Generalizations of Ramanujan’s Continued Fractions
N.D. Bagis Stenimahou 5 Edessa Pellas Greece 58200
[email protected]
abstract In this article we continue a previous work in which we have generalized the Rogers Ramanujan continued fraction (RR) introducing what we call, the Ramanujan-Quantities (RQ). We use the Mathematica package to give several modular equations for certain cases of Ramanujan Quantities-(RQ). We also give the modular equations of degree 2 and 3 for the evaluation of the first derivative of RogersRamanujan continued fraction. More precisely for certain cases of (RQ)’s we show how we can find the corresponding continued fraction expansions-S, in which we are able to evaluate with numerical methods some lower degree modular equations of this fraction and its derivatives. A systematically method for evaluating theoriticaly certain (RQ)’s (not for all) and their derivatives, with elementary functions used by Ramanujan is presented. We give applications and several results. keywords: Ramanujan; Continued Fractions; Modular Equations; Derivatives; Evaluations; Quantities
1
1
Definitions and Introductory Results
In this chapter we will define and study expressions that rise from continued fractions, analogous to that of Rogers-Ramanujan (RR), Ramanujan’s Cubic (RC), Ramanujan-Gollnitz-Gordon (RGG) (see relations (1.53), (1.54) and (1.55) below). The results are new since no work has been done in this area and most of them are experimental observations. The focused quantities are Q∞ (1 − q a q np )(1 − q p−a q np ) −(a−b)/2+(a2 −b2 )/(2p) Qn=0 , (1) R(a, b, p; q) = q ∞ b np p−b q np ) n=0 (1 − q q )(1 − q where a, b, p are positive rationals such that a < b < p. General theorems such (a1 − b1 q1 )(b1 − a1 q1 ) (a1 − b1 q13 )(b1 − a1 q13 ) (a1 − b1 q15 )(b1 − a1 q15 ) q A−B = (1 − a1 b1 )+ (1 − a1 b1 )(q12 + 1)+ (1 − a1 b1 )(q14 + 1)+ (1 − a1 b1 )(q16 + 1) + . . . Q∞ (1 − q a q np )(1 − q p−a q np ) = Qn=0 (2) ∞ b np p−b q np ) n=0 (1 − q q )(1 − q where a1 = q A , b1 = q B , q1 = q A+B , a = 2A + 3p/4, b = 2B + p/4 and p = 4(A + B), |q| < 1, are proved. As someone can see these quantities behave as (RR), √ the (RC) and (RGG) continued fractions identities. For example when q = e−π r , r positive rational, they are algebraic numbers and satisfy modular equations. Their derivatives also are all obey the same nome. The idea of formulation (1.1) is described below. Let k−1 Y (a; q)k := (1 − aq n ) (3) n=0
For |q| < 1, the Rogers Ramanujan continued fraction is (see for example [7],[18],[19]): q 1/5 q q 2 q 3 R(q) := ··· (4) 1+ 1+ 1+ 1+ and satisfies the famous Rogers-Ramanujan identity (see [7],[2]): R∗ (q) := q −1/5 R(q) =
∞ Y (q; q 5 )∞ (q 4 ; q 5 )∞ = (1 − q n )X2 (n) (q 2 ; q 5 )∞ (q 3 ; q 5 )∞ n=1
where X2 (n) is the Legendre symbol n5 . Also hold ϑ4 (3ix/4, e−5x/2 ) R(e−x ) = e−x/5 ,x > 0 ϑ4 (ix/4, e−5x/2 )
2
(5)
(6)
where ϑ4 (a, q) is the 4th elliptic theta function (see [9],[14],[20]). We can rewrite (1.6) into the form (see [14]): ! ∞ 4nx 3nx 2nx nx X e − e − e + e 1 R(e−x ) = exp −x/5 − , x > 0. n e5nx − 1 n=1
(7)
For |q| < 1, the Ramanujan-Gollnitz-Gordon (RGG) continued fraction is H(q) =
q 1/2 q2 q4 q6 ··· (1 + q)+ (1 + q 3 )+ (1 + q 5 )+ (1 + q 7 )+
(8)
Also for this continued fraction holds H(e
−x
∞ X 1 e7nx − e5nx − e3nx + enx ) = exp −x/2 − n e8nx − 1 n=1
H(e−x ) = e−x/2
! ,x > 0
ϑ4 (3ix/2, e−4x ) ,x > 0 ϑ4 (ix/2, e−4x )
(9)
(10)
As we will show it is true that exists generalizations for these expansions, but there is no theory developed in this direction, especially for evaluations and modular equations. An example is the continued fraction (1.2).
2
Theorems on Rogers Ramanujan Quantities
Definition 1. √ In general if q = e−π r where a, p, r > 0 we denote ’Agile’ the quantity [a, p; q] = (q p−a ; q p )∞ (q a ; q p )∞
(11)
Definition 2. If |q| < 1, we call ’Ramanujan Quantity’ the R(a, b, p; q) := q −(a−b)/2+(a
2
−b2 )/(2p) [a, p; q]
[b, p; q]
,
(12)
because many of Ramanujan’s continued fractions can be put in this form. Also [a, p; q] R∗ (a, b, p; q) := (13) [b, p; q] Lemma 2.1 If |2t| < πa, then ∞ X cosh(2tk) = log(P0 ) − log(ϑ4 (it, e−aπ )) k sinh(πak)
k=1
3
(14)
where P0 =
Q∞
n=1 (1
− e−2nπa ) and ϑ4 (u, q) = 1 +
P∞
n=1 (−1)
n n2
q
cos(2nu).
Proof. From ([2] pg.170 relation (13-2-12)) and the definition of theta functions we have ∞ Y ϑ4 (z, q) = (1 − q 2n+2 )(1 − q 2n−1 e2iz )(1 − q 2n−1 e−2iz ) (15) n=0
The above identity is known as Jacobi triple product identity. By taking the logarithm in both sides and expanding the logarithm of the individual terms in a power series it is simple to show (1.14) from (1.15), where q = e−πa and a is positive real. Theorem 1. For all positive reals a, b, p, x, a < b < p we have a − b ϑ4 (p − 2a)ix/4, e−px/2 a2 − b2 −x = (16) +x R(a, b, p; e ) = exp −x 2p 2 ϑ4 (p − 2b)ix/4, e−px/2 " # 2 X ∞ a − b2 a−b 1 eanx + e(p−a)nx − e(p−b)nx − ebnx = exp −x − − 2p 2 n epnx − 1 n=1 (17) Proof. From Definitions 1,2 and the relations (1.14), (1.15) we can rewrite R in the form exp P∞ cosh(nx(p−2b)/2) 2 2 n=1 n sinh(pnx/2) a − b a − b P R(a, b, p; e−x ) = exp −x +x ∞ cosh((p−2a)nx/2) 2p 2 exp n=1
n sinh(pnx/2)
from which as one can see (1.16) and (1.17) follow. Conjecture √ 1. If q = e−π r , a, b, p, r positive rationals then 2
q p/12−a/2+a
/(2p)
[a, p; q] = Algebraic
(18)
Theorem 2. If a, b, p, r are positive rationals and a < b < p, then 2
R(a, b, p; q) := q −(a−b)/2+(a
−b2 )/(2p)
R∗ (a, b, p; q) = Algebraic
(19)
Proof. Equation (1.19) follows easy from the Conjecture 1 and the Definitions 1,2. 4
One example is the Rogers-Ramanujan continued fraction q 1/5 R∗ (1, 2, 5; q) = R∗ (q)q 1/5 = R(q)
(20)
For the continued fraction (8) we give some evaluations using the command ’Recognize’ or ’RootApproximant’ of Mathematica: q √ √ ? H e−π = 4 − 2 2 − 1 − 2 √ ? H e−π 2 = (1 − 8t − 12t2 − 8t3 + 38t4 + 8t5 − 12t6 + 8t7 + t8 )3 , where (P (x))n is the nth root of the equation P (x) = 0 taken with Mathematica program. Theorem 3. (The Rogers Ramanujan Identity of the Quantities) If a, b, p are positive integers and p − a 6= p − b, a < b < p, |q| < 1, then R∗ (a, b, p; q) =
∞ Y
(1 − q n )X(n)
(21)
n=1
where
1, n ≡ (p − a)modp −1, n ≡ (p − b)modp 1, n ≡ amodp X(n) = −1, n ≡ bmodp 0, else
(22)
Proof. Use Theorem 2. Take the logarithms and expand the product (1.21). The proof is easy.
Hence if we set M (q) = M (a, b, p; q) := q Then q
dR(a, b, p; q) 1 dq R(a, b, p; q)
∞ X dR(a, b, p; q) 1 =Q− q n τ (n), dq R(a, b, p; q) n=1
(23)
(24)
where Q = −(a − b)/2 + (a2 − b2 )/(2p) 5
(25)
τ (n) = τ (a, b, p, n) :=
X
X(d)d.
(26)
d|n
Theorem 4. Let |q| < 1, then log(R∗ (a, b; p; q)) = −
∞ X qn X X(d)d n n=1
(27)
d|n
Proof. Follows from Theorem 3. Theorem 5. For every positive integers a, b, p with a < b < p, p−prime, we have τ (n) = τ (pn)
(28)
where n = 1, 2, 3, . . . Proof. For n = 1 we have τ (1) = X(1) andPτ (p) = X(1)1 + X(p)p = X(1), (from 1.26). Let for n = n0 is τ (n0 ) = d|n0 X(d)d. For n1 = n0 p, p−prime: P τ (n0 p) = d|(n0 p) X(d)d. Let (p, n0 ) = 1, then τ (n0 ) = τ (n0 p) or equivalently P d|n0 X(pd)pd = 0, which is true since p > a, b and definition of X. If was k0 (p, n0 ) = s > 1, then s = p and exist k0 , m: n0 = Pmp , (m, p) = 1. Again all the multiplies of p are canceled and we lead to d|mpk0 X(d)d − τ (m) = 0. Hence we get the proof. Proposition 1.(See [7] pg. 24) Suppose that a, b and q are complex numbers with |ab| < 1 and |q| < 1 or that a = b2m+1 for some integer m. Then P (a, b, q) := =
(a2 q 3 ; q 4 )∞ (b2 q 3 ; q 4 )∞ = (a2 q; q 4 )∞ (b2 q; q 4 )∞
(a − bq)(b − aq) (a − bq 3 )(b − aq 3 ) (a − bq 5 )(b − aq 5 ) 1 (29) (1 − ab)+ (1 − ab)(q 2 + 1)+ (1 − ab)(q 4 + 1)+ (1 − ab)(q 6 + 1) + . . .
Theorem 6. If A, B > 0, a = 2A + 3p/4, b = 2B + p/4 and p = 4(A + B), |q| < 1, then R∗ (a, b, p; q) = (1 − q B−A )P (q A , q B , q A+B )
(30)
Proof. One can see that P (q A , q B , q A+B ) =
(q a ; q p )∞ (q 2p−a ; q p )∞ [b, p; q] 6
(31)
where a = 2A + 3p/4, b = 2B + p/4 and p = 4(A + B). Define the q analog of the Gauss hypergeometric function 2 φ1 [a, b; c; q, z] :=
and ψ(a, q, z) :=
∞ X (a; q)n (b; q)n z n (c; q)n (q; q)n n=0
∞ X (a; q)n n z = 2 φ1 [a, 0, 0, q, z] (q, q)n n=0
(32)
(33)
Then ψ(q p , q p , q p−a )R∗ (a, b, p; q) = P [q A , q B , q A+B ]
(34)
The proof of (1.30) follows easily from (1.34) and the q-binomial theorem (see [7]): ∞ Y 1 − azq n ψ(a, q, z) = 1 − zq n n=0 Note. Relation (1.30) is an expansion of a certain Ramanujan Quantity in continued fraction (see relation (1.2)).
Also in view of [7] page 14 Entry 4, we have R∗ (a, b, p; q)
∞ X (q a−b ; q p )n (q b+a−p ; q p )n (p−a)n q = 1. (q p ; q p )n (q a ; q p )n n=0
(35)
and we get the next Theorem 7. For |q| < 1 we have a−b
a2 −b2
q − 2 + 2p R(a, b, p; q) = a−b , q b+a−p ; q a ; q p , q p−a ] 2 φ1 [q
(36)
Theorem 8. If a, b, p ∈ N, with a < b < p, p−prime and ζpm = e−πim/p , then R(a, b, p; q p ) =
p−1 Y
R a, b, p; (−1)m ζpm q
(37)
m=0
Proof. From Theorem 5 holds τ (n) = τ (np), hence log(R(a, b, p; q)) = Q log(q) −
∞ ∞ X X qn qn τ (n) = Q log(q) − p τ (np) n np n=1 n=1
7
(38)
Also holds the general identity: If y(x) =
∞ X
an xn
n=0
then
∞ X
p−1 X
anp xnp = p−1
y (−1)m ζpm x
(39)
m=0
n=0
From (1.38) and (1.39) we get the proof. Note. Another form of Theorem 8 is: If a, b ∈ N and p-positive prime: a, b < p, then R(a, b, p; q p ) =
p−1 Y
R a, b, p; e2πim/p q
(40)
m=0
Conjecture 2. i) If a, b are odd positive integers and p-even, then ?
τ (a, b, p, 2n) = τ (a, b, p, n)
(41)
ii) If a 6= 0(mod3), b 6= 0(mod3) and p positive integer such that (p, 3) > 1, then ?
τ (a, b, p, 3n) = τ (a, b, p, n)
(42)
. . . etc In general if p0 is prime and if a 6= 0(modp0 ), b 6= 0(modp0 ) and p is positive integer such that (p, p0 ) > 1 then ?
τ (a, b, p, p0 n) = τ (a, b, p, n)
(43)
Theorem 9. If a, b, p as in Conjecture 2−(i) and R(q) = R(a, b, p; q), then 2M (q 2 ) = M (q) + M (−q)
(44)
R(q 2 ) = R(q) |R(−q)|
(45)
Proof. From τ (2n) = τ (n) we have, expanding M (q) into series: M (−q) + M (q) = 2Q − 2
∞ X
τ (2n)q 2n = 2 Q −
n=1
∞ X n=1
8
! τ (n)(q 2 )n
= 2M (q 2 )
3
The Programs
In this section we present the programs that we use to find τ -relations and unproved modular equations. The program of τ with Mathematica
Routine 1. Suppose a = 1, b = 4, p = 17. Set Q1[n, k] := If[Mod[n, k] == 0, 1, 0] t[n] :=
n X
X[1, 4, 17, k]kQ1[n, k]
k=1
then use the ’Solve’ routine as 17 X Solve[Table[ c[j]t[jn] == 0, {n, 1, 289}], Table[c[r], {r, 1, 17}]] j=1
We get {{c[1] → −4c[8] − 4c[16] − c[17], c[2] → 4c[8], c[3] → 0, c[4] → −c[8] + 3c[16], , c[5] → 0, c[6] → 0, c[7] → 0, c[9] → 0, c[10] → 0, c[11] → 0, c[12] → 0, , c[13] → 0, c[14] → 0, c[15] → 0}} We change in the above output equalities the symbol → into = Clear[t] Coefficient[
17 X
c[j]t[jn], c[16]]
j=1
−4t[n] + 3t[4n] + t[16n] Coefficient[
17 X
c[j]t[jn], c[8]]
j=1
−4t[n] + 4t[2n] − t[4n] + t[8n]
With the above program one can find for every Ramanujan Quantity relations between the τ ’s. We present such relations in the next two Propositions:
9
Proposition 2. If a = 1, b = 4, p = 17 we have −4τ (n) + 3τ (4n) + τ (16n) = 0
(46)
−4τ (n) + 4τ (2n) − τ (4n) + τ (8n) = 0
(47)
and
Proposition 3. If a = 1, b = 5, p = 26 we have −τ (n) + τ (3n) − τ (5n) + τ (15n) = 0 26 17 17 51 τ (n) − τ (3n) + τ (7n) − τ (11n) + τ (17n) = 0 77 7 7 77 19 19 57 134 τ (n) + τ (3n) − τ (7n) + τ (11n) + τ (19n) = 0 − 77 7 7 77 23 34 − τ (n) + τ (11n) + τ (23n) = 0 11 11 −5τ (n) + 4τ (5n) + τ (25n) = 0 −
A simple program for finding possible modular equations with Mathematica
Routine 2. x = Series[R[a, b, p, q α ], {q, 0, A}] y = Series[R[a, b, p, q β ], {q, 0, A}] s = ...
s s X X c[i, j]x∧ iy ∧ j, q ∧ n == 0, {n, 1, s∧ 2} ; t = Table Coefficient i=0 j=0
r = Table[c[i, j], {i, 0, s}, {j, 0, s}]; Clear[u, v] r1 = Table[u∧ iv ∧ j, {i, 0, s}, {j, 0, s}]; m = Normal[Extract[CoefficientArrays[t//Flatten, r//Flatten], 2]]; m0 = Normal[m]; Take[NullSpace[m0], 1].Flatten[r] Take[NullSpace[m0], 1].Flatten[r1] 10
(48) (49) (50) (51) (52)
4
The first Order Derivatives of Ramanujan’s Quantities
Observe that if q 1/5 q q 2 q 3 (q; q 5 )∞ (q 4 ; q 5 )∞ · · · = q 1/5 2 5 1+ 1+ 1+ 1+ (q ; q )∞ (q 3 ; q 5 )∞
(53)
q 1/3 q + q 2 q 2 + q 4 q 3 + q 6 (q; q 6 )∞ (q 5 ; q 6 )∞ · · · = q 1/3 1+ 1+ 1+ 1+ (q 3 ; q 6 )2∞
(54)
8 7 8 q 1/2 q2 q4 1/2 (q; q )∞ (q ; q )∞ · · · = q (1 + q)+ (1 + q 3 )+ (1 + q 5 )+ (q 3 ; q 8 )∞ (q 5 ; q 8 )∞
(55)
R1 (q) = R2 (q) = R3 (q) =
R1 (q) = R(q) = R(1, 3, 5; q), R2 (q) = V (q) = R(1, 3, 6; q), R3 (q) = H(q) = R(1, 3, 8; q) are respectively the Rogers-Ramanujan the Ramanujan’s Cubic and Ramanujan-Gollnitz-Gordon continued fractions, then have derivatives such that qπ 2 0 = Algebraic (56) R1,2,3 (q) K(kr )2 whenever q = e−π
√
r
and r is a positive rational. This lead us to use the folowing:
Conjecture 3. If a, b, p, r are positive rationals with a < b < p, then 2 d ? K(kr ) R(a, b, p; q) = Algebraic 2 dq qπ
(57)
K(k )2 d p/12−a/2+a2 /(2p) ? r q [a, p; q] = Algebraic dq qπ 2
(58)
where K(x), kr are the complete elliptic integral of the first kind and the elliptic singular modulus (see [20]).
Let for simplicity k = kr . In [12] we have prove the following relation (see also [6] pg. 87]), for real r > 0: √ dr π r = (59) dk K(k)2 k(1 − k 2 ) Hence
dq −qπ 2 = dk 2k(1 − k 2 )K(k)2
These formulas along with Conjecture 3 lead us to the concluding remark dR(a, b, p; q) = dk 11
(60)
=
dR(a, b, p; q) dq −qπ 2 K(k)2 = · Algebraic 2 dq dk qπ 2K(k)2 2k(1 − k 2 )
Hence Theorem 10.√ When q = e−π r , a, b, p, r positive rationals with a < b < p, then dR(a, b, p; q) ? = Algebraic dk
(61)
We give the proof of relations (1.53),(1.54),(1.55) above without using the Conjecture 3 and Theorem 10. Theorem√11. If q = e−π r , then √ −qπ 2 dH(q) 1 − k0 √ √ = 2 2 0 dq 2k(1 − k )K(k) k (k 2 + 2 1 − k 0 )
(62)
Proof. In [10], we have proved that H(q) = −t + which gives
p t2 + 1 , t =
kr (1 − kr0 )
√ dH(q) 1 − k0 √ √ = 0 dk k (k 2 + 2 1 − k 0 )
(63)
(64)
using now (1.60) we get the result. Note. In [12] we have derive the first order derivative for the Cubic Continued fraction: √ Let q = e−π r , r > 0 then V 0 (q) =
dV (q) 4K 2 (kr )kr02 (V (q) + V 4 (q)) = √ p dq 3qπ 2 r 1 − 8V 3 (q)
(65)
Hence we have prove that (1.54) and (1.55) obey (1.57). In [10] we have given a formula for the (RR) and its first derivative involving equations that can not solve in radicals (higher than 4). Also in [13] we have given a formula for (RR), but contains the function k (−1) (x), which is the inverse function of kr . However the general formula for (RR) continued fraction first derivative is p R0 (q) = 5−1 q −5/6 f (−q)4 R(q) 6 R(q)−5 − 11 − R(q)5 (66) 12
where q = e−π
√
r
, r > 0 and
f (−q) =
∞ Y
(1 − q n ) = 21/3 π −1/2 q −1/24 kr1/12 kr01/3 K(kr )1/2
(67)
n=1
From (1.66) and (1.67) we get (1.57). We will return in these functions later. Examples. eπ Γ(1/4)4 d R(1, 2, 4; q) = dq 64 · 25/8 π 3 q=e−π d eπ Γ(1/4)4 = p1 R(1, 2, 5; q) dq 16π 3 q=e−π
(68) (69)
where p1 = (16 − 240t2 + 800t3 − 2900t4 − 6000t5 − 6500t6 + 17500t7 + 625t8 )3 ! r √ d 7√ 64eπ π = 2+ 2− 5− R(1, 3, 8; q) 2 (70) 4 dq 2 Γ − 41 q=e−π √ (6 + 4 2)e2π Γ(5/4)4 d = R(1, 3, 8; q) p2 (71) dq π3 q=e−2π p2 = (16384 − 1720320t2 − 6684672t3 + 143104t4 − 18432t5 − 1664t6 + t8 )3 ! r √ √ 1 −π −2 − 3 + 3 3 + 2 3 V (e ) = (72) 2 p p √ √ √ 64 −26 − 15 3 + 10 3 + 2 3 + 6 9 + 6 3 eπ π V 0 (e−π ) = − q (73) p p √ √ √ 1 4 45 + 26 3 − 18 3 + 2 3 − 10 9 + 6 3 Γ − 4 Hence we have theoretical and numerical verifications of Conjectures 1 and 3.
5
Modular equations and Ramanujan Quantities
With the help of Theorem 2 we can evaluate R(a, b, p; q) in series of q Q : R(a, b, p; q) =
N X
cn q nQ
(74)
n=0
where N is positive integer and Q=
a2 − b2 a−b − 2p 2 13
(75)
Setting as in [11]: RS =
X
aij ui v j
(76)
0≤i+j≤d
and d is suitable large positive integer, we try to solve RS = 0, where u = R(a, b, p; q), v = R(a, b, p; q ν )
(77)
are given from (74) above and ν positive integer. Evaluating the aij , we obtain modular equations for R(a, b, p; q). 1) We present some modular equations for the Ramanujan Quantity R(1, 2, 4; q), which: R(1, 2, 4; q) = 1 +
(1 + q) q 2 (q + q 3 ) q 4 (q 2 ; q 4 )2∞ ... = 1+ 1+ 1+ 1+ (q; q 4 )∞ (q 3 ; q 4 )∞
(78)
a) If u = R(1, 2, 4; q) and v = R(1, 2, 4; q 2 ), then ?
u4 − v 2 + 4u4 v 4 = 0
(79)
b) If u = R(1, 2, 4; q) and v = R(1, 2, 4; q 3 ), then ?
u4 − uv + 4u3 v 3 − v 4 = 0
(80)
c) If u = R(1, 2, 4; q) and v = R(1, 2, 4; q 5 ), then ?
u6 − uv + 5u4 v 2 − 5u2 v 4 + 16u5 v 5 − v 6 = 0
(81)
d) If u = R(1, 2, 4; q) and v = R(1, 2, 4, q 7 ), then ?
u8 − uv + 7u2 v 2 − 28u3 v 3 + 70u4 v 4 − 112u5 v 5 + 112u6 v 6 − 64u7 v 7 + v 8 = 0 (82) 2) For the Ramanujan Quantity R(1, 2, 6; q) we have a) If u = R(1, 2, 6; q) and v = R(1, 2, 6; q 2 ), then ?
u4 − v 2 + 3u4 v 2 + v 4 = 0
(83)
. . .etc One can find with the help of Mathematica many relations such above The 5-degree modular equation of Ramanujan’s Cubic continued fraction: If u = R(1, 3, 6; q) and v = R(1, 3, 6; q 5 ), then ?
u6 −uv+5u4 v+5u2 v 2 −10u5 v 2 −20u3 v 3 +5uv 4 +20u4 v 4 −10u2 v 5 −16u5 v 5 +v 6 = 0 (84) 14
The 7-degree modular equation of Ramanujan’s Cubic continued fraction: If u = R(1, 3, 6; q) and v = R(1, 3, 6; q 7 ), then u8 − uv + 7u4 v + 28u6 v 2 − 56u5 v 3 + 7uv 4 + 21u4 v 4 − 56u7 v 4 − 56u3 v 5 + ?
+28u2 v 6 − 56u4 v 7 − 64u7 v 7 + v 8 = 0
(85)
If a > b then from the definition of the Ramanujan Quantity (RQ) we have R(a, b, p; q) =
1 R(b, a, p; q)
(86)
Suppose that a = aa12 , b = bb21 , p = pp12 , and u(q) = R(a, b, p; q), then u can be writen as 1 1 u q a2 b2 p2 = R a1 b2 p2 , b1 a2 p2 , p1 a2 b2 ; q a2 b2 p2 = w(q), (87) if a1 b2 p2 < b1 a2 p2 , (otherwise we use (1.86)). But w1 := w(q), wν := w(q ν ) are 1 ν related with a modular equation F (w1 , wν ) = 0, or F (u(q a2 b2 p2 ), u(q a2 b2 p2 )) = 0 or equivalently F (u(q), u(q ν )) = 0. Hence Theorem 12. When a = aa12 , b = bb12 , p = pp21 , a1 , a2 , b1 , b2 , p1 , p2 ∈ N and a1 b2 p2 < b1 a2 p2 then the modular equation which relates u1 := R(a, b, p; q) and uν := R(a, b, p; q ν ), ν ∈ N is that of w(q) = R(a1 b2 p2 , b1 a2 p2 , p1 a2 b2 ; q) and w(q ν ).
(88)
Example. The modular equation between z1 = z(q) = R 1, 12 , 2; q and z2 = z(q 2 ) is ?
4 + z24 − z14 z22 = 0
(89)
Proof. We have 1 1 1 = (90) z1 = z(q) = R 1, , 2; q = R(2, 1, 4; q 1/2 ) = 2 R(1, 2, 4; q 1/2 ) u(q 1/2 ) But z2 = z(q 2 ) =
1 u(q 2/2 )
using (1.79) we have
u(q 1/2 )4 − u(q)2 + 4u(q 1/2 )4 u(q)4 = 0 15
from which (1.89) follows.
From Theorem 4 differentiating (1.27) and using Conjecture 3 we have that q
∞ X X dR(a, b, p; q) 1 =Q− qn X(d)d dq R(a, b, p; q) n=1
(91)
d|n
and
N (q) = q −1/6 f (−q)−4 Q −
∞ X n=1
qn
X
X(d)d = Algebraic
(92)
d|n
This is a resulting formula for the first derivative of (RR): p dR(q) R0 (q) = 5 · 21/3 (kk 0 )1/3 = 6 R(q)−5 − 11 − R(q)5 = N (q) R(q) dk (93) √ The function N (q) take algebraic values when q = e−π r , r positive rational and satisfies modular equations. With the same method as in R(a, b, p; q) which we use in the begining of paragraph 5 and Routine 2 of paragraph 3, we have: 5q 5/6 f (−q)−4
The 2-degree Modular equation for the first derivative of RR continued fraction For a = 1, b = 2, p = 5, we have the case of (RR) and a) If u = N (q) and v = N (q 2 ) then 5u6 − u2 v 2 − 125u4 v 4 + 5v 6 = 0
(94)
The 3-degree Modular equation for the first derivative of RR continued fraction b) If u = N (q) and v = N (q 3 ) we have 125u12 + u3 v 3 + 1125u9 v 3 + 1125u3 v 9 + 1953125u9 v 9 − 125v 12 = 0
(95)
Proofs for (1.94) and (1.95) can given if we use the identity (1.66) and Ramanujan’s duplication and triplication formulas of R(q): R(q 2 ) − R(q)2 = R(q)R(q 2 )2 , R(q 2 ) + R(q)2 R(q 3 ) − R(q)3 1 + R(q)R(q 3 )3 = 3R(q)2 R(q 3 )2
(96) (97)
The hard thing is that have very tedious algebraic calculations, but they can verified with a mathematical program.
16
Suppose now that q0 = e−π
√
r0
and we know R(1) (q0 ) =
dR(q) dq
, then q=q0
from equations (1.92),(1.93),(1.94),(1.95),(1.96) and (1.97) we can evaluate in radicals, any high order values of the first derivative of the (RR) in which r = 4n 9m r0 , for n, m integers. Note. If K(kr ) = K[r] then holds: 1 + kr0 K[r] 2
(98)
K[9r] = m3 (r)K[r]
(99)
27m3 (r)4 − 18m3 (r)2 − 8(1 − 2kr2 )m3 (r) − 1 = 0
(100)
K[4r] = and where m3 (r) is solution of
The formulas for evaluation of k4r and k9r are in [7] and [15]. Proposition 4. If R(q) is the Rogers Ramanujan continued fraction and M (q) =
R0 (q) R(q)
(101)
then r M (q 2 ) =
√ p q 5/2 f (−q 2 )8 M (q) 125q 5/2 M (q) − 5 8f (−q)24 + 3125q 5 M (q) √ 2 5q 5/2 f (−q)8 (102)
where f (−q 2 ) = 4
1 + kr0 (2m2 − 1)4 qf (−q)24 , where m2 = m2 (m2 − 1) 2
(103)
Proof. The proof follows from (1.94), the above note and (1.67).
6
Application in a wide class of continued fractions
In this section we give examples and evaluations of a certain class of continued fractions with values a, b, p positive rationals. Under the conditions in parameters of Theorem 6 holds. I. The case of A = 1, B = 2, then a = 11, b = 7, p = 12 and S(q) = S1,2 (q) = R(11, 7, 12; q) = 17
1 R(7, 11, 12; q)
(104)
Hence we get S1,2 (q) = =q
1 − q q 3 (1 − q 2 )(1 − q 4 ) q 3 (1 − q 8 )(1 − q 10 ) q 3 (1 − q 14 )(1 − q 16 ) 1 − q 3 + (1 − q 3 )(1 + q 6 )+ (1 − q 3 )(1 + q 12 )+ (1 − q 3 )(1 + q 18 ) + ... Q∞ −1 (1 − q 12n+11 )(1 − q 12n+1 ) (105) = q Qn=0 ∞ 12n+7 )(1 − q 12n+5 ) n=0 (1 − q
With the above methods we find that continued fraction S1,2 (q) obeys the following modular equations: 1) If we set u = S1,2 (q) and v = S1,2 (q 2 ), then ?
−u2 + v − 2uv + u2 v − v 2 = 0
(106)
2) If u = S1,2 (q) and v = S1,2 (q 3 ), then ?
u3 − v + 3uv − u3 v + v 2 − 3u2 v 2 + u3 v 2 − v 3 = 0
(107)
3) If u = S1,2 (q) and v = S1,2 (q 5 ), then −u5 + v − 5uv + 5u2 v + 5u5 v − 10u3 v 2 − 5u5 v 2 + 10u2 v 3 + 10u4 v 3 − 5uv 4 − ?
−10u3 v 4 + 5uv 5 + 5u4 v 5 − 5u5 v 5 + u6 v 5 − uv 6 = 0
(108)
4) If u = S1,2 (q) and v = S1,2 (q 7 ), then −u7 + v − 7uv + 14u2 v − 7u3 v + 7u5 v − 7u6 v + 7u7 v + 7uv 2 − 28u2 v 2 + +7u3 v 2 − 28u5 v 2 + 28u6 v 2 − 14u7 v 2 − 7uv 3 + 28u2 v 3 − 7u3 v 3 + +35u4 v 3 + 7u5 v 3 − 7u6 v 3 + 7u7 v 3 − 35u3 v 4 − 35u5 v 4 + 7uv 5 − −7u2 v 5 + 7u3 v 5 + 35u4 v 5 − 7u5 v 5 + 28u6 v 5 − 7u7 v 5 − 14uv 6 + +28u2 v 6 − 28u3 v 6 + 7u5 v 6 − 28u6 v 6 + 7u7 v 6 + 7uv 7 − 7u2 v 7 + ?
+7u3 v 7 − 7u5 v 7 + 14u6 v 7 − 7u7 v 7 + u8 v 7 − uv 8 = 0
(109)
II. In the case of A = 1, B = 3, then a = 14, b = 10, p = 16 and S1,3 (q) = R(14, 10, 16; q) =
1 R(10, 14, 16; q)
(110)
and from Theorem 5 we get S1,3 (q) = =q
1 − q 2 q 4 (1 − q 2 )(1 − q 6 ) q 4 (1 − q 14 )(1 − q 10 ) q 4 (1 − q 22 )(1 − q 18 ) = 1 − q 4 + (1 − q 4 )(1 + q 8 )+ (1 − q 4 )(1 + q 12 )+ (1 − q 4 )(1 + q 20 ) + . . . Q∞ −1 (1 − q 16n+10 )(1 − q 16n+6 ) = q Qn=0 (111) ∞ 16n+14 )(1 − q 16n+2 ) n=0 (1 − q 18
With the above methods we find that Continued fraction SA,B (q) obeys the following modular equations: 1) If we set u = S1,3 (q) and v = S1,3 (q 2 ), then ?
u2 − v + u2 v + v 2 = 0
(112)
2) If u = S1,3 (q) and v = S1,3 (q 3 ), then ?
u3 − v + 3u2 v + 3uv 2 − 3u3 v 2 − 3u2 v 3 + u4 v 3 − uv 4 = 0
(113)
3) If u = S1,3 (q) and v = S1,3 (q 5 ), then u5 − v + 5u2 v + 10u3 v 2 − 5u5 v 2 − 10u2 v 3 + 10u4 v 3 + 5uv 4 − 10u3 v 4 − ?
−5u4 v 5 + u6 v 5 − uv 6 = 0
(114)
7
4) If u = S1,3 (q) and v = S1,3 (q ), then u8 − uv + 7u3 v − 7u5 v − 7u7 v + 28u6 v 2 + 7uv 3 − 49u3 v 3 − 7u5 v 3 − −7u7 v 3 + 70u4 v 4 − 7uv 5 − 7u3 v 5 − 49u5 v 5 + 7u7 v 5 + 28u2 v 6 − ?
−7uv 7 − 7u3 v 7 + 7u5 v 7 − u7 v 7 + v 8 = 0
(115)
As a starting value we can get S1,3 (e
−π/2
?
) = −1 −
√
r √ 2+ 2 2+ 2
(116)
from (1.87) we get S1,3 (e s
−π
√ ) = −3 − 2 2 + 2 ?
q q √ √ 2 + 2 + 4 + 2 2−
r q √ √ √ − 30 + 22 2 − 16 2 + 2 − 12 2 2 + 2
. . . etc A value for the derivative according to the Conjecture 2 is q √ √ π/2 64 4 + 2 2 − 2 10 + 7 2 e π dS1,3 (q) ? = 4 dq Γ − 14 q=e−π/2
(117)
(118)
Note. It is clear by this way that one can produce continued fractions such found and studied by Ramanujan. Approaches for the evaluation of Ramanujan Quantities and their derivatives In this section we will try to find theoretical results that can used to recover the Ramanujan quantities and their derivatives. We will also drop the notation 19
of X which we use until now. More precisely we use in place of X(n) a general arithmetic function. The symbol X(a, b, p; n) will be that of relation (22). Lastly Xp (n) will be the characteristic function taking the value 1 when n ≡ 0(modp) and 0 else. The generalized Ramanujan quantity R(q) will be R(q) =
∞ Y
(1 − q n )X(n) , |q| < 1
(119)
n=1
with X(n) arbitrary sequence (as long as (119) converges). Further ∞ Y
f (−q) =
(1 − q n ), |q| < 1
(120)
n=1
will be the Ramanujan’s-Dedekind eta function and ∞ X nq n 1 − qn n=1
L1 (q) :=
(121)
Also L(q) = 1−24L1 (q) will be the classical series attached to that of Eisenstein. The following elementary relations hold (see [7] pg.120) ∞ X (−1)n nq n 1 = (3 − L(q) − 4L(q 2 )) n 1 − q 24 n=1 ∞ ∞ X X n nq n n nq n 1 (−1 − L(q) + 2L(q 2 )) = = n n 4 1 − q 14 1 − q 24 n=1 n=1
1 − (−1)n 2 8 4 2 ∞ ∞ n X n X (3n + 1)q 3n+1 nq 1 3 = (2 − L(q) − 3L(q )) + 2 3 1 − qn 24 1 − q 3n+1 n=1 n=1 n
=
n
,
n
(122)
(123)
=
(124)
We state a first result: Lemma 2. q
∞ X R0 (q) X(n)nq n = M (q) = R(q) 1 − qn n=1
Proof. From the relation R(q) =
∞ Y
(1 − q n )X(n) ,
n=1
taking the logarithmic derivative, we get immediately the result.
20
(125)
Lemma 3. If X(n) = gn2 and g-prime, then f (−q) f (−q g )
(126)
q −1 f (−q) (1 − g − L(q) + gL(q g )). 24 f (−q g )
(127)
R(q) = R0 (q) =
Proof. The proof follows from the above Lemma 1. √
Here we must mention that when q = e−π r , L1 (q) can evaluated from 1 1 6α(r) K 2 [r] 2 √ −1 − k L1 (q) = − √ + + (128) r 24 4π r 6π 2 r where
π K(w) = 2 F1 2
1 1 , ; 1; w2 2 2
(129)
is the complete elliptic integral of the first kind, kr is the elliptic singular modulus and α(r) is the elliptic alpha function (see [7] and [15]). Here K[r] = K(kr ). If g is odd prime number, then g−1
n g2
=
2 X
X(j, 0, g, n) +
j=1
and
g−1 Xg (n) 2
(130)
g−1 2 Y
f (−q) f (−q g )
[j, g; q]∞ =
j=1
R(q) =
∞ Y
(131)
g−1
n
(1 − q )
n g2
?
g
= f (−q )
n=1
g−1 2
2 Y
R(j, 0, g; q) =
j=1 g−1
g−1
g
= f (−q )
g−1 2
2 2 Y Y [j, g; q]∞ = [j, g; q]∞ f (−q g ) j=1 j=1
(132)
But also we can write R(q) =
∞ Y
(1 − q n )
n=1
n g2
=
∞ Y
(1 − q n )
n=1
∞ Y
(1 − q ng )−1 = f (−q)f (−q g )−1 (133)
n=1
Note that R(j, 0, g; q) is the (RQ) which coresponds to (1.21) as defined by (1.22) and it is not proper but gives correct results. That is why we add the 21
”?” in relation (2.14). Trying to Generalize the Problem From the above it became obvious, that we can write for any G-integer with G > 1: ν n X = X(cj , 0, G; n) + cXG (n), (134) G2 j=1 where c−integer constant depending from G, and cj positive integers depending also form G. The function X(a, b, p, n) is that of relation (1.22). Theorem 13. In the case which
n (135) G2 then if g1 , g3 , . . . , gλ are positive primes and G = g1w1 g2w2 . . . gλwλ is the factorization of G, with g1 < g2 < . . . < gλ , then X(n) =
ν Y
?
[cj , G; q]∗∞ = f (−q)
j=1
λ Y
f (−q gi )−1
i=1
×
Y
f (−q gi gj )1
i
