On a continued fraction of order twelve
We present some new relations between a continued fraction $U(q)$ of order twelve (which is established by M. S. M. Naika et al.) and $U(q^n)$ for $n = 7,9,11\;\text{and}\; 13$.
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| Дата: | 2010 |
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2010
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508995777724416 |
|---|---|
| author | Kahtan, Abdulrawf A. A. Sathish, Kumar C. Sharath, G. Vasuki, K. R. Кантан, Абдулрашф А. А. Сатниш, Кумар С. Шаратх, Г. Васюкі, К. Р. |
| author_facet | Kahtan, Abdulrawf A. A. Sathish, Kumar C. Sharath, G. Vasuki, K. R. Кантан, Абдулрашф А. А. Сатниш, Кумар С. Шаратх, Г. Васюкі, К. Р. |
| author_sort | Kahtan, Abdulrawf A. A. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:41:53Z |
| description | We present some new relations between a continued fraction $U(q)$ of order twelve (which is established by
M. S. M. Naika et al.) and $U(q^n)$ for $n = 7,9,11\;\text{and}\; 13$. |
| first_indexed | 2026-03-24T02:34:04Z |
| format | Article |
| fulltext |
UDC 511.72
K. R. Vasuki, Abdulrawf A. A. Kahtan, G. Sharath, C. Sathish Kumar
(Univ. Mysore, India)
ON A CONTINUED FRACTION OF ORDER TWELVE
ПРО ЛАНЦЮГОВИЙ ДРIБ ДВАНАДЦЯТОГО ПОРЯДКУ
We present some new relations between a continued fraction U(q) of order twelve (which is established by
M. S. M. Naika et al.) and U(qn) for n = 7, 9, 11 and 13.
Наведено деякi новi спiввiдношення мiж ланцюговим дробом U(q) дванадцятого порядку (який описано
М. С. М. Найка та iншими авторами) i U(qn) для n = 7, 9, 11 та 13.
1. Introduction. Throughout the paper, we assume |q| < 1 and for a positive integer n,
we use the standard notation
(a)0 := (a; q)0 = 1,
(a)n := (a; q)n =
n−1∏
i=0
(1− aqi)
and
(a)∞ := (a; q)∞ =
∞∏
n=0
(1− aqn).
In Chapter 16 of his Second Notebook [12, p. 197; 3, p. 34] S. Ramanujan develops the
theory of theta functions and his theta function is defined by
f(a, b) :=
∞∑
n=−∞
a
n(n+1)
2 b
n(n−1)
2 = (−a; ab)∞(−b; ab)∞(ab; ab)∞, |ab| < 1.
Following Ramanujan [12, p. 197], we define
φ(q) := f(q, q) = 1 + 2
∞∑
k=1
qk
2
=
(−q;−q)∞
(q;−q)∞
,
ψ(q) := f(q, q3) =
∞∑
k=0
q
k(k+1)
2 =
(q2; q2)∞
(q; q2)∞
and
f(−q) := f(−q,−q2) =
∞∑
k=−∞
(−1)kq
k(3k−1)
2 = (q; q)∞.
For convenience, we denote
f(−qn) = fn.
The celebrated Rogers – Ramanujan continued fraction is defined as
c© K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR, 2010
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12 1609
1610 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR
R(q) :=
q1/5f(−q,−q4)
f(−q2,−q3)
=
q1/5
1 +
q
1 +
q2
1 +
q3
1 + . . .
. (1.1)
On p. 365 of his Lost Notebook [13], Ramanujan recorded five identities, which shows
the relations between R(q) and five continued fractions R(−q), R(q2), R(q3), R(q4)
and R(q5). He also recorded these identities at the scattered places of his Notebooks
[12]. L. J. Rogers [14] found the modular equations relating R(q) and R(qn) for n =
= 2, 3, 5 and 11, the latter equations is not found in Ramanujan’s works. For a proof
of these, one may refer [4]. Recently K. R. Vasuki and S. R. Swamy [19] found the
modular equation relating R(q) with R(q7).
The Ramanujan’s cubic continued fraction G(q) be defined by
G(q) :=
q1/3f(−q,−q5)
f(−q3,−q3)
=
q1/3
1 +
q + q2
1 +
q2 + q4
1 +
q3 + q6
1 + . . .
. (1.2)
The continued fraction (1.2) was first introduced by Ramanujan in his second letter to
G. H. Hardy [10]. He also recorded the continued fraction (1.2) on p. 365 of his Lost
Notebook [13] and claims that there are many results of G(q) which are analogous to
the famous Rogers – Ramanujan continued fraction (1.1).
Motivated by Ramanujan’s claim H. H. Chan [5], N. D. Baruah [1], K. R. Vasuki
and B. R. Srivatsakumar [18] have established the modular relations between G(q) and
G(qn) for n = 2, 3, 5, 7, 11 and 13.
The Ramanujan Göllnitz – Gordon continued fraction [13, p. 44; 8; 9] is defined by
H(q) :=
q1/2f(−q3,−q5)
f(−q,−q7)
=
q1/2
1 +
q2
1 + q3 +
q4
1 + q5 + . . .
. (1.3)
Chan and S. S. Hang [6], K. R. Vasuki and B. R. Srivatsakumar [17] have established
the identities which gives the relations between H(q) and H(qn) with n = 3, 4, 5, 11
by employing the modular equations stated by Ramanujan. Recently B. Cho, J. K. Koo
and Y. K. Park [7] have extended the above results on continued fraction (1.3) to all odd
primes p by computing the affine models of modular curves X(Γ) with Γ = Γ1(8) ∩
∩ Γ0(16p).
Motivated by the above works on the continued fractions (1.1) – (1.3), in this paper,
we establish the relation between the following continued fraction U(q) with U(qn) for
n = 2, 7, 9, 11 and 13
U(q) :=
qf(−q,−q11)
f(−q5,−q7)
=
q(1− q)
(1− q3) +
q3(1− q2)(1− q4)
(1− q3)(1 + q6) +
q3(1− q8)(1− q10)
(1− q3)(1 + q12) + . . .
. (1.4)
The continued fraction (1.4) was established by M. S. M. Naika et al. [11] as a particu-
lar case of fascinating continued fraction identity recorded by Ramanujan in his Second
Notebook [12, p. 74]. Furthermore they have also established the modular relation be-
tween the continued fraction U(q) with U(qn) with n = 3 and 5.
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
ON A CONTINUED FRACTION OF ORDER TWELVE 1611
2. Some preliminary results.
Theorem 2.1 [11]. We have
φ(q)
φ(q3)
=
1 + U(q)
1− U(q)
. (2.1)
Theorem 2.2. We have
φ2(−q)
φ2(−q3)
= 1− 4U(q)
1 + U2(q)
. (2.2)
Proof. From [2], we have(
φ2(q) + φ2(q3)
)2
=
4φ(q3)φ3(−q3)φ(q)
φ(−q)
, (2.3)
and also from [16], we have(
3φ2(q3)− φ2(q)
)2
=
4φ(q3)φ(q)φ3(−q)
φ(−q3)
. (2.4)
From (2.3) and (2.4), we deduce that
φ2(−q)
φ2(−q3)
=
3− φ2(q)
φ(q3)
1 +
φ2(q)
φ2(q3)
.
Employing (2.1) in the right-hand side of the above identity, we obtain (2.2).
Theorem 2.3. We have
ψ2(q2)
qψ2(q6)
= U(q) +
1
U(q)
− 1. (2.5)
Proof. From [2], we obtain
1− φ2(−q)
φ2(−q3)
=
4qf1f
3
12
f4f33
, (2.6)
and also from [2], we have
1 +
ψ2(q2)
qψ2(q6)
=
f33 f4
qf1f312
.
Using (2.6) and the above identity, we find that
1 +
ψ2(q2)
qψ2(q6)
=
4
1− φ2(−q)
φ2(−q3)
,
which upon using (2.2) gives the required result.
Theorem 2.4. We have
ψ2(q)
q1/2ψ2(q3)
=
1 + U(q)
1− U(q)
√
U(q) +
1
U(q)
− 1. (2.7)
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
1612 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR
Proof. Consider,
ψ4(q)
qψ4(q3)
=
φ2(q)ψ2(q2)
qφ2(q3)ψ2(q6)
(2.8)
where, we have used the identity φ(q)ψ(q2) = ψ2(q) from Entry 25 [3, p. 40]. Now
using (2.1) and (2.7) in the right-hand side of the above, we obtain the required result.
Theorem 2.5. We have
ψ4(−q)
qψ4(−q3)
=
[
U2(q) + 1
U(q)
+
4U(q)
1 + U2(q)
− 5
]
. (2.9)
Proof. Changing q to −q in (2.8) and then using (2.2) and (2.5) on the right-hand
side of (2.8), we obtain (2.9).
To prove our main results, we require following easily deducible identities:
φ(q) =
f52
f21 f
2
4
, ψ(q) =
f22
f1
,
φ(−q) =
f21
f2
, ψ(−q) =
f1f4
f2
,
f(q) =
f32
f1f4
, χ(q) =
f22
f1f4
and
χ(−q) =
f1
f2
. (2.10)
And also the following identities analogous to Rogers – Ramanujan forty identities:
Theorem 2.6 [15]. Let
M(q) =
f(−q5,−q7)
f4
and
N(q) =
f(−q,−q11)
f4
.
Then
M(q2)N(q) + qM(q)N(q2) =
f3f24
f4f8
, (2.11)
M(q2)N(q)− qM(q)N(q2) =
f21 f6f24
f2f3f4f8
, (2.12)
M(q3)N(q) + q2M(q)N(q3) =
f1f
5
6 f9f36
f22 f
2
3 f
3
12f18
, (2.13)
M(q3)N(q)− q2M(q)N(q3) =
f1f
2
18
f4f9f12
, (2.14)
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
ON A CONTINUED FRACTION OF ORDER TWELVE 1613
M(q5)N(q)− q4M(q)N(q5) =
f1f
2
6 f30
f2f3f12f20
(2.15)
and
M(q)M(q5)− q6N(q)N(q5) =
f5f6f
2
30
f4f10f15f60
. (2.16)
A proof of (2.11) – (2.16) can be found in [15].
3. Main results. In this section, we derive the relation between the continued
fraction U(q) with U(qn) for n = 2, 3, 5, 7, 9, 11 and 13.
Theorem 3.1. Let u := U(q) and v := U(q2). Then
u2 − v + 2uv − u2v + v2 = 0.
Proof. By (1.4), we have
U(q) =
qN(q)
M(q)
. (3.1)
Dividing (2.11) by (2.12) and on using (2.10), we obtain
M(q2)N(q) + qM(q)N(q2)
M(q2)N(q)− qM(q)N(q2)
=
ϕ(−q3)
ϕ(−q)
.
Squaring on both sides of the above identity and using (3.1) on the left-hand side and
(2.2) on the right-hand side of the above identity and then dividing throughout by 4u,
we deduce the required result.
Theorem 3.2 [11]. Let u = U(q) and v = U(q3). Then
u3 − v3 + v2 − v + 3uv − 3u2v2 + u3v2 − u3v = 0.
Proof. Dividing (2.14) by (2.13) and on using (2.10), we find that
M(q3)N(q)− q2M(q)N(q3)
M(q3)N(q) + q2M(q)N(q3)
=
φ(−q2)φ(−q3)φ(−q18)
φ(−q6)φ(−q9)φ(−q6)
.
Squaring on both sides of the above identity, and using (3.1) on the left-hand side and
employing the identity
φ(q)φ(−q) = φ2(−q2)
on the right-hand side of the above, we obtain(
U(q)− U(q3)
U(q) + U(q3)
)2
=
φ(q)
φ(q3)
φ(−q)
φ(−q3)
φ(−q3)
φ(−q9)
φ(q9)
φ(q3)
.
Squaring again on both sides of the above, employing (2.1) and (2.2) on the right-hand
side of the above identity and then factorizing using maple, we deduce that
8(−v3 + v2 − v + 3uv − 3u2v2 + u3v2 − u3v + u3)×
×(u5v − u4v3 + u4v2 − 3u4v + 5u3v3 − u3v2 − u3v + u3 − u2v5 + u4v2+
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
1614 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR
+u2v3 − 5u2v2 + 3v4u− uv3 + uv2 − v4) = 0.
Now, from the definition of u and v, we have
u = U(q) = 1− q + q5 − q6 + q7 + . . . (3.2)
and
v = U(q3) = 1− q3 + q15 − q18 + q21 + . . . .
Using (3.2) and the above in the above factors, we see that the first factor becomes
q3(3− 6q + 3q2 − 3q3 + 6q4 + 3q5 − 9q6 + . . .),
and the second factor becomes
−q3(10− 22q + 15q2 − 22q3 + 33q4 − 9q5 + . . .).
Thus the second factor does not vanish. Hence by the identity Theorem, we must
have
v3 − v2 + v − 3uv + 3u2v2 − u3v2 + u3v − u3 = 0.
Theorem 3.2 is proved.
Theorem 3.3 [11]. Let u := U(q) and v := U(q5). Then
uv6 − u6v5 + 5u5v5 − 5u4v5 − 5uv5 + 10v4u3 + 5uv4 − 10u4v3 − 10u2v3+
+5u5v2 + 10u3v2 − 5u5v − 5u2v + 5uv − v + u5 = 0.
Proof. Dividing (2.15) by (2.16) and using (2.10), we find that
M(q5)N(q)− q4M(q)N(q5)
M(q)M(q5)− q6N(q)N(q5)
=
qψ(−q)
ψ(−q3)
ψ(−q15)
ψ(−q5)
.
Taking power 4 on both sides of the above and employing (2.9) on the right-hand
side of the above and then factorizing using Maple, we deduce that
(u2 + u2v2 + 1− 4uv + v2)(uv6 − u6v5 + 5u5v5 − 5u4v5 − 5uv5 + 10v4u3 + 5uv4−
−10u4v3 − 10u2v3 + 5u5v2 + 10u3v2 − 5u5v − 5u2v + 5uv − v + u5) = 0.
Changing q to q5 in (3.2), we find that
v = U(q5) = 1− q5 + q25 − q30 + q35 + . . . .
Using (3.2) and the above in the above factors, we see that the first factor becomes
q2(2− 4q4 + 2q5 − 4q6 + 4q8 − 2q9 + 3q10 + . . .),
and the second factor becomes
−q7(40− 40q + 20q2 + 10q3 − 55q4 + 110q5 − 180q6 + . . .).
Thus the first factor does not vanish. Hence by the identity theorem, we must have
uv6 − u6v5 + 5u5v5 − 5u4v5 − 5uv5 + 10v4u3 + 5uv4 − 10u4v3 − 10u2v3+
+5u5v2 + 10u3v2 − 5u5v − 5u2v + 5uv − v + u5 = 0.
Theorem 3.3 is proved.
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
ON A CONTINUED FRACTION OF ORDER TWELVE 1615
Theorem 3.4. Let u = U(q) and v = U(q7). Then
7u3v2 − 35u3v4 − 7u3v − 7u3v3 + 35u4v3 − u7+
+7u5v − 7u6v + 7u7v − 28u5v2 + 28u6v2 − 14v2u7 + 7v3u5 − 7v3u6+
+7v3u7 − 35v4u5 + 7v5u− 7v5u2 + 7v5u3 + 35v5u4 − 7v5u5 + 28v5u6−
−7v5u7 − 14v6u+ 28v6u2 − 28v6u3 + 7v6u5 − 28v6u6 + 7v6u7 + 7v7u−
−7v7u2 + 7v7u3 − 7v7u5 + 14v7u6 − 7v7u7 + v7u8 − v8u+ v − 7uv+
+7uv2 + 14u2v − 28u2v2 − 7uv3 + 28u2v3 = 0.
Proof. If
P =
φ(q)
φ(q3)
and
Q =
φ(q7)
φ(q21)
,
then (
Q
P
)4
−
(
P
Q
)4
+ 14
[(
Q
P
)2
−
(
P
Q
)2
]
=
(PQ)3 +
27
(PQ)3
+ 7
(
PQ+
3
PQ
)[
1−
(
P
Q
)2
−
(
Q
P
)2
]
. (3.3)
We have deduced the identity (3.3) on the lines remarked by N. D. Baruah [1]. Now
employing (2.1) in (3.3), we deduce the required result.
Theorem 3.5. Let u = U(q) and v = U(q9), then
9u4v + 126v4u− 90uv3 + 27v8u3 + v − 9vu+ 45v2u+ 378u7v5+
+135u8v4 + 54u8v2 − 9u8v + 18u7v + 558v4u6 − 558v5u3 + 561v4u3 − 10u9v2−
−333u6v3 + 4u9v − 561u6v5 + 16u9v3 − 19u9v4 + 369v5u2−
−99u5v2 − 378v4u2 + 135v7u2 + 99uv6 + 16u9v5 + 333u3v6+
+243u7v3 − 243u2v6 − 126u8v5 − 27u6v − 54v7u+ 9v8u− 135u7v2+
+10v3 − 198v6u4 + 369u4v5 − 16v6 + 19v5 − 16v4 − u9 − 4v8+
+10v7 − 243v4u4 − 153v2u2 + 99v7u4 + 297v3u2 + 9u8v8 + 243u5v5−
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
1616 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR
−18v8u2 − 297v6u7 + 30u6v8 + 153v7u7 − 30vu3 − 165v7u3 + 198u5v3−
−99u8v3 − 423v3u3 − 9u4v8 − 369u7v4 + 165u6v2+
+423u6v6 − 153u5v6 − 171u6v7 − 369u5v4 − 9u5v8 + 54u5v7+
+27vu2 + 90u8v6 + v9 + 9u5v − 10u9v6 − 45u8v7 + 171v2u3 + 153v3u4+
+4u9v7 − 4v2 − 27u7v8 − 54u4v2 − u9v8 − 135v5u = 0.
Proof. Let w = U(q3) then, by Theorem 3.2, we have
u3 − w3 + w2 − w + 3uw − 3u2w2 + u3w2 − u3w = 0. (3.4)
Changing q to q3 in (3.4), we obtain
w3 − v3 + v2 − v + 3wv − 3w2v2 + w3v2 − w3v = 0. (3.5)
Now eliminating w between (3.4) and (3.5), we deduce the required result.
Theorem 3.6. Let u = U(q) and v = U(q11), then
−363v7u10 − 528v10u8 + 11v10u11 − 1089v9u8 + 1463v4u8 − 1012v7u8+
+11v11u2 + vu− 968v4u4 + 44v3u− 374v3u2 + 77v4u−
−528v4u2 − 11vu2 − 363v5u2 + 55v7u11 + v11u11+
+759v7u9 + 55v5u+ 55v11u7 − 759v3u7 − 55v7u+
+363v7u2 − 77v8u+ 528v8u2 + 176v10u2+
+44v11u9 + 363v2u5 − 44vu9 − 11v2u11 − 11v10u+
+528v10u4 + 528v4u10 + 374v9u2 − 44v9u−
−110v2u2 + 11v2u+ 77vu8 + 374v3u10 − 363v2u7−
−77vu4 + 11vu10 + 1089v4u3 − 374v9u10+
+44vu3 − 759v9u5 + 55vu5 + 759v5u3+
+363v5u10 + 1012v5u8 − 1012v5u4 − 55vu7 + 1496v5u5−
−759v5u9 − 528v2u4 − 704v5u7 − 55v5u11 − 968v8u8+
+1463v8u4 + 1089v3u8 − 528v8u10 − 1089v3u4−
−1089v8u3 − 1012v8u5 + 1496v7u7 + 77v8u11+
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
ON A CONTINUED FRACTION OF ORDER TWELVE 1617
+1089v8u9 − 363v10u5 + 1012v8u7 + u12 − 374v2u9+
+1023v3u3 − 110v10u101012v7u4 − 704v7u5 − 77v11u8+
+1012v4u5 + 11vu11 − 44v11u3 − 1012v4u7 + 924v6u6+
+1089v9u4 + 1023v9u9 − 759v7u3 + 759v3u5 + 44v9u11+
+528v2u8 + 759v9u7 − 803v3u9 − 44v3u11 + 363v10u7+
+1089v4u9 + 11v11u+ 374v2u3 − 77v4u11 + 77v11u4 + 374v10u9−
−55v11u5 + v12 − 374v10u3 − 11v11u10 − 803v9u3 + 176v2u10 = 0.
Proof. Let
P =
ϕ(q)
ϕ(q3)
and
Q =
ϕ(q11)
ϕ(q33)
,
then from [18], we have
(PQ)5 +
35
(PQ)5
− 11
[
(PQ)3 +
33
(PQ)3
]
+ 308
[
PQ+
3
PQ
]
=
=
(
P
Q
)6
+
(
Q
P
)6
+ 22
[(
P
Q
)4
+
(
Q
P
)4
][
3−
(
PQ+
3
PQ
)]
+
+11
[(
P
Q
)2
+
(
Q
P
)2
][
(PQ)3 +
33
(PQ)3
− 15
(
PQ+
3
PQ
)
+ 45
]
+ 924.
Using (2.1) in the above, we deduce the required result.
Theorem 3.7. Let u = U(q) and v = U(q13), then
−8346u8v5 + 611u11v12 + 6318u4v8 + 8723u8v6 − 8346u5v8 − 78u7v5+
+910u10v7 + 1508u8v2 − 780u8v7 + 8476u8v9 − 10322u4v9 − 5148u8v8−
−481u3v12 − 7228u8v10 + 10894u5v9 + 611u3v2 − 208u6v13 + 910u7v10−
−10322u5v10 − 143u12v12 + 6409u3v9 − 1911u5v12 + 1313u10v2−
−832u7v11 − 5317u3v10 − 130u7v12 − 2184u11v11 − 182u13v9 + 2652u3v11−
−5148u6v6 − 5317u10v3 − 91u13v4 + 1508u6v12 + 78uv7 + 8723u6v8+
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
1618 K. R. VASUKI, ABDULRAWF A. A. KAHTAN, G. SHARATH, C. SATHISH KUMAR
+4706u6v3 + 1846u5v2 + 1508u2v8 + v14 + 9607u10v4 − 5317u4v11−
−182uv5 + 78u6v − 1170u6v2 − 2184u3v3 − 7228u4v6 − 6331u11v9+
+195u9v + 10894u9v5 + 6409u9v3 − 7228u10v8 + 6318u10v6 + 78u7v13+
+4706u3v6 − 780u6v7 + 13u13v12 − 130u2v7 + 52u2v13 − 1911u9v2+
+6318u8v4 + 78uv6 + 4407u11v10 + 1846u2v5 − 8346u6v9 − 1768u7v7−
−8346u9v6 − 832u7v3 − 10322u10v5 − 7631u4v4 − 3926u8v3 − 780u7v8+
+910u7v4 − 130u7v2 + 78u7v + 10257u4v5 + u14 − 832u3v7 − 78u7v9−
−78u9v7 + 10257u10v9 + 8476u9v8 − 780u7v6 − 12909u9v9 − 7631u10v10+
+846u12v9 − u13v13 + 156u13v10 − 10322u9v4 + 4407u4v3 + 13uv2+
+10257u9v10 + 6409u11v5 − 6331u5v3 − 1378u4v2 + 13u2v + 4407u10v11−
−3926u11v6 − 3926u3v8 − 143u2v2 − 1911u12v5 + 6409u5v11 + 8476u6v5+
+156uv4 − 65u3v + 1508u12v6 − 208uv8 + 13u12v13 − 65u11v13 + 156u10v13+
+195u13v5 − 1378u10v12 + 1313u4v12 − 91u4v13 − 130u12v7 − 3926u6v11−
−832u11v7 − 65uv3 − 208u13v6 − 1170u12v8 + 9607u4v10 − 13u3v13+
+6318u6v10 + 78u13v8 + 611u12v11 + 156u4v + 611u2v3 + 78u13v7+
+4706u11v8 − 1378u2v4 + 10257u5v4 − 208u8v − 26u13v − 7228u6v4−
−91u10v + 52u12v + 4407u3v4 − 13u11v + 8476u5v6 − 481u11v2 − 1378u12v10−
−1170u8v12 + 78u8v13 + 4706u8v11 − 182u5v − 13uv11 + 195uv9 − 91uv10−
−65u13v11 − uv − 39u2v12 − 481u2v11 − 26uv13 − 1170u2v6 + 52uv12−
−1911u2v9 + 1846u9v12 + 910u4v7 − 6331u9v11 − 13u13v3 + 2652u11v3−
−481u12v3 − 182u9v13 + 52u13v2 + 195u5v13 + 1313u2v10 − 39u12v2−
−6331u3v5 − 5317u11v4 + 1313u12v4 − 12909u5v5 − 78u5v7 = 0.
Proof. Let P =
ϕ(q)
ϕ(q3)
and Q =
ϕ(q13)
ϕ(q39)
, then from [18], we have
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
ON A CONTINUED FRACTION OF ORDER TWELVE 1619
(
Q
P
)7
+
(
P
Q
)7
+ 13
[(
P
Q
)6
+
(
Q
P
)6
]
− 26
[(
P
Q
)5
+
(
Q
P
)5
]
−
−13
[
3(PQ)2 +
27
(PQ)2
+ 10
][(
Q
P
)4
+
(
P
Q
)4
]
+ 13
[
10(PQ)2 +
90
(PQ)2
+ 68
]
×
×
[(
Q
P
)3
+
(
P
Q
)3
]
+ 13
[
10(PQ)2 +
90
(PQ)2
+ 68
][(
Q
P
)3
+
(
P
Q
)3
]
+
+13
[
(PQ)4 +
81
(PQ)4
− 20
(
(PQ)2 +
9
(PQ)2
)
− 115
][(
Q
P
)2
+
(
P
Q
)2
]
−
−13
[
(PQ)4 +
81
(PQ)4
− 10
(
(PQ)2 +
9
(PQ)2
)
− 131
][(
Q
P
)
+
(
P
Q
)]
=
= (PQ)6 +
729
(PQ)6
− 26
[
(PQ)4 +
81
(PQ)4
]
+ 169
[
(PQ)2 +
9
(PQ)2
]
+ 832.
Now using (2.1) in the above identity, we deduce the required result.
Acknowledgement. The authors are thankful to DST, New Delhi for awarding
research project [No. SR/S4/MS:517/08] under which this work has been done. Further,
authors thank the referee for comments to improve the manuscript.
1. Baruah N. D. Modular equations for Ramanujan’s cubic continued fraction // J. Math. Anal. and Appl.
– 2002. – 268. – P. 244 – 255.
2. Baruah N. D., Barman R. Certain Theta-function identities and Ramanujan’s modular equations of degree
3 // Indian J. Math. – 2006. – 48, № 1. – P. 113 – 133.
3. Berndt B. C. Ramanujan notebooks. Pt III. – New York: Springer, 1991.
4. Berndt B. C. Ramanujan Notebooks. Pt V. – New York: Springer, 1998.
5. Chan H. H. On Ramanujan’s cubic continued fraction // Acta arithm. – 1995. – 73, № 4. – P. 343 – 355.
6. Chan H. H., Huang S. S. On the Ramanujan – Göllnitz – Gordan continued fraction // Ramanujan J. –
1997. – 1. – P. 75 – 90.
7. Cho B., Koo J. K., Park Y. K. Arithmetic of the Ramanujan Göllnitz Gordan continued fraction // J.
Number Theory. – 2009. – 4, № 129. – P. 922 – 947.
8. Gollnitz H. Partition mit Diffrenzebedingungen // J. reine und angew. Math. – 1967. – 25. – S. 154 – 190.
9. Gordon B. Some continued fractions of the Rogers Ramanujan type // Duke Math. J. – 1965. – 32. –
P. 741 – 748.
10. Hardy G. H. Ramanujan. – 3rd ed. – New York: Chelsea, 1978.
11. Naika M. S. M., Dharmendra B. N., Shivashankra K. A continued fraction of order twelve // Cent. Eur.
J. Math. – DOI: 10.2478/s 11533-008-0031-y.
12. Ramanujan S. Notebooks (2 volumes). – Tata Inst. Fundam. Research. – Bombay, 1957.
13. Ramanujan S. The “Lost” Notebook and other unpublished papers. – New Delhi: Narosa, 1988.
14. Rogers L. J. On a type of modular relation // Proc. London Math. Soc. – 1920. – 19. – P. 387 – 397.
15. Vasuki K. R., Guruprasad P. S. On certian new modular relations for the Rogres – Ramanujan type
functions of order twelve // Proc. Jangjeon Math. Soc. (to appear).
16. Vasuki K. R., Sharath G., Rajanna K. R. Two modular equations for squares of the cubic functions with
Applications // Note Math. (to appear).
17. Vasuki K. R., Srivastakumar B. R. Certian identities for Ramanujan – Göllnitz – Gordan continued fraction
// J. Comput. Appl. Math. – 2006. – 187. – P. 87 – 95.
18. Vasuki K. R., Srivastakumar B. R. Two identities for Ramanujan’s cubic continued fraction. – Preprint.
19. Vasuki K. R., Swamy S. R. A new identity for the Rogers – Ramanujan continued fraction // J. Appl.
Math. Anal. and Appl. – 2006. – 2, № 1. – P. 71 – 83.
Received 27.10.09,
after revision — 08.07.10
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
|
| id | umjimathkievua-article-2986 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:34:04Z |
| publishDate | 2010 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/f7/8af9b781da61a7e6b6b2d4253ae762f7.pdf |
| spelling | umjimathkievua-article-29862020-03-18T19:41:53Z On a continued fraction of order twelve Про ланцюговий дріб дванадцятого порядку Kahtan, Abdulrawf A. A. Sathish, Kumar C. Sharath, G. Vasuki, K. R. Кантан, Абдулрашф А. А. Сатниш, Кумар С. Шаратх, Г. Васюкі, К. Р. We present some new relations between a continued fraction $U(q)$ of order twelve (which is established by M. S. M. Naika et al.) and $U(q^n)$ for $n = 7,9,11\;\text{and}\; 13$. Наведено деякі нові співвідношення між ланцюговим дробом $U(q)$ дванадцятого порядку (який описано М. С. М. Найка та іншими авторами) і $U(q^n)$ для $n = 7,9,11\;\text{та}\; 13$. Institute of Mathematics, NAS of Ukraine 2010-12-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2986 Ukrains’kyi Matematychnyi Zhurnal; Vol. 62 No. 12 (2010); 1609 - 1619 Український математичний журнал; Том 62 № 12 (2010); 1609 - 1619 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2986/2722 https://umj.imath.kiev.ua/index.php/umj/article/view/2986/2723 Copyright (c) 2010 Kahtan Abdulrawf A. A.; Sathish Kumar C.; Sharath G.; Vasuki K. R. |
| spellingShingle | Kahtan, Abdulrawf A. A. Sathish, Kumar C. Sharath, G. Vasuki, K. R. Кантан, Абдулрашф А. А. Сатниш, Кумар С. Шаратх, Г. Васюкі, К. Р. On a continued fraction of order twelve |
| title | On a continued fraction of order twelve |
| title_alt | Про ланцюговий дріб дванадцятого порядку |
| title_full | On a continued fraction of order twelve |
| title_fullStr | On a continued fraction of order twelve |
| title_full_unstemmed | On a continued fraction of order twelve |
| title_short | On a continued fraction of order twelve |
| title_sort | on a continued fraction of order twelve |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2986 |
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