On a continued fraction of order twelve
We present some new relations between a continued fraction U(q) of order 12 (established by M. S. M. Naika et al.) and U(q n) for n = 7, 9, 11, 13. Наведено деякі нові співвідношення між ланцюговим дробом U(q) дванадцятого порядку (який описано М. С. М. Найка та іншими авторами) і U(qn) для n=7,9,11...
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| Опубліковано в: : | Український математичний журнал |
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Інститут математики НАН України
2010
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| Цитувати: | On a continued fraction of order twelve / K.R. Vasuki, A.A. Kahtan Abdulrawf, C. Satish Kumar // Український математичний журнал. — 2010. — Т. 62, № 12. — С. 1609 - 1619. — Бібліогр.: 19 назв. — англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1859708078078820352 |
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| author | Vasuki, K.R. Abdulrawf, A.A. Kahtan Sathish Kumar, C. |
| author_facet | Vasuki, K.R. Abdulrawf, A.A. Kahtan Sathish Kumar, C. |
| citation_txt | On a continued fraction of order twelve / K.R. Vasuki, A.A. Kahtan Abdulrawf, C. Satish Kumar // Український математичний журнал. — 2010. — Т. 62, № 12. — С. 1609 - 1619. — Бібліогр.: 19 назв. — англ. |
| collection | DSpace DC |
| container_title | Український математичний журнал |
| description | We present some new relations between a continued fraction U(q) of order 12 (established by M. S. M. Naika et al.) and U(q n) for n = 7, 9, 11, 13.
Наведено деякі нові співвідношення між ланцюговим дробом U(q) дванадцятого порядку (який описано М. С. М. Найка та іншими авторами) і U(qn) для n=7,9,11та13.
|
| first_indexed | 2025-12-01T04:15:41Z |
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| 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.
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Received 27.10.09,
after revision — 08.07.10
ISSN 1027-3190. Укр. мат. журн., 2010, т. 62, № 12
|
| id | nasplib_isofts_kiev_ua-123456789-166304 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1027-3190 |
| language | English |
| last_indexed | 2025-12-01T04:15:41Z |
| publishDate | 2010 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Vasuki, K.R. Abdulrawf, A.A. Kahtan Sathish Kumar, C. 2020-02-18T17:08:30Z 2020-02-18T17:08:30Z 2010 On a continued fraction of order twelve / K.R. Vasuki, A.A. Kahtan Abdulrawf, C. Satish Kumar // Український математичний журнал. — 2010. — Т. 62, № 12. — С. 1609 - 1619. — Бібліогр.: 19 назв. — англ. 1027-3190 https://nasplib.isofts.kiev.ua/handle/123456789/166304 511.72 We present some new relations between a continued fraction U(q) of order 12 (established by M. S. M. Naika et al.) and U(q n) for n = 7, 9, 11, 13. Наведено деякі нові співвідношення між ланцюговим дробом U(q) дванадцятого порядку (який описано М. С. М. Найка та іншими авторами) і U(qn) для n=7,9,11та13. en Інститут математики НАН України Український математичний журнал Статті On a continued fraction of order twelve Про ланцюговий дріб дванадцятого порядку Article published earlier |
| spellingShingle | On a continued fraction of order twelve Vasuki, K.R. Abdulrawf, A.A. Kahtan Sathish Kumar, C. Статті |
| 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 |
| topic | Статті |
| topic_facet | Статті |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/166304 |
| work_keys_str_mv | AT vasukikr onacontinuedfractionofordertwelve AT abdulrawfaakahtan onacontinuedfractionofordertwelve AT sathishkumarc onacontinuedfractionofordertwelve AT vasukikr prolancûgoviidríbdvanadcâtogoporâdku AT abdulrawfaakahtan prolancûgoviidríbdvanadcâtogoporâdku AT sathishkumarc prolancûgoviidríbdvanadcâtogoporâdku |