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(1) . . . . . . . . . . On Modular Forms and zeta values. .

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(3) . . On modular forms and zeta values. .

(4). . . StudentRen-Hau Li. . . AdvisorYifan-Yang . A Thesis Submitted to Department of Applied Mathematics College of Science National Chiao Tung University in Partial Fulfillment of the Requirements for the Degree of Master in Applied Mathematics June 2006 Hsinchu, Taiwan, Republic of China. . ! " # $ % &.

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(13) L. K. ò. }. …………………………………………………….. i. ó. K. ò. }. …………………………………………………….. ii. ¾. . …………………………………………………….. iii. ô. õ. …………………………………………………….. iv. 1.. Introduction………………………………………………...1. 2.. Modular group and congruence subgroup………………….3. 3.. Modular functions and modular forms……………………..9. 4.. Beuker’s Proof ……………………………………………13. 5.. Differential equation satisfied by modular forms…………17. 6.. Construction of series converging to ζ (5) …………………21. 7.. Reference………………………………………………….33. iv.

(14) 1. Introduction The Riemann zeta function was first introduced by Euler and is defined by ∞ X 1 ζ(s) = ns n=1. The series is convergent when s is a complex number with Re s > 1. Some special values of ζ(s) are well known. For example ζ(2) = π 2 /6 and ζ(4) = π 4 /90. In general, when s = 2n is a positive even integer we have ζ(s) = π 2n r for some rational numbers r. In fact, the number r can be expressed in terms of the Bernoulli numbers. For odd integer 2n + 1 ≥ 3, however, not much about ζ(2n + 1) is known. It is not even known whether ζ(2n + 1) are rational, except for the case 2n + 1 = 3, which was established relatively recently. In 1970’s, R. Apéry [3] proved that ζ(3) =. P∞ n=1. n−3 is an irrational number by. constructing two sequences an =. ¶2 n µ ¶2 µ X n n+k k=0. k. k.       µ ¶ µ ¶ n n k 2 2  X X n n + k X 1 (−1)m−1 µ ¶µ ¶ bn = + , 3   n n + m k k m 3   m=1 2m k=0 m=1  m m and then showing that bn /an converges to ζ(3) fast enough to ensure irrationality of. ζ(3). Another remarkable discovery of Apéry is that an and bn satisfy the recursive relation (n + 1)3 un+1 = (34n3 + 51n2 + 27n + 5)un − n3 un−1 P P∞ n n Thus, if we set A(t) = ∞ n=0 an t and B(t) = n=0 bn t , then the functions A(t) and B(t) satisfy the differential equations (1 − 34t + t2 )Dt3 A + (3t2 − 51t)Dt2 A + (3t2 − 27t)Dt A + (t2 − 5t)A = 0 and (1 − 34t + t2 )Dt3 B + (3t2 − 51t)Dt2 B + (3t2 − 27t)Dt B + (t2 − 5t)B = 6t, where Dt denotes the differential operator td/dt. It turns out that these differential equations have a modular-function origin. It can be shown that if we choose two linearly independent solutions F1 and F2 appropriately, then t is a modular function. 1.

(15) of τ = F2 /F1 on Γ0 (6), A a modular form of weight 2, and d3 (B/A)/dτ 3 a modular form of weight 4. This connection between ζ(3) and modular forms was discovered by Beukers [1]. (Note that, in general, if F is a modular form of weight k and t a modular function, then F as a function of t, satisfied an (k + 1)-st order linear differential equation. See Section 5 below.) In this thesis we will construct a sequence of rational numbers cn converging to ζ(5) using Beukers’ idea [1]. Our result is as follows. Theorem 1. Let an and bn be two sequences satisfying the recursive relation n5 un = A7 un−1 − A6 un−2 − A5 un−3 − A4 un−4 − A3 un−5 − A2 un−6 + A1 un−7 − A0 un−8 , where A7 = 24n5 + 420n4 + 2960n3 + 10500n2 + 18744n + 13468, A6 = 92n5 + 1380n4 + 8160n3 + 23760n2 + 33968n + 18960, A5 = 600n5 + 7500n4 + 38480n3 + 101100n2 + 135704n + 74260, A4 = 966n5 + 9660n4 + 40800n3 + 90240n2 + 103840n + 49472, A3 = 600n5 + 4500n4 + 14480n3 + 24660n2 + 21944n + 8076, A2 = 92n5 + 460n4 + 800n3 + 560n2 + 48n − 80, A1 = 24n5 + 60n4 + 80n3 + 60n2 + 24n + 4, A0 = n5 with initial values a0 = 1, a1 = 4, a2 = 34, a3 = 308, a4 = 3083, a5 = 32696, a6 = 361428, a7 = 4119288 and b0 = 0, b1 = 144/25, b2 = 333/10, b3 = 217042/675, b4 = 138004123/43200, b5 = 1144320384083/33750000, b6 = 25297127932859/67500000, b7 = 2422896637170749569/567236250000. Then the series {bn /an } converges to ζ(5). More precisely, we have. s¯ ¯ √ ¯ ¯ n ¯ bn lim sup ¯ − ζ(5)¯¯ ≤ ( 3 − 1)4 /4 an n→∞. Here we tabulate bn /an for n = 1 . . . 10 below. We remark that the error between b10 /a10 and ζ(5) is about 10−8 . In order to get the same magnitude of error term P −5 using N n=1 n . One would need N to be about 100.. 2.

(16) ζ(5) = 1.03692775. b1 /a1 = 1.44000000 b2 /a2 = 0.97941176 b3 /a3 = 1.04397306 b4 /a4 = 1.03617900 b5 /a5 = 1.03700113 b6 /a6 = 1.03692095 b7 /a7 = 1.03692836 b8 /a8 = 1.03692770 b9 /a9 = 1.03692776 b10 /a10 = 1.03692775 The rest of the thesis is organized as follows. We first introduce the basic theory of modula groups, congruence subgroups and modular forms in section 2 and section 3. Then we will describe Beukers’ approach to irrationality proof using modular forms in section 4. Next we introduce the result of P. F. Stiller [3] and the method of Y.Yang [5] for determining the differential equation satisfied by modular form in section 5. Finally, we will prove theorem 1 and apply the method of [5] to find the recursive relation given above in the last section. 2. Modular group and congruence subgroup In this section we briefly recall the definition of modular groups and congruence subgroups. For a ring R with unity 1, we denote by R× the group of invertible elements in R. The general linear group GL2 (R) is defined by (Ã ! ) a b GL2 (R) = | a, b, c, d ∈ R and ad − bc ∈ R× c d Here we consider the situations when R = R or R = Z. We set (Ã ! ) a b GL+ | a, b, c, d ∈ R and ad − bc > 0 2 (R) = c d (Ã SL2 (R) =. a b c d. !. ) | a, b, c, d ∈ R and ad − bc = 1. 3.

(17) We call SL2 (Z) and its subgroups of finite index modular groups. The special linear group SL2 (Z) is also called the full modular group. A class of modular groups that are of special interest to number theorists is the congruence subgroups. Their definition is given as follows. Definition 2.1. For a positive integer N , we define the subgroups Γ0 (N ), Γ1 (N ) and Γ(N ) of SL2 (Z) by (Ã Γ0 (N ) = (Ã Γ1 (N ) =. a b. a b. c d !. !. ) ∈ SL2 (Z) | c ≡ 0 mod N ). ∈ SL2 (Z) | c ≡ 0, a ≡ d ≡ 1 mod N c d (Ã ! ) a b Γ(N ) = ∈ SL2 (Z) | b ≡ c ≡ 0, a ≡ d ≡ 1 mod N c d We note that SL2 (Z) = Γ0 (1) = Γ1 (1) = Γ(1) and SL2 (Z) ⊃ Γ0 (1) ⊃ Γ1 (N ) ⊃ Γ(N ) Further, if M | N , then Γ0 (M ) ⊃ Γ0 (N ) ,. Γ1 (M ) ⊃ Γ1 (N ) ,. Γ(M ) ⊃ Γ(N ).. These subgroups are modular groups since [Γ(1) : Γ(N )] < ∞. We call Γ(N ) a principal congruence modular group, and Γ0 (N ) and Γ1 (N ) modular groups of Hecke type. We call N the level of Γ0 (N ), Γ1 (N ) and Γ(N ). A modular group containing a principal congruence modular group is called a congruence modular group. There is another type of congruence subgroup that is of great interest to number theorists. Let N be a positive integer and n be an integer such that gcd(n, N/n) = 1 ( Ã ! ) an b 1 wn = √ : adn2 − bcN = n n cN dn. The elements wn are called Atkin-Lehner involutions. The set of all Γ∗0 (N ) of Γ0 (N ) union all the Atkin-Lehner involutions lies in the normalizer of Γ0 (N ) in SL2 (R). 4.

(18) and we have Γ∗0 (N )/Γ0 (N ) ' Zk2 , where k is the number of distinct prime divisors of N . Group action on the upper half-plane Let H denote that upper half-plane {τ ∈ C : Im τ > 0}. We define a mapping SL2 (R) × H 7→ H by. where τ ∈ H and α =. Ã a b. aτ + b , (α, τ ) 7→ ατ = cτ + d !. ∈ SL2 (R). Then we can check that this mapping is c d a group action. Moreover, this mapping is called a linear fractional transformation and is also a conformal mapping. Cusps and elliptic points We now classify the linear fractional transformation defined above. A nonconstant element α of GL+ 2 (R) is called elliptic, parabolic, or hyperbolic, when it satisfies tr(α)2 < 4 det(α) ,. tr(α)2 = 4 det(α),. or. tr(α)2 > 4 det(α),. respectively. When τ ∈ H∗ is a fixed point of an elliptic, parabolic or hyperbolic element of Γ, we say that τ is an elliptic point, a parabolic point, or a hyperbolic point, respectively. We also call a parabolic point of Γ a cusp of Γ. Remark. In SL2 (Z) the above classification implies that if tr(α) = 0, then α2 = −I, and if tr(α) = ±1, then α is order 3 in P SL2 (Z). Fundamental domains Let G denote any subgroup of the modular group Γ(1). Two points τ and τ 0 in the upper half-plane H are said to be equivalent under G if τ 0 = Aτ for some A ∈ G. This is an equivalence relation since G is a group. Let Γ be a discrete subgroup of SL2 (R). A fundamental domain for Γ is a connected open subset D of H such that no two points of D are equivalent under Γ ¯ where D ¯ is the closure of D. The standard fundamental domain for and H = ∪γ D, SL2 (Z) is shown in Fig 1. Γ := {τ ∈ H |. −1 1 ≤ Re τ ≤ and |τ | ≥ 1} 2 2 5.

(19) Figure 1. Fundamental domain of Γ(1) Next we consider the fundamental domain for congruence subgroups. The following fact is well-known. Proposition 2.1. Let Γ be a discrete subgroup of SL2 (R), and let D be a fundamental domain for Γ. Let Γ0 be a subgroup of Γ of finite index, and write Γ as a disjoint union of right cosets of Γ0 : Γ = Γ0γ1 ∪ · · · ∪ Γ0γm Then D0 = ∪γi D is a fundamental domain for Γ0 . ¯ and γ = γ 0 γi for some γ 0 ∈ Γ0 Proof. Let τ ∈ H. Then τ = γτ 0 for some τ 0 ∈ D, Thus τ = γ 0 γi τ ∈ Γ0 . If γD0 ∩ D0 6= φ, then it would contain a transformation of D. But then γγi D = γj D for some i 6= j, which would imply that γγi = γj and this is a contradiction.. ¤. The Riemann surface Γ\H∗ Let Γ be a discrete subgroup of SL2 (R). We now consider the quotient space Γ\H. It is in general a non-compact Riemann surface. To compactify it, we begin by adding cusps of Γ to H. Let PΓ be the set of all cusps of Γ and put H∗ = H ∪ PΓ . When Γ has no cusps, PΓ = φ and H∗ = H. We put U ∗ = Ul ∪ {∞} l > 0. Ul = {τ ∈ H | Im (τ ) > l}. Now we define the topology on H∗ as follows:. 6.

(20) (i) for τ ∈ H, we take as the fundamental neighborhood system at τ in H∗ that at τ in H. (ii) for x ∈ PΓ , we take as the fundamental neighborhood system at x the family {σ −1 Ul∗ | l > 0} , where σ ∈ SL2 (R) such that σx = ∞. Then H∗ is also a Hausdorff space under this topology. In fact, put σ =. Ã ! a b c d. and x = −d/c. Then we see that σ −1 Ul = {τ ∈ H | Im (τ )/|cτ + d|2 > l} and this is the inside of a circle with the radius (2lc2 )−1 tangent to the real axis at x. For x ∈ PΓ , we call {σ −1 Ul } a neighborhood of x in H. Since the action of Γ on H is a conformal mapping which maps circles or lines to those, Γ also acts on the topological space H∗ . Therefore, the quotient space Γ\H∗ can be defined, and we conclude that Γ\H∗ is compact. The genus of Γ\H∗ Let < be a compact Riemann surface and χ be the Euler-Poincaré characteristic of <. We define the genus of < by χ = 2 − 2g Then g is a non-negative integer. We now compute the genus of the Riemann surface Γ\H∗ introduced above. The main tool we use is the Riemann-Hurwitz formula. Assuming that <0 −→ < is a covering of Riemann surfaces, then the RiemannHurwitz formula states that 2g 0 − 2 = n(2g − 2) +. X. (eb − 1). b∈<0. where g 0 is the genus of <0 , n is the degree of the covering and eb is the ramification index at the point b. Proposition 2.2. Let Γ be a modular group, and g the genus of Γ\H∗ Then g = 1 + m/12 − v2 /4 − v3 /3 − v∞ /2. 7.

(21) where v2 is the number of inequivalent elliptic points of order 2; v3 is the number of inequivalent elliptic points of order 3; v∞ is the number of inequivalent cusps; and ¯ ¯ m = [Γ(1) : Γ]. ¯ ) is denoted the image of Γ(N ) in Γ(1)\{±I}. Remark. Γ(N Proof. Since H∗Γ = H∗Γ(1) , there exists a natural mapping F : <Γ = Γ\H∗Γ −→ <Γ(1) = Γ(1)\H∗Γ(1) We put H∗ = H∗Γ = H∗Γ(1) , and < = <Γ(1) . Let πΓ : H∗ → <Γ , and π : H∗ → < be the natural mappings. For any point b of <Γ , take a point τ ∈ H∗ so that πΓ (τ ) = b. Hence {<Γ , F } is a covering of < of degree m. Let eb = eb,F be the ramification index of the covering at b, and put F (b) = a. Let a2 ,a3 and a∞ be the elliptic points of order 2 and 3,and the cusp on <, respectively. If a 6= a2 , a3 , a∞ , then b is an ordinary point and eb = 1. Suppose a = a2 , then eb = 1 or 2. Put t = # {b ∈ <Γ | F (b) = a2 }. Then m = v2 + 2(t − v2 ). Therefore X (1) (eb − 1) = m − t = (m − v2 )/2, where. 2. P 2. is the summation over the points b such that F (b) = a2 . A similar. argument implies X. (2) where. P 3. (eb − 1) = 2(m − v3 )/3,. 3. is the summation over the points b such that F (b) = a3 . Next assume. F (b) = a∞ . Then b is a cusp in <Γ , and v∞ = # {b ∈ <Γ | F (b) = a∞ }. P Denote by ∞ is the summation over the points b such that F (b) = a∞ . Then P ∞ eb = m, so that X (3) (eb − 1) = m − v∞ . ∞. Consequently, the formula of genus follows from (1), (2), (3) and the RiemannHurwitz formula.. ¤. 8.

(22) We now restrict our attention to Γ0 (N )\H∗ . The geometric data can be easily seen to be [Γ(1) : Γ0 (N )] = N. Y p|N. 1 (1 + ) p.    if 4 | N. 0, µ µ ¶¶ Y v2 (Γ0 (N )) = −1  if 4 6 |N. 1+   p p|N.    if 9 | N. 0, µ ¶¶ µ v3 (Γ0 (N )) = Y −3  if 9 6 |N. 1+   p p|N. v∞ (Γ0 (N )) =. X. φ((d, N/d)). 0<d|N. where φ(n) is the Euler function and (−) denote the Legendre symbol. Therefore we can get the genus of Γ0 (N )\H∗ by Proposition 2.2 . Remark. Using the above formula we see that if Γ0 (N ) is of genus zero, then N = 1, . . . , 10, 12, 13, 16, 18, 25. 3. Modular functions and Modular forms In this section we are concerned with modular functions and modular forms. In particular, we will introduce the classes of modular forms, namely, the Eisenstein series and the η−function. Ã Dedekind ! a b Let γ = ∈ SL2 (Z) and f (τ ) be a function on H∗ with values in C∪{∞}. c d Let k be an integer. The action of γ on f is defined to be ¯ f (τ )¯[γ] = (cτ + d)−k f (γτ ) f or k. More generally, for γ =. Ã ! a b c d. Ã γ=. !. a b c d. ∈ SL2 (Z). ∈GL+ 2 (Q). We define. ¯ f (τ )¯[γ] = (det γ)k/2 (cτ + d)−k f (γτ ) f or k. 9. γ=. Ã ! a b c d. ∈ GL+ 2 (Q).

(23) We now define modular functions, modular forms and cusp forms for a congruence subgroup Γ ⊂ SL2 (Z). Let N be the level of Γ. We set qN = e2πiτ /N . Definition 3.1. Let f (τ ) be a meromorphic function on H, and let Γ be a congruence subgroup of level N , i.e., Γ ⊃ Γ(N ). Let k ∈ Z. We call f (τ ) a modular function of weight k for Γ if it satisfies the following two conditions: ¯ (a) f ¯[γ] = f f or all γ ∈ Γ and, k ¯ P n (b) if for any γ0 ∈ SL2 (Z), f ¯[γ0 ] has the form an qN with an = 0 for n ¿ 0. k. Here an = 0 for n ¿ 0 means that an = 0, for n ≤ −M for some fixed integer M . We call such an f (τ ) a modular form of weight k for Γ if it is holomorphic on H and if for all γ0 ∈ SL2 (Z) we have an = 0 for all n < 0 in condition (b). We call a modular form a cusp-form if in addition a0 = 0 in condition (b) for all γ0 ∈ SL2 (Z). We let Mk (Γ) and Sk (Γ) denote the set of modular forms of weight k for Γ and the set of cusp-forms of weight k for Γ, respectively. Now we illustrate two examples of modular forms which will appear later. Eisenstein series Let k be an even integer greater than 2 and write X (4) Gk (Λ) = ω −k ω∈Λ. define Gk (τ ) = Gk (τ Z + Z) where Λ is denote the lattice spanned by 1 and τ . Because k is at least 4, the double sum (4) is absolutely convergent and uniformly convergent in any compact subset of H. Hence Gk (τ ) is a holomorphic function on H. It is obvious that P Gk (τ ) = Gk (τ + 1), and that the Fourier expansion of Gk (τ ) = n∈Z an q n has no negative terms. Because Gk (τ ) approaches a finite limit as τ → i∞: X lim Gk (τ ) = n−k = 2ζ(k) τ →i∞. n∈Z,n6=0. Finally, we easily check that τ −k Gk (−1/τ ) = Gk (τ ) Thus we have proved that Gk ∈ Mk (SL2 (Z)).. 10.

(24) We now compute the q-expansion coefficients for Gk (τ ). We shall find these coefficients are essentially the arithmetic functions. (5). σk (n) =. X. dk. d|n. of n. Proposition 3.1. Let k be an even integer greater than 2, and let τ ∈ H. Then the modular form Gk (τ ) has q-expansion Ã. ∞ 2k X σk−1 (n)q n Gk (τ ) = 2ζ(k) 1 − Bk n=1. !. where q = e2πiτ , and the Bernoulli number Bk are defined by setting ∞. X xk x = Bk k! ex − 1 k=0. Proof. The logarithmic derivative of the product formula for sine is ∞. 1 X 1 1 π cot(πa) = + ( + ), a∈H a n=1 a + n a − n. (6). If we write the left side as πi(eπa + e−πa )/(eπa + e−πa ) = π + 2πi/(e2πia − 1), multiply both sides by a, replace 2πia by x, and expand both series in powers of x, we obtain the well-known formula for ζ(k): ζ(k) = −2(πi)k. Bk 2k!. f or k > 0 even. Next, if we successively differentiate both sides of (6) with respect to a and then replace a by mτ , we obtain: ∞ X. ∞ ∞ X 1 (2πi)k X k−1 2πimnτ 2k = n e = − ζ(k) dk−1 q dm k (mτ + n) (k − 1)! B k n=−∞ n=1 d=1. Thus. Ã ! 1 2k X k−1 dm Gk (τ ) = 2ζ(k) + 2 = 2ζ(k) 1 − d q (mτ + n)k Bk m,d=1 m=1 n=−∞ ∞ X ∞ X. Collecting coefficient of a fixed power q n in the last double sum, we obtain the sum in (5) as the coefficient of q n . This completes the proof.. 11. ¤.

(25) Because of Proposition 3.1, it is useful to define the normalized Eisenstein series, obtained by dividing Gk (τ ) by the constant 2ζ(k) in: ∞ 2k X Gk σk−1 (n)q n =1− Ek (τ ) = Bk n=1 2ζ(k). Thus, Ek (τ ) is defined so as to have rational q-expansion coefficients.The first few Ek are: E4 (τ ) = 1 + 240 E6 (τ ) = 1 − 504 E8 (τ ) = 1 + 480. ∞ X n=1 ∞ X n=1 ∞ X. σ3 (n)q n ; σ5 (n)q n ; σ7 (n)q n. n=1. Next we consider the second example. Dedekind eta function Definition 3.2. Let τ be a complex number with Im τ > 0.The ordinary Dedekind eta function is defined by πiτ /12. η(τ ) = e. ∞ Y. (1 − e2πiτ ). n=1. This function plays an important role in the study of the theory of modular function and its applications to other areas. One of the most important properties of the eta function is the transformation formula, which will be used in this paper. Ã ! a b ∈ SL2 (Z), the transformation formula for η(τ ) Proposition 3.2. For γ = c d is given by, for c = 0 η(τ + b) = eπib/12 η(τ ), and,for c 6= 0. r η(γτ ) = ²(a, b, c, d). with. cτ + d η(τ ) i. µ ¶ d (1−c)/2 πi(bd(1−c2 )+c(a+d))/12   i e , if c is odd. ²(a, b, c, d) = ³ c´   c eπi(ac(1−d2 )+d(b−c+3))/12 , if d is odd. d 12.

(26) µ ¶ d is the Legendre-Jacobi symbol where c. M. Newman [2] gave criteria for a product of η−function to be modular on Γ0 (N ) Proposition 3.3. If f (τ ) = Q (1) (drd ) is a square, P (2) drd ≡ 0mod 24, PN r ≡ 0 mod 24, (3) d d. Q d|N. η(dτ )rd satisfies. then f (τ ) is a modular function on Γ0 (N ) of weight. 1 2. P. rd .. We will use these results to find modular functions suitable for our purpose. Finally, we discuss the properties of the action of Atkin-Lehner involutions on modular forms. Proposition 3.4. Let n be a positive integer with gcd(n, N/n) = 1. Let f (τ ) be a ¯ modular form of weight k on Γ0 (N ). Then f ¯[γ] for γ ∈ wn is again a modular form ¯ ¯ ¯ of weight k on Γ0 (N ), and that f ¯[γ1 ] = f ¯[γ2 ] for all γ1 ,γ2 ∈ wn .Note that f (τ )¯[γ] is ¯ defined by f (τ )¯ = (det γ)k/2 (cτ + d)−k f (γτ ) [γ]. Proof. Recall that wn normalizes Γ0 (N ). That is, wn Γ0 (N )wn−1 = Γ0 (N ). We have, for all α ∈ Γ0 (N ),. ¯ ¯ ¯ f ¯[γ] ¯[α] = f ¯[α0 γ] ,. where α0 is an element in Γ0 (N ). Since f is a modular form on Γ0 (N ), it follows that. ¯ ¯ ¯ f ¯[γ] ¯[α] = f ¯[γ] .. ¯ ¯ ¯ That is, f ¯[γ] is modular on Γ0 (N ). The assertion that f ¯[γ1 ] = f ¯[γ2 ] for all γ1 , γ2 ∈ wn follows from the fact that γ1−1 γ2 ∈ Γ0 (N ). This proves the proposition.. ¤. 4. Beukers’ Proof In [1], Beukers gave a modular form interpretation of the Apéry sequence. Now we will describe the Beukers’ proof in this section. P n Let t(q) = ∞ n=0 tn q be a power series convergent for all |q| < 1 and W (q) be another analytic function on |q| < 1. Then consider W as function of t. In general it will be a multivalued function over which we have no control. However, we shall. 13.

(27) introduce some assumptions. First, t0 = 0, t1 6= 0. Let q(t) be the local inverse of t(q) with q(0) = 0. Choose W (q(t)) for the value of w around t = 0. Then in P n order to determine the radius of convergence of power series W (q(t)) = ∞ n=0 wn t we introduce branching values of t. We say that t branches above t0 , if either t0 is not in the image of t, or if t0 (q0 ) = 0 for some q0 with t(q0 ) = t0 . Now assume that t has a discrete set of branching values t1 , t2 , · · · where we have exclude zero as a possible value and suppose |t1 | < |t2 | < · · · . It is clear now that the radius of convergence is in general t1 . We shall be interested in cases where the radius of convergence is larger than |t1 |. Let γ be a closed contour in the complex t-plane beginning and ending at the origin, not passing through any ti and which encircles the point t1 exactly once. Suppose that analytic continuation of W (q(t)) along γ again yields the same branch of W (q(t)). Then W (q(t)) can be continued analytically to the disc |t| < |t2 | with exception of possible isolated singularity t1 . If W (q(t)) remains bound around t1 we can conclude that the radius of convergence is at least |t2 |. The construction of the function t(q) and W (q) will proceed using modular forms and functions. The value for which Beukers obtain irrationality results are in fact values at integral points of Dirichlet series associated to modular forms. We recall two propositions of Beukers. For completeness we also include their proofs. Proposition 4.1 (Beukers). 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 [1, . . . , n]r where r, d are certain fixed positive integers and [1, . . . , n] is the lowest common multiple of 1, . . . , 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 nonzero Taylor coefficients. If ρ > der ,then at least one of θ1 , θ2 , · · · θk is irrational. Proof. Choose ε > 0 such that ρ > der(1+ε) . Let fi (t) =. P∞ n=0. ain tn Since the radius. of convergence of f0 (t) + θ1 f1 (t) + θ2 f2 (t) + · · · + θk fk (t) is ρ, we have for sufficiently large n, |a0n + a1n θ1 + · · · + akn θk |. Suppose θ1 θ2 , . . . , θk are all rational and we have common denominator D. Then An = Ddn [1, . . . , n]r |a0n + a1n θ1 + · · · + akn θk | is an integer smaller than Ddn [1, . . . , n]r (ρ − ε)−n . By the prime number theorem. 14.

(28) (1+ε)r. we have [1, . . . , n] < e(1+ε)n for sufficiently large n, hence |An | < D( deρ−ε )n . Since. der(1+ε) (ρ−ε)−1 < 1 this implies that An = 0 for sufficiently large n, in contradiction with our assumption An 6= 0 for infinitely many n. Thus our proposition follows. ¤ Proposition 4.2 (Beukers). Let F (τ ) =. P∞. an q n , q = e2πiτ , be a Fourier series. n=1. convergent for |q| < 1,such that for some k,N ∈ N, √ F (−1/N τ ) = ε(−iτ N )k F (τ ). where ε = ±1. Let f (τ ) be the Fourier series ∞ X an n q f (τ ) = nk−1 n=1. Let L(F, s) =. ∞ X an n=1. and finally,. X. h(τ ) = f (τ ) −. 0≤r< k−2 2. Then. ns. L(F, k − r − 1) (2πiτ )r r!. √ h(τ ) − D = (−1)k−1 ε(−iτ N )k−2 h(−1/N τ ) k. where D=0 if k is odd and D = L(F, k2 )(2πiτ ) 2 −1 /( k2 − 1)! if k is even. Moreover,. L(F, k2 ) = 0 if ε = −1. √ Proof. We apply a lemma of Hecke, see [4] with G(τ ) = εF (τ )/(i N )k. to obtain f (τ ) − ε(−1). k−1. (−iτ. p. k−2. (N )). f (−1/N τ ) =. k−2 X L(F, k − r − 1) r=0. r!. (2πiτ )r. Split the summation on the right hand side into summation over r < k2 −1, r > k2 −1. and, possibly ,r =. k 2. − 1.For the region r >. k 2. − 1 we apply the functional equation. √ L(F, k − r − 1) L(F, r + 1) = ε(−1)k (−i N )k−2 (−1/N )k−r−2 (2πi)k−2r−2 r! (k − r − 2)!. and substitute r by k − 2 − r.. ¤. 15.

(29) Having introduced these two propositions, we start to describe Beukers’ proof. He first defined a modular function t on Γ0 (6) + w6 , and found the branching values √ √ √ √ t(1/2) = ∞ t(2/5+i/5 6) = ( 2+1)4 , t(i∞) = 0, t(i/ 6) = ( 2−1)4 ,. of t. Thus, if one writes a modular form E(τ ) on Γ0 (6) as a series of t, the series in √ general has a radius of convergence ( 2 − 1)4 . Then Beukers found a modular form. F (τ ) of weight 4 such that the conditions in Proposition 4.2 holds with F (−1/6τ ) = −36τ 4 F (τ ) (that is, ² = −1) and L(F, 3) = ζ(3). Thus, choosing E(τ ) to be of weight 2 with E(−1/6τ ) = −6τ 2 E(τ ) and setting ( dτd )3 f (τ ) = (2πi)3 F (τ ), by. Proposition 4.2, we see that E(−1/6τ )(f (−1/6τ ) − ζ(3)) = E(τ )(f (τ ) − ζ(3)) From this Beukers concluded that the radius of convergence of E(t)(f (t) − ζ(3)) equals at least the next branching value. He also checked that the coefficients of P n 3 E(t) ∈ Z[t] and E(t)f (t) = ∞ n=1 bn t ,where bn ∈ Z/[1, 2 . . . n] . Finally, he proved that ζ(3) is irrational by applying Proposition 4.1. Our construction of sequences converging to ζ(5) basically follows Beuker’s approach. However, our result is not strong enough to conclude that ζ(5) is irrational. Here we give a weaker version of Proposition 4.1 applicable to our situation. Proposition 4.3. Let f0 (t) =. P. an tn , f1 (t) =. P. bn tn be power series in t. Suppose. that θ is a real number such that f0 (t) − θf1 (t) has radius of convergence α and lim sup 1/(α|bn |1/n ) < 1, n→∞. then an /bn converges to θ.. 16.

(30) 5. Differential equations satisfied by modular forms In this section, we will give the result of [3] and introduce the the method of [5]. Theorem 2 (Stiller). Let Γ be a discrete subgroup of SL2 (R) commensurable with SL2 (Z). Suppose that t = t(q) is a non-constant (meromorphic) modular function invariant under Γ, and F (t) = F (t(q)) is a (meromorphic) modular form of weight k on the group Γ with respect to a multiplier system χ. Then the functions F (t), τ F (t), . . ., τ k F (t) are linearly independent solutions of a (k + 1)-st order linear differential equation. The idea of Theorem 2 is to consider the vector-valued function V (τ ) = (F (τ ), τ F (τ ), . . . , τ k F (τ )). This function behaves like a modular function (of weight 0), and so do the derivatives dm V /dtm . Thus the coefficients of the linear relation among k + 2 vectors dm V /dtm , m = 0 . . . k + 1, are Γ-invariant, and thus are algebraic functions of t. From the general theory of differential equations, we know that the differential equations in Theorem 2 can be expressed in terms of the Wronskians. In practice, we find the following proof of Y. Yang more suitable for the computational purpose. Theorem 3 (Yang). Setting G1 =. Dq t , t. G2 =. Dq F F. and p1 (t) =. Dq G1 − 2G1 G2 /k , G21. p2 (t) = −. Dq G2 − G22 /k , G21. then the differential equations satisfied by F and t are for k = 1, Dt2 F + p1 Dt F + p2 F = 0, for k = 2, Dt3 F + 3p1 Dt2 F + (2p21 + tp01 + 2p2 )Dt F + (2p1 p2 + tp02 )F = 0, and in general rm (t) are polynomials of t, p1 , p2 , and derivatives of p1 and p2 .. 17.

(31) d Remark. The notation in Theorem 3 Dt and Dq are denoted by Dt = t dtd , Dq = t dq. Y. Yang first prove a lemma showing that the functions p1 and p2 in the statement of Theorem 3 are indeed algebraic functions of t. Lemma 5.1. Let t, F , G1 and G2 be given as in Theorem 3. Then G1 is a meromorphic modular form of weight 2, while Dq G1 − 2G1 G2 /k and Dq G2 − G22 /k are meromorphic modular forms of weight 4. Proof. Throughout the proof of the lemma we let f˙(τ ) denote the derivative of a function f (τ ) with respect to τ . The meromorphic property of the functions concerned is clear. We Ã now show ! a b that the functions have the claimed modular property. Since, for all γ = ∈ c d Γ, t(γτ ) = t(τ ),. F (γτ ) = χ(γ)(cτ + d)k F (τ ),. taking the logarithmic derivatives of the above equalities with respect to τ , we obtain. t˙ t˙ (γτ ) = (cτ + d)2 (τ ), t t. F˙ F˙ (γτ ) = kc(cτ + d) + (cτ + d)2 (τ ), F F. or equivalently, (7). G1 (γτ ) = (cτ + d)2 G1 (τ ),. G2 (γτ ) =. 1 kc(cτ + d) + (cτ + d)2 G2 (τ ). 2πi. This shows that G1 is a meromorphic modular form of weight 2. Differentiating the expressions in (7) with respect to τ again, we obtain G˙ 1 (γτ ) = 2c(cτ + d)3 G1 (τ ) + (cτ + d)4 G˙ 1 (τ ) and. 1 G˙ 2 (γτ ) = kc2 (cτ + d)2 + 2c(cτ + d)3 G2 (τ ) + (cτ + d)4 G˙ 2 (τ ). 2πi It follows that ½ ¾ 4πi 4πi 4 G˙ 1 (γτ ) − G1 (γτ )G2 (γτ ) = (cτ + d) G˙ 1 (τ ) − G1 (τ )G2 (τ ) k k. and. ½ ¾ 2πi 2πi 2 4 2 G˙ 2 (γτ ) − G2 (γτ ) = (cτ + d) G˙ 2 (τ ) − G2 (τ ) . k k. This shows that Dq G1 −2G1 G2 /k and Dq G2 −G22 /k are meromorphic modular forms of weight 4, and the proof of the lemma is completed.. 18. ¤.

(32) We are now ready to prove our Theorem 3. Proof of Theorem 3. With a slight abuse of notation we will alternate the use of F (τ ), F (q) and F (t) freely. We first show that F satisfies a (k+1)-st order differential equation. By the definitions of Dt , Dq , G1 and G2 we have (8). Dt F = t. G2 F G2 Dq F , =F =t G1 tG1 Dq t. and Dt. Dq G2 G2 G2 Dq G1 t Dq G2 G2 − 3 Dq G1 . = −t 2 = G1 G21 G1 Dq t G1 Dq t G1. By Lemma 5.1 the functions Dq G1 − 2G1 G2 /k and Dq G2 − G22 /k are meromorphic modular forms of weight 4, and so is G21 . Therefore we can write (Dq G1 − 2G1 G2 /k)/G12 and (Dq G2 − G22 /k)/G21 as algebraic functions of t, say, Dq G1 − 2G1 G2 /k , G21. p1 (t) =. p2 (t) = −. Dq G2 − G22 /k . G21. Thus, we have (9). Dt. G2 G2 G2 = − 22 − p1 − p2 . G1 kG1 G1. Using (8) and (9) we can now compute higher order derivatives of F inductively. We have. µ Dt2 F. = Dt. and Dt3 F. G2 F G1. ¶. ½ =F. G2 G2 (1 − 1/k) 22 − p1 − p2 G1 G1. ½ ¾ G22 G2 (1 − 1/k) 2 − p1 − p2 G1 G1 ½ ¾ G2 G2 G2 0 G2 0 + F 2(1 − 1/k) Dt − tp1 − p1 Dt − tp2 G1 G1 G1 G1. G2 =F G1. (. ¾. = F (1 − 1/k)(1 − 2/k). G32 G22 + (3/k − 3)p 1 2 G31 G1. ) ¢ G 2 + (2/k − 3)p2 − tp01 + p21 + p1 p2 − tp02 . G1 ¡. It follows that, for k = 1, Dt2 F = −p1 F. G2 − p2 F = −p1 Dt F − p2 F, G1 19.

(33) and, for k = 2, Dt3 F. =. −3p1 Dt2 F. ¾ ½ G2 0 2 0 − 2p1 p2 − tp2 + F (−2p1 − tp1 − 2p2 ) G1. = −3p1 Dt2 F + (−2p21 − tp01 − 2p2 )Dt F − (2p1 p2 + tp02 )F. In general, the n-th derivative takes the form ) ( n−2 n−1 n n−1 Y G G G 2 (1 − j/k) + sn,n−1 2n−1 + sn,n−2 2n−2 + · · · , Dtn F = F Gn1 j=1 G1 G1. where sn,j are polynomials of t, p1 , p2 and their derivatives. When n = k + 1, the k+1 term involving Gk+1 is annihilated, and we see that Dtk+1 F is equal to a linear 2 /G1. sum of lower order derivatives of F (with algebraic functions of t as coefficients). ¤ In the final section, we will apply the method of [5] to find the differential equations for n = 5. Moreover, we will obtain the recursive relation of sequences an and bn from these differential equations.. 20.

(34) 6. Construction of series converging to ζ(5) Now we start to construct a sequence {cn } of rational numbers converging to ζ(5) in this section. According to our idea described in section 4, we have to find three functions F (τ ), E(τ ) and t(τ ) appropriately and consider the power series E(t)(f (t) − ζ(5)). Now we will divide this section into two parts. We will find the sequence cn and prove it converges to ζ(5) in first part. Then we will apply the method of [5] to find the recursive relations in the second part. 6.1 To find suitable t, E and F , our first task is to determine the congruence subgroup that they should be modular on. First of all, we notice that the zeta functions appear naturally in the L-function associated with an Eisenstien series. Namely, we have ∞ X σk−1 (n) n=1. ns. =. ∞ ∞ ∞ X 1 X k−1 X 1 X 1 = ζ(s + 1 − k)ζ(s). d = s s+1−k s n d n n=1 n=1 d=1 d|n. From Proposition 4.2 we see that in order to get ζ(5) we require the weight k to be 6. Furthermore, taking account that the Atkin-Lehner involution wN identitify cusps by pairs and that we have two conditions L(F, 4) = 0, L(F, 3) = 0 in Proposition 4.2 that must be fulfilled, the congruence subgroup Γ0 (N ) must have at least 6 cusps. The smallest positive integer N with this property is 12. That is, we shall choose t, E, F to be modular on Γ0 (12). We first consider the E function. To make the computation of the differential Q equation satisfied by E easier, we choose E to be a product d|12 η(dτ )ed of ηP functions, where ed = 8. Thus, if E is to satisfy E(−1/12τ ) = ±122 τ 4 E(τ ), then the sign must be positive. In other words, E satisfies (1) E(τ ) ∈ M4 (Γ0 (12)), (2) E(−1/12τ ) = 122 τ 4 E(τ ). Accordingly, by Proposition 3.3, the exponents ed should satisfy P (i) ed = 8, (ii) e1 = e12 , e2 = e6 , e3 = e4 , Q (iii) d|12 is a square, P (iv) ded , (v) E(τ ) is holomorphic at τ = 1/2 and τ = 1/3,. 21.

(35) Then from these conditions we find that we have the following five choices for Efunction: (e1 , e2 , e3 ) = (−4, −4, 12) (−5, 2, 7) (−6, 8.2) (−7, 14, −3) (−8, 20, −8). We now consider the possible choices of the modular form F . Let f (τ ) be determined by d5 f (τ )/dτ 5 with the constant term in the Fourier expansion being 0. In order for f (τ ) to satisfy E(−1/12τ )(f (−1/12τ ) − ζ(5)) = E(τ )(f (τ ) − ζ(5)), by Proposition 4.2, the function F should meet the conditions (i) F (τ ) ∈ M6 (Γ0 (12)), F (i∞) = 0, (ii) F (−1/12τ ) = 123 τ 6 F (τ ), (iii) L(F, 3) = L(F, 4) = 0 and L(F, 5) = ζ(5), where L denotes the Dirichlet series associated with F (τ ). The first condition means that the q-expansion of F start from q. That is F (q) = a1 q + a2 q 2 + · · · . Meanwhile, to fulfill condition (ii), the function F should be a linear combination of f1 (τ ) = E6 (τ )+123 E6 (12τ ) , f2 (τ ) = E6 (2τ )+33 E6 (6τ ) , f3 (τ ) = 33 E6 (3τ )+26 E6 (4τ ) where fi (1/12τ ) = 123 τ 6 fi (τ ), for i = 1, 2, 3. That is, we have xF (τ ) = A[E6 (τ ) + 123 E6 (12τ )] + B[E6 (2τ ) + 33 E6 (6τ )] + C[33 E6 (3τ ) + 26 E6 (4τ )] where A, B, C, x are constants. Then from the conditions (i) and (iii) we have the following equations:.   13A + B + 7C = 0    4 145A + 5B + 25C = x  7    1729A + 28B + 91C = 0. =⇒ A = A , B = −104A , C = 13A , x = −175A/2 where A is a constant. Thus the F function can be decided. Finally, we consider the t function. Referring to Beukers’ proof, now we want to define a t-function which is modular with respect to Γ0 (12) + w12 . The choice of. 22.

(36) t-function is similarly with E-function. We also construct the t-function by Proposition 3.3 and Proposition 3.4 as follows: Y t(τ ) = η(dτ )ed d|12. where d and ed satisfy the following conditions: P (i) ed = o, (ii) e1 = e12 , e2 = e6 , e3 = e4 , P (iii) ded = 24( mod 24), (iv) t has only one simple zero at τ = i∞, From these above conditions we choose the t-function to be ¶4 µ η(τ )η(12τ ) t= η(3τ )η(4τ ). We now determine the branching values of t, which occurs at either the elliptic points or the cusps. In other words, we need to evaluate the values of t at τ = √ √ 1/2, 1/3, i/ 12, and (2+ −12)/5. We first note that t and E are modular on Γ0 (12),. which has six inequivalent cusps 1, 1/2, 1/3, 1/4 and 1/6, 1/12. Furthermore, the function field of modular functions is generated by µ ¶ η(τ )η(12τ )3 g= η(4τ )η(3τ )3. and the value of g at cusps are given by g(0) = 1/4 g(1/2) = −1/2 g(1/3) = ∞ g(1/4) = 1 g(1/6) = 1/2 g(1/12) = 0 The function y(−1/12τ ) is again invariant on Γ0 (12) and one easily checks that (10). y(τ ) − 1/4 y(τ ) − 1. y(−1/12τ ) =. Moreover, the modular function t and g have the following relation (11). t(τ ) = g(τ ). 1 − 4g(τ ) 1 − g(τ ). That t(i∞) = 0, t(1/3) = ∞ can been seen from the values g(i∞) = 0 , g(1/3) = ∞. √ √ From (10) it follows that for τ = i/ 12 and g0 = g(i/ 12) we have g0 = (g0 − √ √ √ 1/4)/(g0 − 1), hence g0 = 1 ± 5/2 and correspondingly , t(i/ 12) = ( 3 ± 1)4 /4.. 23.

(37) On other hand, the next branching value is t(1/6) = −1. Therefore, we have the branching values of t as follows: √ √ √ √ ( 3 + 1)4 ( 3 − 1)4 , t(1/6) = −1, t((2 + −12)/5) = t(i∞) = 0, t(i/ 12) = 4 4 Now we are ready to give the proof of Theorem 1. Proof of Ã Theorem 1. Let −175 F (τ ) 2. !. E6 (τ ) + 123 E6 (12τ ). =. E(τ ) =. Ã. !. − 104 E6 (2τ ) + 27E6 (6τ ). Ã + 13 27E6 (3τ ) + 26 E6 (4τ ). 12. (η(3τ )η(4τ )) 4 (η(2τ )η(6τ )η(τ )η(12τ )). Notice that F (τ ) ∈ M6 (Γ0 (12)) and F (−1/12τ ) = 123 τ 6 F (τ ), F (i∞) = 0 and E(τ ) ∈ M4 (Γ0 (12)), E(−1/12τ ) = 122 τ 4 E(τ ) . Then the Dirichlet series corresponding to F (τ ) reads:. L(F, s) =. ∞ X n=1. =. (µ. ¶ −504σ5 (n) −504σ5 (n) + − 104 (6n)s (2n)s µ ¶) −504σ5 (n) 6 −504σ5 (n) +13 27 +2 (3n)s (4n)s. −504σ5 (n) −504σ5 (n) + 123 s (12n)s n. ¶. µ. ¢ 1008 ¡ 1 + 123−s − 104(2−s + 33 6−s ) + 13(33−s + 26−2s ) ζ(s)ζ(s − 5) 175. Define f (τ ) by ( dτd )5 f (τ ) = (2πi)5 F (τ ) , f (i∞) = 0 . From Proposition 4.2 and. the fact that F (−1/12τ ) = 123 τ 6 F (τ ) follows 122 τ 4 [f (−1/12τ ) − L(F, 5)] = [f (τ ) − L(F, 5)] and since L(F, 5) = ζ(5) , E(−1/12τ ) = 122 τ 4 E(τ ), we have 122 τ 4 [f (−1/12τ ) − ζ(5)] = [f (τ ) − ζ(5)] Multiplication with E(−1/12τ ) = 122 τ 4 E(τ ) yields (12). E(−1/12τ )[f (−1/12τ ) − ζ(5)] = E(τ )[f (τ ) − ζ(5)] The function E(τ )[f (τ ) − ζ(5)] can be considered as a multivalued function of. t(τ ). We choose it at t = 0 as follows. From the expansion t = q − 4q 2 + 2q 3 + · · · one infers the inverse expansion q = t + 4t2 + 30t3 + · · · . Then, from E(τ ) =. 24. !.

(38) 1 + 4q + 18q 2 + · · · one finds E(t) = 1 + 4t + 34t2 + 308t3 + · · · and similarly , E(t)f (t) = 144/25t + 1665/50t2 + · · · . Since the inverse function t 7→ τ branches at t =. √ ( 3−1)4 4. one expects the radius. √ 4 . However, by (12), the function of convergence of E(t)[f (t) − ζ(5)] to be ( 3−1) 4 √ ( 3−1)4 , and its radius of convergence t 7→ E(t)[f (t)−ζ(5)] has no branch point at t = 4. equals at least the next branching value, which is −1 . Then we can conclude that p 1 where β = 1 lim sup n | bn − an ζ(5)| ¿ β n→∞. where bn and an are the coefficients of E(t)f (t) and E(t). On other hand, we know the coefficients an of E(t) from the branching values. That is p √ 1 where α = ( 3 − 1)4 /4 as n −→ ∞ lim sup n |an | ≤ α n→∞. Thus,. s¯ ¯ ¯ α ¯ n ¯ bn lim sup ¯ − ζ(5)¯¯ ≤ β an n→∞. √ since α/β = ( 3 − 1)4 /4 < 1. Then (α/β)n tends to 0 as n−→∞ . Thus bn /an. converges to ζ(5). This completes the proof of Theorem 1.. ¤. 6.2 In this subsection, we will find the sequences an and bn by applying the method of Y.Yang. The main purpose of this part is to prove the following proposition. Proposition 6.1. Let µ ¶4 η(τ )η(12τ ) t= η(3τ )η(4τ ). E(τ ) =. (η(3τ )η(4τ ))12 (η(2τ )η(6τ )η(τ )η(12τ ))4. Then we have (13). Dt5 E + r4 (t)Dt4 E + r3 (t)Dt3 E + r2 (t)Dt2 E + r1 (t)Dt E + r0 (t)E = 0, r4 (t) = 20. (t2 − 10t − 3)t (1 + t)(t2 − 14t + 1). r3 (t) = 80. (2t5 − 41t4 + 182t3 + 158t2 + 12t − 1)t (t + 1)2 (t2 − 14t + 1)2. r2 (t) = 20. (32t6 − 557t5 + 1745t4 + 3310t3 + 1202t2 + 31t − 3)t (t + 1)3 (t2 − 14t + 1)2. 25.

(39) (160t7 − 2343t6 + 4246t5 + 16963t4 + 12980t3 + 2743t2 + 6t − 3)t r1 (t) = 8 (t + 1)4 (t2 − 14t + 1)2. r0 (t) = 4. (256t7 − 3367t6 + 4740t5 + 18565t4 + 12368t3 + 2019t2 − 20t − 1)t (t + 1)4 (t2 − 14t + 1)2. If we set E(t) =. P∞ n=0. an tn which satisfy the differential equation (13), then we. can easily obtain the recursive relation with the initial conditions: n5 an = A7 an−1 − A6 an−2 − A5 an−3 − A4 an−4 − A3 an−5 − A2 an−6 + A1 an−7 − A0 an−8 a0 = 1 , a1 = 4 , a2 = 34 , a3 = 308 , a4 = 3083 , a5 = 32696 , a6 = 361428 a7 = 4119288 . Thus we get the sequence an from the recursive relation. On other hand, we introduce the following proposition to find the sequence bn . Proposition 6.2. Suppose that t(τ ) and E(τ ) are given in Proposition 6.1, and that F (τ ) and f (τ ) be defined as in the proof of Theorem 1. Then A(τ ) = E(τ )f (τ ) satisfies the inhomogeneous differential equations Dt5 A + r4 (t)Dt4 A + r3 (t)Dt3 A + r2 (t)Dt2 A + r1 (t)Dt A + r0 (t)A = H(t). (14) where. 144 9504 2 3744 4 5616 5 t− t + t + t 25 25 25 25 P∞ n and G1 = Dq t/t. Moreover, if we set E(t)f (t) = n=0 bn t . Then bn satisfy H(t) =. the recursive relation given in the statement of Theorem 1 with the initial values b0 = 0. b1 = 144/25. b2 = 333/10. b3 = 217042/675 b4 = 138004123/43200 b6 = 25297127932859/67500000. b5 = 1144320384083/33750000. b7 = 2422896637170749569/567236250000.. Before we prove the proposition 6.1 and 6.2, we introduce the formula for the number of zeros of modular form with respect to Γ0 (12). This result will be used in the proof of Proposition 6.1. Lemma 6.1. Let f be a nonzero modular function of weight k for Γ0 (12). For p ∈ H, let vp (f ) denote the order of zero of f (τ ) at the point p. Let v∞ (f ) denote the index of the first nonvanishing term in the q − expansion of f (τ ). Then we have the formula. 26.

(40) K. A. 0 B. 1/6 1/5 1/4 C D E. 1/3 F. 1/2 G. 2/3 H. 3/4 I. 1 J. Figure 2. Fundamental domain of Γ0 (12). (15). X. v∞ (f ) +. vp (f ) = 2k. p∈Γ0 (12)\H. Proof. The idea of the proof is to count the zeros and poles in Γ0 (12)\H by integrating the logarithmic derivative of f (τ ) around the boundary of the fundamental domain F . More precisely, let L be the contour in Figure 2. The top of L is a horizontal line from K = 1 + iT to A = 0 + iT , where T is taken larger than the imaginary part of any of the zeros or poles of f (τ ). The rest contour follows around the boundary of F , except that it detours around any zeros or poles on the boundary along circular arcs of small radius ε. According to the residue theorem, we have Z 0 X 1 f (τ ) (16) dτ = vp (f ) 2πi L f (τ ) p∈Γ\H. On the other hand, we evaluate the integral in (16) section by section. First of all, the integral from A to B cancels the integral from J to K. Next, we evaluate the integral over KA. To do this we make the change of variables q = e2πiτ . P Let f˜(q) = f (τ ) = an q n be the q-expansion. Since f 0 (τ ) = d f˜(q) dq , we find that dq. dτ. this section of the integral in (15) is equal to the following integral over the circle of radius e−2πT centered at zero : 1 2πi. Z. df˜/dq dq. f˜(q). Since the circle is traversed in a clockwise direction as τ goes from K to A, it follows that this integral is minus the order of zero or pole of f˜(q) at 0, and this is what we mean by −v∞ (f ).. 27.

(41) Finally, we consider the integral from B to J. Since f (τ ) is a nonzero modular function of weight k for Γ0 (12), we have f (A(τ )) = (cτ + d)k f (τ ), where A(τ ) =. aτ +b . cτ +d. Differentiation of this equation gives us. f 0 (A(τ ))(A0 (τ )) = (cτ + d)k f 0 (τ ) + kc(cτ + d)k−1 f (τ ) From this we find. kc f 0 (τ ) f 0 (A(τ ))(A0 (τ )) + = cτ + d f (τ ) f (A(τ )) Consequently, for any path γ not through a zero we have Z Z 0 Z kc 1 f (τ ) 1 f 0 (u) 1 dτ dτ + du = (17) 2πi γ cτ + d 2πi γ f (τ ) 2πi A(γ) f (u). Therefore the integrals of semicircles along the B to J in Figure do not cancel unless k = 0. Now thereÃare eight ! semicircles between B and J. We note that 7 −1 the transformation γ1 = takes BC to DC, i.e., γ1 τ goes from D to C 36 −5 along the contour goes from B to C along Ãthe contour. Similarly, we can Ã as τ ! ! −19 4 −17 5 find that γ2 = takes DE to JI, γ3 = takes EF to IH and −24 5 −24 7 Ã ! −7 3 γ4 = takes F G to HG. So the integral from B to J can be evaluated as −12 5 follows: (Z Z Z Z Z Z Z Z Z ) 1 f 0 (τ ) 1 = + + + + + + + 2πi BJ f (τ ) 2πi BC CD DE DE FG GH HI IJ. 1 = 2πi. 1 = 2πi. (Z. (Z BC. Z. Z. − BC. Z. + γ1 (BC). 36k dτ + 36τ − 5. Z DE. Z. − DE. Z. + γ2 (DE). −24k dτ + −24τ + 5. − FG. Z FG. + γ3 (F G). ). Z − HI. −12k dτ + −12τ + 5. Therefore, we have the following conclusion: Z 1 f 0 (τ ) = 2k 2πi BJ f (τ ) 28. Z. γ4 (HI). Z EF. −24k dτ −24τ + 7. ).

(42) This completes the proof.. ¤. Proof of Proposition 6.1. Referring to the proof of Y.Yang, by using (8) and (9) we can now compute higher order derivatives of F inductively. We have. µ Dt2 F. = Dt. G2 F G1. ¶. ½ =F. G2 G2 − p2 (1 − 1/k) 22 − p1 G1 G1. ¾. and. Dt3 F. ½ ¾ G22 G2 (1 − 1/k) 2 − p1 − p2 G1 G1 ½ ¾ G2 G2 G2 0 G2 0 + F 2(1 − 1/k) Dt − tp1 − p1 Dt − tp2 G1 G1 G1 G1. G2 =F G1. (. = F (1 − 1/k)(1 − 2/k). G32 G22 + (3/k − 3)p 1 2 G31 G1. ) ¢ G 2 + (2/k − 3)p2 − tp01 + p21 + p1 p2 − tp02 . G1 ¡. (. Dt4 F = F (1 − 1/k)(1 − 2/k)(1 − 3/k). G42 G32 + 6p (1 − 1/k)(1 − 2/k) + 1 G41 G31. (7p1 − 6p2 − 4p01 t + (−7p21 + 14p2 + 4p01 t)/k − 8p2 /k 2 ). G22 − G21. G2 (2) (p31 − 10p1 p2 + p01 t + p1 t2 + 4p02 t − 3p01 tp1 + (8p1 p2 − 2p02 t)/k) − G1 ) (2). (p02 t + p2 t2 + p21 p2 − p1 p02 t − 2p01 tp2 − 3p21 + (2p22 )/k) .. 29.

(43) ( Dt5 F = F. − p01 t. G2 G2 (2) − 15/4p01 t − 3p2 t2 + p1 p02 t + 3p01 tp2 + − 9/2p02 t G1 G1. (4) G2. p1. G1. − 8p1 p22 − 15/16p1. G32 G42 0 G2 + p31 p2 + − 15/8p − 15/4p t 2 1 3 4 G1 G1 G1. 75/8p21. 2 G2 G32 (3) 3 2 0 3 G2 − p1 p02 t+ − p t + 17/2p tp − 45/4p 2 2 2 1 2 + 39/2p1 p2 3 G1 G1 G1. 17/2p22. G2 G2 G2 (2) G2 − 45/8p02 t 22 − + 4p01 tp1 − p02 t − p02 t − 3p1 t2 G1 G1 G1 G1. (2) G2 9/2p2 t2 G1. −. 2 (2) 2 G2 15/4p1 t 2 G1. G2 G2 G22 + 25/2p1 p02 t + + 75/4p1 p2 2 + 16p01 tp2 G1 G1 G1. G2 G2 G22 (2) (2) (2) + 4p1 t2 p1 + + 3p01 t2 p02 + p1 p2 t2 + 3p1 t2 p2 − p1 (3)t3 2 G1 G1 G1 ) G2 G2 − 5p01 tp1 p2 − 6p01 tp21 3(p01 )2 t2 G1 G1. 75/4p01 tp1. (2). = −10p1Dt4 F − (5p2 + 35p21 + 10p01 t)Dt3 F − (5p01 t + 50p31 + 15/2p02 t + 5p1 t2 + (2). 45p01 tp1 + 30p1 p2 )Dt2 F − (p01 t + 4p22 + 7p1 02 t2 + 46p01 tp21 + 11p1 t2 p1 + (3). (2). (2). 24p41 + p1 t3 + 9/2p2 t2 + 30p1 p02 t + 3p1 t2 + 14p01 tp2 + 52p21 p2 + 11p01 tp1 + (3). (2). (2). 9/2p02 t)Dt F − (24p31 p2 + 4p02 tp2 + p2 t3 + 7p01 t2 p02 + 9p1 p2 t2 + 3p2 t2 + (2). 20p01 tp1 p2 + 2p1 t2 p2 + 26p21 p02 t + 9p1 p02 t + 2p01 tp2 + p02 t + 8p1 p22 ) Then we find that Dt5 F can be represented as follows: (18). Dt5 F + r4 (t)Dt4 F + r3 (t)Dt3 F + r2 (t)Dt2 F + r1 (t)Dt F + r0 (t)F = 0,. where rm (t) are polynomials of t, p1 , p2 and their derivatives. Next we want to express the modular functions p1 = (Dq G1 − G1 G2 )/G21 and p2 = (Dq G2 − G22 )/G21 as a rational functions of t. By Lemam5.1 we know that G1 is a meromorphic modular form of weight 2, so G21 is a modular form of weight 4 on Γ0 (12). On the other hand, by Lemma 6.1 we know that a modular form of weight 4 on Γ0 (12) has eight zeros and by (11), the expression for (Dq G1 − G1 G2 )/G21 in terms of t takes the form (a0 + a1 t + a2 t2 + a3 t3 + a4 t4 )/(b0 + b1 t + b2 t2 + b3 t3 + b4 t4 ). Comparing the first terms of (Dq G1 − G1 G2 )/G21 with (a0 + a1 t + a2 t2 + a3 t3 + a4 t4 )/(b0 + b1 t + b2 t2 + b3 t3 + b4 t4 ), we obtain p1 as follows: p1 (t) =. 2t4 − 18t3 − 26t2 − 6t (t2 − 14t + 1)(t + 1)2. 30.

(44) Similarly, we have. 4t4 − 28t3 − 36t2 − 4t (t2 − 14t + 1)(t + 1)2 Finally, from (18) we complete the proof. p2 (t) =. ¤. Proof of Proposition 6.2. Recall the the method of variation goes as follows. If u1 , u2 , u3 , u4 , u5 are solutions of Dt5 A + r4 (t)Dt4 A + r3 (t)Dt3 A + r2 (t)Dt2 A + r1 (t)Dt A + r0 (t)A = 0 then a solution v of Dt5 A + r4 (t)Dt4 A + r3 (t)Dt3 A + r2 (t)Dt2 A + r1 (t)Dt A + r0 (t)A = H(t) can be solved by assuming v = p1 u1 + p2 u2 + p3 u3 + p4 u4 + p5 u5 with    (Dt p1 )u1 + (Dt p2 )u2 + (Dt p3 )u3 + (Dt p4 )u4 + (Dt p5 )u5 = 0       D p D u + Dt p2 Dt u2 + Dt p3 Dt u3 + Dt p4 Dt u4 + Dt p5 Dt u5 = 0    t 1 t 1 Dt p1 Dt2 u1 + Dt p2 Dt2 u2 + Dt p3 Dt2 u3 + Dt p4 Dt2 u4 + Dt p5 Dt2 u5 = 0     Dt p1 Dt3 u1 + Dt p2 Dt3 u2 + Dt p3 Dt3 u3 + Dt p4 Dt3 u4 + Dt p5 Dt3 u5 = 0      Dt p1 D4 u1 + Dt p2 D4 u2 + Dt p3 D4 u3 + Dt p4 D4 u4 + Dt p5 D4 u5 = H(t). t t t t t and then solving Dt p1 , Dt p2 , Dt p3 , Dt p4 , Dt p5 . Now, by Stiller’s theorem, we have u1 = E, u2 = τ E, u3 = τ 2 E, u4 = τ 3 E, u5 = τ 4 E. Thus, using the definition of G1 and G2 , it follows that Dt u1 = EG2 /G1 , Dt u2 = E(1 + τ G2 )/G1 , Dt u3 = τ E(2 + τ G2 )/G1 , Dt u4 = τ 2 E(3 + τ G2 )/G1 ,Dt u5 = τ 3 E(4 + τ G2 )/G1 , where G2 = Dq E/E. The higher derivatives can be computed analogously. At the end, we find Dt p1 =. Hτ 4 G41 −Hτ 3 G41 Hτ 2 G41 −Hτ G41 HG41 , Dt p2 = , Dt p3 = , Dt p4 = , Dt p5 = 24E 6E 4E 6E 24E. Now let Dt5 A + r4 (t)Dt4 A + r3 (t)Dt3 A + r2 (t)Dt2 A + r1 (t)Dt A + r0 (t)A = H(t) be a differential equation satisfied by Ef . Reversing the procedure above, we have p1 u1 + p2 u2 + p3 u3 + p4 u4 + p5 u5 = Ef Then. Z Z Z Z Z 2. 3. 4. 5. p1 + τ p2 + τ p3 + τ p4 + τ p5 = f = (2πi). F τ. 31. τ. τ. τ. τ.

(45) Dt p1 + τ Dt p2 + τ 2 Dt p3 + τ 3 Dt p4 + τ 4 Dt p5. Z Z Z Z 1 F + p2 /G1 + 2τ p3 /G1 + 3τ p4 /G1 + 4τ p5 /G1 = − G1 τ τ τ τ Z Z Z Z 2 3 =⇒ p2 + 2τ p3 + 3τ p4 + 4τ p5 = − F 2. Then. 3. τ 3. τ. τ. τ. 3. Dt p2 + 2τ Dt p3 + 3τ Dt p4 + 4τ Dt p5. Z Z Z 1 F + 2p3 /G1 + 6τ p4 /G1 + 12τ p5 /G1 = G1 τ τ τ Z Z Z 2 =⇒ 2p3 + 6τ p4 + 12τ p5 = F 2. τ. τ. τ. 1 2Dt p3 + 6τ Dt p4 + 12τ Dt p5 + 6p4 /G1 + 24τ p5 /G1 = − G1 Z Z =⇒ 6p4 + 24τ p5 = − F τ τ Z Z 1 F 6Dt p4 + 24τ Dt p5 + 24p5 /G1 = G1 τ τ Z =⇒ 24p5 = F. Z Z. 2. F τ. τ. τ. 24Dt p5 =. HG41. E. =. −1 F G1. Thus. 144 9504 2 3744 4 5616 5 t− t + t + t. 25 25 25 25 where the expression of H in terms of t is determined in the same way as in PropoH(t) = −EF/G51 =. sition 6.1. This completes the proof of Proposition 6.2.. 32. ¤.

(46) References [1] F. Beukers. Irrationality proofs using modular forms. Astérisque, (147-148):271–283, 345, 1987. Journées arithmétiques de Besan¸con (Besan¸con, 1985). [2] Morris Newman. Construction and application of a class of modular functions. Proc. London. Math. Soc. (3), 7:334–350, 1957. [3] P. Stiller. Special values of Dirichlet series, monodromy, and the periods of automorphic forms. Mem. Amer. Math. Soc., 49(299):iv+116, 1984. [4] André Weil. Remarks on Hecke’s lemma and its use. In Algebraic number theory (Kyoto Internat. Sympos., Res. Inst. Math. Sci., Univ. Kyoto, Kyoto, 1976), pages 267–274. Japan Soc. Promotion Sci., Tokyo, 1977. [5] Yifan Yang. On differential equations satisfied by modular forms. Math. Z., 246(1-2):1–19, 2004.. 33.

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