THE IRRATIONALITY OF ζ(3) AND APÉRY NUMBERS

Transcription

1 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-202

2 To Maa and Dadi i

3 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 wor 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

4 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 lie 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 nowledge. 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 thans 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 bacground in algebra, analysis and number theory. I would lie to than my classmate Ashish Kumar Pandey who as a good friend was always willing to help and give his best suggestions. Many thans to all my friends for their encouragement and constant support. Last but by no means the least I than my parents and my sister and brother for being there for me always and bestowing their best wishes on me forever. iii

5 Contents Introduction 2 Algebraic and Transcendental numbers 3 2. Algebraic numbers Properties of Algebraic numbers Transcendence of e and π Transcendence of e Transcendence of π Rational Approximations Apéry s Proof of Irrationality of ζ(3) 2 3. Preliminaries Irrationality of ζ(3) Theory of Modular forms SL(2, Z) and its congruence subgroups Modular functions and modular forms q-expansion of a modular form Estimates for the coefficients in q-expansion The Eisenstein series of weight L-functions and Modular forms Mellin Transform Irrationality Proofs using Modular Forms Preliminaries The group Γ (6) and the irrationality of ζ(3) Irrationality of L-functions using modular forms Congruence Properties of Apéry numbers Congruences of Apéry numbers Congruences of Apéry Numbers relating modular forms Conclusion 62 References 65 iv

6 Chapter Introduction The Riemann zeta function, ζ(s) = n s 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 ζ(s) = (2πis)s B s 2s! for s > 0 and s is even and B s are the Bernoulli numbers. This was proved by Euler in 735. 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 Beuer 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 Γ (6). The sequences a n and b n 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 Atin Swinnerton-Dyer congruence relations for certain congruence subgroups [T2]. Apéry s proof is also significant as their relation with

7 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 978. Chapter 4 introduces modular forms. It also discusses the Dedeind 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 Hece s lemma by Weil and Razar. Chapter 5 discusses Beuer s proof of irrationality of ζ(3) using modular forms. This proof is taen from F. Beuers paper on irrationality proofs using modular forms. The last section of the chapter proves the irrationality of certain combinations of L-functions using Hece 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 Beuers of irrationality of ζ(3) and the relation of irrationality proof of ζ(3) with the theory of differential equations. 2

8 Chapter 2 Algebraic and Transcendental numbers 2. Algebraic numbers Definition 2... (Algebraic numbers) A complex number α is called an algebraic number if there exists p(x) Z[x], p 0 with p(α) = 0. Definition (Transcendental numbers) A complex number α is called transcendental if it is not algebraic. Remar 2... 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 set in [0, ] and hence essentially all numbers are transcendental. Example 2... (Algebraic numbers) All rational numbers, square root of any rational number, the cube root of any rational number, numbers of the form r p q where r, p, q Q, i =, Example (Transcendental numbers) e, π, e π, 2 2, ζ(3) Remar The transcendence of the number γ = lim n ( ) n log e(n) 3

9 2 Algebraic and Transcendental numbers (called Euler s constant) is a well nown open question in transcendental number theory. It is not even nown if γ is rational or not. 2.. Properties of Algebraic numbers Theorem 2... The set of algebraic numbers forms a countable set. Theorem (Cantor) The set of all real numbers is uncountable. The proof of the above theorem is a consequence of the following result. Lemma Let S be the set of all binary sequences, i.e. sequences {y i } i Z with y i {0, }. 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, ] whose decimal expansion consists entirely of 0 s and 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 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

10 2 Algebraic and Transcendental numbers Lemma 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 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 If α is an algebraic integer and α is rational, then α is an integer. 2.2 Transcendence of e and π We now 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 0 n! to be transcendental, and this was one of the first numbers proven to be transcendental. Lambert (76), Legendre (794), Hermite (873) and others proved π is irrational. Lindemann (882) proved π transcendental. What about e? Euler(737) proved that e and e 2 are irrational. Liouville(844) proved e is not an algebraic number of degree 2, and Hermite(873) proved e is transcendental. Liouville (85) 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 mae use of the function, I(t) = t 0 e t u f(u)du 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

11 2 Algebraic and Transcendental numbers gives: I(t) = f(u)e t u t 0 + t 0 f (u)e t u du = f(t) + f(0)e t ( f (t) + f (0)e t ) + = e t d f j=0 f j (0) d f j=0 f j (t) t 0 f (u)e t u du where d f is the degree of f also note that if f(x) d f j=0 a jx j, we set f(x) = d f j=0 a j x j. Then, I(t) t 0 e t u f(u) du t max e t u max( f(u) ) t e t f( t ). Hence we get an upper bound for I(t) Transcendence of e Theorem The number e is transcendental Proof. Assume e is algebraic. There is a minimal polynomial g(x) = r b x Z[x], =0 with r g(e) = b e = 0. =0 Let p be a large prime greater than max {r, b 0 } to be chosen suitably and define r f(x) = x p (x j) p. j= 6

12 2 Algebraic and Transcendental numbers The degree of f is n = (r + )p. Consider J = b 0 I(0) + b I() b r I(r) = = = ( r b =0 e n f j (0) j=0 = r =0 j=0 n f j (0) j=0 r b e =0 n b f j () n = b 0 f j (0) j=0 r = r r b I() =0 ) n f j () j=0 =0 j=0 ( p n b f j () b f j () j=0 ) n b f j (). Here we have used the fact that the degree of f is (r + )p. computation we get, 0 if j p 2 f j (0) = (p )!( ) p...( n) p if j = p 0 (mod p!) if j p. For r, f j () = j=p { 0 if j p 0 (mod p!) if j p. With a simple Now since p > b 0 the first summand is completely divisible by (p )! and the second summand is divisible by p! hence by (p )! thus we find that J is an integer which is divisible by (p )! but not p and Since f(x) = x p r j= (x j)p we get f() p J (p )!. r ( + j) p (2r) n < (2r) 2rp j= for 0 r. From this we deduce that, r r J b j I(j) b j je j f(j) r(r + )Ce r (2r) 2rp < K p, j=0 j=0 7

13 2 Algebraic and Transcendental numbers where C = max r { b }, and K = r(r + )Ce r (2r) 2r, which is a constant not depending on p. But J is bounded below by (p )!, thus (p )! J K p, but p has been chosen arbitrarily large and this inequality bounds p hence we get a contradiction. Thus, e is transcendental Transcendence of π Theorem 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 = θ, θ 2,..., θ r denote the algebraic conjugates of θ. Let b be the leading coefficient of g(x) (so bθ j is an algebraic integer). Since e θ = e iπ =, called the Euler s identity for e, so ( + e θ )( + e θ 2 )...( + e θr ) = 0. Rewriting this expression into an expression containing 2 r number of terms we get, r ( + e θ j ) = j= ρ= P r j= b jθ j b j {0,} Let θ,..., θ n denote the non-zero expressions of the form e ρ so that e ρ. 2 r n + e θ + e θ e θn = 0. Let f(x) = b np x p n i= (x θ i ) p where p is an arbitrarily chosen large prime. Consider θ, θ 2,..., θ 2 r first n of which are non zero. Use 2 r (x θ j ) = x 2r n j= j= n (x θ j ) 8

14 2 Algebraic and Transcendental numbers is symmetric in θ,..., θ n thus by the fundamental theorem of elementary symmetric functions and with the use of Lemmas 2..5 and 2..6 we can chec f(x) Z. Define n J = I(θ ) where I(t) = e t m = f j (0) j=0 j=0 m f j (t) where m = (n + )p is the degree of f(x). Substituting the value of I(θ ) in J we get J = ( n = = m e θ() j=0 = = (2 r n) m j=0 f j (0) n f j (θ ) + ) m f j (θ ) j=0 m f j (0) j=0 m f j (0) m j=0 j=0 = n = e θ n f j (θ ). Observe that n = f j (θ ) is a symmetric polynomial in bθ,..., bθ n with integer coefficients and thus a symmetric polynomial with integer coefficients in the 2 r numbers, bθ i. By the fundamental theorem of symmetric functions, this sum is a rational number and so is an integer. Since, f j (θ ) = 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 (0) = b np ( ) np (p )!(θ θ 2...θ n ) p. From the fundamental theorem of elementary symmetric functions and Lemmas 2..5 and 2..6, we find f p (0) is an integer divisible by (p )!, if p is sufficiently large, then f p (0) is not divisible by p. If p > 2 r n, then J (p )! On the other hand using the upper bound for I(t), we get J n θ e θ f(θ ) C p = 9

15 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 < and that p q is irreducible, i.e. it is expressed in its lowest terms) i.e., given ξ and ɛ > 0, there exists r = p such that q r ξ = p q ξ ɛ. Dirichlet (842) proved that for any real irrational number, there are infinitely many rational numbers approximating it very closely. Theorem (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 0 < q n, such that ξ p q. q 2 Remar If ξ is a rational and is equal to a b where a and b are integers. If r ξ, then From the theorem we get, hence r ξ = p q a b = bp aq bq bq. bq ξ p q < q 2 p q a b < q. 2 From this we get, q < b and hence there are only finite number of solutions to the inequality, ξ p q < q. 2 0

16 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. Consider [ ] the Q + numbers, 0, {ξ}, {2ξ}, {3ξ},..., {Qξ} and the Q intervals, where i, i+ Q Q i {0,,..., Q }. Hence there must be an interval, say [ j, j+ ] containing two of Q Q these Q + numbers, say aξ and bξ with 0 a < b Q. Hence, bξ aξ 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, This is true for any Q and as q Q, to the inequality, thus proving the theorem. qξ p Q ξ p q pq. qq, we get an infinite number of solutions q2 ξ p q < q 2 From the Dirichlet s theorem we get a criterion for proving a number irrational. Corollary A real number is irrational if there exists a δ > 0 and infinite pairs p, q Z and (p, q) = such that ξ p q < q. +δ

17 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. Preliminaries We begin with stating an identity involving the lowest common multiple of first n natural numbers Lemma 3... where d n = lcm {, 2,..., n}. d n = p n p is prime p log n log p Proof. Suppose p is a prime number and x is a real number. Then, n = p x x = log n log p so, x = is the highest integer power of p for which p x < n. Now suppose log n log p {A, A 2,..., A n } is a set of positive integers where A i = p y i p 2 y i2...p α y iα for i =,..., n and p i is prime and (y i will be 0 for many i). As lcm {A, A 2,..., A n } = p m p 2 m 2...p α m α where m j = max{y i, i =,..., n}. In the special case where A i = i, log n m j = log p j. 2

18 3 Apéry s Proof of Irrationality of ζ(3) Lemma Given n, p n p is prime p log n log p = Proof. To prove the lemma, it suffices to show p n p log n log p p n p is prime = n. log n Let p log p log n = y log p = log y log n = log y n = y. Thus, log p p n p log n log p = n. 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 ζ(3) =: n n 3 = 5 2 Proof. We will prove the lemma step by step. Step : ( ) n n 3( ) 2n. n K = where a s are integers. Put Now, a a 2...a (x + a )(x + a 2 )...(x + a ) = x a a 2...a K x(x + a )...(x + a K ) A 0 = x and A K = a a 2...a K x(x + a )...(x + a K ). a a 2...a (x + a )(x + a 2 )...(x + a ) = a a 2...a x(x + a )(x + a 2 )...(x + a ) a a 2...a (x + a )(x + a 2 )...(x + a ) = A A 3

19 3 Apéry s Proof of Irrationality of ζ(3) Hence, K = a a 2...a (x + a )(x + a 2 )...(x + a ) = K A A = A 0 A K. = Step 2: Putting x = n 2 ; a = 2 and letting K n we have K = ( ) 2 (n 2 )...(n 2 2 ) = K = ( ) ( ) 2 (n 2 2 )...(n 2 2 ). n = ( ) ( )! 2 (n 2 2 )...(n 2 2 ) = n 2 ( )n (n )! 2 n2n n 2 (2n)! 2 = n 2 ( )n 2n!n! n 2 (2n)! = n ( )n 2 2 n 2( ) 2n. n Write Since upon expanding we get, ɛ n, = 2 (!) 2 (n )!. 3 (n + )! ( ) n(ɛ n, ɛ n, ) = ( ) ( )! 2 (n 2 2 )...(n 2 2 ), N n ( ) (ɛ n, ɛ n, ) = = N ( ) (ɛ N, ɛ, ) = = N = ( ) 2 3( N+ )( N )+ 2 N ( ) 3( ). 2 = From above, the right hand side of the equation = N = n 2 N ( ) n 3 n 3( ) 2n. n Hence N = n 2 N ( ) n 3 n 3( ) 2n = n N = ( ) 2 3( N+ )( N ) + 2 N ( ) 3( ). 2 = 4

20 3 Apéry s Proof of Irrationality of ζ(3) Tae N, the first term vanishes and we get n = 5 ( ) 3 2 3( ). 2 = = Thus, define the double sequence, One may remar that c n, c n, := n m= m 3 + ( ) m 2m 3( n m m= )( n+m m ζ(3) as n, uniformly in i.e., the sequence c n,n ζ(3) but the following lemma will show that our conclusion will be false and we cannot apply the irrationality criterion. ). Lemma where d n = lcm(, 2,..., n). ( ) n + 2d 3 n c n, is an integer Proof. It suffices to show, ( ) n + 2c n, is a rational number whose denominator expressed in lowest terms divides d n 3. Furthermore to prove this it suffices to show that the number of times any given prime ( p divides the denominator of 2c n+ ) n, is less then the number of times that prime divides d 3 n. Let ord p (x) mean the highest exponent of p that divides x. Observe, ) ) Now, ord p ( n m ) ( n+ ( n+m m ) = ( n+ m ( m). [ ] ( ) n(n + )...(n m ) n! = ord p ord p m! m! = ord p (n!) ord p (m!) ord p (d n ) ord p (m) log n = ord p (m). log p 5

21 3 Apéry s Proof of Irrationality of ζ(3) ( Next, we chec that the denominator of each term in the expression for 2c n+ ) n, divides d n 3. Now, ( ) ( ) n ( ) n + n + n + 2c n, = 2 m The denominator of m= ( ) n n + 2 m 3 m= m= ( ) m 2m 3( n m )( n+m m divides d n 3. Hence we will chec for the denominator of the second term, whose denominator is Hence, As m n, and therefore ( ( m 3 n m ord p )( n+m m ( n+ ) ( ) n + ( ) m 2 2m 3( n m 2m 3( n )) m= )( n+m m m ( n+ )( n+m m ) ) = 2m3( n ) m)( m ). ( n+ m ( ( m 3 n )) = ord m)( m p ) ( n+ m log n 3ord p (m) + ord p (m) log p ( ) log n + + ord p (m) ord p log p m log n log + + ord p (m). log p log p ord p (m) ord p (d m ) ord p (d ) ord p (d n ) log n log + + ord p (m) log p log p log m But ord p (d m ) = 3 log p log n + log p log n = ord p (d 3 n ). log p ) log + ord p (d m ). log p ). 6

22 3 Apéry s Proof of Irrationality of ζ(3) Corollary The Denominator of a n divides d n 3. Proof. From the definition of a n, where a n = c n, = n ( ) 2 ( ) 2 n n + c n, =0 n m= m 3 + ( ) m 2m 3( n m m= )( n+m m and from the above lemma, denominator of c n, divides d n 3 and thus denominator of a n divides d n Irrationality of ζ(3) ) Lemma The equation, n 3 u n + (n ) 3 u n 2 = (34n 3 5n n 5)u n, n 2, is equivalent to (n + ) 3 u n+ (34n 3 + 5n n + 5)u n + n 3 u n = 0, n. Proof. The recurrence relation, (n + ) 3 u n+ (34n 3 + 5n n + 5)u n + n 3 u n = 0 can be obtained by replacing n by n in n 3 u n + (n ) 3 u n 2 = (34n 3 5n n 5)u n. And hence we get (n + ) 3 u n+ (34n 3 + 5n n + 5)u n + n 3 u n = 0, n. 7

23 3 Apéry s Proof of Irrationality of ζ(3) Lemma a n = n ( n ) 2 ( n+ ) 2cn, =0 and b n = n ( n ) 2 ( n+ ) 2 =0 satisfy the recurrence relation, n 3 u n + (n ) 3 u n 2 = (34n 3 5n n 5)u n, n 2. Proof. In order to show that the sequence a n satisfies the recurrence relation, we define a double sequence, ( ) 2 ( ) 2 n n + B n, = 4(2n + )[(2 + ) (2n + ) 2 ]. Step : We observe ( ) 2 ( ) 2 ( ) 2 ( n + n + + n n + B n, B n, = (n + ) 3 (34n 3 + 5n n + 5) ( ) 2 ( ) 2 n n + +n 3. Step 2: Using the definition of c n, = n m= This can be obtained as follows: n c n, c n, = m + 3 m= = n 3 + m 3 + m= ( ) m 2m 3( n m )( n+m m c n, c n, = ( )! 2 (n )!. n 2 (n + )! m= m= ( ) m 2m 3( n m The fraction ( )m (m )! 2 (n m )! (n + m)! ( ) m (m )! 2 (n m )! (n + m)! )( n+m m ) n m= m 3 + ( ) m (m )! 2 (n m )!. (n + m)! ) then ( ) m 2m 3( n m m= can be split into two parts, )( n +m m = ( )m (m )! 2 (n m )!(m 2 + n 2 m 2 ) n 2 (n + m)! = ( )m m! 2 (n m )! ( ) m+ (m )! 2 (n m)!(n + m) n 2 (n + m)! = ( )m (m!) 2 (n m )! n 2 (n + m)! 8 ( )m+ (m )! 2 (n m)!. n 2 (n + m )! ) ) 2

24 3 Apéry s Proof of Irrationality of ζ(3) Thus, c n, c n, = n + ( ) m (m!) 2 (n m )! 3 n 2 (n + m)! m= = ( ) ( (n )! n + 3 n 2 n! ( ) m (m )! 2 (n m)! n 2 (n + m )! ) m= ( ) (!) 2 (n )! n 2 (n + )! Step 3: If b n, = ( ) n 2 ( n+ ) 2, define Bn, = 4(2n )((2 + ) (2n + ) 2 )b n, then. (B n, B n, )c n, +(n + ) 3 b n, (c n, c n, ) n 3 b n, (c n, c n, ) = A n, A n, where Step 4: A n, = B n, c n, + 5(2n + )( ) n(n + ) ( n )( n + ). Using the previous definition of a n and c n, we will show a n satisfies: n 3 u n + (n ) 3 u n 2 = 34n 3 5n n 5)u n, n 2. From Lemma 3.2., the expression n 3 u n + (n ) 3 u n 2 = 34n 3 5n n 5)u n, n 2, is equivalent to (n + ) 3 u n+ (34n 3 + 5n n + 5)u n + n 3 u n = 0, n. Substituting a n for u n in the above identity, n (n + ) 3 b n+, c n+, P (n) = n b n, c n, + n 3 = n b n, c n, = 0, n, where b n, = ( ) n 2 ( n+ ) 2 and P (n) = 34n 3 + 5n n + 5. Since for r > n, ( n r) = 0, n n n (n + ) 3 b n+, c n+, P (n) b n, c n, + n 3 b n, c n, = 0, n, = = 9 = =

25 3 Apéry s Proof of Irrationality of ζ(3) can be written as n+ { (n + ) 3 } b n+, c n+, P (n)b n, c n, + n 3 b n, c n, = 0, n. () =0 We will now prove this expression. From Step, B n, B n, = (n + ) 3 b n+, P (n)b n, + n 3 b n, where B n, = 4(2n )[(2 + ) (2n + ) 2 ]b n, and therefore P (n)b n, c n, = (B n, B n, )c n, (n + ) 3 b n+, c n, n 3 b n, c n,. Substituting in () n+ { (n + ) 3 b n+, c n+, + (B n, B n, )c n, (n + ) 3 } b n+, c n, n 3 b n, c n, + n 3 b n, c n, =0 n+ { = (n + ) 3 b n+, (c n+, c n, ) + (B n, B n, )c n, + n 3 b n, (c n, c n, ) }. =0 From the previous lemma this becomes where Now But n+ (A n, A n, ) =0 A n, = B n, c n, + 5(2n + )9 ) ( n n(n + ) n+ A n, A n, = A n,n+ A n,. =0 )( n+ A n,n+ = B n,n+ c n,n+ + 5(2n + )( )n (n + ) ( n n+ n(n + ) ). )( 2n+ n+ (as B n,n+ has a factor ( n n+) ). Similarly An, = 0. Thus, a n satisfies the recurrence n 3 u n + (n ) 3 u n 2 = 34n 3 5n n 5)u n, n ) = 0

26 3 Apéry s Proof of Irrationality of ζ(3) Step 5: Finally we will show b n = satisfies the recurrence relation, n ( ) 2 ( ) 2 n n + = =0 n =0 b n, n 3 u n + (n ) 3 u n 2 = 34n 3 5n n 5)u n, n 2. From earlier, ( ) 2 ( ) 2 n n + B n, = 4(2n+)((2+) (2n + ) 2 ) = 4(2n+)((2+) (2n + ) 2 )b n,. From Step (), B n, B n, = (n + ) 3 b n+, P (n)b n, n 3 b n, := G n,. So, B n, = G n, + B n, = G n, + G n, + B n, 2 = G n, + G n, + G n, Continuing in this manner and using the fact that B n,r = 0 for r < 0, we get Thus, B n, = G n,i. i=0 B n,n+ = = n+ G n,i i=0 n+ ( (n + ) 3 ) b n+,i P (n)b n,i n 3 b n,i i=0 n+ n+ n+ = (n + ) 3 b n+,i P (n) b n,i n 3 i=0 n+ = (n + ) 3 b n+,i P (n) i=0 i=0 i=0 i=0 i=0 2 b n,i n n b n,i n 3 b n,i.

27 3 Apéry s Proof of Irrationality of ζ(3) Now since b n,n+ = 0, b n,n+ = 0, b n,n = 0, B n,n+ = (n + ) 3 b n+ P (n)b n n 3 b n. But B n,r = 0 for r > n. Thus, (n + ) 3 b n+ P (n)b n n 3 b n = 0. And hence b n satisfies the recurrence relation. Lemma Given that a n and b n satisfy the recurrence relation n 3 u n (n ) 3 u n 2 = P (n)u n, n 2 where P (n) = 34n 3 5n n 5 then a n b n a n b n = 6 n 3 for a n and b n as defined. Proof. From the above lemma since a n and b n satisfy the recurrence relation, we substitute a n and b n in the recurrence relation to get, n 3 a n + (n ) 3 a n 2 = P (n)a n, n 3 b n + (n ) 3 b n 2 = P (n)b n. Multiplying the first equation by b n and the second equation by a n and subtracting gives, n 3 (a n b n b n a n ) = (n ) 3 (a n 2 b n b n 2 a n ). 22

28 3 Apéry s Proof of Irrationality of ζ(3) a n b n b n a n = = =. = (n )3 (a n 3 n b n 2 b n a n 2 ) (n )3 (n 2) 3 n 3 (n ) 3 (a n 2b n 3 a n 3 b n 2 ) (n (n ))3 n 3 ( an (n ) b n n a n n b n (n ) ) = n 3 (a b 0 a 0 b ) = n 3. Now from definition a = 6, b 0 =, a 0 = 0, b = 5 and so, a b 0 a 0 b = 6 hence we get a n b n a n b n = 6 n 3. Lemma If c n, = n m= m 3 + ( ) m 2m 3( n m m= then c n, ζ(3) as n uniformly in. )( n+m m ) n Proof. c n, ζ(3) = Since m= Next we see that ( ) m 2m 3( n m m= m=n+ m 3 ( ) m 2m 3( n m m= )( n+m m ) m 3 m=n+ + m is convergent, it follows that given ɛ > 0, n > N, 3 < ɛ. )( n+m m ) 2n(n + ) m 3 m=n+ ( ) m 2m 3( n m m= )( n+m m n 2n(n + ) = 2(n + ) 2n < ɛ, n > ɛ. 23 ).

29 3 Apéry s Proof of Irrationality of ζ(3) Let N 2 = 2ɛ, N = max(n, N 2 ) then for n > N, c n, ζ(3) 2ɛ + ɛ = 3ɛ since this convergence is independent of so the convergence is uniform. Corollary Proof. Using the fact that a n b n ζ(3) as n. a n = b n = n b n, c n,, = and the result that if x n,, y n, R and y n, L as n (uniformly in n) then n = b n, n = x n,y n, n = x n, L as n. We get a n b n ζ(3) as n. Lemma If a n b n a n b n = 6 n 3, then ζ(3) a n b n = =n+ 6 3 b b. Proof. We have Let ζ(3) an b n = x n, then a n b n a n b n = 6 n 3 b n b n. x n x n+ = a n+ b n+ a n b n = 24 6 (n + ) 3 b n+ b n.

30 3 Apéry s Proof of Irrationality of ζ(3) Hence ζ(3) a n b n = and proceeding in this manner we get, 6 (n + ) 3 b n+ b n + x n+, ζ(3) a n b n = = 6 (n + ) 3 + b n+ b n m =n+ 6 3 b b + x n+m. 6 (n + 2) 3 b n+2 b n (n + m) 3 b n+m b n+m + x m+n Now, ζ(3) an b n 0 as n and thus, ζ(3) a n b n = =n+ 6 3 b b. Theorem ζ(3) is irrational Proof. From the above estimation, we have ζ(3) a n b n = =n+ 6 3 b b ζ(3) a n b n = O(b n 2 ). The recurrence relation satisfied by the b ns can be rewritten as b n (34 5n + 27n 2 5n 3 )b n + ( 3n + 3n 2 n 3 )b n 2 = 0. From this relation we get the generating polynomial for b n which is x 2 34x+ and it has zeros 7±2 2 = ( ± 2) 4. From this we get b n = O(α n ), where α = ( + 2) 4 Now, we have already seen that a ns are rationals with denominator dividing d n 3 where d n = lcm(, 2,.., n) we define two new sequences p n and q n such that ζ(3) p n n. Define q n as p n = 2d n 3 a n, q n = 2d n 3 b n. 25

31 3 Apéry s Proof of Irrationality of ζ(3) Clearly p n, q n Z and q n = O(α n e 3n ), ζ(3) p n q n = ζ(3) a n b n = O(b n 2 ) = O(α 2n ). Choosing we get log α = δ = log α 3 log α + 3 = > 0, 3( + δ) ( δ) α +δ = e 3(+δ) α 2 α +δ = e 3(+δ) α 2 = α δ e 3(+δ) α 2n = (α n e 3 n) (+δ). Thus, O(α 2n ) = O(q n (+δ) ). Hence we have This implies ζ(3) is irrational. ζ(3) p n q n = O(q n (+δ) ). 26

32 Chapter 4 Theory of Modular forms In this chapter we discuss the theory on modular forms, in particular Eisenstein series, Dadeind eta function and L- functions associated to modular forms. 4. SL(2, Z) and its congruence subgroups Let {( ) } a b SL(2, Z) = a, b, c, d Z, ad bc =. c d We give an action of SL 2 (Z) on the extended complex plane C = C { } by { az+b cz+d a c if z C γz = if z = ( ) a b where γ =. The upper half plane H = {z C Im z > 0} is stable under c d the given action of SL 2 (Z). Let S and T be two elements of SL(2, Z) given by ( ) ( ) 0 S = and T =. Note that SL(2, Z) is generated by S and T. For 0 0 any z H we have, Sz = z, T z = z +. A fundamental domain D is a subset of H satisfying the properties:. z H, g G s.t. gz D, 2. If z, z 2 int(d), then g G s.t. gz = z 2. Theorem 4... The region D = {z H : z, 2 Re(z) 2 } is a fundamental domain for the action of modular group SL(2, Z) on H. Definition 4... The principal congruence subgroup of SL(2, Z) of level N is {( ) ( ) ( ) } a b a b 0 Γ(N) = SL(2, Z) : (mod N). c d c d 0 27

33 4 Theory of Modular forms Definition 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 Γ 0 (N) = SL(2, Z) : c d c d ( ) 0 } (mod N) (where * means unspecified )and {( ) ( ) a b a b Γ (N) = SL(2, Z) : c d c d ( ) 0 } (mod N), satisfying Γ(N) Γ (N) Γ 0 (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 be an integer. Suppose that f(z) satisfies the relation ( ) f(γz) = (cz + d) a b f(z) γ = SL(2, Z). c d In particular, for the elements γ = T = gives, ( ) and S = 0 ( ) f(z + ) = f(z); f = ( z) f(z). z ( ) 0 above condition 0 Hence f is a -periodic function. f(z) is called a modular function of weight for SL(2, Z) if there are finitely many non zero coefficients a n with n < 0 in the fourier series expansion of f, f(z) = n Z a n q n, where q = e 2πiz. 28

34 4 Theory of Modular forms Definition For Z, a function f is a modular form of weight for SL(2, Z) if f is holomorphic on H and holomorphic at infinity (a n = 0 n < 0) and satisfies f(γz) = (cz + d) f(z) γ SL ( 2, Z), z H. The space of all modular forms of weight is denoted by M (SL(2, Z)). Definition A cusp form is a modular form with the constant term a 0 = 0 in its fourier series expansion. The space of all cusp forms of weight are denoted by S (SL(2, Z)). ( ) 0 Remar As the element SL(2, Z), f(z) = ( ) f(z). Hence if 0 is odd, there are no nonzero modular functions of weight for SL(2, Z). Definition Let γ SL 2 (Z), then slash operator of weight on functions on H is defined as (f γ)(z) := (cz + d) f(γz). Definition Let Γ be a congruence subgroup of SL 2 (Z) of level N and let be an integer. A function f : H C is a modular form of weight with respect to Γ if. f is holomorphic on H, 2. f α = f f Γ, 3. f α is holomorphic at for all α SL 2 (Z). If in addition, we have a 0 = 0 in the Fourier expansion of f α for all α SL 2 (Z), then we call f a cusp form of weight with respect to Γ. Example Now we will explicitly construct a class of modular forms of all even weights > 2, the Eisenstein series of weight. 29

35 4 Theory of Modular forms Definition Let V be a real vector space. A subgroup Γ of V is a lattice if there exists a R basis {e,...e n } of V, which is a Z basis for Γ, i.e. Γ = Ze... Ze n. Lemma Let Γ be a lattice in C. Then the series γ Γ,γ 0 σ > 2. Definition The Eisenstein series of index is defined by G (z) = m,n Z (m,n) (0,0), z H. (mz + n) 2 Proposition G is a modular form of weight. Also G ( ) = 2ζ(). Proof. First we see that G (z + ) = G (z) ( ) G = z 2 G (z) z γ σ is convergent for Next we show G 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 > and Re(z) > 2, mz + n 2 = (mz + n)(m z + n) = m 2 z 2 +2mnRe(z) + n 2 m 2 mn + n 2 = mω + n 2 mz + n 2 mw + n 2 mz + n 2 m,n Z (m,n) (0,0) m,n Z (m,n) (0,0) mω + n 2. But the latter series runs over all the lattice points of the lattice generated by ω and, hence by Lemma 4.2. it is convergent for 2 > 2 i.e. > and hence the 30

36 4 Theory of Modular forms Eisenstein series G. Since for any other point z H there exists g G such that gz D. From the relation G (gz) = z 2 G (z), we show G converges for all z H and hence it is holomorphic in H. lim G (z) = lim z i z i = m,n Z (m,n) (0,0) m,n Z (m,n) (0,0) lim z i (mz + n) 2 (mz + n) 2. For m 0 it vanishes but for m = 0 it contributes n Z = 2ζ(2). Thus n 2 G ( ) = 2ζ(2). Hence G (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) = a n q n where q = e i2πz. n=0 We compute the q-expansion coefficients for G. Proposition For even integer > 2, z H. Then, ( ) G (z) = 2ζ() 2 σ (n)q n B where, σ (n) = d n d and the Bernoulli numbers are defined by the expansion, =0 x e x = x B!. We normalize the value of G at infinity by defining E i.e., E (z) = G (z) 2ζ() = + c 3 σ (n)q n

37 4 Theory of Modular forms where c = (2iπ) ( )! ζ() = 2. B We shall be mainly interested in the standard Eisenstein series, E 4 (z) = σ 3 (n)q n Estimates for the coefficients in q-expansion In this section, we give an estimate on the growth of the coefficients a n in q-expansion of modular form f. Proposition If f = G, the a n = O(n 2 ). Proof. From the q-expansion of G in Proposition 4.3., for some constant A, a n = A( ) σ 2 (n) a n n 2 = A d n where B = Aζ(2 ), we have a n = Aσ 2 (n) An 2 d 2 n 2 = A d n d A 2 d= d 2 = B < An 2 a n Bn 2 Thus, a n = O(n 2 ). Hece. The next theorem is on estimates of coefficients of cusp forms which is due to Theorem If f is a cusp form of weight 2, then a n = O(n ). 32

38 4 Theory of Modular forms Proof. Since for a cusp form a 0 = 0, ( ) f(z) = q a n q n. Hence, if q 0, f(z) = O(q) = O(e 2πy ), where y = Im(z). Define a function φ(z) = f(z) y. 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) My z H. Using the formula to compute Fourier coefficients a n = 0 f(z)e i2πnz dx. a n My e 2πny y > 0. Putting y = we get, n an Mn e 2π an = O(n ) The Eisenstein series of weight 2 Define the Eisenstein series G 2, G 2 (z) = c Z d Z c (cz + d) 2 where Z c = Z {0} if c = 0 and Z c = Z otherwise. The normalized Eisenstein series is, E 2 E 2 (z) = 24 σ (n)q n = 24q 72q 2... satisfies all the properties of a modular form except the transformation law E ( /z) = z 2 E (z). In fact the Eisenstein series of weight 2 satisfies the following transformation law: 33

39 ( ) a b Proposition For z H and SL c d 2 (Z), we have ( ) az + b G 2 = (cz + d) 2 G 2 (z) πic(cz + d). cz + d 4 Theory of Modular forms Definition The Dedeind eta function η : H C is defined by an infinite product η(τ) = q 24 ( q n ) for q = e 2πiz. The transformation law for the Dedeind eta function is Theorem Let denote the branch of the square root having non negative real part. Then ( ) η = z z i η(z). Proof. Compute the derivative of the logarithm of the eta function. d dτ πi log(η(τ)) = 2 + 2πi dq d q d d= = πi 2 + 2πi d d= = πi 2 + 2πi m= m m= d= q dm dq dm = πi 2 + 2πi d q n = πi 2 E 2(τ). 0<d n Thus, and d dτ log(η(τ)) = πi 2 τ 2 E 2 ( ) d dτ log( iτη(τ)) = 2τ + πi 2 E 2(τ) = πi ( E 2 (τ) + 2 ). 2 2πiτ τ 34

40 4 Theory of Modular forms Now πi 2 τ 2 E 2 ( τ ) = πi ( E 2 (τ) + 2 ). 2 2πiτ Hence the desired relation holds up to a multiplicative constant. Setting τ = i shows that the constant is. Remar η(τ) 24 is a cusp form of weight 2 of SL(2, Z) and also of Γ (6) and so η(τ) 24 S 2 (Γ (6)). 4.4 L-functions and Modular forms Each modular form f M (Γ (N)) has an associated Dirichlet series, its L-function. Definition The L-function associated to a modular form, f(z) = n=0 a nq n for a complex variable s C is L(s, f) = a n n s. Theorem If f M (Γ (N)) is a cusp form then L(s, f) converges absolutely for all s with Re(s) > +. If f is not a cusp form then L(s, f) converges absolutely 2 for all s with Re(s) >. L(s, f) satisfies the functional equation, (2π) s Γ(s)L(s, f) = ( ) 2 (2π) ( s) Γ( s)l( s, f) Mellin Transform If f(τ) = a nq n S (Γ (N). As T Γ (N), we have a q-expansion rather than q N, we set g(s) := i 0 f(z)z s dz. 35

41 4 Theory of Modular forms This integral is nown as the Mellin transform of the function f. It converges for Re s > c + since for f(z) = a nq n, if a n = O(n c ) then an n c < as n. We conclude this section by stating the Hece s Lemma for Dirichlet L-functions. Theorem (J2). (Hece s Lemma by Weil s and Razar) Let F (τ) = n=0 a n qn λ, G(τ) = b n qn λ, f(τ) = a 0z ( )! + (2πi λ ) ( ) a n n e qn λ, g(τ) = b ( ) 0τ ( ) 2πi ( )! + b n λ n e qn λ. Let L(F, s) = ( an ) n s be the associated L function for F. Here is a positive integer and γ C such that F (z) = γ(τ) G( ), then τ ( ) f(τ) γτ 2 g = τ 2 j=0 L(F, j) j! ( ) ( j) 2πi τ j. λ 36

42 Chapter 5 Irrationality Proofs using Modular Forms In this chapter we discuss Beuer s proof of irrationality of ζ(3) using modular forms. Let t(q) = n=0 t nq n be a power series convergent for all q <. Let w(q) be another analytic function on q <. Our aim is to study w as a function of t. In order to mae it a single-valued function, assume t 0 = 0, t 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 power series w(q(t)) = n=0 w nt n. 5. Preliminaries Definition 5... Let t(q) = n=0 t nq n. x is a branching value of t, if either x is not in the image of t, or if t (y) = 0 for some y with t(y) = x. Let us assume, that t has a discrete set of branching values, say t, t 2,... where t i 0 i and t < t 2 <... and so the radius of convergence is t. 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 < t 2 with exception of the possible isolated singularity t and it remains bounded around t we can conclude that the radius of convergence is at least t 2. 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

43 5 Irrationality Proofs using Modular Forms Theorem 5... Let f 0 (t), f (t),..., f (t) be power series in t. Suppose that for any n N, i = 0,,..., the n-th coefficient in the Taylor series of f i is rational and has denominator dividing d n d n r where r, d are certain fixed positive integers and d n is the lowest common multiple of, 2,..., n. Suppose there exist real numbers θ, θ 2,..., θ such that f 0 (t) + θ f (t) + θ 2 f 2 (t) θ f (t) has radius of convergence ρ and infinitely many non zero Taylor coefficients. If ρ > de r, then at least one of θ,..., θ is irrational. Proof. As ρ > de r, there exists an ɛ > 0 such that ρ ɛ > de r(+ɛ). Let f i (t) = n=0 a int n. From the hypothesis, if a in = p in q in, then q in d n d n r. Since the radius of convergence of f o (t) + θ f (t) θ f (t) is ρ, we have for sufficiently large n, a 0n + a n θ a n θ < (ρ ɛ) n. Suppose θ,..., θ are all rational and have common denominator D. Then, is an integer and A n = Dd n d n r a 0n + a n θ a n θ A n < Dd n d n r (ρ ɛ) n for sufficiently large n. But for such n d n < e (+ɛ)n. Hence we get, Since, A n < Dd n e (+ɛ)nr (ρ ɛ) n. de r(+ɛ)r (ρ ɛ) < implies that A n = 0 for sufficiently large n, which is a contradiction that A n 0 for infinitely many n. Thus, at least one of θ,..., θ is irrational. The values for which irrationality results are obtained are integral points of Dirichlet series associated to modular forms. 38

44 5 Irrationality Proofs using Modular Forms Proposition Let F (τ) = a nq n be a Fourier series convergent for q < such that for some, N N, F ( ) = ɛ( iτ N) F (τ), Nτ where ɛ = ±. Let f(τ) be the Fourier series f(τ) = a n n qn. Let and Then h(τ) = f(τ) L(F, s) = 0 r 2 2 a n n s L(F, r ) (2πiτ) r. r! h(τ) D = ( ) ɛ( iτ N) 2 h ( ) Nτ where D = 0 if is odd and D = L(F, 2 ) (2πiτ)( 2 ) ( 2 )! if is even. Moreover L(F, 2 ) = 0 if ɛ =. Proof. Applying the lemma of Hece, with G(τ) = ɛ F (τ) (i N) to obtain, f(τ) ɛ( ) ( iτ N) 2 f ( ) 2 = Nτ r=0 L(F, r ) (2πiτ) r. r! Splitting the summation on the right hand side into summations over r < 2, r > 2 and r = 2. We thus write, 2 r=0 L(F, r ) (2πiτ) r = r! 4 2 r=0 2 + r= 2 L(F, r ) r! (2πiτ) r + L(F, 2 ) ( 2 2 )! L(F, r ) (2πiτ) r. r! 39

45 5 Irrationality Proofs using Modular Forms Now using the functional equation of the Dirichlet L-function for F ( ) s N Λ(s) = CΛ( s), where Λ(s) = Γ(s)L(f, s). 2π For the region r > 2, putting s = r + we get, ( ) r+ ( ) r N N r!l(f, r + ) = C ( r )!L(F, r ). 2π 2π L(F, r ) r! Now substitute r by 2 r. = ɛ( ) ( i ( ) r 2 N) 2 (2πi) 2r 2 L(F, r + ) N ( r 2)!. 5.2 The group Γ (6) and the irrationality of ζ(3) Define a modular function on Γ (6), y(τ) = η(6τ)8 η(τ) 4 η(2τ) 8 η(3τ) 4 where, Using η(τ) = q 24 ( q n ). y(0) = 9, y( 3 ) =, y( ) =, y( ) = 0, 2 we find the function y( ) is invariant on Γ 6τ (6). From this one derives ( ) y = y(τ) 9 6τ y(τ). (5.2.) Remar t(τ) = y(τ) 9y(τ) y(τ) is invariant under the involution τ 6τ. Since t ( ) = y 6τ ( ) 9y( ) 6τ 6τ y( ). 6τ Now replace y( ) by the identity described to get the stated result. 6τ 40

46 5 Irrationality Proofs using Modular Forms Remar t(τ) = ( ) (6τ) (τ) 2. (3τ) (2τ) Proof. Write, ( ) ( ) ( ) η(6τ) 8 η(τ) 4 η( τ t(τ) = 9y(τ)y = 9 )8 η( 6τ )4 6τ η(2τ) 8 η(3τ) 4. η( 2 6τ )8 η( 3 6τ )4 Use, η ( ) τ = τ η(τ). Hence, we have i ( ) y = η(τ) 8 η(6τ) 4 6τ 9 η(3τ) 8 η(2τ) 4. Now, (τ) = η(τ) 24 so t(τ) = η(6τ)8 η(τ) 4 η(2τ) 8 η(3τ) 4 η(τ)8 η(6τ) 4 ( ) 2 η(6τ)η(τ) η(3τ) 8 η(2τ) 4 =. η(2τ)η(3τ) Remar The function, defined above is modular with respect to Γ (6), invariant under τ 6τ and its zeros and poles coincide with those of t(τ). Proposition The function ( 2 + ) 4, t( 2 ) =. t(i ) = 0, t(i 6) = ( 2 ) 4, t ( ) 2 + i 5 5 = 6 Proof. By definition t(τ) = y(τ) 9y(τ). So, y( ) = 0 t(i ) = 0. And, y ( y(τ) 2) = t ( ) ( ) 2 =. For τ = i i 6 and y 0 = y 6 we find, ( ) i y = 6 ( hence y 0 = ± 2 2 and correspondingly t 3 ( ) i y 6 9 ( ) i y 6 ) i 6 = ( 2 ± ) 4. To decide which sign should be taen, one can estimate by obtaining the values numerically. Using the above functions we show the irrationality of ζ(3). 4

47 5 Irrationality Proofs using Modular Forms Theorem ζ(3) is irrational. Proof. Construct functions F (τ) = 40 (E 4(τ) 36E 4 (36τ) 7 (4E 4 (2τ) 9E 4 (3τ))), where E(τ) = 24 ( 5 (E 2(τ) 6E 2 (6τ)) + 2E 2 (2τ) 3E 2 (3τ)) E 4 (τ) = σ 3 (n)q n ; E 2 (τ) = 24 σ n q n. We note that F (τ) M 4 (τ(6)) and ( ) ( F = ɛ iτ ) ( ) F N (τ) F = 36τ 4 F (τ). Nτ 6τ Also E(τ) M 2 (Γ (6)), The Dirichlet series corresponding to F (τ) is ( 6σ3 (n) L(F, s) = 36 6σ 3(n) n s (6n) s E ( ) = 6τ 2 E(τ). 6τ 28 6σ 3(n) (2n) s σ ) 3(n) (3n) s. F (τ) = 40 (E 4(τ) 36E 4 (36τ) 7 (4E 4 (2τ) 9E 4 (3τ))) ( ( )) = σ 3 (n)q n σ 3 (n)q 36n 40 ( ( ) ( )) σ 3 (n)q 2n σ 3 (n)q 3n 40 = 6 σ 3 (n)q n 36 6 σ 3 (n)q 36n 28 6 σ 3 (n)q 2n σ 3 (n)q 3n. Dirichlet series in the form of zeta function is L(F, s) = 6( 6 2 s s s ) σ 3 (n) n s = 6( 6 2 s s s )ζ(s)ζ(s 3). 42

48 5 Irrationality Proofs using Modular Forms Define f(τ) by ( ) 3 d f(τ) = (2πi) 3 F (τ) ; f(i ) = 0. dτ Comparing with the hypothesis of Proposition 5..4 here = 4 and r = 0. h(τ) = f(τ) L(F, 3) L(F, 2)(2πiτ) = ɛ( iτ ( ) ) 6) (f 2 L(F, 3). 6τ ( ) ) f(τ) L(F, 3) = 6τ (f 2 L(F, 3). 6τ Since L(F, 3) = 6 ( ) 3 ζ(3)ζ(0) = ζ(3) ( ) ) 6τ (f 2 ζ(3) = (f(τ) ζ(3)). 6τ Multiplying the above equation by E ( ) 6τ = 6τ 2 E(τ) we get, ) E ( ) ( f 6τ ( ) ζ(3) 6τ = E(τ) (f(τ) ζ(3)). () In general, the function E(τ)(f(τ) ζ(3)) can be a multivalued function of t = t(τ). t(τ) = q ( q 6n+ ) 2 ( q 6n+5 ) 2 = q 2q q 3 220q q 5 The inverse function expansion is q = t + 2t t 3 + We get E(t) = + 5t + 73t t 3 +. and E(t) Z[t]. We also have E(t)f(t) = a nt n where a n Z/d 3 n. The function t τ has a branch point at t = ( ( ) 2 ) 4 i since t 6 = ( ( ) 2 ) 4 and t i 6 = 0. We can expect the radius of convergence of E(t)(f(t) ζ(3)) to be ( 2 ) 4 but by equation (), the function t E(t)(f(t) ζ(3)) has no branch point at ( + 2) 4, 43

49 5 Irrationality Proofs using Modular Forms and thus its radius of convergence, ρ equals at least the next branching value, which is t = ( + ( ) 2) 4 2 (since t + i 5 5 = ( + 2) 4 ). Thus the radius of convergence 6 ρ > e 3. Remar 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..3 with the two Taylor series, f 0 (t) = E(t)f(t) and f (t) = ζ(3)f(t). The n-th coefficient in the Taylor series of f i is rational with denominator dividing d 3 n. Also, the radius of convergence is larger than e 3 and since E(t)(f(t) ζ(3)) is not a polynomial, it has infinitely many nonzero Taylor coefficients. Therefore from Theorem 5..3 we have at least one of θ = or θ 2 = ζ(3) is irrational. But 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 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, is irrational. π 2 L(F, 2) and L(F, 3) + 47L(F, 2)ζ(3) 48π 2 44

50 5 Irrationality Proofs using Modular Forms Proof. The function F (τ) M 4 (Γ (6)) and it is a cusp form. From the functional equation satisfied by the Dadeind-eta function we get, F ( ) = 36τ 4 F (τ). 6τ Let f(τ) be defined as the Fourier series such that ( d dτ ) 3f(τ) = (2πi) 3 F (τ) and f(i ) = 0. Apply Proposition 5..4 for h(τ) = f(τ) L(F, 3). Observe that f(τ) L(F, 3) L(F, 2)(2πiτ) = ( iτ ( ) ) 6) (f 2 L(F, 3). 6τ ( ) ) 6τ (f 2 L(F, 3) = f(τ) L(F, 3) L(F, 2)(2πiτ). 6τ Define the function G(τ) such that 240G(τ) = 3(E 4 (τ) + 36E 4 (6τ) 37(4E 4 (2τ) + 9E 4 (3τ)). Note that this function has the properties G(i ) = 0, G ( ) 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) = 3 σ 3 (n)q n σ 3 (n)q 6n 48 σ 3 (n)q 2n 333 σ 3 (n)q 3n. Thus the L-series is Hence, From Proposition L(G, s) = ( s s s) σ 3 (n) n s. L(G, s) = ( s s s) ζ(s)ζ(s 3) 6τ 2 (g L(G, 3) = 47 ζ(3) ; L(G, 2) = 48ζ(2). 6 ( ) L(G, 3)) = g(τ) L(G, 3) L(G, 2)(2πiτ), (2) 6τ 45

51 5 Irrationality Proofs using Modular Forms and so 6τ 2 (g ( ) 47 ) 6τ 6 ζ(3) = g(τ) 47 ζ(3) + 48ζ(2)(2πiτ). (3) 6 The function, satisfies the equation, ( h(τ) = 48ζ(2)(f(τ) L(F, 3)) + L(F, 2) g(τ) 47 ) 6 ζ(3) 6τ 2 h ( ) = h(τ). 6τ This is obtained by eliminating 2πiτ from equations (2) and (3). Consider the function E(τ) = E 2 (τ) 2E 2 (2τ) + 6E 2 (6τ) 3E 2 (3τ). We see, E(τ) M 2 (Γ (6)) and E ( ) 6τ = 6τ 2 E(τ). Consequently, E ( 6τ ) 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 6 ζ(3))e(t) is a power series in t with radius of convergence ( 2 + ) 4. In order to apply Theorem 5..3 we have the Taylor series, f 0 (t) = E(t)f(t), f (t) = E(t)g(t), f 2 (t) = E(t) where the n-th coefficient divides d n 3. Thus the hypothesis of Theorem 5..3 are satisfied. On substituting the value of ζ(2) we get, at least one of the numbers is irrational. ( ) 47L(F, 2)ζ(3) π 2 L(F, 2) ; L(F, 3) + 48π 2 46

52 Chapter 6 Congruence Properties of Apéry numbers The Apéry numbers a n, b n are defined by the finite sums, a n = n ( )( ) 2 n + n, b n = =0 n ( ) 2 ( n + n =0 ) 2 for every integer n 0. These numbers were introduced by R. Apéry[978] in proving irrationality of ζ(3) and ζ(2). Hence these numbers are called Apéry numbers. The first few Apéry numbers, a n s and b n s are, 5, 73, 445, 3300 and, 3, 9, 47, 25 respectively. The Apéry numbers satisfy some congruence properties. Chowla, Cowles and Cowles[J4] were the first to consider congruences for a ns. They proved. for n 0, a 5n+ 0 (mod 5) and a 5n+3 0 (mod 5). 2. for all primes p, a p a (mod p 3 ). Beuer further found some congruences for a ns. The significance of these congruence properties and the reason for the Apéry numbers to satisfy these congruences is still unnown 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