
−1
Nτ
√ k
= (−iτ N ) F (τ ),
where = ±1. Let f (τ ) be the Fourier series
f (τ ) =
∞
X
an n
q .
k−1
n
n=1
Let
L(F, s) =
∞
X
an
n=1
ns
and
h(τ ) = f (τ ) −
L(F, k − r − 1)
(2πiτ )r .
r!
X
0≤r≤ k−2
2
Then
k−1
h(τ ) − D = (−1)
√
k−2
(−iτ N )
(k
2 −1)
)
where D = 0 if k is odd and D = L(F, k2 ) (2πiτk −1
( 2 )!
if = −1.
f (τ ) − (−1)
√ k−2
(−iτ N ) f
−1
Nτ
=
h
−1
Nτ
if k is even. Moreover L(F, k2 ) = 0
Proof. Applying the lemma of Hecke, with G(τ ) = k−1
F (τ )
√ k
(i N )
to obtain,
k−2
X
L(F, k − r − 1)
r=0
r!
(2πiτ )r .
Splitting the summation on the right hand side into summations over r <
r>
k
2
− 1 and r =
k
2
r!
− 1,
− 1. We thus write,
k−2
X
L(F, k − r − 1)
r=0
k
2
k−4
(2πiτ )r =
2
X
L(F, k − r − 1)
r=0
k−2
X
+
r= k2
39
r!
(2πiτ )r +
L(F, k − r − 1)
(2πiτ )r .
r!
L(F, k2 )
( k−2
)!
2
5 Irrationality Proofs using Modular Forms
Now using the functional equation of the Dirichlet L-function for F
√ !s
N
Λ(s) = CΛ(k − s), where Λ(s) =
Γ(s)L(f, s).
2π
For the region r >
k
2
− 1, putting s = r + 1 we get,
√ !r+1
√ !k−r−1
N
N
r!L(F, r + 1) = C
(k − r − 1)!L(F, k − r − 1).
2π
2π
√ k−2 −1 k−r−2
L(F, k − r − 1)
L(F, r + 1)
k
= (−1) (−i N )
.
⇒
(2πi)k−2r−2
r!
N
(k − r − 2)!
Now substitute r by k − 2 − r.
5.2
The group Γ1(6) and the irrationality of ζ(3)
Define a modular function on Γ1 (6),
η(6τ )8 η(τ )4
η(2τ )8 η(3τ )4
y(τ ) =
where,
η(τ ) = q
1
24
∞
Y
(1 − q n ).
n=1
Using
1
1
1
y(0) = , y( ) = 1, y( ) = ∞, y(∞) = 0,
9
3
2
we find the function y( −1
) is invariant on Γ1 (6). From this one derives
6τ
y
−1
6τ
=
y(τ ) − 19
.
y(τ ) − 1
)
Remark 5.2.1. t(τ ) = y(τ ) 1−9y(τ
is invariant under the involution τ →
1−y(τ )
Since
t
−1
6τ
=y
−1
6τ
1 − 9y( −1
)
6τ
−1 .
1 − y( 6τ )
Now replace y( −1
) by the identity described to get the stated result.
6τ
40
(5.2.1)
−1
.
6τ
5 Irrationality Proofs using Modular Forms
Remark 5.2.2.
t(τ ) =
∆(6τ )∆(τ )
∆(3τ )∆(2τ )
12
.
Proof. Write,
t(τ ) = 9y(τ )y
Use, η
−1
τ
=
pτ
i
−1
6τ
η(6τ )8 η(τ )4
η(2τ )8 η(3τ )4
=9
!
η( −1
) η( −1
)
τ
6τ
8
4
8
4
) η( −3
)
η( −2
6τ
6τ
!
.
η(τ ). Hence, we have
y
−1
6τ
=
1 η(τ )8 η(6τ )4
.
9 η(3τ )8 η(2τ )4
Now, ∆(τ ) = η(τ )24 so
η(6τ )8 η(τ )4
η(τ )8 η(6τ )4
t(τ ) =
×
=
η(2τ )8 η(3τ )4 η(3τ )8 η(2τ )4
η(6τ )η(τ )
η(2τ )η(3τ )
12
.
Remark 5.2.3. The function, defined above is modular with respect to Γ1 (6), invariant
under τ →
1
6τ
and its zeros and poles coincide with those of t(τ ).
√
√
4
Proposition 5.2.1. The function t(i∞) = 0, t(i 6) = ( 2 − 1) , t 52 +
√
4
( 2 + 1) , t( 12 ) = ∞.
i
√
5 6
)
Proof. By definition t(τ ) = y(τ ) 1−9y(τ
. So, y(∞) = 0 ⇒ t(i∞) = 0. And, y
1−y(τ )
∞ ⇒ t 12 = ∞. For τ = √i6 and y0 = y √i6 we find,
y
hence y0 = 1 ±
√
2 2
3
i
√
6
y
√i
6
−
y
√i
6
−1
=
and correspondingly t
√i
6
1
2
=
=
1
9
√
4
= ( 2 ± 1) . To decide which sign
should be taken, one can estimate by obtaining the values numerically.
Using the above functions we show the irrationality of ζ(3).
41
5 Irrationality Proofs using Modular Forms
Theorem 5.2.2. ζ(3) is irrational.
Proof. Construct functions
F (τ ) =
E(τ ) =
1
(E4 (τ ) − 36E4 (36τ ) − 7 (4E4 (2τ ) − 9E4 (3τ ))) ,
40
1
(−5 (E2 (τ ) − 6E2 (6τ )) + 2E2 (2τ ) − 3E2 (3τ ))
24
where
E4 (τ ) = 1 + 240
∞
X
σ3 (n)q n ; E2 (τ ) = 1 − 24
n=1
∞
X
σn q n .
n=1
We note that F (τ ) ∈ M4 (τ (6)) and
√ k
−1
−1
F
= −iτ N F (τ ) ⇒ F
= −36τ 4 F (τ ).
Nτ
6τ
Also
E(τ ) ∈ M2 (Γ1 (6)), E
−1
6τ
= −6τ 2 E(τ ).
The Dirichlet series corresponding to F (τ ) is
∞ X
6σ3 (n)
6σ3 (n)
6σ3 (n)
6σ3 (n)
− 36
− 28
+ 63
L(F, s) =
.
ns
(6n)s
(2n)s
(3n)s
n=1
1
(E4 (τ ) − 36E4 (36τ ) − 7 (4E4 (2τ ) − 9E4 (3τ )))
40
!!
∞
∞
X
X
1
=
1 + 240
σ3 (n)q n − 36 1 + 240
σ3 (n)q 36n
40
n=1
n=1
!
!!
∞
∞
X
X
−7
4 1 + 240
σ3 (n)q 2n − 9 1 + 240
σ3 (n)q 3n
40
n=1
n=1
F (τ ) =
= 6
∞
X
n=1
n
σ3 (n)q − 36 × 6
∞
X
σ3 (n)q
36n
− 28 × 6
n=1
∞
X
σ3 (n)q
2n
+ 63 × 6
n=1
n=1
Dirichlet series in the form of zeta function is
L(F, s) = 6(1 − 62−s − 7.22−s + 7.32−s )
∞
X
σ3 (n)
1
ns
= 6(1 − 62−s − 7.22−s + 7.32−s )ζ(s)ζ(s − 3).
42
∞
X
σ3 (n)q 3n .
5 Irrationality Proofs using Modular Forms
Define f (τ ) by
d
dτ
3
f (τ ) = (2πi)3 F (τ ) ; f (i∞) = 0.
Comparing with the hypothesis of Proposition 5.1.4 here k = 4 and r = 0.
h(τ ) = f (τ ) − L(F, 3) − L(F, 2)(2πiτ )
√ 2
−1
− L(F, 3) .
= −(−iτ 6) f
6τ
−1
2
⇒ f (τ ) − L(F, 3) = −6τ f
− L(F, 3) .
6τ
ζ(3)ζ(0) = ζ(3)
Since L(F, 3) = 6 × −1
3
−1
6τ f
− ζ(3) = − (f (τ ) − ζ(3)) .
6τ
Multiplying the above equation by E −1
= −6τ 2 E(τ ) we get,
6τ
2
E
−1
6τ
−1
f
− ζ(3) = E(τ ) (f (τ ) − ζ(3)) .
6τ
(1)
In general, the function E(τ )(f (τ ) − ζ(3)) can be a multivalued function of t = t(τ ).
t(τ ) = q
∞
Y
−12
12
(1 − q 6n+1 ) (1 − q 6n+5 )
= q − 12q 2 + 66q 3 − 220q 4 + 495q 5 − · · ·
n=1
The inverse function expansion is
q = t + 12t2 + 222t3 + · · ·
We get
E(t) = 1 + 5t + 73t2 + 1445t3 + · · · .
P
3
n
and E(t) ∈ Z[t]. We also have E(t)f (t) = ∞
n=1 an t where an ∈ Z/dn . The function
√
√
4
4
i
0 √i
√
t → τ has a branch point at t = ( 2 − 1) since t 6 = ( 2 − 1) and t
= 0.
6
√
4
We can expect the radius of convergence of E(t)(f (t) − ζ(3)) to be ( 2 − 1) but by
√ 4
equation (1), the function t → E(t)(f (t) − ζ(3)) has no branch point at (1 + 2) ,
43
5 Irrationality Proofs using Modular Forms
and thus its radius of convergence, ρ equals at least the next branching value, which
√ 4
√ 4
2
i
√
is t = (1 + 2) (since t 5 + 5 6 = (1 + 2) ). Thus the radius of convergence
ρ > e3 .
Remark 5.2.4. The function,
E(t)(f (t) − ζ(3))
is not a polynomial since otherwise weight of E(τ ) being 2, weight of f (τ ) − ζ(3)
would be −2 which is not possible.
Apply Theorem 5.1.3 with the two Taylor series,
f0 (t) = E(t)f (t) and f1 (t) = ζ(3)f (t).
The n-th coefficient in the Taylor series of fi is rational with denominator dividing
dn 3 . Also, the radius of convergence is larger than e3 and since E(t)(f (t) − ζ(3)) is
not a polynomial, it has infinitely many nonzero Taylor coefficients. Therefore from
Theorem 5.1.3 we have at least one of θ1 = 1 or θ2 = ζ(3) is irrational. But 1 is
rational, hence ζ(3) is irrational.
5.3
Irrationality of L-functions using modular forms
In this section we prove the irrationality of some combinations of L-functions.
Theorem 5.3.1. Let F (τ ) = η(τ )2 η(2τ )2 η(3τ )2 η(6τ )2 and L(F, s) the corresponding
Dirichlet series. Then at least one of the two numbers,
π −2 L(F, 2) and L(F, 3) +
is irrational.
44
47L(F, 2)ζ(3)
48π 2
5 Irrationality Proofs using Modular Forms
Proof. The function F (τ ) ∈ M4 (Γ1 (6)) and it is a cusp form. From the functional
equation satisfied by the Dadekind-eta function we get,
−1
6τ
F
= 36τ 4 F (τ ).
d 3
f (τ )
dτ
Let f (τ ) be defined as the Fourier series such that
= (2πi)3 F (τ ) and
f (i∞) = 0. Apply Proposition 5.1.4 for h(τ ) = f (τ ) − L(F, 3). Observe that
√
−1
f (τ ) − L(F, 3) − L(F, 2)(2πiτ ) = (−iτ 6) f
− L(F, 3) .
6τ
−1
2
⇒ 6τ f
− L(F, 3) = f (τ ) − L(F, 3) − L(F, 2)(2πiτ ).
6τ
2
Define the function G(τ ) such that
240G(τ ) = 13(E4 (τ ) + 36E4 (6τ ) − 37(4E4 (2τ ) + 9E4 (3τ )).
Note that this function has the properties G(i∞) = 0, G
−1
6τ
= 36τ 4 G(τ ). In order
to evaluate the corresponding Dirichlet series for G(τ ) write the q-expansion of G(τ ).
G(τ ) = 13
∞
X
n
σ3 (n)q + 13 × 36
n=1
∞
X
σ3 (n)q
6n
− 148
n=1
∞
X
σ3 (n)q
2n
− 333
n=1
∞
X
n=1
Thus the L-series is
2−s
L(G, s) = 13 + 13.6
2−s
− 37.2
2−s
− 37.3
∞
X
σ3 (n)
.
ns
n=1
⇒ L(G, s) = 13 + 13.62−s − 37.22−s − 37.32−s ζ(s)ζ(s − 3)
Hence,
L(G, 3) =
47
ζ(3) ; L(G, 2) = −48ζ(2).
6
From Proposition 5.0.4
2
6τ (g
−1
6τ
− L(G, 3)) = g(τ ) − L(G, 3) − L(G, 2)(2πiτ ),
45
(2)
σ3 (n)q 3n .
5 Irrationality Proofs using Modular Forms
and so
−1
47
47
6τ g
− ζ(3) = g(τ ) − ζ(3) + 48ζ(2)(2πiτ ).
6τ
6
6
2
(3)
The function,
47
h(τ ) = 48ζ(2)(f (τ ) − L(F, 3)) + L(F, 2) g(τ ) − ζ(3)
6
satisfies the equation,
2
6τ h
−1
6τ
= h(τ ).
This is obtained by eliminating 2πiτ from equations (2) and (3). Consider the function
E(τ ) = E2 (τ ) − 2E2 (2τ ) + 6E2 (6τ ) − 3E2 (3τ ).
We see, E(τ ) ∈ M2 (Γ1 (6)) and E
E
−1
6τ
−1
6τ
= 6τ 2 E(τ ). Consequently,
−1
h
= E(τ )h(τ ).
6τ
Repeating the arguments of the previous theorem we find
48ζ(2)f (t)E(t) + L(F, 2)g(t)E(t) − (48ζ(2)L(F, 3) + L(F, 2)
47
ζ(3))E(t)
6
√
4
is a power series in t with radius of convergence ( 2 + 1) .
In order to apply Theorem 5.1.3 we have the Taylor series, f0 (t) = E(t)f (t), f1 (t) =
E(t)g(t), f2 (t) = E(t) where the n-th coefficient divides dn 3 . Thus the hypothesis of
Theorem 5.1.3 are satisfied. On substituting the value of ζ(2) we get, at least one of
the numbers
−2
π L(F, 2) ; L(F, 3) +
is irrational.
46
47L(F, 2)ζ(3)
48π 2
Chapter 6
Congruence Properties of Apéry numbers
The Apéry numbers an , bn are defined by the finite sums,
an =
2
n X
n+k
n
k=0
k
k
, bn =
2 2
n X
n+k
n
k=0
k
k
for every integer n ≥ 0. These numbers were introduced by R. Apéry[1978] in proving
irrationality of ζ(3) and ζ(2). Hence these numbers are called Apéry numbers. The
first few Apéry numbers, an 0 s and bn 0 s are
1, 5, 73, 1445, 33001 · · ·
and
1, 3, 19, 147, 1251 · · ·
respectively. The Apéry numbers satisfy some congruence properties. Chowla, Cowles
and Cowles[J4] were the first to consider congruences for a0n s. They proved
1. for n ≥ 0, a5n+1 ≡ 0 (mod 5) and a5n+3 ≡ 0 (mod 5).
2. for all primes p, ap ≡ a1 (mod p3 ).
Beuker further found some congruences for a0n s. The significance of these congruence
properties and the reason for the Apéry numbers to satisfy these congruences is still
unknown and is believed to be due to the interplay between the Apéry numbers and
the ζ function of a certain algebraic K-3 surface. In this chapter we will discuss the
following congruence properties of the Apéry numbers.
47
6 Congruence Properties of Apéry numbers
1. For m, r ∈ N and p > 3 prime. Then for p ≥ 3 prime,
ampr −1 ≡ ampr−1 −1
(mod p3r ), bmpr −1 ≡ bmpr−1 −1
(mod p3r ).
2. Let
4
4
f (τ ) = η(2τ ) η(2τ ) =
∞
X
γn q n
n=1
where η(τ ) is the Dedekind eta function. f (τ ) is a modular form of weight 4
for Γ0 (4). Then
a (p−1) ≡ γp
(mod p).
2
Remark 6.0.1. Beuker conjectured that the congruence
a (p−1) ≡ γp
(mod p2 )
2
also holds true for all primes p ≥ 2. This conjecture is true and is proved by Ahlgren
and Ono[J9].
6.1
Congruences of Apéry numbers
Theorem 6.1.1. Let m, r ∈ N and p > 3 prime. Then
ampr −1 ≡ ampr−1 −1
(mod p3r ).
Before proving this we will prove a weaker version of Theorem 6.1.1 which holds
for all primes.
Theorem 6.1.2. Let m, r ∈ N and let p be any prime. Then
bmpr −1 ≡ bmpr−1 −1
(mod pr ), ampr −1 ≡ ampr−1 −1
In order to prove this we require the congruences
48
(mod pr ).
6 Congruence Properties of Apéry numbers
Lemma 6.1.3.
2.
mpr
kp
3.
mpr −1
k
≡
1.
mpr−1
k
≡
mpr
k
≡ 0 (mod pr ) ∀m ∈ Z, k, r ∈ N, k 6≡ 0 (mod p),
(−1)k(p−1) (mod pr ) ∀m ∈ Z, k, r ∈ N,
mpr−1 −1
[k/p]
(−1)k−[k/p] (mod pr ) ∀m ∈ Z, k, r ∈ N.
r
r
r mpr −1
. Now mp
and
Proof. Congruence (1) is easy to show since mp
= mp
k
k−1
k
k
r
mpr −1
are integers thus mp
is also an integer but p does not divide k and so k|m
k−1
k
hence (pr k)|(mpr ). Finally we get
mpr
k
(mod pr ).
≡0
To prove the second congruence one proceeds by proving first
r
(1 − X)mp ≡ (1 − X p )mp
r−1
(mod pr )
by induction on r. For r = 1 it is trivial since
m
(1 − X)mp = ((1 − X)p )
m
p(p + 1) 2
p
X − ... − X
=
1 − pX +
2
≡ (1 − X p )m
(mod p).
Let the congruence be true for r = k. Now for r = k + 1,
(1 − X)mp
k+1
k
p
= ((1 − X)mp )
k−1 p
≡ (1 − X p )mp
= (1 − X p )mp
k
(mod pk+1 )
(mod pk+1 ),
where the equivalence in Step 2 follows from induction. Thus we have,
r
(1 − X)mp ≡ (1 − X p )mp
49
r−1
(mod pr ).
6 Congruence Properties of Apéry numbers
We then equate the coefficient of X kp and get
r r−1 mp
kp mp
(−1)
≡
(−1)k
kp
k
(mod pr ).
Hence we get congruence (2). In order to prove congruence (3), we compare the
coefficients of X k on both sides
(1 − X)mp
r −1
≡ (1 − X p )mp
r−1 −1
≡ (1 − X p )mp
r−1 −1
1 − Xp
1−X
(1 + X + ... + X p−1 )
(mod pr ).
We get
r−1
mpr − 1
mp
−1
(−1)
≡
(mod pr )
k
[ kp ]
r
r−1
mp − 1
mp
−1
k−[ kp ]
⇒
≡
(−1)
(mod pr ).
k
k
[p]
k
Now we shall prove the congruence established in Theorem 6.1.2 using the lemma
above.
Proof. Notice that
n+k
k
= (−1)k
−n−1
k
we get,
mpr −1 ampr −1
r
2
mpr − 1 + k
mp − 1
=
k
k
k=0
mpr −1 X mpr − 12 −mpr =
(−1)k .
k
k
k=0
X
−mpr
k
Since by congruence(1) for p - k,
≡ 0 (mod pr ), we have for p - k,
ampr −1 ≡ 0
(mod pr ).
The remaining sum is over all k from 0 to mpr − 1 for which p|k and so putting k = lp
we get
mpr−1 −1 ampr −1 ≡
X
l=0
2 mpr − 1
−mpr
(−1)lp
lp
lp
50
(mod pr ).
6 Congruence Properties of Apéry numbers
Using congruences (2) and (3) we obtain
mpr−1 −1 ampr −1
2 mpr−1 − 1
−mpr−1
(−1)lp (−1)l(p−1) (mod pr )
≡
l
l
l=0
2 mpr−1 −1 X
mpr−1 − 1
−mpr−1
(−1)l (mod pr )
≡
l
l
l=0
2 r−1
mpr−1 −1 X
mpr−1 − 1
mp
−1+l
(mod pr )
≡
l
l
l=0
X
≡ ampr−1 −1
(mod pr ).
Similarly we prove congruences for bn ’s.
In order to prove Theorem 6.1.1, we shall make use of the following congruences
which we list down in the form of lemmas and propositions. The notations used in
the lemma are:
p is a fixed prime ≥ 5
r is a fixed natural number
[x] is the largest integer not exceeding x
{x} is x − [x]
Q0
P0
k∈V or
k∈V means that we take the product or sum over those values k ∈ V for
which p - k
vp (n) is the number of prime factors p in n.
Lemma 6.1.4. For any l ∈ Z we have
0
X 1
≡0
λ
−r
[λp
(mod p2r ).
]=l
Proof. Observe that
0
pr (l+1)−1 1 X
1
1
1 X
1
1
+
=
+
.
2 −r
λ (2l + 1)pr − λ
2 λ=pr l
λ (2l + 1)pr − λ
[λp
]=l
p-λ
51
6 Congruence Properties of Apéry numbers
On expanding the summand term by term we get
pr (l+1)−1
X
λ=pr l
p-λ
1
=
λ
But
pr (l+1)−1
X
λ=pr l
p-λ
1
.
(2l + 1)pr − λ
r
0
X 1 p (l+1)−1
X 1
=
.
λ
λ
r
r
λ=p l
[λp ]=l
p-λ
Thus we have,
0
0
X 1
1 X
1
1
=
+
λ
2 −r
λ (2l + 1)pr − λ
[λpr ]=l
[λp ]=l
!!
0
1 X
1
1
1
=
−
2 r
λ λ 1 − (2l+1)pr
λ
[λp ]=l
0
1 X 1
1
(2l + 1)pr
=
−
+ ...
1+
2 r
λ λ
λ
[λp ]=l
0
!
1 (2l + 1)pr (2l + 1)2 p2r
1
− −
−
.....
λ λ
λ2
λ3
1 X
=
2 r
[λp ]=l
r
p −1
(2l + 1)pr X 1
≡ −
≡0
2
λ2
λ=1
(mod p2r ).
p-λ
Hence
0
X 1
≡0
λ
−r
[λp
(mod p2r ).
]=l
Lemma 6.1.5. For any m ∈ Z, k ∈ N we have
1.
r−1
k
mpr − 1
mp
− 1 Y mpr − λ
=
k
[k/p]
λ
λ=1
p-λ
52
6 Congruence Properties of Apéry numbers
2. If p | k, then
mpr
k
=
k
mpr−1 Y mpr − λ
.
k
λ
p
λ=1
p-λ
Proof. (Proof of (1)) The idea is to write the binomial coefficient
mpr −1
k
in terms of
products and then split the product into products with p - λ and one with λ = µp,
i.e.,
r
mp − 1
(mpr − 1)(mpr − 2)...(mpr − k)
=
k
k!
k
Y mpr − λ
=
λ
λ=1
[k]
p
k
Y
mpr − λ Y mpr − µp
=
λ
µp
µ=1
λ=1
p-λ
λ=µp
[k]
=
p
k
Y
mpr − λ Y mpr−1 − µ
λ=1
p-λ
λ
µ
µ=1
k
Y
mpr − λ mpr−1 − 1
.
=
λ
[ kp ]
λ=1
p-λ
Hence part (1) is proved.
Proof. (Proof of (2)) We will again proceed by a similar method as in part (1) by
first breaking the summand into products
r
mp
mpr mpr − 1
=
k
k
k−1
k−1
mpr−1 mpr−1 − 1 Y mpr − λ
=
.
k
[ k−1
λ
]
p
p
λ=1
p-λ
]=
Note that [ k−1
p
k
p
− 1 and
k−1
Y
λ=1
p-λ
=
k
Y
λ=1
p-λ
53
.
6 Congruence Properties of Apéry numbers
Hence we get,
k−1
r−1 Y
k
mpr − λ
mpr−1 mpr−1 − 1 Y mpr − λ
mp
=
.
k
k
]
[ k−1
λ
λ
p
p
p
λ=1
λ=1
p-λ
p-λ
Thus we have proved part (2) of the lemma
r r−1 Y
k
mp
mp
mpr − λ
.
=
k
λ
k
p
λ=1
p-λ
Lemma 6.1.6. Let ak ∈ Zp (k = 0, 1, 2, 3...) be such that
X
ak ≡ 0
(mod ps ) for any s, l ∈ Z≥0 .
[kp−s ]=l
Let e ∈ N. Then,
1.
X
[kp−r ]=l
r
2 e
mp − 1
−mpr − 1
ak
(−1)ke ≡ 0
k
k
(mod pr ) ∀ l, m ∈ Z≥0 .
(1)
If in addition ak = 0 for all k ≡ 0 (mod p), then
2.
X
[kp−r ]=l
r
2 e
mp − 1
−mpr − 1
ak
(−1)ke ≡ 0
k
k−1
(mod pr ) ∀ l, m ∈ Z≥0 .
(2)
Proof. (Proof of (1)) We prove using induction on r. For r = 0 it is trivial. Assume
it to be true for r − 1. We then show that the right hand side of (1) which we denote
by A, is zero mod pr . To prove this we will apply
r
r−1
k
mp − 1
mp
−1
=
(−1){ p }p ,
k
k
[p]
54
m ∈ Z.
6 Congruence Properties of Apéry numbers
This equality can be derived from Lemma (1.2.3) part (3). Using this we get,
r
2 e
X
mp − 1
−mpr − 1
ak
(−1)ke
k
k
−r
[kp
X
=
[kp−r ]=l
]=l
2 e
r
k
−mpr−1 − 1
mp − 1
ak
(−1)ke (−1)(k−[ p ])e
k
k
[p]
[p]
(mod pr )
Collecting all such terms which have the same [ kp ] from the summand above we get,


r−1
2 e
X
X
mp
−1
−mpr−1 − 1


A≡
ak
(−1)en (mod pr )
n
n
−r+1
−1
[np
]=l
[kp
]=n
Applying induction hypothesis for r − 1 with the new coefficients
a¯n =
1
p
X
ak .
[kp−1 ]=n
a¯n satisfy the hypothesis of our lemma since
X
a¯n =
[np−s ]=l
1
p
X
ak ≡ 0
(mod ps ) ∀ s, l ∈ Z≥0 .
[kp−s−1 ]=l
We have
X
an =
[np−s ]=l
1
p
X
ak =
[kp−1 ]=n
1 X
1
ak =
p k
p
[
]=l
ps+1
X
≡0
(mod ps ) ∀ s, l ∈ Z
[kp−s−1 ]=l
and so we obtain,
A=
X
[np−r+1 ]=l
2 e
mpr−1 − 1
−mpr−1 − 1
pa¯n
(−1)en ≡
n
n
(mod pr ).
Proof. (Proof of (2)) To prove part (2) we apply induction to the left hand side of
the congruence which is
X
[kp−r ]=l
r
2 e
mp − 1
−mpr − 1
ak
(−1)ke
k
k−1
and put [ k−1
] = [ kp ] since the sum is over (k, p) = 1. Then we can proceed similarly
p
as in the proof of Lemma 6.1.6 (part 1).
55
6 Congruence Properties of Apéry numbers
We state some results used to prove Theorem 6.1.1 in the form of a lemma.
1. Let s, t ∈ Z≥0 . Then,
Lemma 6.1.7.
(a)

0
X
X

[np−t ]=i
[kp−s ]=n


n
X
1 
1
≡0
k
λ
0
(mod pt+2s ) ∀ i ∈ Z.
λ=1
(b)
s
0
kp
X 1X
1
≡0
k
λ
0
−t
[kp
]=i
(mod pt+s ) ∀ i ∈ Z.
λ=1
2. For any l ∈ N ∪ 0, m ∈ N we have,
0
X 1 mpr − 12 −mpr − 1
(−1)k ≡ 0
k
k
k−1
−r
[kp
(mod p2r ).
]=l
With the aid of the above lemmas we will now prove Theorem 6.1.1 on congruences satisfied by Apéry numbers:
ampr −1 ≡ ampr−1 −1
(mod p3r ).
Proof. (Proof of Theorem 6.1.1)
We begin by writing
r −1 2 2 mp
2 2
X
mpr − 1
−mpr
mpr − 1
−mpr
+
.
lp
lp
k
k
k=0
mpr−1 −1 ampr −1 =
X
l=0
p-k
Then we observe
mpr −1 X
k=0
p-k
r −1
2 2
2 2 mp
X
−mpr − 1
mpr − 1
−mpr
m2 p2r mpr − 1
.
=
2
k
k
−
1
k
k
k
k=0
p-k
Since,
0
X
[kp−s ]=n
1
≡0
k2
(mod ps ) ∀ n ∈ Z, s ∈ N,
56
(3)
6 Congruence Properties of Apéry numbers
we apply Lemma 6.1.6 part (2) with ak = k −2 if p - k and ak = 0 otherwise and
conclude that expression (3) is zero mod p3r . Hence
mpr−1 −1 ampr −1 ≡
X
l=0
2 2
mpr − 1
−mpr
lp
lp
(mod p3r ).
We then apply Lemma 6.1.3 part (2) to get,
mpr−1 −1 bmpr −1 ≡
X
l=0
2 2 lp 2
m2 p2r
mpr−1 − 1
−mpr−1 Y
1−
l
l
λ
λ=1
(mod p3r ).
p-λ
Notice that
lp Y
λ=1
p-λ
m2 p2r
1−
λ
2
lp
X
1
≡ 1 − 2m p
λ2
λ=1
2 2r
(mod p3r ) ≡ 1
(mod pmin(3r,2r+vp (lp) )
p-λ
and
2
2
2r−2
−mpr−1
−mpr−1 − 1
2p
=m
≡0
l
l2
l−1
(mod p2max(r−vp (lp),0) ).
Thus
mpr−1 −1 ampr −1 ≡
X
l=0
2 2
mpr−1 − 1
−mpr−1
≡ ampr−1 −1
l
l
(mod p3r ).
The proof of
bmpr −1 ≡ bmpr−1 −1
(mod p3r )
is also established likewise.
6.2
Congruences of Apéry Numbers relating modular forms
The congruences proved in the preceding section appears to be proven by a brute
force method. The congruences in this section appears to be more interesting as the
proof relates the generating function of a0n s to a certain modular form.
57
6 Congruence Properties of Apéry numbers
Proposition 6.2.1. Let p be a prime and
w(t) =
∞
X
fn tn−1 dt
n=1
a differential form with fn ∈ Zp , the p-adic integers, ∀n. Let t(u) =
Zp , ∀n and suppose
w(t(u)) =
∞
X
P∞
n=1
cn un−1 du.
n=1
Suppose there exist αp , βp ∈ Zp with p | βp such that
fmpr − αp fmpr−1 + βp fmpr−2 ≡ 0
(mod pr ) ∀m, r ∈ N
(4)
cmpr − αp cmpr−1 + βp cmpr−2 ≡ 0
(mod p) ∀m, r ∈ N.
(5)
Then
Moreover, if A1 is a p-adic unit then the congruences (5) imply (4).
Remark 6.2.1. We will use the convention bmps = 0 if mps ∈
/ Z.
Proof. We first observe that the congruence (2) is equivalent to
w(t) −
αp
βp
2
w(tp ) + 2 w(t(up )) = dv(t), dv(t) ∈ Zp [[t]]
p
p
Since,
w(t) =
∞
X
fn tn−1 dt
n=1
⇒ w(tp ) =
∞
X
fn (tp )(n−1) d(tp )
n=1
1
⇒ w(tp ) =
p
∞
X
n=1
∞
X bn
fn np−p p
t
d(t ) =
d(tpn ).
p
np
n=1
But we also have
t(u)np = t(up )n + nphn (u)
58
hn (u) ∈ Zp [[u]],
(6)
An un , An ∈
6 Congruence Properties of Apéry numbers
so substituting t(u)np in the expression for p1 w(tp ) we get,
∞
X
1
fn
p
w(t ) =
d(t(u)np )
p
np
n=1
∞
X
fn
=
d(t(up )n ) + fn dhn (u)
np
n=1
=
1
w(t(up )) + dψ(u),
p
ψ(u) ∈ Zp [[u]].
So now substituting the expression of p1 w(tp ) in (6) we have,
w(t(u)) −
αp
βp
2
w(t(up )) + 2 w(t(up )) = dv(t) + αp dψ(u) := dϕ(u).
p
p
And hence equating the coefficient of mpr from both sides we get,
fmpr − αp fmpr−1 + βp fmpr−2 ≡ 0
(mod pr ) ∀m, r ∈ N
Thus we showed that congruence (4) is equivalent to (6).
The generating function of the Apéry numbers, an satisfies an identity in terms
of products of η(τ ).
Proposition 6.2.2. Let
λ(τ ) =
η(2τ )η(12τ )
η(4τ )η(6τ )
6
Then,
A(λ2 )d(λ) = (η(2τ ))4 (η(4τ ))4 − 9(η(6τ ))4 (η(12τ ))4 dq.
where A(λ) is the generating function of the Apéry numbers an .
Theorem 6.2.3. Let
∞
X
n
γn q = q
n=1
∞
Y
4
4
(1 − q 2n ) (1 − q 4n ) .
n=1
Then for any odd prime p and any m, r ∈ N, m odd, we have
a mpr − 1 − γp a mpr−1 − 1 + p3 a mpr−2 − 1 ≡ 0
2
2
2
59
(mod pr ).
6 Congruence Properties of Apéry numbers
Proof. The function
∞
X
n
γn q = q
n=1
∞
Y
4
4
(1 − q 2n ) (1 − q 4n )
n=1
is the unique cusp form in S4 (Γ0 (8)), its corresponding Dirichlet series has an Euler
product,
∞
X
γn
n=1
ns
=
1
Y
p odd
1 − γp
p−s
+ p3−2s
.
Let
4
4
4
4
η(2τ ) η(4τ ) − 9η(6τ ) η(12τ ) =
∞
X
γ˜n q n .
n=1
Then
∞
X
γ˜n
n=1
ns
=
∞
X
γn − 9γn/3
n=1
ns
X
∞
γn
9
=
1− s
3 n=1 ns
9 Y
1
=
1− s
.
−s
3 p odd 1 − γp p + p3−2s
Multiplying both sides of the above equation by (1 − γp p−s + p3−2s ) and then equating
mpr -th coefficient we get,
For any prime p ≥ 5,
γ̃mpr − γp γ̃mpr−1 + p3 γ̃mpr−2 = 0 ∀ m, r ∈ N.
For p = 3,
γ̃m3r − γ3 γ̃mpr−1 + 27γ̃m3r−2 =
0
if 9 k m3r
−9γm if 3 k m3r , ∀ m, r ∈ N
Hence for all odd primes p,
γ̃mpr − γp γ̃mpr−1 + p3 γ̃mpr−2 ≡ 0
60
(mod pr ) ∀m, r ∈ N
6 Congruence Properties of Apéry numbers
From Proposition 6.1.9,
∞
X
2n
an λ (q)dλ(q) =
n=0
∞
X
γ̃n q
n−1
dq, where λ(q) = q
n=1
6
∞ Y
1 − q 12n+2
n=0
1 − q 12n+10
and Proposition 6.1.8 implies the congruence.
a mpr − 1 − γp a mpr−1 − 1 + p3 a mpr−2 − 1 ≡ 0
2
2
2
61
(mod pr ).
.
Chapter 7
Conclusion
The n-th Apéry numbers
an =
2 2
n X
n+k
n
k=0
k
k
satisfy the recurrence relation
(n + 1)3 un+1 = 34n3 + 51n2 + 27n + 5)un − n3 un−1
with initial values u0 = 1, u1 = 5. Consider the generating function of the Apéry
P
n
0
numbers, A(t) = ∞
n=0 an t . From the recurrence relation for an s, the generating
function A(t) satisfies the third order differential equation.
000
00
0
(t4 − 34t3 + t2 )y + (6t3 − 153t2 + 3t)y + (7t2 − 112t + 1)y + (t − 5)y = 0.
One observes that this differential equation is the symmetric square of a second order
differential equation with coefficients in C(t). A third order differential equation is the
symmetric square of a second order differential equation if the third order equation
is spanned by squares of solutions of second order differential equation.
To test whether a third differential equation is a symmetric square or not, we can
find the second order differential equation whose symmetric square is the third order
equation. This can be achieved by taking a1 =
c2
3
0
and a0 =
c1 −a1 −2a1 2
4
and test if
0
4a0 a1 + 2a0 = c0 .
Using this we find that the differential equation of the generating function of the
Apéry numbers,
000
00
0
(t4 − 34t3 + t2 )y + (6t3 − 153t2 + 3t)y + (7t2 − 112t + 1)y + (t − 5)y = 0
62
7 Conclusion
is the symmetric square of a second order differential equation given by
1
00
0
(t3 − 34t2 + t)y + (2t2 − 51t + 1)y + (t − 10)y = 0.
4
(1)
p
A(t) is the solution of (1). Equation (1) has four singularities at t =
√ 4
0, (1 ± 2) , ∞. Notice that all these singular points are regular singular points and
Thus,
hence the second order differential equation is a Fuchsian equation which is a linear
ordinary differential equation whose coefficients are rational functions having only
regular singular points. The local exponents of a singular point is the solution of the
indicial equation corresponding to the singular point. For the singular point t = 0,
√ 4
the exponents are 0, 0, for (1 ± 2) , the exponents are 0, 12 and for the singular point
1 1
x(1 − 9x)
t = ∞, the local exponents are , . Consider the rational function f (x) =
.
2 2
(1 − x)
√
√ 4
3±2 2
. Moreover every point except
Now, f (x) takes the value (1 ± 2) at x =
3
√ 4
the points (1 ± 2) has two pre images. Thus, we can conclude that the rational
√
√ 4
(3 ± 2 2)
.
function has degree two and ramifies above the points (1 ± 2) at x =
3
1
x(1 − 9x)
Replacing t in equation (2) by
and then a replacing y by (1 − x2 ) 2 y we
1−x
get,
00
0
x(x − 1)(9x − 1)y + (27x2 − 20x + 1)y + (9x − 3)y = 0.
This equation is the Picard-Fuchs equation of the modular family of elliptic curves
associated to Γ1 (6)[J3]. This gives a relation between the proof of irrationality of ζ(3)
and modular forms on Γ1 (6). The relation of irrationality proof with modular forms is
also visible from some of the observations we make from Beukers proof of irratinality
of ζ(3) using modular forms. The coefficients of the Taylor series expansion of E(t)
are indeed the Apéry numbers themselves. Moreover the radius of convergence of the
√
4
Taylor series expansion of E(t)f (t) − ζ(3)E(t) is ( 2 + 1) = α but recall that the
order of the Apéry number bn = O(αn ). Some of these observations have found an
explanation in Beuker’s and Steinstra’s paper [J7].
63
7 Conclusion
The most remarkable results for the arithmetic nature of Riemann zeta function at
odd integers greater than 3 has been obtained by Tanguay Rivoal and others where it
is proved that there are infinitely many irrational values of the Riemann zeta function
at odd positive integers. Infact it was proved by Rivoal that at least one of the
nine numbers ζ(5), ζ(7), ..., ζ(21) is irrational. Later Zudilin improved upon this and
obtained that at least one of the four numbers ζ(5), ζ(7), ζ(9) and ζ(11) is irrational.
The example of ζ(3) in irrationality proofs of Riemann zeta function at other odd
integer values using modular forms has been used extensively without much success.
It will be challenging and interesting to investigate more on the relation of modular
forms with irrationality of ζ(3) and ζ(2) in particular and use it to say something on
the arithmetic nature of other zeta values, ζ(5), ζ(7) · · · in general.
64
References
Journals and Papers (J)
1. F. Beukers, Some congruences for the Apèry Numbers. J. Number Theory 21,
141-155 (1985)
2. F. Beukers, Irrationality proofs using modular forms. Journees arithmetiques
Besancon, Asteriaque, (1985)
3. F. Beukers, C.A.M. Peters, A family of K3 surfaces and ζ(3). J. Reine Angew
Math 351, 188-190 (1980)
4. S. Chowla, J. Cowles, M. Cowles, Congruence properties of Apèry Numbers. J.
Number Theory 12, 188-190 (1980)
5. I. Gessel, Some congruences for the Apèry Numbers. J. number Theory 14,
362-368 (1982)
6. A. J. Van der Poorten, A Proof that Euler missed...Apèry’s proof of irrationality
of ζ(3). Math. Intelligencer 1, 195-203 (1979)
7. J. Steinstra, F. Beukers, On the Picard-Fuchs equation and the formal Brauer
group of certain elliptic K3 surfaces. Math. Annals 271, 269-304 (1985)
8. F. Beukers, Another Congruence for the Apèry Numbers. J. Number Theory
25, 201-210 (1987)
9. S. Ahlgren and K. Ono, A Gaussian hypergeometric series evaluation and Apery
number congruences, J. reine angew. Math. 518, 187-212 (2000)
Books (B)
1. Neal Koblitz, Introduction to Elliptic Curves and modular Forms (SpringerVerlag, New York, 1993)
2. Fred Diamond, Jerry Shurman, A First Course in Modular Forms (SpringerVerlag, New York, 2005)
3. Jean-Pierre Serre, A Course in Arithmetic (Presses Universitaires de France,
Paris, 1970)
65
REFERENCES
4. Bruinie Hendrick Jan. et al., 1-2-3 of modular forms (the): Lectures at a
summer school in Nordfjordeid, Norway (Springer-Verlag, New York, 2008)
5. Mollin A. richard, Advanced number theory with applications (CRC Press, New
York, 2009)
6. Carl Ludwig Siegel, Transcendental numbers (hindustan Book Agency, 2012)
Technical reports (T)
1. Gert Almkvist, Wadim Zudilin ”Differential Equations, Mirror Maps and Zeta
Values“ (E-print math. NT/0402386, December 2004)
2. Frits Beukers, Consequences of Apèry’s work on ζ(3) (May 9, 2003). [the easiest
access to this source is by Internet]
Theses and personal communications (C)
1. Victor Legge, Master Thesis, San Jose State University (2001).
66