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The Rogers–Ramanujan continued fraction and its level 13 analogue
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The Rogers–Ramanujan continued fraction and its level 13 analogue Shaun Cooper ∗ , Dongxi Ye Institute of Natural and Mathematical Sciences, Massey University-Albany, Private Bag 102904, North Shore Mail Centre, Auckland, New Zealand Received 18 July 2013; received in revised form 14 January 2014; accepted 22 January 2014
Communicated by Special Issue Guest Editor Dedicated to Richard Askey in celebration of his 80th birthday
Abstract One of the properties of the Rogers–Ramanujan continued fraction is its representation as an infinite product given by
(q) = q
1/5
∞
(1 − q ) j
j 5
j=1
where
j p
is the Legendre symbol. In this work we study the level 13 function
R(q) = q
∞
(1 − q j )
j 13
j=1
and establish many properties analogous to those for the fifth power of the Rogers–Ramanujan continued fraction. Many of the properties extend to other levels ℓ for which ℓ − 1 divides 24, and a brief account of these results is included. c 2014 Elsevier Inc. All rights reserved. ⃝
Keywords: Dedekind eta function; Eisenstein series; Hypergeometric function; Modular form; Pi; Ramanujan’s theories of elliptic functions to alternative bases
∗ Corresponding author.
E-mail addresses: [email protected] (S. Cooper), [email protected] (D. Ye). http://dx.doi.org/10.1016/j.jat.2014.01.008 c 2014 Elsevier Inc. All rights reserved. 0021-9045/⃝
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1. Introduction The Rogers–Ramanujan continued fraction (q) is defined for |q| < 1 by q 1/5 q
(q) = 1+ 1+
.
q2 q3 1+ 1 + ···
One of its main properties, due to Rogers [28], is the infinite product formula given by (q) = q
1/5
∞
(1 − q ) j
j=1
j 5
= q 1/5
∞ (1 − q 5 j−4 )(1 − q 5 j−1 ) (1 − q 5 j−3 )(1 − q 5 j−2 ) j=1
where pj is the Legendre symbol. This work is about the level 13 analogue defined by R(q) = q
∞
(1 − q j )
j 13
j=1 ∞ (1 − q 13 j−12 )(1 − q 13 j−10 )(1 − q 13 j−9 )(1 − q 13 j−4 )(1 − q 13 j−3 )(1 − q 13 j−1 ) =q . (1 − q 13 j−11 )(1 − q 13 j−8 )(1 − q 13 j−7 )(1 − q 13 j−6 )(1 − q 13 j−5 )(1 − q 13 j−2 ) j=1
Our goal is to show that although R(q) does not have a simple expansion as a continued fraction, it has many other properties similar to the fifth power of the Rogers–Ramanujan continued fraction. Let us illustrate with two examples. First, if r (q) = 5 (q) then it is well-known that ∞ 1 1 (1 − q j )6 − 11 − r (q) = . r (q) q j=1 (1 − q 5 j )6
Ramanujan found an analogous property for R(q), namely ∞ 1 1 (1 − q j )2 − 3 − R(q) = . R(q) q j=1 (1 − q 13 j )2
This is one of five identities in Entry 8(i) of Chapter 20 in Ramanujan’s second notebook [26]. One of these identities is notable for being the last result in the 21 chapters of the notebook to be proved; see the paper by Evans [18] for more information. Here is the second example. If r = r (q) = 5 (q), then it was shown in [9] that n ∞ d r r (1 − 11r − r 2 ) log (1) q = a(n) dq 1 − 11r − r 2 (1 + r 2 )2 n=0 where a(n) =
2n n
n 2 n n+ j . j j j=0
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A general class series for 1/π involving the coefficients a(n) was derived in [9]. A specific example of such a series is ∞ 1705 71 −1 n 1 = √ a(n) n + . (2) π 682 15228 81 47 n=0 We will show that an analogue of (1) is given by n ∞ d R R(1 − 3R − R 2 ) log = q A(n) dq 1 − 3R − R 2 (1 + R 2 )2 n=0
(3)
where the coefficients A(n) satisfy a 6-term recurrence relation. We will also derive a general class of series for 1/π involving the coefficients A(n) and exhibit the example ∞ 1 161 9 −1 n A(n) n + =√ , (4) π 750 124 31 n=0 which is an analogue of (2). The objective of this work is to provide a systematic development of the main properties of the level 13 analogue R(q). To emphasize the analogy between the level 5 function r (q) and the level 13 function R(q), most of the main results will be presented in pairs: the level 5 result will be stated first followed by its level 13 analogue. The level 5 results are generally known, so references to proofs are provided. Many of the level 13 results are new, in which case proofs are given. 2. Definitions This section is organized in three parts. The first part contains some standard definitions and results. In the second part, some functions related to the Rogers–Ramanujan continued fraction are defined. The level 13 analogues are defined in the third part. To emphasize the analogy, lower case letters are used for functions related to the Rogers–Ramanujan continued fraction, while capital letters are used for the level 13 analogues. 2.1. Standard definitions and results Let τ be a complex number with positive imaginary part and put q = exp(2πiτ ). Ramanujan’s Eisenstein series L, M and N , the Dedekind eta function η(τ ) and the j-invariant are defined by L(q) = 1 − 24
∞
jq j , 1−qj
j=1 ∞
N (q) = 1 − 504
j=1
η(τ ) = q 1/24
∞
M(q) = 1 + 240
∞ j 3q j , 1−qj j=1
j 5q j , 1−qj
(1 − q j )
j=1
and
j (τ ) =
1728M 3 (q) . M 3 (q) − N 2 (q)
We will sometimes make use of the compact notation for infinite products defined by (x1 , x2 , . . . , xm ; q)∞ =
∞ j=0
(1 − x1 q j )(1 − x2 q j ) · · · (1 − xm q j ).
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Let ℓ be a positive integer, called the level. The modular group Γ and the congruence subgroup Γ0 (ℓ) are defined by a b Γ = : a, b, c, d ∈ Z, ad − bc = 1 c d and Γ0 (ℓ) =
a c
b d
∈Γ :c≡0
(mod ℓ) .
The results in the following lemmas are well known. Lemma 2.1. Ramanujan’s Eisenstein series satisfy the differential equations q
L2 − M dL = , dq 12
q
dM LM − N = , dq 3
and
q
dN L N − M2 = dq 2
(5)
as well as the identity M 3 (q) − N 2 (q) = 1728q
∞
(1 − q j )24 .
(6)
j=1
Proof. See [29, pp. 18,36] for proofs and more information. Lemma 2.2. Let ac db ∈ Γ , q = exp(2πiτ ) and q ′ = exp 2πi η24
aτ + b cτ + d
aτ +b cτ +d
. Then
= (cτ + d)12 η24 (τ )
(7)
and L(q ′ ) = (cτ + d)2 L(q) + In the special case ac db = 10
−1 η τ
=
τ η(τ ) i
6c(cτ + d) . πi −1 these transformation formulas become 0
and
L(q ′ ) = τ 2 L(q) +
6τ . πi
(8)
(9)
Proof. The first result is standard, e.g., see [2, pp. 50–51]. The transformation formula (8) follows from (7) by logarithmic differentiation with respect to τ . Lemma 2.3. Let m be a fixed positive integer and let ω = exp(2πi/m). Let j and k be real numbers. Suppose Imτ > 0, q = exp(2πiτ ) and p = exp(−2πi/mτ ). Then ∞ ∞ sin πmj (1 − ω j q n )(1 − ω− j q n ) (1 − p mn− j )(1 − p mn−m+ j ) E = p (1 − ωk q n )(1 − ω−k q n ) (1 − p mn−k )(1 − p mn−m+k ) sin πk n=1 m n=1
where the exponent E is given by E=
k(m − k) − j (m − j) (k − j)(m − j − k) = . 2m 2m
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Proof. This is obtained by applying the classical transformation formula for the theta function, e.g., [3, p. 36, Entry 20], to a quotient of two theta functions. 2.2. Definitions pertaining to the Rogers–Ramanujan continued fraction Let r , s and t be the modular functions defined by 5 ∞ ∞ j (1 − q 5 j−4 )5 (1 − q 5 j−1 )5 (1 − q j )( 5 ) = q r = r (q) = 5 (q) = q (1 − q 5 j−3 )5 (1 − q 5 j−2 )5 j=1 j=1 s = s(q) = q
∞ (1 − q 5 j )6 (1 − q j )6 j=1
and t = t (q) =
s . 1 + 22s + 125s 2
Let u and v be the weight 2 modular forms defined by ∞ ∞ ∞ d j j jq j u = u(q) = q = 1 − 5 log r (q) = 1 − 5 jq jk j dq 5 1 − q 5 j=1 k=1 j=1 and v = v(q) =
∞ j j=1
5
∞ ∞ qj j kq jk . = 5 (1 − q j )2 j=1 k=1
Let y and z be the weight 2 modular forms defined by y = y(q) =
∞ (1 − q j )5 (1 − q 5 j ) j=1
and z = z(q) = q
d 1 log s = 5L(q 5 ) − L(q) . dq 4
2.3. Definitions pertaining to the level 13 analogues Now we will define level 13 analogues of the functions defined in Section 2.2. Each function defined by a capital letter is an analogue of the function defined using the corresponding lower case letter in Section 2.2. Let R, S and T be the modular functions defined by R = R(q) = q
∞
(1 − q j )
j 13
j=1
=q
(q, q 3 , q 4 , q 9 , q 10 , q 12 ; q 13 )∞ (q 2 , q 5 , q 6 , q 7 , q 8 , q 11 ; q 13 )∞
∞ (1 − q 13 j−12 )(1 − q 13 j−10 )(1 − q 13 j−9 )(1 − q 13 j−4 )(1 − q 13 j−3 )(1 − q 13 j−1 ) = q , (1 − q 13 j−11 )(1 − q 13 j−8 )(1 − q 13 j−7 )(1 − q 13 j−6 )(1 − q 13 j−5 )(1 − q 13 j−2 ) j=1
S = S(q) = q
∞ (1 − q 13 j )2 (1 − q j )2 j=1
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and S . 1 + 6S + 13S 2 Let U and V be the weight 2 modular forms defined by ∞ ∞ ∞ d j j jq j U = U (q) = q = 1 − jq jk log R(q) = 1 − j dq 13 1 − q 13 j=1 k=1 j=1 T = T (q) =
and V = V (q) =
∞ ∞ ∞ j j qj = kq jk . j )2 13 13 (1 − q j=1 j=1 k=1
Let Y and Z be the modular forms of weights 6 and 2, respectively, defined by Y = Y (q) =
∞ (1 − q j )13 (1 − q 13 j ) j=1
and Z = Z (q) = q
d 1 log S = 13L(q 13 ) − L(q) . dq 12
Although y and Y are clearly analogues, y is a modular form of weight 2 while Y has weight 6. Therefore, in formulas that involve these functions, Y will occur as an analogue of y 3 . 3. Identities involving modular functions In this section we outline some basic properties of the functions r and R. Theorem 3.1. The following identities hold: 1 1 − 11 − r = r s
(10)
1 1 −3− R = . R S
(11)
and
Proof. The identity (10) was given by Ramanujan in his paper [25]. A simple proof using only the Jacobi triple product identity has been given by Hirschhorn [20]. More information, and references to other proofs, can be found in the book by Andrews and Berndt [1, pp. 11–12]. The identity (11) was known to Ramanujan [26, Ch. 20, Entry 8(i)]. Proofs have been given by Evans [18, (5.9)], Berndt [3, p. 375, (8.15)] and Horie and Kanou [22, Theorem 2.2]. Corollary 3.2. t=
r (1 − 11r − r 2 ) s = 1 + 22s + 125s 2 (1 + r 2 )2
and T =
S R(1 − 3R − R 2 ) = . 1 + 6S + 13S 2 (1 + R 2 )2
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Proof. The first equality in each identity is just the definition; the second equality in each identity follows from the results in Theorem 3.1. The next result states how r and R transform under the congruence subgroups Γ0 (5) and Γ0 (13), respectively. Theorem 3.3. Let e(τ ) = exp(2πiτ ). d a b d aτ +b = 5 r (e(τ )) 5 . If c d ∈ Γ0 (5) then r e cτ +d d a b aτ +b d If c d ∈ Γ0 (13) then R e cτ +d = 13 R(e(τ )) 13 . Proof. See [22, Theorem 1.1].
The next result describes how r and R transform under the Fricke involutions τ → −1/5τ and τ → −1/13τ , respectively. Theorem 3.4. Let e(τ )√ = exp(2πiτ ), q = e(τ ), q5 = e(−1/5τ ) and q13 = e(−1/13τ ). Let √ 1+ 5 β = 2 and δ = 3+2 13 . The following transformation formulas hold: r (q5 ) =
1 − β 5r (q) β 5 + r (q)
(12)
and R(q13 ) =
1 − δ R(q) . δ + R(q)
(13)
Proof. The identity (12) may be rearranged to the symmetric form √ 5 √ 5+1 5 5 10 . r (q) + β r (q5 ) + β = 1 + β = 5 5 2 This was stated by Ramanujan in the Lost Notebook; see [1, pp. 91–92] for a discussion and proof. It remains to prove (13). For brevity, write R = R(q) and R ′ = R(q13 ). By Theorem 3.1 we have 1 η2 (τ ) −3− R = 2 . (14) R η (13τ ) If we replace τ with −1/13τ and apply the modular transformation for the Dedekind eta function given in (9), we get 1 η2 (−1/13τ ) η2 (13τ ) = 13 2 . − 3 − R′ = 2 ′ R η (−1/τ ) η (τ ) On multiplying (14) by (15) we get 1 1 ′ −3− R − 3 − R = 13. R R′ This may be expanded and rearranged to give 2 3 13 1 − (R + R ′ ) − R R ′ = (R + R ′ )2 2 4
(15)
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and it follows that √ 13 3 ′ ′ (R + R ′ ). 1 − (R + R ) − R R = ± 2 2
(16) √ The sign√may be determined by substituting the value τ = i/ 13. This gives R = R ′ = R(e−2π/ 13 ), so (16) reduces to √ 1 − 3R − R 2 = ± 13 R. (17) Using the first two terms in the q-expansion R(q) = q − q 2 + O(q 3 ) we obtain the approximation √
R(e−2π/
√ 13
) ≈ e−2π/
13
√ 13
− e−4π/
≈ 0.14.
It follows that we must select the plus sign in (17) and hence in (16). On rearranging (16) we complete the proof of (13) and also obtain the symmetric form √ √ 13 2 3 + 13 . (R(q) + δ) (R(q13 ) + δ) = 1 + δ = 2 Corollary 3.5. The following evaluations hold: √ r exp(−2π/ 5) = β 10 + 1 − β 5 and
√ R exp(−2π/ 13) = δ 2 + 1 − δ,
where the values of β and δ are as given in Theorem 3.4. √ √ Proof. The values τ √ = i/ 5 and τ = i/ 13 give √fixed points for (12) and (13), namely q = q5 = exp(−2π/ exp(−2π/ 13), respectively. On solving (12) and √ 5) and q = q13 = √ (13) for r (exp(−2π/ 5)) and r (exp(−2π/ 13)), respectively, and choosing the solutions that are positive, we obtain the claimed evaluations. The results in the next theorem provide factorizations of the identities in Theorem 3.1. Theorem 3.6. Let ζ = exp(2πi/5), α =
√ 1− 5 2
and β =
√ 1+ 5 2 .
Then
∞ √ 1 1 1 √ − α 5 r = 5 1/12 j )5 (1 − ζ 4 q j )5 (q s) (1 − ζ q r j=1
(18)
∞ √ 1 1 1 . √ − β 5 r = 5 1/12 2 q j )5 (1 − ζ 3 q j )5 (q s) (1 − ζ r j=1
(19)
and
If ξ = exp(2πi/13), γ =
√ 3− 13 2
and δ =
√ 3+ 13 2 ,
then
√ 1 1 1 × √ −γ R = 1/4 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q; q) (q S) (ξ q, ξ R ∞
(20)
–
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√ 1 1 1 × 2 . √ −δ R = 1/4 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q; q) (q S) (ξ q, ξ R ∞
(21)
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√ √ Proof. √ The identities (18) and (19), as well as infinite product representations for 1/ − α √ and 1/ − β , were given by Ramanujan in the Lost Notebook [27, p. 206]. Proofs of these identities have been given in [1, pp. 21–24], [5,6,8,13,21,23]. We will prove (20) and (21). By Theorem 3.4 we have 4 , q 9 , q 10 , q 12 ; q 13 ) (q13 , q 3 , q13 1 − δ R(q) 13 13 13 13 ∞ = R(q13 ) = q13 2 13 . 5 , q 6 , q 7 , q 8 , q 11 ; q 13 ) δ + R(q) (q13 , q13 13 13 13 13 13 ∞
By three applications of Lemma 2.3 with m = 13, p = q13 = exp(−2πi/13τ ) we deduce 3π 4π π sin 13 sin 13 sin 13 (ξ q, ξ 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q; q) 1 − δ R(q) ∞ = . 2π 5π 6π (ξ 2 q, ξ 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q; q)∞ δ + R(q) sin 13 sin 13 sin 13 Since 3π 4π √ sin 13 sin 13 13 − 3 1 = = −γ = 2 δ sin 5π sin 6π sin 2π 13 13 13
π sin 13
we may multiply both sides by δ to get 1 − δ R(q) (ξ q, ξ 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q; q)∞ . = 2 1 − γ R(q) (ξ q, ξ 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q; q)∞
(22)
On the other hand, the identity (11) in Theorem 3.1 may be written as ∞ (1 − γ R(q))(1 − δ R(q)) 1 (1 − q j )2 = . R(q) q j=1 (1 − q 13 j )2
(23)
On dividing (23) by (22) we get ∞ (ξ 2 q, ξ 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q; q)∞ 1 (1 − γ R(q))2 (1 − q j )2 = × R(q) q j=1 (1 − q 13 j )2 (ξ q, ξ 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q; q)∞
=
∞ 1 1 (1 − q j ) × . 3 4 9 10 12 2 q j=1 (1 − q 13 j ) (ξ q, ξ q, ξ q, ξ q, ξ q, ξ q; q)∞
On taking square roots we obtain (20). The identity (21) may be obtained in a similar way, starting by multiplying (22) by (23). The identities (20) and (21) are equivalent to the pair of identities (q 2 , q 5 , q 6 , q 7 , q 8 , q 11 , q 13 , q 13 ; q 13 )∞ − γ q (q, q 3 , q 4 , q 9 , q 10 , q 12 , q 13 , q 13 ; q 13 )∞ = (ξ 2 q, ξ 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q, q, q; q)∞
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and (q 2 , q 5 , q 6 , q 7 , q 8 , q 11 , q 13 , q 13 ; q 13 )∞ − δ q (q, q 3 , q 4 , q 9 , q 10 , q 12 , q 13 , q 13 ; q 13 )∞ = (ξ q, ξ 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q, q, q; q)∞ . It would be interesting to have direct proofs of these results. 4. Identities involving modular forms In this section we outline the basic properties of the modular forms u, v, y and z and their level 13 analogues U , V , Y and Z . All of these forms have weight 2 except for Y which has weight 6. We begin by noting some known transformation formulas involving S, U and V . Similar transformation formulas hold for the functions s, u and v and the group Γ0 (5) but, as they will not be required, we do not state them here. Lemma 4.1. Let S, U and V be as defined in Section 2.3. a b Let c d ∈ Γ0 (13). −2πi +b and q = exp . Let q = exp(2πiτ ), q ′ = exp 2πi aτ 13 cτ +d 13τ Then 1 (i) S(q ′ ) = S(q), (ii) S(q13 ) = , 13S(q) d 1 (iii) U (q ′ ) = (cτ + d)2 U (q), (iv) U (q13 ) = √ (13τ )2 V (q), 13 13 √ d ′ 2 (v) V (q ) = (cτ + d) V (q), (vi) V (q13 ) = 13 τ 2 U (q). 13 Proof. For (i), see [2, p. 87, Th. 4.9]. The identity (ii) follows immediately from the definition of S and the transformation formula for the Dedekind eta function in (9). For (iii)–(vi), see [11, Th. 6.1]. Theorem 4.2. The modular forms u, v, y and z (respectively, U , V , Y and Z ) are interrelated by 1 (i) y = u (ii) v = su, (iii) z = r + v (24) r and U3 (i) Y = , (1 + 5S + 13S 2 )2
(ii) V = SU,
Proof of (24). Two classical identities are ∞ ∞ j jq j (1 − q j )5 1−5 = 5 1−qj (1 − q 5 j ) j=1 j=1 and ∞ j j=1
5
∞ qj (1 − q 5 j )5 = q . (1 − q j ) (1 − q j )2 j=1
(iii) Z =
1 R+ R
V.
(25)
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These were known to Ramanujan and many proofs exist; see [1, pp. 406–407] for links to Ramanujan’s work and references to proofs. They imply infinite product formulas for u and v, respectively, and on comparing them with the infinite products that occur in the definitions of s and y, we immediately deduce the identities (24)(i) and (ii). The identity (24)(iii) was known d to the to Ramanujan; see [1, p. 88, Entry 3.2.6]. It can be proved by applying the operator q dq identity (10) and appealing the definitions of v and z. This completes the proof of the identities in (24). (q) and g(τ ) = S(q) and note that f and g are analytic in the Proof of 25(i). Let f (τ ) = UY 2 (q) region Imτ > 0. By the definitions of η, S and Y we have 6
f (τ ) = U 6 (q)
∞ U 6 (q)S(q) (1 − q 13 j )2 = . (1 − q j )26 η24 (τ ) j=1
By Lemma 2.2 and Parts (i) and (iii) of Lemma 4.1 it follows that f (τ ) and g(τ ) are each invariant under Γ0 (13). Let us examine the behavior at τ = 0. By Lemma 2.2 and Parts (ii) and (iv) of Lemma 4.1 we have −1 U 6 (q13 )S(q13 ) f = 24 13τ η (−1/13τ ) V 6 (q) × S(q)η24 (13τ ) 134 1 26 76 267 600 1 4 = 4 + 7+ 6+ 5+ 4 + 3 8 q q q q q 13 q 1448 2552 + 2 + + 4601 + O(q) q q =
1
and g
−1 13τ
1 13S(q) 1 1 = − 2 − q + 2q 2 + q 3 + 2q 4 − 2q 5 − 2q 7 + O(q 8 ) . 13 q
=
Thus, f (−1/13τ ) and g(−1/13τ ) have poles of orders 8 and 1, respectively, at q = 0. It follows that there are unique constants, a8 , a7 , . . . , a0 , that can be determined by successively comparing coefficients of q −8 , q −7 , . . . , q −1 , q 0 , such that −1 −1 −1 −1 8 7 f − a8 g + a7 g + · · · + a1 g + a0 = O(q). 13τ 13τ 13τ 13τ On solving for a8 , a7 , . . . , a0 , we conclude that f
−1 13τ
4 −1 −1 − 13g 2 + 5g + 1 = O(q). 13τ 13τ
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Replacing τ with −1/13τ we deduce that the function h(τ ) defined by 4 h(τ ) := f (τ ) − 13g 2 (τ ) + 5g(τ ) + 1 4 U 6 (q) = 2 − 13S 2 (q) + 5S(q) + 1 Y (q) is analytic at the vertex τ = 0. Clearly, h(τ ) is also analytic in the region Imτ > 0, and 4 U 6 (0) h(i∞) = 2 − 13S 2 (0) + 5S(0) + 1 = 0. Y (0) It follows that h is bounded. Since h is invariant under Γ0 (13), Theorem 4.4 in [2, p. 79] implies that h is constant. Since h(i∞) = 0, it follows that h is identically zero and so U6 . (1 + 5S + 13S 2 )4 On taking square roots and evaluating both sides when q = 0 to determine the sign, we complete the proof of (25)(i). Y2 =
(q) and g(τ ) = S(q) and follow the steps in the proof of (25)(i) Proof of 25(ii). Let f (τ ) = VY 2 (q) to deduce V3 . Y = 3 S (1 + 5S + 13S 2 )2 On comparing this result with the result from (25)(i) we obtain (25)(ii). 6
d Proof of 25(iii). Apply the operator q dq to the identity (11) to get 1 dR 1 dS 1+ 2 q = 2q . dq R S dq
By the definitions of U and Z , this becomes 1 1 R+ U = Z. R S On applying (25)(ii) we complete the proof.
The essence of the above proof of (25)(i) is that S(q) is a Hauptmodul for Γ0 (13) and U 3 /Y is modular on Γ0 (13) with no poles in its fundamental domain. Therefore U 3 /Y must be a polynomial in S(q). 5. Eisenstein series In the lost notebook [27, pp. 50,51] Ramanujan noted that the Eisenstein series M(q), M(q 5 ), N (q) and N (q 5 ) may be expressed as algebraic functions of η(τ ) and η(5τ ). The results may be stated as: Theorem 5.1. Let M and N be the Eisenstein series defined in Section 2.1 and let s and y be the eta-quotients defined in Section 2.2. Then M(q) = y 2 (1 + 250s + 3125s 2 ), M(q 5 ) = y 2 (1 + 10s + 5s 2 ), N (q) = y 3 (1 + 22s + 125s 2 )1/2 (1 − 500s − 15625s 2 )
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13
and N (q 5 ) = y 3 (1 + 22s + 125s 2 )1/2 (1 + 4s − s 2 ). Ramanujan also gave analogous results for M(q), M(q 7 ), N (q) and N (q 7 ) in terms of η(τ ) and η(7τ ). The goal of this section is to prove the level 13 analogues given by Theorem 5.2. Let M and N be the Eisenstein series defined in Section 2.1 and let S and Y be the eta-quotients defined in Section 2.3. Then M(q) = Y 2/3 (1 + 5S + 13S 2 )1/3 × (1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 ), M(q 13 ) = Y 2/3 (1 + 5S + 13S 2 )1/3 (1 + 7S + 20S 2 + 19S 3 + S 4 ), N (q) = Y (1 + 6S + 13S 2 )1/2 × (1 − 494S − 20618S 2 − 237276S 3 − 1313806S 4 − 3712930S 5 − 4826809S 6 ) and N (q 13 ) = Y (1 + 6S + 13S 2 )1/2 ×(1 + 10S + 46S 2 + 108S 3 + 122S 4 + 38S 5 − S 6 ). The method of proof will be to adapt the procedure developed in [14] where the analogous results for levels 5 and 7 were derived. By Theorem 4.2, we may work with any of U , V or Z in place of Y . To simplify the calculations we will work in terms of S and U and then use Theorem 4.2 to re-express the final results in terms of the eta-quotients S and Y . We will not list any level 5 results in this section; for these, as well as the level 7 results, the reader is referred to [14]. Lemma 5.3. The following identity holds: 1 13L(q 13 ) − L(q) = U 1 + 6S + 13S 2 . 12 Moreover, the following relations among differential operators hold: q
d d d = UR = U 1 + 6S + 13S 2 S . dq dR dS
Proof. By the definition of Z in Section 2.3 we have d 1 Z =q log S = 13L(q 13 ) − L(q) . dq 12 On the other hand, by (11), (25)(ii) and (iii), and some algebraic manipulation, we have 1 Z = SU R + R 2 1 = SU − R +4 R
(26)
(27)
(28)
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= SU
)
–
2 1 +3 +4 S
= U 1 + 6S + 13S 2 .
(29)
On combining (28) and (29) we obtain 1 13L(q 13 ) − L(q) = Z = U 1 + 6S + 13S 2 12 which proves (26). Next, by the chain rule and the definition of U in Section 2.3 we have d d d d = q log R × R . q = UR dq dq dR dR Finally, by the chain rule, the definition of Z in Section 2.3 and (29) we have d d d d d q = q log S × S =Z× S = U 1 + 6S + 13S 2 S . dq dq dS dS dS Lemma 5.4. The following identities hold: q
q
∞
(1 − q j )24 = Y 2 S = U 6
j=1 ∞ 13
S , (1 + 5S + 13S 2 )4
(1 − q 13 j )24 = Y 2 S 13 = U 6
j=1
S 13 . (1 + 5S + 13S 2 )4
Proof. These are immediate consequences of the definitions of S and Y in Section 2.3 and the identity (25)(i). Lemma 5.5. The following identities hold: S (1 + 5S + 13S 2 )4 S 13 M 3 (q 13 ) − N 2 (q 13 ) = 1728 U 6 . (1 + 5S + 13S 2 )4 M 3 (q) − N 2 (q) = 1728 U 6
Proof. These are immediate consequences of the identity (6) and Lemma 5.4. Lemma 5.6. The following identities hold: 1 − 5S − 91S 2 dU U + 6S L(q) = 1 + 6S + 13S 2 dS 1 + 5S + 13S 2
(30)
and 1 L(q ) = 1 + 6S + 13S 2 13 13
13 + 45S + 65S 2 dU U + 6S 2 dS 1 + 5S + 13S
.
(31)
S. Cooper, D. Ye / Journal of Approximation Theory (
Hence,
1 + 6S + 13S 2 S
)
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15
dU dS
2 = 1 + 6S + 13S 2 3
5S + 26S 2 1 + 5S + 13S 2
U−
13 L(q 13 ) − L(q) . 72
(32)
Proof. Take the logarithmic derivative of the first identity in Lemma 5.4, using (27), to get q
∞ d log q (1 − q j )24 dq j=1 d = U 1 + 6S + 13S 2 S log S − 4 log(1 + 5S + 13S 2 ) + 6 log U . dS
On calculating the derivatives we obtain (30). The identity (31) may be obtained similarly, starting from the second identity in Lemma 5.4. Finally, the identity (32) may be obtained by subtracting (31) from (30). Lemma 5.7. The following identities hold: 169M(q 13 ) − M(q) = 24 U 2
(1 − 13S 2 )(7 + 39S + 91S 2 ) (1 + 5S + 13S 2 )
(33)
and 2197N (q 13 ) − N (q) = U 1 + 6S + 13S 3 × 2197M(q 13 ) − M(q) −
48U 2 p(S) , (1 + 5S + 13S 2 )2
(34)
where p(S) = S(76 + 481S − 1014S 2 − 19773S 3 − 74698S 4 − 107653S 5 ). d to the identity (26), and make use of (27) to get Proof. We apply the operator q dq
q d d 13L(q 13 ) − L(q) = U 1 + 6S + 13S 2 S U 1 + 6S + 13S 2 . 12 dq dS The derivative on the left hand side can be computed using (5); the derivative on the right hand side can be computed by straightforward calculation and using (32) to express dU d S in terms of U , 13 S, L(q ) and L(q). This gives 1 169(L 2 (q 13 ) − M(q)) − (L 2 (q) − M(q)) 144 2 2 5S + 26S 2 2 = U (1 + 6S + 13S ) 3 1 + 5S + 13S 2 13 − U 1 + 6S + 13S 2 L(q 13 ) − L(q) + U 2 S(3 + 13S), 72
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which may be rearranged to give (1 − 13S 2 )(7 + 39S + 91S 2 ) 169M(q 13 ) − M(q) − 24 U 2 (1 + 5S + 13S 2 ) = 13L(q 13 ) − L(q) − 12U 1 + 6S + 13S 2 × 13L(q 13 ) + L(q) + 14U 1 + 6S + 13S 2 .
(35)
By (26), the right hand side of (35) is identically zero. Hence, we deduce (33). d Now we will prove (34). Applying q dq to (33) and making use of (27) we get q
d 169M(q 13 ) − M(q) dq d (1 − 13S 2 )(7 + 39S + 91S 2 ) . = 24 U 1 + 6S + 13S 2 S U2 dS (1 + 5S + 13S 2 )
The derivative on the left hand side may be computed using (5), while the derivative on the right hand side may be computed by direct calculation and using (32) to express dU d S in terms of U , S, L(q 13 ) and L(q). The result simplifies to 1 13L(q 13 ) − L(q) 2197M(q 13 ) − M(q) 2197N (q 13 ) − N (q) = 12 p(S) − 48U 3 1 + 6S + 13S 2 2 . 1 + 5S + 13S 2 On applying (26), we complete the proof of (34).
Theorem 5.8. M(q) = U 2
1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 , 1 + 5S + 13S 2
(36)
1 + 7S + 20S 2 + 19S 3 + S 4 , 1 + 5S + 13S 2 √ 1 + 6S + 13S 2 2 3 3 N (q) = U 2 1 − 494S − 20618S − 237276S 1 + 5S + 13S 2 − 1313806S 4 − 3712930S 5 − 4826809S 6
(38)
√ 1 + 6S + 13S 2 N (q ) = U 2 1 + 5S + 13S 2 × 1 + 10S + 46S 2 + 108S 3 + 122S 4 + 38S 5 − S 6 .
(39)
M(q 13 ) = U 2
(37)
and 13
3
Proof. Let a and b be defined by a = 1 + 6S + 13S 2
and
b = 1 + 5S + 13S 2 .
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17
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Let u 1 , u 2 , u 3 and u 4 be defined by u1 = b
M(q) , U2
u2 = b
M(q 13 ) , U2
b2 N (q) u3 = √ a U3
and
b2 N (q 13 ) u4 = √ . (40) a U3
When q = 0 we have S = 0, U = 1, a = b = 1 and u 1 = u 2 = u 3 = u 4 = 1. The results of Lemmas 5.5 and 5.7 may be written as bu 31 − au 23 = 1728S, bu 32 − au 24 = 1728S 13 , −u 1 + 169u 2 = 24 1 − 13S 2 7 + 39S + 91S 2 , bu 1 − 2197bu 2 − u 3 + 2197u 4 = −48 p(S), where p(S) = S(76 + 481S − 1014S 2 − 19773S 3 − 74698S 4 − 107653S 5 ). This non-linear system may be solved to give u 1 , u 2 , u 3 and u 4 in terms of S by the method in [14, Theorem 2.6]. The results are u 1 = 1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 , u 2 = 1 + 7S + 20S 2 + 19S 3 + S 4 , u 3 = 1 − 494S − 20618S 2 − 237276S 3 − 1313806S 4 − 3712930S 5 − 4826809S 6 , u 4 = 1 + 10S + 46S 2 + 108S 3 + 122S 4 + 38S 5 − S 6 . On using these in (40) we deduce (36)–(39).
We are now ready for: Proof of Theorem 5.2. Use (25)(i) to eliminate U from each of (36)–(39).
Corollary 5.9. The j-invariant defined in Section 2.1 may be expressed in terms of the etaquotient S defined in Section 2.3, as follows: (1 + 5S + 13S 2 )(1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 )3 S (1 + 5S + 13S 2 )(1 + 7S + 20S 2 + 19S 3 + S 4 )3 j (13τ ) = . S 13 j (τ ) =
Proof. Substitute the results of Theorem 5.2 into the definition of the j-invariant.
It should be noted that Corollary 5.9 is the level 13 analogue of the results j (τ ) =
(1 + 250s + 3125s 2 )3 , s
j (5τ ) =
(1 + 10s + 5s 2 )3 , s5
and that the identities [14, Theorem 2.7] and [16, (2.5)] may be obtained from these on using (10). Many transformation formulas of degree 13 for hypergeometric functions can be deduced from Theorem 5.2 and Corollary 5.9. The next result gives two of the simplest transformation formulas.
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)
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Theorem 5.10. The following hypergeometric transformation formulas hold in a neighborhood of x = 0: 1 5 1728x 2 F1 12 , 12 ; 1; (1+5x+13x 2 )(1+247x+3380x 2 +15379x 3 +28561x 4 )3 √ 4 1 + 247x + 3380x 2 + 15379x 3 + 28561x 4 1 5 1728x 13 2 F1 12 , 12 ; 1; (1+5x+13x 2 )(1+7x+20x 2 +19x 3 +x 4 )3 = √ 4 1 + 7x + 20x 2 + 19x 3 + x 4 and 1 5 1 1728x F , , ; 1, 1; 3 2 6 6 2 (1+5x+13x 2 )(1+247x+3380x 2 +15379x 3 +28561x 4 )3 √ 1 + 247x + 3380x 2 + 15379x 3 + 28561x 4 1 5 1 1728x 13 3 F2 6 , 6 , 2 ; 1, 1; (1+5x+13x 2 )(1+7x+20x 2 +19x 3 +x 4 )3 = . √ 1 + 7x + 20x 2 + 19x 3 + x 4 Proof. From [10, (2.5) and Th. 4.11] we have 1 5 1 N (q) M 1/4 (q) = 2 F1 , ; 1; 1 − 3/2 6 6 2 M (q) and from [17, p. 111, (2)] we have 1 5 1 5 , ; 1; x = 2 F1 , ; 1; 4x(1 − x) . 2 F1 6 6 12 12 Therefore, M 1/4 (q) = 2 F1
1 5 M 3 (q) − N 2 (q) , ; 1; 12 12 M 3 (q)
= 2 F1
1 5 1728 , ; 1; 12 12 j (τ )
and so 2 F1
1728 1 5 12 , 12 ; 1; j (τ ) M 1/4 (q)
2 F1
=1=
1 5 1728 12 , 12 ; 1; j (13τ ) M 1/4 (q 13 )
.
Now substitute the results of Theorem 5.2 and Corollary 5.9, and then replace S with x to obtain the first result. The second result can be obtained by squaring the first result and applying Clausen’s formula [17, p. 185] 2 1 1 = 3 F2 2a, 2b, a + b; 2a + 2b, a + b + ; x . 2 F1 a, b; a + b + ; x 2 2 6. Differential equations The goal of this section is to show that Z satisfies a third order linear differential equation with respect to T . We begin by constructing a second order nonlinear differential equation for U in terms of R.
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Theorem 6.1. The following differential equation holds: d dU 1 dU 2 f (R) U R R = R + dR dR 2U dR 2(1 − 3R − R 2 )2 (1 − R + 5R 2 + R 3 + R 4 )2 where f (R) = R(−2 + 13R − 94R 2 + 7R 3 + 124R 4 − 23R 5 − 124R 6 + 7R 7 + 94R 8 + 13R 9 + 2R 10 ). Proof. We may use (11) and (27) to rewrite (30) and (36) in the forms L(q) = U
(1 − 21R − 39R 2 + 21R 3 + R 4 ) dU + 6R 2 3 4 2 dR (1 − R + 5R + R + R )(1 − 3R − R )
(41)
and M(q) = U
2
1 + 235R + 1207R 2 + 955R 3 + 3840R 4 − 955R 5 + 1207R 6 − 235R 7 + R 8 , (42) (1 − R + 5R 2 + R 3 + R 4 )(1 − 3R − R 2 )2
d respectively. From (27) we have q dq = U R ddR . Therefore, on solving (41) for R dU d R and applying d , we obtain the operator q dq d dU q dL UR d U (1 − 21R − 39R 2 + 21R 3 + R 4 ) UR . R = − dR dR 6 dq 6 d R (1 − R + 5R 2 + R 3 + R 4 )(1 − 3R − R 2 )
The derivative of the first term on the right hand side may be obtained using (5) and the result may be expressed in terms of U , R and dU/d R, using (41) and (42). The derivative of the second term on the right hand side may be found by direct calculation. On simplifying and dividing both sides by U we complete the proof. The next result gives a third order linear differential equation for U in terms of R. Theorem 6.2. The following differential equation holds: d d dU g(R) dU R R R = R dR dR dR dR (1 − R + 5R 2 + R 3 + R 4 )2 (1 − 3R − R 2 )2 +
h(R) U, (1 − R + 5R 2 + R 3 + R 4 )3 (1 − 3R − R 2 )3
where g(R) = R −2 + 13R − 94R 2 + 7R 3 + 124R 4 − 23R 5 − 124R 6 + 7R 7 + 94R 8 + 13R 9 + 2R 10
and h(R) = R 1 + R 2 −1 + 9R − 119R 2 + 37R 3 + 1019R 4 − 2446R 5 − 2192R 6 + 3206R 7 + 2192R 8 − 2446R 9 − 1019R 10 + 37R 11 + 119R 12 + 9R 13 + R 14 .
(43)
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Proof. Apply the operator R ddR to both sides of the differential equation in Theorem 6.1. The key is to note that dU 2 dU 1 dU 2 1 1 dU d d R R − R = R R R d R 2U dR U dR dR dR 2U dR and then Theorem 6.1 can be applied to simplify the expression in braces. The remaining details are purely computational and we omit them. Theorem 6.3. Let S, T and Z be as defined in Section 2.3, that is, ∞ S (1 − q 13 j )2 , T = S=q j 2 (1 − q ) 1 + 6S + 13S 2 j=1 and Z =q
∞ d jq j log S = 1 + 2 . dq 1−qj j=1 13- j
The following differential equation holds: d3 Z 2 3 2 d Z T 2 (1 − T )3 1 − 12T − 16T 2 + 3T − T 1 − 18T − 32T (1 ) dT 3 dT 2 dZ + (1 − T ) 1 − 42T − 43T 2 + 168T 3 − 108T 4 dT = 2 1 + 11T − 9T 2 + 15T 3 − 6T 4 Z . Proof. By Corollary 3.2 and Theorem 4.2 we have R(1 − 3R − R 2 ) 1 + R2 and Z = U . (1 + R 2 )2 1 − 3R − R 2 On applying this change of variables to the result of Theorem 6.2 we obtain the claimed result. T =
The next result gives an expansion of Z in powers of T as well as representations of Z in terms of the 3 F2 hypergeometric function. Theorem 6.4. Let S, T and Z be as for Theorem 6.3. Let A( j) be the sequence defined by the recurrence relation ( j + 1)3 A( j + 1) = 15 j 3 + 18 j 2 + 10 j + 2 A ( j) − 23 j 3 − 63 j 2 + 27 j − 9 A ( j − 1) − 11 j 3 + 24 j 2 − 118 j + 70 A ( j − 2) + 6 (2 j − 5) 3 j 2 − 9 j + 5 A ( j − 3) − 4 (2 j − 5) (2 j − 7) ( j − 3) A ( j − 4) and initial conditions A(0) = 1,
A(−1) = A(−2) = A(−3) = A(−4) = 0.
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)
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21
Then Z =
∞
A( j)T j =
j=0
=
=
∞
A( j)
j=0
S 1 + 6S + 13S 2
j
(1 + 5S + 13S 2 )1/2 (1 + 6S + 13S 2 )1/2 (1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 )1/2 1728S 1 5 1 × 3 F2 , , ; 1, 1; 3 6 6 2 1 + 5S + 13S 2 1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 (1 + 5S + 13S 2 )1/2 (1 + 6S + 13S 2 )1/2 (1 + 7S + 20S 2 + 19S 3 + S 4 )1/2 1 5 1 1728S 13 ×3 F2 , , ; 1, 1; 3 . 6 6 2 1 + 5S + 13S 2 1 + 7S + 20S 2 + 19S 3 + S 4
Proof. The equalities in the first line of the identity follow immediately from Theorem 6.3. Let us prove the formulas that involve the hypergeometric function. By (29), (36) and the formulas in the proof of Theorem 5.10, we have Z = (1 + 6S + 13S 2 )1/2 U (1 + 5S + 6S 2 )1/2 (1 + 6S + 13S 2 )1/2 = M 1/2 (q) (1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 )1/2 (1 + 5S + 6S 2 )1/2 (1 + 6S + 13S 2 )1/2 1 5 1 1728 = , , ; 1, 1; . 3 F2 6 6 2 j (τ ) (1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 )1/2 Now apply Corollary 5.9 to obtain the first result involving the hypergeometric function. The other result involving the hypergeometric function may be proved in a similar way using (37) instead of (36), or by appealing to Theorem 5.10. 7. Series for 1/π In this section we develop some series for 1/π. The prototypes for such series were given by Ramanujan [24] and they include examples such as √ ∞ 2 1 2 2 4j 2 j (1103 + 26390 j) = . (44) π 9801 j=0 2 j j 3964 j The key to proving such formulas is [7, Theorem 2.1]: Theorem 7.1 (H. H. Chan, S. H. Chan and Z.-G. Liu). Let ℓ be a positive integer. Suppose T = T (q), Z = Z (q) and w = w(q) satisfy the properties: √ √ x Z e−2π x/ℓ = Z e−2π/ xℓ for all x > 0, (45) Z (q) =
∞ j=0
A ( j) T j (q)
(46)
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)
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and q
d log T (q) = w (q) Z (q) . dq
(47)
For any integer N ≥ 2, let m (q) =
Z (q) . Z qN
Let λ N , TN and W N be defined by T dm λN = , 2N dT q=e−2π/√ N ℓ √ TN = T e−2π N /ℓ
(48)
(49) (50)
and √ W N = w e−2π N /ℓ .
(51)
Then
∞ ℓ 1 j = WN A ( j) ( j + λ N ) TN . N 2π j=0
(52)
The integers ℓ and N are called the level and degree, respectively. We are now ready to deduce the following class of level 13 series for 1/π : Theorem 7.2. Let N√≥ 2 be an integer and let T = T (q) be the function defined in Section 2.3. Let q N = exp −2π N /13 and TN = T (q N ). Then the identity ∞ 1 13 j 2 1 − 12TN − 16TN A ( j) ( j + λ N ) TN = (53) 2π N j=0 holds for the values of N , TN and λ N given in Table 1. Proof. Let T and Z be the functions defined in Section 2.3 and let ℓ = 13. The hypotheses (45) and (46) of Theorem 7.1 are satisfied because of (8) and Theorem 6.4, respectively. By the definitions of T and Z and a calculation we have d d S q log T = q log dq dq 1 + 6S + 13S 2 1 − 13S 2 d q log S 2 dq 1 + 6S + 13S = 1 − 12T − 16T 2 Z , =
so the hypothesis (47) of Theorem 7.1 holds with w(q) = 1 − 12T − 16T 2 .
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)
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23
Table 1 Data for Theorems 7.2 and 7.3. q
TN
λN √ 11− 13 27 √ 19 13−67 6 √ 766 − 111 2875 11500 13 √ 3415 1136 14739 − 73695 13
3 7
√ 7 13−25 6 √ 19−5 13 36 √ 15−2 26 363 √ 17−4 13 648 √ − 13−1 6 − 17
15
65 − 27−3 72
31
1 − 124
55
13 − 319−88 1089
79
13−1151 − 320 6399
N 3
√ exp −2π N /13
4 6 10
√ − exp −π N /13
Series does not converge Series does not converge
√
√ √
√
25 − 3 64 320 65 161 750 √ 379 209 1215 − 4860 13 √ 40177 93589 163056 − 326112 13
Therefore, identities of the form (53) hold by Theorem 7.1. In principle, values of TN and λ N can be determined using modular equations; see [8, pp. 408, 409] and [9, Proof of Th. 2.1] for some examples of the technique. In practice, the values are computed and identified numerically. The results that appear to be rational or quadratic irrational are recorded in Table 1. Some details concerning the evaluation of TN will appear in forthcoming work [15]. We also have: Theorem 7.3. Let ≥ 11 bean integer and let T = T (q) be the function defined in Section 2.3. N√ Let q N = − exp −π N /13 and TN = T (q N ). Then the identity ∞ 1 13 j 2 1 − 12TN − 16TN A ( j) ( j + λ N ) TN = (54) π N j=0 holds for the values of N , TN and λ given in Table 1. Proof. Replace q with −q and write the result of Theorem 6.4 in the form Z (−q) =
∞ (−1) j A( j)(−T (−q)) j . j=0
Now apply Theorem 7.1 along with the fact that Z (−q) and −T (−q) are modular forms of level 52. The details are similar to those in the proof of Theorem 7.2, so we omit them. We end this section with some remarks. It should be noted that for the series (54) to converge, √ it is necessary that |TN | < 1/(6 + 2 13). Since N is an integer, this requires N ≥ 11. On the other hand, the series (53) converges for every integer N ≥ 2. In fact, from Corollaries 3.2 and 3.5 it follows that √ R(1 − 3R − R 2 ) 1 T1 = T (exp(−2π/ 13)) = = √ . √ 2 2 (1 + R ) 6 + 2 13 q=exp(−2π/ 13)
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If an explicit combinatorial formula for A( j) was known, then it would be possible to prove the level 13 series for 1/π, without using modular forms, from the method shown in [19]. Finally, the series (4) in the introduction corresponds to the data for Theorem 7.3 in Table 1 for degree N = 31. 8. Other levels: Ramanujan’s theories of elliptic functions to alternative bases, and beyond Most of the analyses in the previous sections can be carried out for other levels ℓ for which ℓ − 1 is a divisor of 24, namely ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25}. The theory in the case ℓ = 4 corresponds to Jacobi’s original theory of theta functions. The levels ℓ = 2, 3 are often referred to as Ramanujan’s theories of elliptic functions to alternative bases. Ramanujan, in fact, had three such theories. The other one corresponds to the level ℓ = 1 which does not fit the general theory to be developed below. All of Ramanujan’s results on elliptic functions to alternative bases have been proved in [4]. A different approach, emphasizing the level, has been given in [10]. For other levels, an analogue of r or R is not the natural starting point (see the column for w in [9, Table 1] for some analogues of r and R for other levels). A more natural and appropriate starting point is an eta-quotient which is an analogue of s and S. We briefly sketch the details for these other levels. Let ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25}. For each level ℓ we associate the integer e defined by (ℓ − 1)e = 24. For each ℓ, let sℓ , yℓ and z ℓ be defined by ∞ (1 − q j )ℓ (1 − q ℓj ) j=1
sℓ = q
∞ (1 − q ℓj )e , (1 − q j )e j=1
zℓ = q
d ℓL(q ℓ ) − L(q) log sℓ = . dq ℓ−1
yℓ =
and
Let tℓ be defined by s2 s3 s4 , t3 = , t4 = , 2 2 (1 + 64s2 ) (1 + 27s3 ) (1 + 16s4 )2 s5 s7 s9 t5 = , t7 = , t9 = , 2 2 1 + 22s5 + 125s5 1 + 13s7 + 49s7 1 + 9s9 + 27s92 s13 s25 t13 = and t25 = . 2 2 1 + 6s13 + 13s13 1 + 2s25 + 5s25 t2 =
That is, tℓ =
sℓ 1 + bℓ sℓ + cℓ sℓ2
where cℓ = ℓ12/(ℓ−1) and bℓ = 128, 54, 32, 22, 13, 9, 6 or 2 corresponding to ℓ = 2, 3, 4, 5, 7, 9, 13 or 25, respectively. Each sℓ and tℓ is a modular function, while z ℓ is a modular form of weight 2 and yℓ is a modular form of weight (ℓ − 1)/2.
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25
In terms of the functions s, t, y and z in Section 2.2 and S, T , Y and Z in Section 2.3, we have s5 = s,
t5 = t,
and
y5 = y
z5 = z
and s13 = S,
t13 = T,
and
y13 = Y
z 13 = Z .
We state the theorems in this section mostly without proof. The results for level 13 have already been proved in detail in Sections 3–7. The other results may be proved by similar methods. Theorem 8.1. Let ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25} and suppose (ℓ − 1)e = 24. Then z ℓ6 /yℓe is a polynomial in sℓ of degree ℓ + 1. Precisely, we have z 26 y224 z 46 y48 z 76 y74 6 z 13 2 y13
z 36
= (1 + 64s2 )3 ,
= (1 + 27s3 )4 ,
y312 z 56
= (1 + 16s4 )5 ,
y56 z 96
4 = 1 + 13s7 + 49s72 ,
y93
3 = 1 + 22s5 + 125s52 , 5 = 1 + 9s9 + 27s92 ,
3 4 2 2 = 1 + 6s13 + 13s13 1 + 5s13 + 13s13
and 6 3 5 z 25 2 2 3 4 = 1 + 2s25 + 5s25 . 1 + 5s25 + 15s25 + 25s25 + 25s25 y25
Theorem 8.2. The following identities, involving the j-invariant, hold: j (τ ) =
(1 + 256s2 )3 , s2
j (τ ) =
(1 + 27s3 )(1 + 243s3 )3 , s3
j (3τ ) =
j (τ ) =
(1 + 256s4 + 4096s42 )3 , s4 (1 + 16s4 )
j (2τ ) =
j (4τ ) =
j (2τ ) =
(1 + 16s4 + 16s42 )3 s44 (1 + 16s4 )
(1 + 16s2 )3 , s22 (1 + 27s3 )(1 + 3s3 )3 s33 (1 + 16s4 + 256s42 )3 s42 (1 + 16s4 )2
,
j (τ ) =
(1 + 250s5 + 3125s52 )3 , s5
j (τ ) =
(1 + 13s7 + 49s72 )(1 + 245s7 + 2401s72 )3 s7
j (7τ ) = j (τ ) =
(1 + 10s5 + 5s52 )3
(1 + 13s7 + 49s72 )(1 + 5s7 + s72 )3
(1 + 9s9
j (3τ ) =
j (5τ ) =
s77 3 ) (1 + 243s
9
+ 2187s92 + 6561s93 )3
s9 (1 + 9s9 + 27s92 ) (1 + 9s9 )3 (1 + 3s9 )3 (1 + 27s92 )3 s93 (1 + 9s9 + 27s92 )3
s55
,
,
,
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S. Cooper, D. Ye / Journal of Approximation Theory ( j (9τ ) = j (τ ) =
)
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(1 + 3s9 )3 (1 + 9s9 + 27s92 + 3s93 )3 s99 (1 + 9s9 + 27s92 )
2 )(1 + 247s + 3380s 2 + 15379s 3 + 28561s 4 )3 (1 + 5s13 + 13s13 13 13 13 13 s13
j (13τ ) = j (τ ) = j (5τ ) =
2 )(1 + 7s + 20s 2 + 19s 3 + s 4 )3 (1 + 5s13 + 13s13 13 13 13 13 13 s13
3
2 +35000s 3 +178125s 4 +631250s 5 1+250s25 +4375s25 25 25 25 6 +3125000s 7 +4296875s 8 +3906250s 9 +1953125s 10 +1640625s25 25 25 25 25
2 + 25s 3 + 25s 4 ) s25 (1 + 5s25 + 15s25 25 25 2 )3 (1 + 5s + 25s 2 )3 (1 + 5s + 20s 2 + 25s 3 + 25s 4 )3 (1 + 5s25 + 5s25 25 25 25 25 25 25 5 (1 + 5s + 15s 2 + 25s 3 + 25s 4 )5 s25 25 25 25 25
j (25τ ) =
2 + 200s 3 + 525s 4 + 1010s 5 + 1425s 6 + 1400s 7 + 875s 8 + 250s 9 + 5s 10 1 + 10s25 + 55s25 25 25 25 25 25 25 25 25 25 (1 + 5s + 15s 2 + 25s 3 + 25s 4 ) s25 25 25 25 25
3 .
Theorem 8.3. For each ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25}, z ℓ satisfies a third order linear differential equation with respect to tℓ , namely: t22 (1 − 256t2 )
d 3 z2 dt23
+ 3t2 (1 − 384t2 )
dz 2 d 2 z2 = 24z 2 , + (1 − 816t2 ) 2 dt2 dt2
d 2 z3 dz 3 + 3t3 (1 − 162t3 ) 2 + (1 − 348t3 ) = 12z 3 , 3 dt3 dt3 dt3 d 2 z4 d 3 z4 dz 4 = 8z 4 , t42 (1 − 64t4 ) 3 + 3t4 (1 − 96t4 ) 2 + (1 − 208t4 ) dt4 dt4 dt4 d 3 z5 d 2 z5 t52 (1 − 44t5 − 16t52 ) 3 + 3t5 (1 − 66t5 − 32t52 ) 2 dt5 dt5 t32 (1 − 108t3 )
d 3 z3
+ (1 − 144t5 − 108t52 ) t72 (1 − 27t7 )(1 + t7 )
d 3 z7 dt73
+ (1 − 86t7 − 186t72 ) t92 (1 − 18t9 − 27t92 )
+ 3t7 (1 − 39t7 − 54t72 )
d 2 z7 dt72
dz 7 = 4(1 + 6t7 )z 7 , dt7
d 3 z9 dt93
dz 5 = 6(1 + 2t5 )z 5 , dt5
+ 3t9 (1 − 27t9 − 54t92 )
d 2 z9 dt92
dz 9 = 3(1 + 9t9 )z 9 , dt9 d3z d2z 13 13 2 2 3 2 t13 + 3t − t 1 − 18t − 32t (1 − t13 )3 1 − 12t13 − 16t13 (1 ) 13 13 13 13 3 2 dt13 dt13 dz 13 2 3 4 + (1 − t13 ) 1 − 42t13 − 43t13 + 168t13 − 108t13 dt13 2 3 4 = 2 1 + 11t13 − 9t13 + 15t13 − 6t13 z 13 + (1 − 60t9 − 189t92 )
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and 3 d3z 25 2 2 2 t25 1 + t25 − t25 1 − 4t25 − 16t25 3 dt25 3 d2z 25 2 2 + 3t25 1 + t25 − t25 1 − 6t25 − 32t25 2 dt25 dz 25 6 4 5 2 2 3 − 108t25 + 77t25 + 192t25 + 1 + t25 − t25 1 − 12t25 − 142t25 − 224t25 dt25 2 3 2 3 4 = 1 + 3t25 + 6t25 − 2t25 1 + 17t25 + 9t25 + 3t25 + 6t25 z 25 . Theorem 8.4. For each ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25} an identity of the form zℓ =
∞
Aℓ (n)tℓn
n=0
holds, where the coefficients satisfy the recurrence relations (n + 1)3 A2 (n + 1) = 8(2n + 1)(16n 2 + 16n + 3)A2 (n), (n + 1)3 A3 (n + 1) = 6(2n + 1)(9n 2 + 9n + 2)A3 (n), (n + 1)3 A4 (n + 1) = 8(2n + 1)3 A4 (n), (n + 1)3 A5 (n + 1) = 2(2n + 1)(11n 2 + 11n + 3)A5 (n) + 4n(2n + 1)(2n − 1)A5 (n − 1), (n + 1)3 A7 (n + 1) = (2n + 1)(13n 2 + 13n + 4)A7 (n) + 3n(3n + 1)(3n − 1)A7 (n − 1), (n + 1)3 A9 (n + 1) = 3(2n + 1)(3n 2 + 3n + 1)A9 (n) + 27n 3 A9 (n − 1); the coefficients A13 (n) satisfy a 6-term recurrence relation given by Theorem 6.4; and the coefficients A25 (n) satisfy a 9-term recurrence relation that can be deduced from the differential equation in Theorem 8.3 (we mercifully suppress the details). Moreover, 2 2 3 2n 4n 2n 3n 2n A2 (n) = , A3 (n) = , A4 (n) = , n 2n n n n n 2 n+ j n 2n , A5 (n) = j n j=0 j A7 (n) =
⌊n/2⌋ j=0
n j
2
2n − j n
2n − 2 j n
,
and A9 (n) =
n 2 n j j + k . j k k n j k
Similar formulas are not yet known for A13 (n) or A25 (n). Finally, we can give two general classes of series for 1/π that contain Theorems 7.2 and 7.3 as special cases.
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Theorem 8.5. Let ℓ be a positive for which ℓ − 1 divides 24. Suppose N ≥ 2 be an √ integer integer and let q N = exp −2π N /ℓ . Let TN = tℓ (q N ). Then, provided the series converges, the identity ∞ 1 ℓ j 2 2 Aℓ ( j) ( j + λ N ) TN = 1 − 2bℓ TN + (bℓ − 4cℓ )TN 2π N j=0 holds, where the value of λ N is given by (49). Theorem 8.6. 4 ϵ= 2 1
Let ℓ be a positive integer for which ℓ − 1 divides 24. Let ϵ be defined by if ℓ ≡ 1 (mod 2), if ℓ ≡ 2 (mod 4), if ℓ ≡ 0 (mod 4).
√ Suppose N ≥ 2 be an integer and let q N = − exp −2π N /ϵ ℓ . Let TN = tℓ (q N ). Then, provided the series converges, the identity ∞ 1 ϵℓ j 2 2 1 − 2bℓ TN + (bℓ − 4cℓ )TN Aℓ ( j) ( j + λ N ) TN = 2π N j=0 holds, where the value of λ N is given by (49). Theorems 7.2 and 7.3 are the special cases of Theorems 8.5 and 8.6 for which ℓ = 13. The instances ℓ = 2, 3, 4, 5 and 9 of Theorems 8.5 and 8.6 were studied in [9] (where the levels 1, 6 and 8 were also studied), and the results for level ℓ = 7 were studied in [12]. An example of a √ series for level ℓ √ = 25 in Theorem √ 8.5 is given by the values N = 4, T4 = t25 (exp(−2π 4/25)) = (7 − 3 5)/4 and λ4 = ( 5 + 1)/10. The series (2) and (4) correspond to the cases (ℓ, N ) = (5, 47) and (ℓ, N ) = (13, 31), respectively, of Theorem 8.6. Ramanujan’s series (44) is the instance (ℓ, N ) = (2, 29) of Theorem 8.5. References [1] G.E. Andrews, B.C. Berndt, Ramanujan’s Lost Notebook, Part I, Springer, New York, 2005. [2] T.M. Apostol, Modular functions and Dirichlet Series in Number Theory, second ed., Springer-Verlag, New York, 1990. [3] B.C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, New York, 1991. [4] B.C. Berndt, S. Bhargava, F.G. Garvan, Ramanujan’s theories of elliptic functions to alternative bases, Trans. Amer. Math. Soc. 347 (1995) 4163–4244. [5] B.C. Berndt, S.-S. Huang, J. Sohn, S.H. Son, Some theorems on the Rogers–Ramanujan continued fraction in Ramanujan’s lost notebook, Trans. Amer. Math. Soc. 352 (2000) 2157–2177. [6] H.-C. Chan, S. Ebbing, Factorization theorems for the Rogers–Ramanujan continued fraction in the lost notebook, Preprint (Oct. 8, 2008), https://edocs.uis.edu/hchan1/www/. [7] H.H. Chan, S.H. Chan, Z.-G. Liu, Domb’s numbers and Ramanujan–Sato type series for 1/π, Adv. Math. 186 (2004) 396–410. [8] H.H. Chan, S.H. Chan, Z.-G. Liu, The Rogers–Ramanujan continued fraction and a new Eisenstein series identity, J. Number Theory 129 (2009) 1786–1797. [9] H.H. Chan, S. Cooper, Rational analogues of Ramanujan’s series for 1/π, Math. Proc. Cambridge Philos. Soc. 153 (2012) 361–383. [10] S. Cooper, Inversion formulas for elliptic functions, Proc. Lond. Math. Soc. (3) 99 (2009) 461–483. [11] S. Cooper, Construction of Eisenstein series for Γ0 ( p), Int. J. Number Theory 5 (2009) 765–778. [12] S. Cooper, Sporadic sequences, modular forms and new series for 1/π, Ramanujan J. 29 (2012) 163–183.
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[13] S. Cooper, M.D. Hirschhorn, Factorizations that involve Ramanujan’s function k(q) = r (q)r 2 (q 2 ), Acta Math. Sin. (Engl. Ser.) 27 (2011) 2301–2308. [14] S. Cooper, P.C. Toh, Quintic and septic Eisenstein series, Ramanujan J. 19 (2009) 163–181. [15] S. Cooper, D. Ye, Explicit evaluations of a level 13 analogue of the Rogers–Ramanujan continued fraction, J. Number Theory (2014) http://dx.doi.org/10.1016/j.jnt.2013.12.015. [16] W. Duke, Continued fractions and modular functions, Bull. Amer. Math. Soc. 42 (2005) 137–162. [17] A. Erd´elyi, Higher Transcendental Functions, vol. 1, McGraw-Hill, New York, 1953. [18] R. Evans, Theta function identities, J. Math. Anal. Appl. 147 (1990) 97–121. [19] J. Guillera, W. Zudilin, Ramanujan-type formulae for 1/ 41 : the art of translation, in: B.C. Berndt, D. Prasad (Eds.), The Legacy of Srinivasa Ramanujan, in: Ramanujan Math. Soc. Lecture Notes Series, vol. 20, 2013, pp. 181–195. [20] M. Hirschhorn, An identity of Ramanujan, and applications, in: M.E.H. Ismail, D. Stanton (Eds.), q-Series from a Contemporary Perspective, American Mathematical Society, Providence, RI, 2000, pp. 229–234. [21] M.D. Hirschhorn, Ramanujan’s “most beautiful identity”, Amer. Math. Monthly 118 (2011) 839–845. [22] T. Horie, N. Kanou, Certain modular functions similar to the Dedekind eta function, Abh. Math. Sem. Univ. Hamburg 72 (2002) 89–117. [23] Z.-G. Liu, A theta function identity and the Eisenstein series on Γ0 (5), J. Ramanujan Math. Soc. 22 (2007) 283–298. [24] S. Ramanujan, Modular equations and approximations to π, Q. J. Math. 45 (1914) 350–372. [25] S. Ramanujan, Some properties of p(n), the number of partitions of n, Proc. Cambridge Phil. Soc. 19 (1919) 207–210. [26] S. Ramanujan, Notebooks (2 volumes), Tata Institute of Fundamental Research, Bombay, 1957. [27] S. Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa, New Delhi, 1988. [28] L.J. Rogers, Second memoir on the expansion of certain infinite products, Proc. Lond. Math. Soc. 25 (1893) 318–343. [29] K. Venkatachaliengar, S. Cooper, Development of Elliptic Functions According to Ramanujan, World Scientific, Singapore, 2012.
ScienceDirect Journal of Approximation Theory (
)
– www.elsevier.com/locate/jat
Full length article
The Rogers–Ramanujan continued fraction and its level 13 analogue Shaun Cooper ∗ , Dongxi Ye Institute of Natural and Mathematical Sciences, Massey University-Albany, Private Bag 102904, North Shore Mail Centre, Auckland, New Zealand Received 18 July 2013; received in revised form 14 January 2014; accepted 22 January 2014
Communicated by Special Issue Guest Editor Dedicated to Richard Askey in celebration of his 80th birthday
Abstract One of the properties of the Rogers–Ramanujan continued fraction is its representation as an infinite product given by
(q) = q
1/5
∞
(1 − q ) j
j 5
j=1
where
j p
is the Legendre symbol. In this work we study the level 13 function
R(q) = q
∞
(1 − q j )
j 13
j=1
and establish many properties analogous to those for the fifth power of the Rogers–Ramanujan continued fraction. Many of the properties extend to other levels ℓ for which ℓ − 1 divides 24, and a brief account of these results is included. c 2014 Elsevier Inc. All rights reserved. ⃝
Keywords: Dedekind eta function; Eisenstein series; Hypergeometric function; Modular form; Pi; Ramanujan’s theories of elliptic functions to alternative bases
∗ Corresponding author.
E-mail addresses: [email protected] (S. Cooper), [email protected] (D. Ye). http://dx.doi.org/10.1016/j.jat.2014.01.008 c 2014 Elsevier Inc. All rights reserved. 0021-9045/⃝
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1. Introduction The Rogers–Ramanujan continued fraction (q) is defined for |q| < 1 by q 1/5 q
(q) = 1+ 1+
.
q2 q3 1+ 1 + ···
One of its main properties, due to Rogers [28], is the infinite product formula given by (q) = q
1/5
∞
(1 − q ) j
j=1
j 5
= q 1/5
∞ (1 − q 5 j−4 )(1 − q 5 j−1 ) (1 − q 5 j−3 )(1 − q 5 j−2 ) j=1
where pj is the Legendre symbol. This work is about the level 13 analogue defined by R(q) = q
∞
(1 − q j )
j 13
j=1 ∞ (1 − q 13 j−12 )(1 − q 13 j−10 )(1 − q 13 j−9 )(1 − q 13 j−4 )(1 − q 13 j−3 )(1 − q 13 j−1 ) =q . (1 − q 13 j−11 )(1 − q 13 j−8 )(1 − q 13 j−7 )(1 − q 13 j−6 )(1 − q 13 j−5 )(1 − q 13 j−2 ) j=1
Our goal is to show that although R(q) does not have a simple expansion as a continued fraction, it has many other properties similar to the fifth power of the Rogers–Ramanujan continued fraction. Let us illustrate with two examples. First, if r (q) = 5 (q) then it is well-known that ∞ 1 1 (1 − q j )6 − 11 − r (q) = . r (q) q j=1 (1 − q 5 j )6
Ramanujan found an analogous property for R(q), namely ∞ 1 1 (1 − q j )2 − 3 − R(q) = . R(q) q j=1 (1 − q 13 j )2
This is one of five identities in Entry 8(i) of Chapter 20 in Ramanujan’s second notebook [26]. One of these identities is notable for being the last result in the 21 chapters of the notebook to be proved; see the paper by Evans [18] for more information. Here is the second example. If r = r (q) = 5 (q), then it was shown in [9] that n ∞ d r r (1 − 11r − r 2 ) log (1) q = a(n) dq 1 − 11r − r 2 (1 + r 2 )2 n=0 where a(n) =
2n n
n 2 n n+ j . j j j=0
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A general class series for 1/π involving the coefficients a(n) was derived in [9]. A specific example of such a series is ∞ 1705 71 −1 n 1 = √ a(n) n + . (2) π 682 15228 81 47 n=0 We will show that an analogue of (1) is given by n ∞ d R R(1 − 3R − R 2 ) log = q A(n) dq 1 − 3R − R 2 (1 + R 2 )2 n=0
(3)
where the coefficients A(n) satisfy a 6-term recurrence relation. We will also derive a general class of series for 1/π involving the coefficients A(n) and exhibit the example ∞ 1 161 9 −1 n A(n) n + =√ , (4) π 750 124 31 n=0 which is an analogue of (2). The objective of this work is to provide a systematic development of the main properties of the level 13 analogue R(q). To emphasize the analogy between the level 5 function r (q) and the level 13 function R(q), most of the main results will be presented in pairs: the level 5 result will be stated first followed by its level 13 analogue. The level 5 results are generally known, so references to proofs are provided. Many of the level 13 results are new, in which case proofs are given. 2. Definitions This section is organized in three parts. The first part contains some standard definitions and results. In the second part, some functions related to the Rogers–Ramanujan continued fraction are defined. The level 13 analogues are defined in the third part. To emphasize the analogy, lower case letters are used for functions related to the Rogers–Ramanujan continued fraction, while capital letters are used for the level 13 analogues. 2.1. Standard definitions and results Let τ be a complex number with positive imaginary part and put q = exp(2πiτ ). Ramanujan’s Eisenstein series L, M and N , the Dedekind eta function η(τ ) and the j-invariant are defined by L(q) = 1 − 24
∞
jq j , 1−qj
j=1 ∞
N (q) = 1 − 504
j=1
η(τ ) = q 1/24
∞
M(q) = 1 + 240
∞ j 3q j , 1−qj j=1
j 5q j , 1−qj
(1 − q j )
j=1
and
j (τ ) =
1728M 3 (q) . M 3 (q) − N 2 (q)
We will sometimes make use of the compact notation for infinite products defined by (x1 , x2 , . . . , xm ; q)∞ =
∞ j=0
(1 − x1 q j )(1 − x2 q j ) · · · (1 − xm q j ).
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Let ℓ be a positive integer, called the level. The modular group Γ and the congruence subgroup Γ0 (ℓ) are defined by a b Γ = : a, b, c, d ∈ Z, ad − bc = 1 c d and Γ0 (ℓ) =
a c
b d
∈Γ :c≡0
(mod ℓ) .
The results in the following lemmas are well known. Lemma 2.1. Ramanujan’s Eisenstein series satisfy the differential equations q
L2 − M dL = , dq 12
q
dM LM − N = , dq 3
and
q
dN L N − M2 = dq 2
(5)
as well as the identity M 3 (q) − N 2 (q) = 1728q
∞
(1 − q j )24 .
(6)
j=1
Proof. See [29, pp. 18,36] for proofs and more information. Lemma 2.2. Let ac db ∈ Γ , q = exp(2πiτ ) and q ′ = exp 2πi η24
aτ + b cτ + d
aτ +b cτ +d
. Then
= (cτ + d)12 η24 (τ )
(7)
and L(q ′ ) = (cτ + d)2 L(q) + In the special case ac db = 10
−1 η τ
=
τ η(τ ) i
6c(cτ + d) . πi −1 these transformation formulas become 0
and
L(q ′ ) = τ 2 L(q) +
6τ . πi
(8)
(9)
Proof. The first result is standard, e.g., see [2, pp. 50–51]. The transformation formula (8) follows from (7) by logarithmic differentiation with respect to τ . Lemma 2.3. Let m be a fixed positive integer and let ω = exp(2πi/m). Let j and k be real numbers. Suppose Imτ > 0, q = exp(2πiτ ) and p = exp(−2πi/mτ ). Then ∞ ∞ sin πmj (1 − ω j q n )(1 − ω− j q n ) (1 − p mn− j )(1 − p mn−m+ j ) E = p (1 − ωk q n )(1 − ω−k q n ) (1 − p mn−k )(1 − p mn−m+k ) sin πk n=1 m n=1
where the exponent E is given by E=
k(m − k) − j (m − j) (k − j)(m − j − k) = . 2m 2m
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Proof. This is obtained by applying the classical transformation formula for the theta function, e.g., [3, p. 36, Entry 20], to a quotient of two theta functions. 2.2. Definitions pertaining to the Rogers–Ramanujan continued fraction Let r , s and t be the modular functions defined by 5 ∞ ∞ j (1 − q 5 j−4 )5 (1 − q 5 j−1 )5 (1 − q j )( 5 ) = q r = r (q) = 5 (q) = q (1 − q 5 j−3 )5 (1 − q 5 j−2 )5 j=1 j=1 s = s(q) = q
∞ (1 − q 5 j )6 (1 − q j )6 j=1
and t = t (q) =
s . 1 + 22s + 125s 2
Let u and v be the weight 2 modular forms defined by ∞ ∞ ∞ d j j jq j u = u(q) = q = 1 − 5 log r (q) = 1 − 5 jq jk j dq 5 1 − q 5 j=1 k=1 j=1 and v = v(q) =
∞ j j=1
5
∞ ∞ qj j kq jk . = 5 (1 − q j )2 j=1 k=1
Let y and z be the weight 2 modular forms defined by y = y(q) =
∞ (1 − q j )5 (1 − q 5 j ) j=1
and z = z(q) = q
d 1 log s = 5L(q 5 ) − L(q) . dq 4
2.3. Definitions pertaining to the level 13 analogues Now we will define level 13 analogues of the functions defined in Section 2.2. Each function defined by a capital letter is an analogue of the function defined using the corresponding lower case letter in Section 2.2. Let R, S and T be the modular functions defined by R = R(q) = q
∞
(1 − q j )
j 13
j=1
=q
(q, q 3 , q 4 , q 9 , q 10 , q 12 ; q 13 )∞ (q 2 , q 5 , q 6 , q 7 , q 8 , q 11 ; q 13 )∞
∞ (1 − q 13 j−12 )(1 − q 13 j−10 )(1 − q 13 j−9 )(1 − q 13 j−4 )(1 − q 13 j−3 )(1 − q 13 j−1 ) = q , (1 − q 13 j−11 )(1 − q 13 j−8 )(1 − q 13 j−7 )(1 − q 13 j−6 )(1 − q 13 j−5 )(1 − q 13 j−2 ) j=1
S = S(q) = q
∞ (1 − q 13 j )2 (1 − q j )2 j=1
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and S . 1 + 6S + 13S 2 Let U and V be the weight 2 modular forms defined by ∞ ∞ ∞ d j j jq j U = U (q) = q = 1 − jq jk log R(q) = 1 − j dq 13 1 − q 13 j=1 k=1 j=1 T = T (q) =
and V = V (q) =
∞ ∞ ∞ j j qj = kq jk . j )2 13 13 (1 − q j=1 j=1 k=1
Let Y and Z be the modular forms of weights 6 and 2, respectively, defined by Y = Y (q) =
∞ (1 − q j )13 (1 − q 13 j ) j=1
and Z = Z (q) = q
d 1 log S = 13L(q 13 ) − L(q) . dq 12
Although y and Y are clearly analogues, y is a modular form of weight 2 while Y has weight 6. Therefore, in formulas that involve these functions, Y will occur as an analogue of y 3 . 3. Identities involving modular functions In this section we outline some basic properties of the functions r and R. Theorem 3.1. The following identities hold: 1 1 − 11 − r = r s
(10)
1 1 −3− R = . R S
(11)
and
Proof. The identity (10) was given by Ramanujan in his paper [25]. A simple proof using only the Jacobi triple product identity has been given by Hirschhorn [20]. More information, and references to other proofs, can be found in the book by Andrews and Berndt [1, pp. 11–12]. The identity (11) was known to Ramanujan [26, Ch. 20, Entry 8(i)]. Proofs have been given by Evans [18, (5.9)], Berndt [3, p. 375, (8.15)] and Horie and Kanou [22, Theorem 2.2]. Corollary 3.2. t=
r (1 − 11r − r 2 ) s = 1 + 22s + 125s 2 (1 + r 2 )2
and T =
S R(1 − 3R − R 2 ) = . 1 + 6S + 13S 2 (1 + R 2 )2
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Proof. The first equality in each identity is just the definition; the second equality in each identity follows from the results in Theorem 3.1. The next result states how r and R transform under the congruence subgroups Γ0 (5) and Γ0 (13), respectively. Theorem 3.3. Let e(τ ) = exp(2πiτ ). d a b d aτ +b = 5 r (e(τ )) 5 . If c d ∈ Γ0 (5) then r e cτ +d d a b aτ +b d If c d ∈ Γ0 (13) then R e cτ +d = 13 R(e(τ )) 13 . Proof. See [22, Theorem 1.1].
The next result describes how r and R transform under the Fricke involutions τ → −1/5τ and τ → −1/13τ , respectively. Theorem 3.4. Let e(τ )√ = exp(2πiτ ), q = e(τ ), q5 = e(−1/5τ ) and q13 = e(−1/13τ ). Let √ 1+ 5 β = 2 and δ = 3+2 13 . The following transformation formulas hold: r (q5 ) =
1 − β 5r (q) β 5 + r (q)
(12)
and R(q13 ) =
1 − δ R(q) . δ + R(q)
(13)
Proof. The identity (12) may be rearranged to the symmetric form √ 5 √ 5+1 5 5 10 . r (q) + β r (q5 ) + β = 1 + β = 5 5 2 This was stated by Ramanujan in the Lost Notebook; see [1, pp. 91–92] for a discussion and proof. It remains to prove (13). For brevity, write R = R(q) and R ′ = R(q13 ). By Theorem 3.1 we have 1 η2 (τ ) −3− R = 2 . (14) R η (13τ ) If we replace τ with −1/13τ and apply the modular transformation for the Dedekind eta function given in (9), we get 1 η2 (−1/13τ ) η2 (13τ ) = 13 2 . − 3 − R′ = 2 ′ R η (−1/τ ) η (τ ) On multiplying (14) by (15) we get 1 1 ′ −3− R − 3 − R = 13. R R′ This may be expanded and rearranged to give 2 3 13 1 − (R + R ′ ) − R R ′ = (R + R ′ )2 2 4
(15)
8
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)
–
and it follows that √ 13 3 ′ ′ (R + R ′ ). 1 − (R + R ) − R R = ± 2 2
(16) √ The sign√may be determined by substituting the value τ = i/ 13. This gives R = R ′ = R(e−2π/ 13 ), so (16) reduces to √ 1 − 3R − R 2 = ± 13 R. (17) Using the first two terms in the q-expansion R(q) = q − q 2 + O(q 3 ) we obtain the approximation √
R(e−2π/
√ 13
) ≈ e−2π/
13
√ 13
− e−4π/
≈ 0.14.
It follows that we must select the plus sign in (17) and hence in (16). On rearranging (16) we complete the proof of (13) and also obtain the symmetric form √ √ 13 2 3 + 13 . (R(q) + δ) (R(q13 ) + δ) = 1 + δ = 2 Corollary 3.5. The following evaluations hold: √ r exp(−2π/ 5) = β 10 + 1 − β 5 and
√ R exp(−2π/ 13) = δ 2 + 1 − δ,
where the values of β and δ are as given in Theorem 3.4. √ √ Proof. The values τ √ = i/ 5 and τ = i/ 13 give √fixed points for (12) and (13), namely q = q5 = exp(−2π/ exp(−2π/ 13), respectively. On solving (12) and √ 5) and q = q13 = √ (13) for r (exp(−2π/ 5)) and r (exp(−2π/ 13)), respectively, and choosing the solutions that are positive, we obtain the claimed evaluations. The results in the next theorem provide factorizations of the identities in Theorem 3.1. Theorem 3.6. Let ζ = exp(2πi/5), α =
√ 1− 5 2
and β =
√ 1+ 5 2 .
Then
∞ √ 1 1 1 √ − α 5 r = 5 1/12 j )5 (1 − ζ 4 q j )5 (q s) (1 − ζ q r j=1
(18)
∞ √ 1 1 1 . √ − β 5 r = 5 1/12 2 q j )5 (1 − ζ 3 q j )5 (q s) (1 − ζ r j=1
(19)
and
If ξ = exp(2πi/13), γ =
√ 3− 13 2
and δ =
√ 3+ 13 2 ,
then
√ 1 1 1 × √ −γ R = 1/4 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q; q) (q S) (ξ q, ξ R ∞
(20)
–
9
√ 1 1 1 × 2 . √ −δ R = 1/4 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q; q) (q S) (ξ q, ξ R ∞
(21)
S. Cooper, D. Ye / Journal of Approximation Theory (
)
and
√ √ Proof. √ The identities (18) and (19), as well as infinite product representations for 1/ − α √ and 1/ − β , were given by Ramanujan in the Lost Notebook [27, p. 206]. Proofs of these identities have been given in [1, pp. 21–24], [5,6,8,13,21,23]. We will prove (20) and (21). By Theorem 3.4 we have 4 , q 9 , q 10 , q 12 ; q 13 ) (q13 , q 3 , q13 1 − δ R(q) 13 13 13 13 ∞ = R(q13 ) = q13 2 13 . 5 , q 6 , q 7 , q 8 , q 11 ; q 13 ) δ + R(q) (q13 , q13 13 13 13 13 13 ∞
By three applications of Lemma 2.3 with m = 13, p = q13 = exp(−2πi/13τ ) we deduce 3π 4π π sin 13 sin 13 sin 13 (ξ q, ξ 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q; q) 1 − δ R(q) ∞ = . 2π 5π 6π (ξ 2 q, ξ 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q; q)∞ δ + R(q) sin 13 sin 13 sin 13 Since 3π 4π √ sin 13 sin 13 13 − 3 1 = = −γ = 2 δ sin 5π sin 6π sin 2π 13 13 13
π sin 13
we may multiply both sides by δ to get 1 − δ R(q) (ξ q, ξ 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q; q)∞ . = 2 1 − γ R(q) (ξ q, ξ 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q; q)∞
(22)
On the other hand, the identity (11) in Theorem 3.1 may be written as ∞ (1 − γ R(q))(1 − δ R(q)) 1 (1 − q j )2 = . R(q) q j=1 (1 − q 13 j )2
(23)
On dividing (23) by (22) we get ∞ (ξ 2 q, ξ 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q; q)∞ 1 (1 − γ R(q))2 (1 − q j )2 = × R(q) q j=1 (1 − q 13 j )2 (ξ q, ξ 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q; q)∞
=
∞ 1 1 (1 − q j ) × . 3 4 9 10 12 2 q j=1 (1 − q 13 j ) (ξ q, ξ q, ξ q, ξ q, ξ q, ξ q; q)∞
On taking square roots we obtain (20). The identity (21) may be obtained in a similar way, starting by multiplying (22) by (23). The identities (20) and (21) are equivalent to the pair of identities (q 2 , q 5 , q 6 , q 7 , q 8 , q 11 , q 13 , q 13 ; q 13 )∞ − γ q (q, q 3 , q 4 , q 9 , q 10 , q 12 , q 13 , q 13 ; q 13 )∞ = (ξ 2 q, ξ 5 q, ξ 6 q, ξ 7 q, ξ 8 q, ξ 11 q, q, q; q)∞
10
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)
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and (q 2 , q 5 , q 6 , q 7 , q 8 , q 11 , q 13 , q 13 ; q 13 )∞ − δ q (q, q 3 , q 4 , q 9 , q 10 , q 12 , q 13 , q 13 ; q 13 )∞ = (ξ q, ξ 3 q, ξ 4 q, ξ 9 q, ξ 10 q, ξ 12 q, q, q; q)∞ . It would be interesting to have direct proofs of these results. 4. Identities involving modular forms In this section we outline the basic properties of the modular forms u, v, y and z and their level 13 analogues U , V , Y and Z . All of these forms have weight 2 except for Y which has weight 6. We begin by noting some known transformation formulas involving S, U and V . Similar transformation formulas hold for the functions s, u and v and the group Γ0 (5) but, as they will not be required, we do not state them here. Lemma 4.1. Let S, U and V be as defined in Section 2.3. a b Let c d ∈ Γ0 (13). −2πi +b and q = exp . Let q = exp(2πiτ ), q ′ = exp 2πi aτ 13 cτ +d 13τ Then 1 (i) S(q ′ ) = S(q), (ii) S(q13 ) = , 13S(q) d 1 (iii) U (q ′ ) = (cτ + d)2 U (q), (iv) U (q13 ) = √ (13τ )2 V (q), 13 13 √ d ′ 2 (v) V (q ) = (cτ + d) V (q), (vi) V (q13 ) = 13 τ 2 U (q). 13 Proof. For (i), see [2, p. 87, Th. 4.9]. The identity (ii) follows immediately from the definition of S and the transformation formula for the Dedekind eta function in (9). For (iii)–(vi), see [11, Th. 6.1]. Theorem 4.2. The modular forms u, v, y and z (respectively, U , V , Y and Z ) are interrelated by 1 (i) y = u (ii) v = su, (iii) z = r + v (24) r and U3 (i) Y = , (1 + 5S + 13S 2 )2
(ii) V = SU,
Proof of (24). Two classical identities are ∞ ∞ j jq j (1 − q j )5 1−5 = 5 1−qj (1 − q 5 j ) j=1 j=1 and ∞ j j=1
5
∞ qj (1 − q 5 j )5 = q . (1 − q j ) (1 − q j )2 j=1
(iii) Z =
1 R+ R
V.
(25)
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)
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11
These were known to Ramanujan and many proofs exist; see [1, pp. 406–407] for links to Ramanujan’s work and references to proofs. They imply infinite product formulas for u and v, respectively, and on comparing them with the infinite products that occur in the definitions of s and y, we immediately deduce the identities (24)(i) and (ii). The identity (24)(iii) was known d to the to Ramanujan; see [1, p. 88, Entry 3.2.6]. It can be proved by applying the operator q dq identity (10) and appealing the definitions of v and z. This completes the proof of the identities in (24). (q) and g(τ ) = S(q) and note that f and g are analytic in the Proof of 25(i). Let f (τ ) = UY 2 (q) region Imτ > 0. By the definitions of η, S and Y we have 6
f (τ ) = U 6 (q)
∞ U 6 (q)S(q) (1 − q 13 j )2 = . (1 − q j )26 η24 (τ ) j=1
By Lemma 2.2 and Parts (i) and (iii) of Lemma 4.1 it follows that f (τ ) and g(τ ) are each invariant under Γ0 (13). Let us examine the behavior at τ = 0. By Lemma 2.2 and Parts (ii) and (iv) of Lemma 4.1 we have −1 U 6 (q13 )S(q13 ) f = 24 13τ η (−1/13τ ) V 6 (q) × S(q)η24 (13τ ) 134 1 26 76 267 600 1 4 = 4 + 7+ 6+ 5+ 4 + 3 8 q q q q q 13 q 1448 2552 + 2 + + 4601 + O(q) q q =
1
and g
−1 13τ
1 13S(q) 1 1 = − 2 − q + 2q 2 + q 3 + 2q 4 − 2q 5 − 2q 7 + O(q 8 ) . 13 q
=
Thus, f (−1/13τ ) and g(−1/13τ ) have poles of orders 8 and 1, respectively, at q = 0. It follows that there are unique constants, a8 , a7 , . . . , a0 , that can be determined by successively comparing coefficients of q −8 , q −7 , . . . , q −1 , q 0 , such that −1 −1 −1 −1 8 7 f − a8 g + a7 g + · · · + a1 g + a0 = O(q). 13τ 13τ 13τ 13τ On solving for a8 , a7 , . . . , a0 , we conclude that f
−1 13τ
4 −1 −1 − 13g 2 + 5g + 1 = O(q). 13τ 13τ
12
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)
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Replacing τ with −1/13τ we deduce that the function h(τ ) defined by 4 h(τ ) := f (τ ) − 13g 2 (τ ) + 5g(τ ) + 1 4 U 6 (q) = 2 − 13S 2 (q) + 5S(q) + 1 Y (q) is analytic at the vertex τ = 0. Clearly, h(τ ) is also analytic in the region Imτ > 0, and 4 U 6 (0) h(i∞) = 2 − 13S 2 (0) + 5S(0) + 1 = 0. Y (0) It follows that h is bounded. Since h is invariant under Γ0 (13), Theorem 4.4 in [2, p. 79] implies that h is constant. Since h(i∞) = 0, it follows that h is identically zero and so U6 . (1 + 5S + 13S 2 )4 On taking square roots and evaluating both sides when q = 0 to determine the sign, we complete the proof of (25)(i). Y2 =
(q) and g(τ ) = S(q) and follow the steps in the proof of (25)(i) Proof of 25(ii). Let f (τ ) = VY 2 (q) to deduce V3 . Y = 3 S (1 + 5S + 13S 2 )2 On comparing this result with the result from (25)(i) we obtain (25)(ii). 6
d Proof of 25(iii). Apply the operator q dq to the identity (11) to get 1 dR 1 dS 1+ 2 q = 2q . dq R S dq
By the definitions of U and Z , this becomes 1 1 R+ U = Z. R S On applying (25)(ii) we complete the proof.
The essence of the above proof of (25)(i) is that S(q) is a Hauptmodul for Γ0 (13) and U 3 /Y is modular on Γ0 (13) with no poles in its fundamental domain. Therefore U 3 /Y must be a polynomial in S(q). 5. Eisenstein series In the lost notebook [27, pp. 50,51] Ramanujan noted that the Eisenstein series M(q), M(q 5 ), N (q) and N (q 5 ) may be expressed as algebraic functions of η(τ ) and η(5τ ). The results may be stated as: Theorem 5.1. Let M and N be the Eisenstein series defined in Section 2.1 and let s and y be the eta-quotients defined in Section 2.2. Then M(q) = y 2 (1 + 250s + 3125s 2 ), M(q 5 ) = y 2 (1 + 10s + 5s 2 ), N (q) = y 3 (1 + 22s + 125s 2 )1/2 (1 − 500s − 15625s 2 )
S. Cooper, D. Ye / Journal of Approximation Theory (
)
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13
and N (q 5 ) = y 3 (1 + 22s + 125s 2 )1/2 (1 + 4s − s 2 ). Ramanujan also gave analogous results for M(q), M(q 7 ), N (q) and N (q 7 ) in terms of η(τ ) and η(7τ ). The goal of this section is to prove the level 13 analogues given by Theorem 5.2. Let M and N be the Eisenstein series defined in Section 2.1 and let S and Y be the eta-quotients defined in Section 2.3. Then M(q) = Y 2/3 (1 + 5S + 13S 2 )1/3 × (1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 ), M(q 13 ) = Y 2/3 (1 + 5S + 13S 2 )1/3 (1 + 7S + 20S 2 + 19S 3 + S 4 ), N (q) = Y (1 + 6S + 13S 2 )1/2 × (1 − 494S − 20618S 2 − 237276S 3 − 1313806S 4 − 3712930S 5 − 4826809S 6 ) and N (q 13 ) = Y (1 + 6S + 13S 2 )1/2 ×(1 + 10S + 46S 2 + 108S 3 + 122S 4 + 38S 5 − S 6 ). The method of proof will be to adapt the procedure developed in [14] where the analogous results for levels 5 and 7 were derived. By Theorem 4.2, we may work with any of U , V or Z in place of Y . To simplify the calculations we will work in terms of S and U and then use Theorem 4.2 to re-express the final results in terms of the eta-quotients S and Y . We will not list any level 5 results in this section; for these, as well as the level 7 results, the reader is referred to [14]. Lemma 5.3. The following identity holds: 1 13L(q 13 ) − L(q) = U 1 + 6S + 13S 2 . 12 Moreover, the following relations among differential operators hold: q
d d d = UR = U 1 + 6S + 13S 2 S . dq dR dS
Proof. By the definition of Z in Section 2.3 we have d 1 Z =q log S = 13L(q 13 ) − L(q) . dq 12 On the other hand, by (11), (25)(ii) and (iii), and some algebraic manipulation, we have 1 Z = SU R + R 2 1 = SU − R +4 R
(26)
(27)
(28)
14
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= SU
)
–
2 1 +3 +4 S
= U 1 + 6S + 13S 2 .
(29)
On combining (28) and (29) we obtain 1 13L(q 13 ) − L(q) = Z = U 1 + 6S + 13S 2 12 which proves (26). Next, by the chain rule and the definition of U in Section 2.3 we have d d d d = q log R × R . q = UR dq dq dR dR Finally, by the chain rule, the definition of Z in Section 2.3 and (29) we have d d d d d q = q log S × S =Z× S = U 1 + 6S + 13S 2 S . dq dq dS dS dS Lemma 5.4. The following identities hold: q
q
∞
(1 − q j )24 = Y 2 S = U 6
j=1 ∞ 13
S , (1 + 5S + 13S 2 )4
(1 − q 13 j )24 = Y 2 S 13 = U 6
j=1
S 13 . (1 + 5S + 13S 2 )4
Proof. These are immediate consequences of the definitions of S and Y in Section 2.3 and the identity (25)(i). Lemma 5.5. The following identities hold: S (1 + 5S + 13S 2 )4 S 13 M 3 (q 13 ) − N 2 (q 13 ) = 1728 U 6 . (1 + 5S + 13S 2 )4 M 3 (q) − N 2 (q) = 1728 U 6
Proof. These are immediate consequences of the identity (6) and Lemma 5.4. Lemma 5.6. The following identities hold: 1 − 5S − 91S 2 dU U + 6S L(q) = 1 + 6S + 13S 2 dS 1 + 5S + 13S 2
(30)
and 1 L(q ) = 1 + 6S + 13S 2 13 13
13 + 45S + 65S 2 dU U + 6S 2 dS 1 + 5S + 13S
.
(31)
S. Cooper, D. Ye / Journal of Approximation Theory (
Hence,
1 + 6S + 13S 2 S
)
–
15
dU dS
2 = 1 + 6S + 13S 2 3
5S + 26S 2 1 + 5S + 13S 2
U−
13 L(q 13 ) − L(q) . 72
(32)
Proof. Take the logarithmic derivative of the first identity in Lemma 5.4, using (27), to get q
∞ d log q (1 − q j )24 dq j=1 d = U 1 + 6S + 13S 2 S log S − 4 log(1 + 5S + 13S 2 ) + 6 log U . dS
On calculating the derivatives we obtain (30). The identity (31) may be obtained similarly, starting from the second identity in Lemma 5.4. Finally, the identity (32) may be obtained by subtracting (31) from (30). Lemma 5.7. The following identities hold: 169M(q 13 ) − M(q) = 24 U 2
(1 − 13S 2 )(7 + 39S + 91S 2 ) (1 + 5S + 13S 2 )
(33)
and 2197N (q 13 ) − N (q) = U 1 + 6S + 13S 3 × 2197M(q 13 ) − M(q) −
48U 2 p(S) , (1 + 5S + 13S 2 )2
(34)
where p(S) = S(76 + 481S − 1014S 2 − 19773S 3 − 74698S 4 − 107653S 5 ). d to the identity (26), and make use of (27) to get Proof. We apply the operator q dq
q d d 13L(q 13 ) − L(q) = U 1 + 6S + 13S 2 S U 1 + 6S + 13S 2 . 12 dq dS The derivative on the left hand side can be computed using (5); the derivative on the right hand side can be computed by straightforward calculation and using (32) to express dU d S in terms of U , 13 S, L(q ) and L(q). This gives 1 169(L 2 (q 13 ) − M(q)) − (L 2 (q) − M(q)) 144 2 2 5S + 26S 2 2 = U (1 + 6S + 13S ) 3 1 + 5S + 13S 2 13 − U 1 + 6S + 13S 2 L(q 13 ) − L(q) + U 2 S(3 + 13S), 72
16
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)
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which may be rearranged to give (1 − 13S 2 )(7 + 39S + 91S 2 ) 169M(q 13 ) − M(q) − 24 U 2 (1 + 5S + 13S 2 ) = 13L(q 13 ) − L(q) − 12U 1 + 6S + 13S 2 × 13L(q 13 ) + L(q) + 14U 1 + 6S + 13S 2 .
(35)
By (26), the right hand side of (35) is identically zero. Hence, we deduce (33). d Now we will prove (34). Applying q dq to (33) and making use of (27) we get q
d 169M(q 13 ) − M(q) dq d (1 − 13S 2 )(7 + 39S + 91S 2 ) . = 24 U 1 + 6S + 13S 2 S U2 dS (1 + 5S + 13S 2 )
The derivative on the left hand side may be computed using (5), while the derivative on the right hand side may be computed by direct calculation and using (32) to express dU d S in terms of U , S, L(q 13 ) and L(q). The result simplifies to 1 13L(q 13 ) − L(q) 2197M(q 13 ) − M(q) 2197N (q 13 ) − N (q) = 12 p(S) − 48U 3 1 + 6S + 13S 2 2 . 1 + 5S + 13S 2 On applying (26), we complete the proof of (34).
Theorem 5.8. M(q) = U 2
1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 , 1 + 5S + 13S 2
(36)
1 + 7S + 20S 2 + 19S 3 + S 4 , 1 + 5S + 13S 2 √ 1 + 6S + 13S 2 2 3 3 N (q) = U 2 1 − 494S − 20618S − 237276S 1 + 5S + 13S 2 − 1313806S 4 − 3712930S 5 − 4826809S 6
(38)
√ 1 + 6S + 13S 2 N (q ) = U 2 1 + 5S + 13S 2 × 1 + 10S + 46S 2 + 108S 3 + 122S 4 + 38S 5 − S 6 .
(39)
M(q 13 ) = U 2
(37)
and 13
3
Proof. Let a and b be defined by a = 1 + 6S + 13S 2
and
b = 1 + 5S + 13S 2 .
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)
17
–
Let u 1 , u 2 , u 3 and u 4 be defined by u1 = b
M(q) , U2
u2 = b
M(q 13 ) , U2
b2 N (q) u3 = √ a U3
and
b2 N (q 13 ) u4 = √ . (40) a U3
When q = 0 we have S = 0, U = 1, a = b = 1 and u 1 = u 2 = u 3 = u 4 = 1. The results of Lemmas 5.5 and 5.7 may be written as bu 31 − au 23 = 1728S, bu 32 − au 24 = 1728S 13 , −u 1 + 169u 2 = 24 1 − 13S 2 7 + 39S + 91S 2 , bu 1 − 2197bu 2 − u 3 + 2197u 4 = −48 p(S), where p(S) = S(76 + 481S − 1014S 2 − 19773S 3 − 74698S 4 − 107653S 5 ). This non-linear system may be solved to give u 1 , u 2 , u 3 and u 4 in terms of S by the method in [14, Theorem 2.6]. The results are u 1 = 1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 , u 2 = 1 + 7S + 20S 2 + 19S 3 + S 4 , u 3 = 1 − 494S − 20618S 2 − 237276S 3 − 1313806S 4 − 3712930S 5 − 4826809S 6 , u 4 = 1 + 10S + 46S 2 + 108S 3 + 122S 4 + 38S 5 − S 6 . On using these in (40) we deduce (36)–(39).
We are now ready for: Proof of Theorem 5.2. Use (25)(i) to eliminate U from each of (36)–(39).
Corollary 5.9. The j-invariant defined in Section 2.1 may be expressed in terms of the etaquotient S defined in Section 2.3, as follows: (1 + 5S + 13S 2 )(1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 )3 S (1 + 5S + 13S 2 )(1 + 7S + 20S 2 + 19S 3 + S 4 )3 j (13τ ) = . S 13 j (τ ) =
Proof. Substitute the results of Theorem 5.2 into the definition of the j-invariant.
It should be noted that Corollary 5.9 is the level 13 analogue of the results j (τ ) =
(1 + 250s + 3125s 2 )3 , s
j (5τ ) =
(1 + 10s + 5s 2 )3 , s5
and that the identities [14, Theorem 2.7] and [16, (2.5)] may be obtained from these on using (10). Many transformation formulas of degree 13 for hypergeometric functions can be deduced from Theorem 5.2 and Corollary 5.9. The next result gives two of the simplest transformation formulas.
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)
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Theorem 5.10. The following hypergeometric transformation formulas hold in a neighborhood of x = 0: 1 5 1728x 2 F1 12 , 12 ; 1; (1+5x+13x 2 )(1+247x+3380x 2 +15379x 3 +28561x 4 )3 √ 4 1 + 247x + 3380x 2 + 15379x 3 + 28561x 4 1 5 1728x 13 2 F1 12 , 12 ; 1; (1+5x+13x 2 )(1+7x+20x 2 +19x 3 +x 4 )3 = √ 4 1 + 7x + 20x 2 + 19x 3 + x 4 and 1 5 1 1728x F , , ; 1, 1; 3 2 6 6 2 (1+5x+13x 2 )(1+247x+3380x 2 +15379x 3 +28561x 4 )3 √ 1 + 247x + 3380x 2 + 15379x 3 + 28561x 4 1 5 1 1728x 13 3 F2 6 , 6 , 2 ; 1, 1; (1+5x+13x 2 )(1+7x+20x 2 +19x 3 +x 4 )3 = . √ 1 + 7x + 20x 2 + 19x 3 + x 4 Proof. From [10, (2.5) and Th. 4.11] we have 1 5 1 N (q) M 1/4 (q) = 2 F1 , ; 1; 1 − 3/2 6 6 2 M (q) and from [17, p. 111, (2)] we have 1 5 1 5 , ; 1; x = 2 F1 , ; 1; 4x(1 − x) . 2 F1 6 6 12 12 Therefore, M 1/4 (q) = 2 F1
1 5 M 3 (q) − N 2 (q) , ; 1; 12 12 M 3 (q)
= 2 F1
1 5 1728 , ; 1; 12 12 j (τ )
and so 2 F1
1728 1 5 12 , 12 ; 1; j (τ ) M 1/4 (q)
2 F1
=1=
1 5 1728 12 , 12 ; 1; j (13τ ) M 1/4 (q 13 )
.
Now substitute the results of Theorem 5.2 and Corollary 5.9, and then replace S with x to obtain the first result. The second result can be obtained by squaring the first result and applying Clausen’s formula [17, p. 185] 2 1 1 = 3 F2 2a, 2b, a + b; 2a + 2b, a + b + ; x . 2 F1 a, b; a + b + ; x 2 2 6. Differential equations The goal of this section is to show that Z satisfies a third order linear differential equation with respect to T . We begin by constructing a second order nonlinear differential equation for U in terms of R.
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Theorem 6.1. The following differential equation holds: d dU 1 dU 2 f (R) U R R = R + dR dR 2U dR 2(1 − 3R − R 2 )2 (1 − R + 5R 2 + R 3 + R 4 )2 where f (R) = R(−2 + 13R − 94R 2 + 7R 3 + 124R 4 − 23R 5 − 124R 6 + 7R 7 + 94R 8 + 13R 9 + 2R 10 ). Proof. We may use (11) and (27) to rewrite (30) and (36) in the forms L(q) = U
(1 − 21R − 39R 2 + 21R 3 + R 4 ) dU + 6R 2 3 4 2 dR (1 − R + 5R + R + R )(1 − 3R − R )
(41)
and M(q) = U
2
1 + 235R + 1207R 2 + 955R 3 + 3840R 4 − 955R 5 + 1207R 6 − 235R 7 + R 8 , (42) (1 − R + 5R 2 + R 3 + R 4 )(1 − 3R − R 2 )2
d respectively. From (27) we have q dq = U R ddR . Therefore, on solving (41) for R dU d R and applying d , we obtain the operator q dq d dU q dL UR d U (1 − 21R − 39R 2 + 21R 3 + R 4 ) UR . R = − dR dR 6 dq 6 d R (1 − R + 5R 2 + R 3 + R 4 )(1 − 3R − R 2 )
The derivative of the first term on the right hand side may be obtained using (5) and the result may be expressed in terms of U , R and dU/d R, using (41) and (42). The derivative of the second term on the right hand side may be found by direct calculation. On simplifying and dividing both sides by U we complete the proof. The next result gives a third order linear differential equation for U in terms of R. Theorem 6.2. The following differential equation holds: d d dU g(R) dU R R R = R dR dR dR dR (1 − R + 5R 2 + R 3 + R 4 )2 (1 − 3R − R 2 )2 +
h(R) U, (1 − R + 5R 2 + R 3 + R 4 )3 (1 − 3R − R 2 )3
where g(R) = R −2 + 13R − 94R 2 + 7R 3 + 124R 4 − 23R 5 − 124R 6 + 7R 7 + 94R 8 + 13R 9 + 2R 10
and h(R) = R 1 + R 2 −1 + 9R − 119R 2 + 37R 3 + 1019R 4 − 2446R 5 − 2192R 6 + 3206R 7 + 2192R 8 − 2446R 9 − 1019R 10 + 37R 11 + 119R 12 + 9R 13 + R 14 .
(43)
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Proof. Apply the operator R ddR to both sides of the differential equation in Theorem 6.1. The key is to note that dU 2 dU 1 dU 2 1 1 dU d d R R − R = R R R d R 2U dR U dR dR dR 2U dR and then Theorem 6.1 can be applied to simplify the expression in braces. The remaining details are purely computational and we omit them. Theorem 6.3. Let S, T and Z be as defined in Section 2.3, that is, ∞ S (1 − q 13 j )2 , T = S=q j 2 (1 − q ) 1 + 6S + 13S 2 j=1 and Z =q
∞ d jq j log S = 1 + 2 . dq 1−qj j=1 13- j
The following differential equation holds: d3 Z 2 3 2 d Z T 2 (1 − T )3 1 − 12T − 16T 2 + 3T − T 1 − 18T − 32T (1 ) dT 3 dT 2 dZ + (1 − T ) 1 − 42T − 43T 2 + 168T 3 − 108T 4 dT = 2 1 + 11T − 9T 2 + 15T 3 − 6T 4 Z . Proof. By Corollary 3.2 and Theorem 4.2 we have R(1 − 3R − R 2 ) 1 + R2 and Z = U . (1 + R 2 )2 1 − 3R − R 2 On applying this change of variables to the result of Theorem 6.2 we obtain the claimed result. T =
The next result gives an expansion of Z in powers of T as well as representations of Z in terms of the 3 F2 hypergeometric function. Theorem 6.4. Let S, T and Z be as for Theorem 6.3. Let A( j) be the sequence defined by the recurrence relation ( j + 1)3 A( j + 1) = 15 j 3 + 18 j 2 + 10 j + 2 A ( j) − 23 j 3 − 63 j 2 + 27 j − 9 A ( j − 1) − 11 j 3 + 24 j 2 − 118 j + 70 A ( j − 2) + 6 (2 j − 5) 3 j 2 − 9 j + 5 A ( j − 3) − 4 (2 j − 5) (2 j − 7) ( j − 3) A ( j − 4) and initial conditions A(0) = 1,
A(−1) = A(−2) = A(−3) = A(−4) = 0.
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Then Z =
∞
A( j)T j =
j=0
=
=
∞
A( j)
j=0
S 1 + 6S + 13S 2
j
(1 + 5S + 13S 2 )1/2 (1 + 6S + 13S 2 )1/2 (1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 )1/2 1728S 1 5 1 × 3 F2 , , ; 1, 1; 3 6 6 2 1 + 5S + 13S 2 1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 (1 + 5S + 13S 2 )1/2 (1 + 6S + 13S 2 )1/2 (1 + 7S + 20S 2 + 19S 3 + S 4 )1/2 1 5 1 1728S 13 ×3 F2 , , ; 1, 1; 3 . 6 6 2 1 + 5S + 13S 2 1 + 7S + 20S 2 + 19S 3 + S 4
Proof. The equalities in the first line of the identity follow immediately from Theorem 6.3. Let us prove the formulas that involve the hypergeometric function. By (29), (36) and the formulas in the proof of Theorem 5.10, we have Z = (1 + 6S + 13S 2 )1/2 U (1 + 5S + 6S 2 )1/2 (1 + 6S + 13S 2 )1/2 = M 1/2 (q) (1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 )1/2 (1 + 5S + 6S 2 )1/2 (1 + 6S + 13S 2 )1/2 1 5 1 1728 = , , ; 1, 1; . 3 F2 6 6 2 j (τ ) (1 + 247S + 3380S 2 + 15379S 3 + 28561S 4 )1/2 Now apply Corollary 5.9 to obtain the first result involving the hypergeometric function. The other result involving the hypergeometric function may be proved in a similar way using (37) instead of (36), or by appealing to Theorem 5.10. 7. Series for 1/π In this section we develop some series for 1/π. The prototypes for such series were given by Ramanujan [24] and they include examples such as √ ∞ 2 1 2 2 4j 2 j (1103 + 26390 j) = . (44) π 9801 j=0 2 j j 3964 j The key to proving such formulas is [7, Theorem 2.1]: Theorem 7.1 (H. H. Chan, S. H. Chan and Z.-G. Liu). Let ℓ be a positive integer. Suppose T = T (q), Z = Z (q) and w = w(q) satisfy the properties: √ √ x Z e−2π x/ℓ = Z e−2π/ xℓ for all x > 0, (45) Z (q) =
∞ j=0
A ( j) T j (q)
(46)
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and q
d log T (q) = w (q) Z (q) . dq
(47)
For any integer N ≥ 2, let m (q) =
Z (q) . Z qN
Let λ N , TN and W N be defined by T dm λN = , 2N dT q=e−2π/√ N ℓ √ TN = T e−2π N /ℓ
(48)
(49) (50)
and √ W N = w e−2π N /ℓ .
(51)
Then
∞ ℓ 1 j = WN A ( j) ( j + λ N ) TN . N 2π j=0
(52)
The integers ℓ and N are called the level and degree, respectively. We are now ready to deduce the following class of level 13 series for 1/π : Theorem 7.2. Let N√≥ 2 be an integer and let T = T (q) be the function defined in Section 2.3. Let q N = exp −2π N /13 and TN = T (q N ). Then the identity ∞ 1 13 j 2 1 − 12TN − 16TN A ( j) ( j + λ N ) TN = (53) 2π N j=0 holds for the values of N , TN and λ N given in Table 1. Proof. Let T and Z be the functions defined in Section 2.3 and let ℓ = 13. The hypotheses (45) and (46) of Theorem 7.1 are satisfied because of (8) and Theorem 6.4, respectively. By the definitions of T and Z and a calculation we have d d S q log T = q log dq dq 1 + 6S + 13S 2 1 − 13S 2 d q log S 2 dq 1 + 6S + 13S = 1 − 12T − 16T 2 Z , =
so the hypothesis (47) of Theorem 7.1 holds with w(q) = 1 − 12T − 16T 2 .
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Table 1 Data for Theorems 7.2 and 7.3. q
TN
λN √ 11− 13 27 √ 19 13−67 6 √ 766 − 111 2875 11500 13 √ 3415 1136 14739 − 73695 13
3 7
√ 7 13−25 6 √ 19−5 13 36 √ 15−2 26 363 √ 17−4 13 648 √ − 13−1 6 − 17
15
65 − 27−3 72
31
1 − 124
55
13 − 319−88 1089
79
13−1151 − 320 6399
N 3
√ exp −2π N /13
4 6 10
√ − exp −π N /13
Series does not converge Series does not converge
√
√ √
√
25 − 3 64 320 65 161 750 √ 379 209 1215 − 4860 13 √ 40177 93589 163056 − 326112 13
Therefore, identities of the form (53) hold by Theorem 7.1. In principle, values of TN and λ N can be determined using modular equations; see [8, pp. 408, 409] and [9, Proof of Th. 2.1] for some examples of the technique. In practice, the values are computed and identified numerically. The results that appear to be rational or quadratic irrational are recorded in Table 1. Some details concerning the evaluation of TN will appear in forthcoming work [15]. We also have: Theorem 7.3. Let ≥ 11 bean integer and let T = T (q) be the function defined in Section 2.3. N√ Let q N = − exp −π N /13 and TN = T (q N ). Then the identity ∞ 1 13 j 2 1 − 12TN − 16TN A ( j) ( j + λ N ) TN = (54) π N j=0 holds for the values of N , TN and λ given in Table 1. Proof. Replace q with −q and write the result of Theorem 6.4 in the form Z (−q) =
∞ (−1) j A( j)(−T (−q)) j . j=0
Now apply Theorem 7.1 along with the fact that Z (−q) and −T (−q) are modular forms of level 52. The details are similar to those in the proof of Theorem 7.2, so we omit them. We end this section with some remarks. It should be noted that for the series (54) to converge, √ it is necessary that |TN | < 1/(6 + 2 13). Since N is an integer, this requires N ≥ 11. On the other hand, the series (53) converges for every integer N ≥ 2. In fact, from Corollaries 3.2 and 3.5 it follows that √ R(1 − 3R − R 2 ) 1 T1 = T (exp(−2π/ 13)) = = √ . √ 2 2 (1 + R ) 6 + 2 13 q=exp(−2π/ 13)
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If an explicit combinatorial formula for A( j) was known, then it would be possible to prove the level 13 series for 1/π, without using modular forms, from the method shown in [19]. Finally, the series (4) in the introduction corresponds to the data for Theorem 7.3 in Table 1 for degree N = 31. 8. Other levels: Ramanujan’s theories of elliptic functions to alternative bases, and beyond Most of the analyses in the previous sections can be carried out for other levels ℓ for which ℓ − 1 is a divisor of 24, namely ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25}. The theory in the case ℓ = 4 corresponds to Jacobi’s original theory of theta functions. The levels ℓ = 2, 3 are often referred to as Ramanujan’s theories of elliptic functions to alternative bases. Ramanujan, in fact, had three such theories. The other one corresponds to the level ℓ = 1 which does not fit the general theory to be developed below. All of Ramanujan’s results on elliptic functions to alternative bases have been proved in [4]. A different approach, emphasizing the level, has been given in [10]. For other levels, an analogue of r or R is not the natural starting point (see the column for w in [9, Table 1] for some analogues of r and R for other levels). A more natural and appropriate starting point is an eta-quotient which is an analogue of s and S. We briefly sketch the details for these other levels. Let ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25}. For each level ℓ we associate the integer e defined by (ℓ − 1)e = 24. For each ℓ, let sℓ , yℓ and z ℓ be defined by ∞ (1 − q j )ℓ (1 − q ℓj ) j=1
sℓ = q
∞ (1 − q ℓj )e , (1 − q j )e j=1
zℓ = q
d ℓL(q ℓ ) − L(q) log sℓ = . dq ℓ−1
yℓ =
and
Let tℓ be defined by s2 s3 s4 , t3 = , t4 = , 2 2 (1 + 64s2 ) (1 + 27s3 ) (1 + 16s4 )2 s5 s7 s9 t5 = , t7 = , t9 = , 2 2 1 + 22s5 + 125s5 1 + 13s7 + 49s7 1 + 9s9 + 27s92 s13 s25 t13 = and t25 = . 2 2 1 + 6s13 + 13s13 1 + 2s25 + 5s25 t2 =
That is, tℓ =
sℓ 1 + bℓ sℓ + cℓ sℓ2
where cℓ = ℓ12/(ℓ−1) and bℓ = 128, 54, 32, 22, 13, 9, 6 or 2 corresponding to ℓ = 2, 3, 4, 5, 7, 9, 13 or 25, respectively. Each sℓ and tℓ is a modular function, while z ℓ is a modular form of weight 2 and yℓ is a modular form of weight (ℓ − 1)/2.
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In terms of the functions s, t, y and z in Section 2.2 and S, T , Y and Z in Section 2.3, we have s5 = s,
t5 = t,
and
y5 = y
z5 = z
and s13 = S,
t13 = T,
and
y13 = Y
z 13 = Z .
We state the theorems in this section mostly without proof. The results for level 13 have already been proved in detail in Sections 3–7. The other results may be proved by similar methods. Theorem 8.1. Let ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25} and suppose (ℓ − 1)e = 24. Then z ℓ6 /yℓe is a polynomial in sℓ of degree ℓ + 1. Precisely, we have z 26 y224 z 46 y48 z 76 y74 6 z 13 2 y13
z 36
= (1 + 64s2 )3 ,
= (1 + 27s3 )4 ,
y312 z 56
= (1 + 16s4 )5 ,
y56 z 96
4 = 1 + 13s7 + 49s72 ,
y93
3 = 1 + 22s5 + 125s52 , 5 = 1 + 9s9 + 27s92 ,
3 4 2 2 = 1 + 6s13 + 13s13 1 + 5s13 + 13s13
and 6 3 5 z 25 2 2 3 4 = 1 + 2s25 + 5s25 . 1 + 5s25 + 15s25 + 25s25 + 25s25 y25
Theorem 8.2. The following identities, involving the j-invariant, hold: j (τ ) =
(1 + 256s2 )3 , s2
j (τ ) =
(1 + 27s3 )(1 + 243s3 )3 , s3
j (3τ ) =
j (τ ) =
(1 + 256s4 + 4096s42 )3 , s4 (1 + 16s4 )
j (2τ ) =
j (4τ ) =
j (2τ ) =
(1 + 16s4 + 16s42 )3 s44 (1 + 16s4 )
(1 + 16s2 )3 , s22 (1 + 27s3 )(1 + 3s3 )3 s33 (1 + 16s4 + 256s42 )3 s42 (1 + 16s4 )2
,
j (τ ) =
(1 + 250s5 + 3125s52 )3 , s5
j (τ ) =
(1 + 13s7 + 49s72 )(1 + 245s7 + 2401s72 )3 s7
j (7τ ) = j (τ ) =
(1 + 10s5 + 5s52 )3
(1 + 13s7 + 49s72 )(1 + 5s7 + s72 )3
(1 + 9s9
j (3τ ) =
j (5τ ) =
s77 3 ) (1 + 243s
9
+ 2187s92 + 6561s93 )3
s9 (1 + 9s9 + 27s92 ) (1 + 9s9 )3 (1 + 3s9 )3 (1 + 27s92 )3 s93 (1 + 9s9 + 27s92 )3
s55
,
,
,
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(1 + 3s9 )3 (1 + 9s9 + 27s92 + 3s93 )3 s99 (1 + 9s9 + 27s92 )
2 )(1 + 247s + 3380s 2 + 15379s 3 + 28561s 4 )3 (1 + 5s13 + 13s13 13 13 13 13 s13
j (13τ ) = j (τ ) = j (5τ ) =
2 )(1 + 7s + 20s 2 + 19s 3 + s 4 )3 (1 + 5s13 + 13s13 13 13 13 13 13 s13
3
2 +35000s 3 +178125s 4 +631250s 5 1+250s25 +4375s25 25 25 25 6 +3125000s 7 +4296875s 8 +3906250s 9 +1953125s 10 +1640625s25 25 25 25 25
2 + 25s 3 + 25s 4 ) s25 (1 + 5s25 + 15s25 25 25 2 )3 (1 + 5s + 25s 2 )3 (1 + 5s + 20s 2 + 25s 3 + 25s 4 )3 (1 + 5s25 + 5s25 25 25 25 25 25 25 5 (1 + 5s + 15s 2 + 25s 3 + 25s 4 )5 s25 25 25 25 25
j (25τ ) =
2 + 200s 3 + 525s 4 + 1010s 5 + 1425s 6 + 1400s 7 + 875s 8 + 250s 9 + 5s 10 1 + 10s25 + 55s25 25 25 25 25 25 25 25 25 25 (1 + 5s + 15s 2 + 25s 3 + 25s 4 ) s25 25 25 25 25
3 .
Theorem 8.3. For each ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25}, z ℓ satisfies a third order linear differential equation with respect to tℓ , namely: t22 (1 − 256t2 )
d 3 z2 dt23
+ 3t2 (1 − 384t2 )
dz 2 d 2 z2 = 24z 2 , + (1 − 816t2 ) 2 dt2 dt2
d 2 z3 dz 3 + 3t3 (1 − 162t3 ) 2 + (1 − 348t3 ) = 12z 3 , 3 dt3 dt3 dt3 d 2 z4 d 3 z4 dz 4 = 8z 4 , t42 (1 − 64t4 ) 3 + 3t4 (1 − 96t4 ) 2 + (1 − 208t4 ) dt4 dt4 dt4 d 3 z5 d 2 z5 t52 (1 − 44t5 − 16t52 ) 3 + 3t5 (1 − 66t5 − 32t52 ) 2 dt5 dt5 t32 (1 − 108t3 )
d 3 z3
+ (1 − 144t5 − 108t52 ) t72 (1 − 27t7 )(1 + t7 )
d 3 z7 dt73
+ (1 − 86t7 − 186t72 ) t92 (1 − 18t9 − 27t92 )
+ 3t7 (1 − 39t7 − 54t72 )
d 2 z7 dt72
dz 7 = 4(1 + 6t7 )z 7 , dt7
d 3 z9 dt93
dz 5 = 6(1 + 2t5 )z 5 , dt5
+ 3t9 (1 − 27t9 − 54t92 )
d 2 z9 dt92
dz 9 = 3(1 + 9t9 )z 9 , dt9 d3z d2z 13 13 2 2 3 2 t13 + 3t − t 1 − 18t − 32t (1 − t13 )3 1 − 12t13 − 16t13 (1 ) 13 13 13 13 3 2 dt13 dt13 dz 13 2 3 4 + (1 − t13 ) 1 − 42t13 − 43t13 + 168t13 − 108t13 dt13 2 3 4 = 2 1 + 11t13 − 9t13 + 15t13 − 6t13 z 13 + (1 − 60t9 − 189t92 )
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and 3 d3z 25 2 2 2 t25 1 + t25 − t25 1 − 4t25 − 16t25 3 dt25 3 d2z 25 2 2 + 3t25 1 + t25 − t25 1 − 6t25 − 32t25 2 dt25 dz 25 6 4 5 2 2 3 − 108t25 + 77t25 + 192t25 + 1 + t25 − t25 1 − 12t25 − 142t25 − 224t25 dt25 2 3 2 3 4 = 1 + 3t25 + 6t25 − 2t25 1 + 17t25 + 9t25 + 3t25 + 6t25 z 25 . Theorem 8.4. For each ℓ ∈ {2, 3, 4, 5, 7, 9, 13, 25} an identity of the form zℓ =
∞
Aℓ (n)tℓn
n=0
holds, where the coefficients satisfy the recurrence relations (n + 1)3 A2 (n + 1) = 8(2n + 1)(16n 2 + 16n + 3)A2 (n), (n + 1)3 A3 (n + 1) = 6(2n + 1)(9n 2 + 9n + 2)A3 (n), (n + 1)3 A4 (n + 1) = 8(2n + 1)3 A4 (n), (n + 1)3 A5 (n + 1) = 2(2n + 1)(11n 2 + 11n + 3)A5 (n) + 4n(2n + 1)(2n − 1)A5 (n − 1), (n + 1)3 A7 (n + 1) = (2n + 1)(13n 2 + 13n + 4)A7 (n) + 3n(3n + 1)(3n − 1)A7 (n − 1), (n + 1)3 A9 (n + 1) = 3(2n + 1)(3n 2 + 3n + 1)A9 (n) + 27n 3 A9 (n − 1); the coefficients A13 (n) satisfy a 6-term recurrence relation given by Theorem 6.4; and the coefficients A25 (n) satisfy a 9-term recurrence relation that can be deduced from the differential equation in Theorem 8.3 (we mercifully suppress the details). Moreover, 2 2 3 2n 4n 2n 3n 2n A2 (n) = , A3 (n) = , A4 (n) = , n 2n n n n n 2 n+ j n 2n , A5 (n) = j n j=0 j A7 (n) =
⌊n/2⌋ j=0
n j
2
2n − j n
2n − 2 j n
,
and A9 (n) =
n 2 n j j + k . j k k n j k
Similar formulas are not yet known for A13 (n) or A25 (n). Finally, we can give two general classes of series for 1/π that contain Theorems 7.2 and 7.3 as special cases.
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Theorem 8.5. Let ℓ be a positive for which ℓ − 1 divides 24. Suppose N ≥ 2 be an √ integer integer and let q N = exp −2π N /ℓ . Let TN = tℓ (q N ). Then, provided the series converges, the identity ∞ 1 ℓ j 2 2 Aℓ ( j) ( j + λ N ) TN = 1 − 2bℓ TN + (bℓ − 4cℓ )TN 2π N j=0 holds, where the value of λ N is given by (49). Theorem 8.6. 4 ϵ= 2 1
Let ℓ be a positive integer for which ℓ − 1 divides 24. Let ϵ be defined by if ℓ ≡ 1 (mod 2), if ℓ ≡ 2 (mod 4), if ℓ ≡ 0 (mod 4).
√ Suppose N ≥ 2 be an integer and let q N = − exp −2π N /ϵ ℓ . Let TN = tℓ (q N ). Then, provided the series converges, the identity ∞ 1 ϵℓ j 2 2 1 − 2bℓ TN + (bℓ − 4cℓ )TN Aℓ ( j) ( j + λ N ) TN = 2π N j=0 holds, where the value of λ N is given by (49). Theorems 7.2 and 7.3 are the special cases of Theorems 8.5 and 8.6 for which ℓ = 13. The instances ℓ = 2, 3, 4, 5 and 9 of Theorems 8.5 and 8.6 were studied in [9] (where the levels 1, 6 and 8 were also studied), and the results for level ℓ = 7 were studied in [12]. An example of a √ series for level ℓ √ = 25 in Theorem √ 8.5 is given by the values N = 4, T4 = t25 (exp(−2π 4/25)) = (7 − 3 5)/4 and λ4 = ( 5 + 1)/10. The series (2) and (4) correspond to the cases (ℓ, N ) = (5, 47) and (ℓ, N ) = (13, 31), respectively, of Theorem 8.6. Ramanujan’s series (44) is the instance (ℓ, N ) = (2, 29) of Theorem 8.5. References [1] G.E. Andrews, B.C. Berndt, Ramanujan’s Lost Notebook, Part I, Springer, New York, 2005. [2] T.M. Apostol, Modular functions and Dirichlet Series in Number Theory, second ed., Springer-Verlag, New York, 1990. [3] B.C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, New York, 1991. [4] B.C. Berndt, S. Bhargava, F.G. Garvan, Ramanujan’s theories of elliptic functions to alternative bases, Trans. Amer. Math. Soc. 347 (1995) 4163–4244. [5] B.C. Berndt, S.-S. Huang, J. Sohn, S.H. Son, Some theorems on the Rogers–Ramanujan continued fraction in Ramanujan’s lost notebook, Trans. Amer. Math. Soc. 352 (2000) 2157–2177. [6] H.-C. Chan, S. Ebbing, Factorization theorems for the Rogers–Ramanujan continued fraction in the lost notebook, Preprint (Oct. 8, 2008), https://edocs.uis.edu/hchan1/www/. [7] H.H. Chan, S.H. Chan, Z.-G. Liu, Domb’s numbers and Ramanujan–Sato type series for 1/π, Adv. Math. 186 (2004) 396–410. [8] H.H. Chan, S.H. Chan, Z.-G. Liu, The Rogers–Ramanujan continued fraction and a new Eisenstein series identity, J. Number Theory 129 (2009) 1786–1797. [9] H.H. Chan, S. Cooper, Rational analogues of Ramanujan’s series for 1/π, Math. Proc. Cambridge Philos. Soc. 153 (2012) 361–383. [10] S. Cooper, Inversion formulas for elliptic functions, Proc. Lond. Math. Soc. (3) 99 (2009) 461–483. [11] S. Cooper, Construction of Eisenstein series for Γ0 ( p), Int. J. Number Theory 5 (2009) 765–778. [12] S. Cooper, Sporadic sequences, modular forms and new series for 1/π, Ramanujan J. 29 (2012) 163–183.
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