Sumudu transform of Dixon elliptic functions with non-zero
modulus as Quasi C fractions and its Hankel determinants
Adem Kilicman
1,?
and Rathinavel Silambarasan
2
1
Department of Mathematics, Faculty of Science, Universiti Putra Malaysia, 43400 UPM Serdang, Selangor, Malaysia.
Email : akilic@upm.edu.my
2
Department of Information Technology, School of Information Technology and Engineering, Vellore Institute of
Technology, Vellore, 632014, Tamilnadu, India. Email : silambu_vel@yahoo.co.in
?
Corresponding Author.
Abstract
Sumudu transform of the Dixon elliptic function with non zero modulus α , 0 for arbitrary powers sm
N
(x, α) ; N ≥
1 , sm
N
(x, α)cm(x, α) ; N ≥ 0 and sm
N
(x, α)cm
2
(x, α) ; N ≥ 0 is given by product of Quasi C fractions. Next by
assuming denominators of Quasi C fraction to 1 and hence applying Heliermann correspondance relating formal power
series (Maclaurin series of Dixon elliptic functions) and regular C fraction, Hankel determinants are calculated and
showed by taking α = 0 gives the Hankel determinants of regular C fraction. The derived results were back tracked
to the Laplace transform of sm(x, α) , cm(x, α) and sm(x,α)cm(x, α).
Keywords : Dixon elliptic functions, non-zero modulus, Sumudu transform, Hankel determinants, Continued
fractions, Quasi C fractions.
Mathematics subject classification : 33E05, 44A10, 11A55, 11C20.
1 Introduction
To determine the coeffecients in the Maclaurin series of Jacobi elliptic functions, continued fractions and the Heilermann
correspondence the relation employing Formal Power Series (FPS) and its continued fraction to calculate Hankel
determinants are used in [2], also determinants of Bernoulli numbers were calculated from the correspondence in [2].
By using continued fraction and Fourier series expansions of Jacobi elliptic functions in [13] obtained orthogonal
polynomials which are related to each other through multiplication formulas of Jacobi elliptic functions in [13]. Laplace
transform of Jacobi elliptic functions expanded as continued fractions and shown their coeffecients are orthogonal
polynomials and derived dual Hahn polynomials in [19]. Fourier series and continued fractions expansions of ratis of
Jacobi elliptic functions and their Hankel determinants are given in [25] from which different ways of representing
sum of square numbers derived in determinant forms in [25]. Laplace transform of bimodular Jacobi elliptic functions
expanded as continued fractions in [14] and by modular transformation results were back tracked to unimodular Jacobi
elliptic functions in [14].
A. C. Dixon studied the cubic curve x
3
+ y
3
− 3αxy = 1 ; α , −1 for the orthogonal polynomials, where the curve
has double period in [16] which then give raise to two set of elliptic functions sm(x, α) and cm(x, α) now known
as Dixon Elliptic Functions (DEF). The examples, its relation to hypergeometric series, modular transformation and
formulae for their ratio given in [17]. When α = 0 in the above cubic curve, their series expansions and transformations
studied in [18]. DEF were used in the study of conformal mapping and geographical structure of world maps in [1]
addition and multiplication formulae for DEF are derived in [1]. Laplace transform applied for DEF for both the cases
of α = 0 and α , 0 to expand as set of continued fractions in [14]. The above cubis curve and its relation to Fermat
curve is studied for the Urn representation and combinatorics in [15]. Number theory related results followed by [25]
for factorial of numbers using DEF given in [4]. DEF relation to trefoil curves and relation to Weierstrass and its
derivative functions shown in [23].
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Fractional heat equations are solved using Sumudu transform in [3]. Sumudu transform embedded in decomposition
method in [27] and in homotopy perturbation method to solve Klein-Gordon equations in [26]. Fractional Maxwell’s
equations solved with Sumudu transform in [28] and some differential equations with Sumudu transform in [29].
Fractional gas dynamics differential equations using Sumudu transform is solved in [5]. Sumudu transform definition
for trigonometric functions and its infinite series expansions proved with examples comprising tables and properties
in [6]. Maxwell’s coupled equations solved with Sumudu transform for magnetic field solutions in TEMP waves given
in [7]. Without using any of decomposition, perturbation (or) analysis techniques Sumudu transform of functions
calculated by differentiating the function in [8]. Symbolic C++ program for Sumudu transform given in [8]. Sumudu
transform applied for bimodular Jacobi elliptic functions [14] for arbitrary powers in [9] as associated continued fraction
and their Hankel determinants. Applying modular transformation, Sumudu transform of tan(x ) and sec(x) derived in [9].
Sumudu transform of f (x) defined in the set A = { f (x)|∃M, τ
1
, τ
2
> 0, | f (x )| < Me
|x|
τ
j
, if x ∈ (−1)
j
× [0, ∞)} given by
integral equation.
S[ f (x)](u)
def
= F(u)
def
=
Z
∞
0
e
−x
f (ux)dx =
1
u
Z
∞
0
e
−
1
u
f (x)dx ; u ∈ (−τ
1
, τ
2
). (1)
In this work Sumudu transform applied for DEF of arbitrary powers sm
N
(x, α) ; N ≥ 1, sm
N
(x, α)cm(x, α) ; N ≥ 0
and sm
N
(x, α)cm
2
(x, α) ; N ≥ 0 and expanded as Quasi C Fractions (QCF). Using the numerator coeffecients of QCF,
Hankel determinants are calculated by the correspondence connecting FPS and Regular C fractions through Sumudu
transform.
2 Preliminaries
Cubic curve x
3
+y
3
−3αxy = 1 ; (α , −1) studied for its orthogonal polynomials in [16] derived the two set of elliptic
functions namely sm(x, α) and cm(x, α) which are having double period. Derivative of DEF (equations (1) and (3),
page 171, [16] and equations (1.18) and (1.19), page 9, [14]) takes the following,
d
dx
sm(x, α) = cm
2
(x, α) − αsm(x, α) and
d
dx
cm(x, α) = −sm
2
(x, α) + αcm(x, α). (2)
and have (equation (1.21), page 10, [14]),
sm(0, α) = 0 and cm(0,α) = 1. (3)
these functions satisfies the cubic curve mentioned above what is called Pythagorean theorem (equation (2), page
171, [16] and equation (1.22), page 10, [14]).
sm
3
(x, α) + cm
3
(x, α) − 3αsm(x, α)cm(x, α) = 1. (4)
Continued fraction notation is followed from (equation (2.1.4b), page 18, [20] and (equation (1.2.5
0
), page 8, [22]).
∞
K
n=1
a
n
b
n
def
=
a
1
b
1
+
a
2
b
2
+
a
3
b
3
+ ·· ·
def
=
a
1
b
1
+
a
2
b
2
+
a
3
b
3
+
a
4
b
4
+
.
.
.
.
Definition 1. Let a = {a
n
} , b = {b
n
} and u is an indeterminate, then the continued fraction of following form is called
C-fraction (equation (7.1.1), page 221, [20]), [30] and (equation (54.2), page 208, [31]).
1 +
∞
K
n=1
a
n
u
β (n)
1
.
2
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When the sequence β (n) is constant then C-fraction is called Regular C fraction. And QCF has the following form.
a
0
b
0
(u)+
∞
K
n=1
a
n
u
b
n
(u)
.
Sometimes the coeffecient a
n
= a
n
(u) thus the coeffecients are functions of u which can be seen in the main results of
this work.
Definition 2. Let c = {c
v
}
∞
v=1
be a sequence in C. Then the following m × m matrices are defined [14, 20,25], whose
determinants are denoted by respective H
(n)
m
and χ
m
.
H
(n)
m
def
= H
(n)
m
(c
v
)
def
= det
c
n
c
n+1
··· c
m+n−2
c
n+m−1
c
n+1
c
n+2
··· c
m+n−1
c
m+n
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
c
m+n−1
c
m+n
··· c
2m+n−3
c
2m+n−2
.
χ
m
def
= χ
m
(c
v
)
def
= det
c
1
c
2
··· c
m−1
c
m+1
c
2
c
3
··· c
m
c
m+2
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
c
m
c
m+1
··· c
2m−2
c
2m
.
Remark 1. The matrix for χ
m
is obtained from the matrix for H
(1)
m+1
by deleting the last row and next to last column
[14, 20, 25]. For n = 1, H
(1)
1
= c
1
and χ
1
= c
2
. Determinants H
(n)
m
and χ
m
are named as persymmetric determinants
(or) Turanian determinants (or) Hankel determinants.
The relation between FPS and Regular C fraction is given as lemma [14], (Theorem 7.2, pp 223-226, [20]), [25].
Lemma 1. When the Regular C fraction converges to FPS.
1 +
∞
∑
v=1
c
v
z
v
= 1 +
∞
K
n=1
a
n
u
1
; (a
n
, 0). (5)
then,
H
(1)
m
([c
v
]) , 0 , H
(2)
m
([c
v
]) , 0 and a
1
= H
(1)
1
([c
v
]) ; (m ≥ 1). (6)
a
2m
= −
H
(1)
m−1
H
(2)
m
H
(1)
m
H
(2)
m−1
and a
2m+1
= −
H
(1)
m+1
H
(2)
m−1
H
(1)
m
H
(2)
m
; (m ≥ 1). (7)
where H
(1)
0
= H
(2)
0
= 1. Conversely if Eqs (6) and (7) holds then Eq (5) holds true. Also,
H
(2)
m
([c
v
]) = (−1)
m
H
(1)
m
([c
v
])
m
∏
j=1
a
2 j
= (−1)
m
H
(1)
m+1
([c
v
])
m
∏
j=1
1
a
2 j+1
; (m ≥ 1). (8)
3 Main results 1 : Sumudu transform of Dixon elliptic functions (α , 0)
Laplace transform of DEF sm(x, α) , cm(x, α) and sm(x, α)(cm(x, α)) given as QCF in [14]. In this work Sumudu
transform Eq (1) of DEF sm
N
(x, α) ; N ≥ 1, sm
N
(x, α)cm(x, α) ; N ≥ 0 and sm
N
(x, α)cm
2
(x, α) ; N ≥ 0 for arbitrary
powers derived as QCF. Followed by assuming the denominator of QCF be 1, using Lemma 1, Hankel determinants
are calculated. The following three theorems are main results of this work.
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Theorem 1. Sumudu transform of DEF sm
N
(x, α) ; N ≥ 1 as QCF given by the following enumerates:
(i) For j ≥ 1.
S[sm(x, α)] =
u
(1 − 2αu)(1 + αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j − 2)(3 j − 1)
2
b
2 j
(u) = (1 − (6 j − 1)αu)
a
2 j+1
= (3 j)
2
(3 j + 1)
b
2 j+1
(u) = (1 − 2(3 j + 1)αu)(1 + (3 j + 1)αu)
(9)
(ii) For j ≥ 1.
S[sm
2
(x, α)] =
1
(1 − αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j − 2)
2
(3 j − 1)
b
2 j
(u) = (1 − 2(3 j − 1)αu)(1 + (3 j − 1)αu)
a
2 j+1
= (3 j − 1)(3 j)
2
b
2 j+1
(u) = (1 − (6 j + 1)αu)
×
2u
2
(1 − 4αu)(1 + 2αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j − 1)(3 j)
2
b
2 j
(u) = (1 − (6 j + 1)αu)
a
2 j+1
= (3 j + 1)
2
(3 j + 2)
b
2 j+1
(u) = (1 − 2(3 j + 2)αu)(1 + (3 j + 2)αu)
(10)
(iii) Let N = 3, 6, 9, 12, ··· and j ≥ 1.
S[sm
N
(x, α)] =
N
3
∏
i=1
(3i − 2)u
(1 − (6i − 3)αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j + 3i −4)
2
(3 j + 3i − 3)
b
2 j
(u) = (1 − 2(3 j + 3i − 3)αu)(1 + (3 j + 3i −3)αu)
a
2 j+1
= (3 j + 3i −3)(3 j + 3i −2)
2
b
2 j+1
(u) = (1 − (6 j + 6i − 3)αu)
×
N
3
∏
i=1
(3i − 1)(3i)u
2
(1 − 2(3i)αu)(1 + (3i)αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j + 3i −3)(3 j + 3i −2)
2
b
2 j
(u) = (1 − (6 j + 6i − 3)αu)
a
2 j+1
= (3 j + 3i −1)
2
(3 j + 3i)
b
2 j+1
(u) = (1 − 2(3 j + 3i)αu)(1 + (3 j + 3i)αu)
(11)
(iv) Let N = 4, 7, 10, 13,· · · and j ≥ 1.
S[sm
N
(x, α)] =
u
(1 − 2αu)(1 + αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j − 2)(3 j − 1)
2
b
2 j
(u) = (1 − (6 j − 1)αu)
a
2 j+1
= (3 j)
2
(3 j + 1)
b
2 j+1
(u) = (1 − 2(3 j + 1)αu)(1 + (3 j + 1)αu)
×
N−1
3
∏
i=1
(3i − 1)u
(1 − (6i − 1)αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j + 3i −3)
2
(3 j + 3i − 2)
b
2 j
(u) = (1 − 2(3 j + 3i − 2)αu)(1 + (3 j + 3i −2)αu)
a
2 j+1
= (3 j + 3i −2)(3 j + 3i −1)
2
b
2 j+1
(u) = (1 − (6 j + 6i − 1)αu)
×
N−1
3
∏
i=1
(3i)(3i +1)u
2
(1 − 2(3i + 1)αu)(1 + (3i + 1)αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j + 3i −2)(3 j + 3i −1)
2
b
2 j
(u) = (1 − (6 j + 6i − 1)αu)
a
2 j+1
= (3 j + 3i)
2
(3 j + 3i + 1)
b
2 j+1
(u) = (1 − 2(3 j + 3i + 1)αu)(1 + (3 j + 3i +1)αu)
(12)
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(v) Let N = 5, 8, 11, 14, ··· and j ≥ 1.
S[sm
N
(x, α)] =
1
(1 − αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j − 2)
2
(3 j − 1)
b
2 j
(u) = (1 − 2(3 j − 1)αu)(1 + (3 j − 1)αu)
a
2 j+1
= (3 j − 1)(3 j)
2
b
2 j+1
(u) = (1 − (6 j + 1)αu)
×
2u
2
(1 − 4αu)(1 + 2αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j − 1)(3 j)
2
b
2 j
(u) = (1 − (6 j + 1)αu)
a
2 j+1
= (3 j + 1)
2
(3 j + 2)
b
2 j+1
(u) = (1 − 2(3 j + 2)αu)(1 + (3 j + 2)αu)
×
N−2
3
∏
i=1
(3i)u
(1 − (6i + 1)αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j + 3i −2)
2
(3 j + 3i − 1)
b
2 j
(u) = (1 − 2(3 j + 3i − 1)αu)(1 + (3 j + 3i −1)αu)
a
2 j+1
= (3 j + 3i −1)(3 j + 3i)
2
b
2 j+1
(u) = (1 − (6 j + 6i + 1)αu)
×
N−2
3
∏
i=1
(3i + 1)(3i + 2)u
2
(1 − 2(3i + 2)αu)(1 + (3i + 2)αu)+
∞
K
n=2
a
n
u
3
b
n
(u)
a
2 j
= (3 j + 3i −1)(3 j + 3i)
2
b
2 j
(u) = (1 − (6 j + 6i + 1)αu)
a
2 j+1
= (3 j + 3i +1)
2
(3 j + 3i + 2)
b
2 j+1
(u) = (1 − 2(3 j + 3i + 2)αu)(1 + (3 j + 3i +2)αu)
(13)
Proof. Defining the Sumudu transform of DEF by integral equations, Let N = 0, 1, 2, ··· .
S[sm
N
(x, α)] = A
N
=
Z
∞
0
e
−x
sm
N
(xu, α)dx. (14)
S[sm
N
(x, α)cm(x, α)] = B
N
=
Z
∞
0
e
−x
sm
N
(xu, α)cm(xu, α)dx. (15)
S[sm
N
(x, α)cm
2
(x, α)] = C
N
=
Z
∞
0
e
−x
sm
N
(xu, α)cm
2
(xu, α)dx. (16)
By parts method, using Eqs (2) - (4), with A
0
= 1 leads to the following:
A
1
= uC
0
− αuA
1
.
A
2
= 2uC
1
− 2αuA
2
.
A
3
= 3uC
2
− 3αuA
3
.
A
N
= NuC
N−1
− NαuA
N
.
Solving with the recurrences of Eqs (15) and (16) yields the following QCF:
A
N
B
N−2
=
(N − 1)Nu
2
(1 − 2Nαu)(1 + Nαu) + N(N + 1)u
2
B
N+1
A
N
; (N ≥ 2). (17)
B
N
A
N−1
=
Nu
(1 − (2N + 1)αu) + (N + 1)u
A
N+2
B
N
; (N ≥ 2). (18)
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