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Page 97n_Multi-Series of the Generalized Difference Equations to Circular Functions
G.Britto Antony Xavier1, S.Sathya2 and S.U.Vasantha Kumar3
1,2,3
Department of Mathematics, Sacred Heart College,
Tirupattur - 635601, Vellore District Tamil Nadu, S.India.
ABSTRACT
We investigate the numerical-complete relation to certain type of higher order generalized difference equation to find the value of n_ multi-series to circular functions in the field of finite difference methods. We also give an example to illustrate the n_ multi-series.
Key words: Complete solution, Circular function, Generalized difference operator, Numerical solution.
AMS Subject classification: 39A70, 47B39, 39A10.
1. INTRODUCTION
The Fractional Calculus is currently a very important research field in several different areas: physics (including classical and quantum mechanics and thermodynamics), chemistry, biology, economics and control theory ([9], [10], [11]). In 1989, K.S.Miller and Ross [12] introduced the discrete analogue of the Riemann-Liouville fractional derivative and proved some properties of the fractional difference operator. The main definition of fractional difference equation (as done in [12]) is the fractional sum of f(t) by
), ( 1
= ) (
1)) ( (
) ( = ) (
s f t
f
s t
s t t
a
s
(1)
where >0. On the other hand when =m is a positive integer, if we replace f(t) by u(k) and by , defined by u(k)=u(k)u(k), (1) becomes
). (
1)! (
1) ( = ) ( = ) ( u
1) (
= )
(
n u k r
r k
u k
n k
n r n
n
(2)Let i >0, u(k) be real valued function on [0,), u(k)=0 for all k(,0], [k/i] be the integer part of k/ , i i(k)=k[k/i]i for i=1,2,,n and 0(k)=k.
), (
u =
) ( u series n_multi
induces (2)
2, n for Then
1] [1, 1 = ]
[1,
n n n
n r n
r k k
n k
(3)where u ( )= ( 1 1)
=1 1 [1,1]
1
r k u k
k
r
(1_ series with respect to 1),
1
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Page 98) (
u = ) (
u 2 2
[1,1] =1 2 [1,2]
2
r k k
k
r
( 2_ multi-series with respect to 1,2),
Substituting u
[1,1]
, u
[1,2]
, , u
1] [1,n
in (3), we get
), (
= ) ( u
1 = 1
= 1 2
1 1
1 = 2 1
1 = 1 1 = ]
[1,
1 2 =
i i n
i r
n n n r n n r k
n r n
n n r k
n r n k
n r n
r k u k
i i r n i k
(4)
which is a numerical solution of the generalized difference equation
0. ), ( = ) )) ( ( (
) (
2 1 ]
[1,
v k v k u k k
n n
(5)
We denote R.H.S of (4) as (~)
] [1,
k u
n
, which depends on 1, 2, , n, k and u(k). By this
notation, (3) and (4)can be expressed as u ( )= (~).
] [1, ]
[1,
k u k
n n
1, = n when ar,
Inparticul u( )= ( )| = (~).
[1,1] ) ( 1 1
[1,1] [1,1]
k u k
u
k k k
(6)
Remark 1.1 u ( 1(~))
1] [1, ] , [
k m m n m
)) (
( u =
= 1 1] [1, 1 = 2
1 1
1 = 2 1
1 = 1 1 =
1 =
i i n
m i m m m
r n
n n r n n r k
n r n
n n r k
n r n k
n r
r k
m i i r n m i k
.
, )
( = ) (
where 1
1 = =
=
1
mm i i r n
m i k i i n
m i i
i n
m i
m k r k r
When 1 =2 =n =, the above n_multi-series (~)
] [1,
k u
n
becomes u ( )
) (
k n
given
in (2). We find that, by expanding the terms, u ( )
] [1,
k n
is independent of the order of the parameters
1
, 2, , n. There are direct formula to find the n_series when u(k)=km, k(m), ak,
k m
a
k etc and 1 =2 ==n = [?, ?].
There is no direct formula to find the value of n_multi-series in the existing literature. We find that the n_multi-series (~)
] [1,
k u
n
is the numerical solution of the generalized difference
equation
), ( = )} (
{ 1) ( ) (
) ( 0
= ]
[1,
k u k
v k
v
A n
L r A r n n
r n
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Page 99 where the operator ] [1,n
is given in (5) and r(Ln) is the set of all subsets of the size t from the set
n
n
L = 1,2,, . The complete solution of equation (7) is denoted by u ( )
] [1,
k n
. Hence, in this
paper we obtain numerical-complete relation of the equation (7) and arrive the n_multi-series to the circular functions.
2. Preliminaries
In this section, we present some notations, basic definitions and preliminary results. Let },
, {1,2,
= n
Jn 0(Jn) = {}, is empty set, 1(Jn) = {{1},{2}, , {n}}, 2(Jn) = {1,3}
{{1,2}, , , {1,n},{2,3}, , {2,n}, ,{n2,n1}}. In general, t(Jn) = set of all subsets of size t in ascending order from the set Jn, ( )= ( )
0 =
n n
t
n t J
J
is power set of Jn.
Let ( )=0
1 =
t f n
t
for all integers n<1, ( )=12 =
i f t
i
if t1and for 1 p<qn , 1 ( )] , [
k u q p
) )) ( ( (
= 1 1
1 1
p p q u k
, u ( )
] [1,
k i
= k k
i i
i k 1( ) 1]
[1, 1
| ) ( u
= u ( ) u ( 1( ))
1] [1, 1 1]
[1, 1
k
k i
i i i
i
for i=2,,n, u ( )= 1 ( )
1 [1,1]
k u
k
and u ( )= ( )
[1,0]
k u k
.
Now we consider following lemma on circular functions.
Lemma 2.1 [1] Let p and q be any real numbers. Then,
. )
cos 2(1
sin ) ( sin = sin
1
j c p
pk k
p
pk
(8)
. )
cos 2(1
cos ) ( cos = cos
and 1 cj
q qk k
q
qk
(9)
Remark 2.2 (i) Hereafter, we take P= p(n12r1)q(n22r2) andP= p(n12r1)q(n22r2)
and hence P and P are varying with respect to n , 1 n , 2 r , 1 r , 2 p and q,
1)) ( ( 2) 1)( ( =
) (
n n r
n n
nr .
(ii) Pi Pi P P i P Pi
2 , 2 ,
, are not multiple of 2 , for i=1,2,,n.
Corollary 2.3 [1] (i) If n and 1 n are odd positive integers, then 2
sin sin
.! ! 1) ( 2
1) ( = cos sin
2 ) 2 ( 2 1
) 1 ( 1 1 2
1 2
0 = 2 2
1 1
0 = 1 1 2 1
2 1 1 2
1 Pk Pk
r n r n qk
pk
r r r n
r n
r n n
n
n n
(10)
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Page 100
Pk Pk
r n r
n qk
pk
r n
r r r n n
r n n n
n
cos cos
! !
1) ( 2
1 = cos sin
2 ) 2 ( 2 2
2 2
0 = 2 1
) 1 ( 1 1 2
1 2
2 1
0 = 1 1 2 1 2
1
. ! 2 ! 2 2 1 2
cos ! 2 2
cos ! 2
2 2 2 2 1
2 1 1 2
2 2 2 1
2 1 1
n n n n k
P P n
n k P P n
n
n n n
n
(11)
Lemma 2.4 [3] If s and rn S are the Stirling numbers of the first and second kinds, and rn
) 1) ( ( ) 2 )( ( =
)
(
k k k k n
k n
, then
. 1) ( = and
= , =
1) ( ) ( 1 )
( 1
= 1
= ) (
k
k k
S k
k s
k rn n r r
n
r n r r n n r n
r n
(12)
3. Main Results
Here we introduce Stirling numbers of third kind and express the polynomial factorial (n)
a k in
terms of (r)
b
k , r=1,2,n and Stirling numbers of third kind. Also we derive the Bernoulli’s multi-series and n_multi-series to circular functions.
Definition 3.1 Let 1 pn, The Stirling number of third kind for the positive reals a and b is defined by
. =
= _ _
p t b t n a t p n t n
p t b n
a
p s S
S
(13)
Lemma 3.2 The expression of (n) a
k in terms of (p) b
k is given by
.
= __ ( )
1 = )
( p
b b n
a p n
p n
a S k
k
(14)Proof. The proof follows from (13) and first, second terms of (12).
Theorem 3.3 Let p1=1, 1,2,n be a set of positive reals and 1 1
2 1 1 1
] [1,
=
n
n
. Then
1 (1)
1 (0) 1
1 =
k k
and .
) (1
) (1
=
1 1
) 1 (1
1 1 _ 1 1
_ 1 1
=1 1
2 = (0) 1
]
[1, n n
n p n
r r
r r p
r r p r p
r p n
r
n p
k
p S k
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Page 101Proof. Since (0) (0)
1 =
=
1 k k and (1)
2 (1)
1 =
k
k from (12), we get
1 (1)
1 (0)
1 1
1 =
k k
. Again taking 1
2
,
we get .
2 =
2 1 (2)
2 (0) 1
[1,2]
k k
Again taking 13 on both sides of the above and applying (14), we
obtain .
) (1 2
= 2
1 = 2
1 =
2 3 2 1
) 2 (1
3 3
2_ 2 _ 2 2
=1 2 ) 2 (
3 3 2_
2 _ 2 2
=1 2 1
3 2 1 (2)
2 1
3 2 1 (0) 1
[1,3] p
k S
k S k
k
p
p p p p
p
Now the proof is completed by taking 1
i
and applying third relation of (12) for n
i=4,5,, respectively.
The following theorem gives the complete solution of the equation (7).
Theorem 3.4 Consider the functions u ( )
] [1,
k i
, i(k) for i=1,2,,n, given in the notations and above. Assume that for each i, 1in, 1 ( )
] [1,
k u i
be any closed form solution of the
difference equation ( )= ( )
] [1,
k u k v i
. Then, for i
n i
k max
1
,
)) ( ( 1)
( ) ( =
| ) (
u 1 1
] 1 [1, ) ( =1 } { =1 1
] [1, ) ( ] [1,
k u k
u
k m
m n J t t s s m t n
t n
k k n n
(0) 1
] , [1 (0) 1 1
] 1 , [1 1 =
)) (
( k k
n t m i
m i m i m t
i
(16)
is the complete solution of equation (7).
Proof. Since 1=k(0), applying the limit from 1(k) to k for ( )
1 1u k
, we have
, )) ( ( )
( = | )
( 1 1 (0)
1 1
1 ) ( 1 1
1u k u k u k k
k
k
which is a complete form solution of equation (7) for n=1.
Taking 1
2
on both sides and applying the limits from 2(k) to k and keeping 1 ( 1( ))
1u k
as
a constant, we obtain
k k k
k k
k k
k u k u k k
k
u ( )
2 (0) 1
2 1
1 1 ) ( 2 1
[1,2] ) ( 2 ) ( 1 1 1 1
2( ( )| )| = ( )| ( ( )) |
which is the complete solution of the equation (7) and it can be expressed as
(0) 1
2 1
1 1 1
[1,2] ) ( 2 [1,2]
)) ( ( )
( = | ) (
u k k k u k u k k
( ( )) ( ( )) 1 ( 2( ))(0).
2 1
1 1 2
1 [1,2]
k k
u k
u
In the right hand side of the above expression, second term is associated to {m1}={1}1(J2), third term to {m1}={2}1(J2) and the fourth term to {m1,m2}={1,2}2(J2). Taking
1 3
on )
(
2 k
u , applying the limits 3(k) and k, and as ( 1( ))
1 1u k
, 2 (0)
1
2( (k))
and 1 ( 2( ))
[1,2]
k u
are constants, we get k k3(k) [1,3]
| ) (
u
= u ( ) u ( 3( ))
[1,2] 1 3 [1,2]
1
3 k k
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Page 102)) ( ( 1)
( ) ( = | ) ( u
1 1
] 1 [1, ) ( 1 } { 3
=1 1
[1,3] ) ( 3 [1,3]
k u k
u
k m
m n J t t s m t
t k
k
1 (0),3] [1 (0) 1 1
] 1 , [1 =1
)) (
( k k
t m i
m i m i m t
i
which is a complete solution of the equation (7) for n=3.
As all the lower limit values are constants, the proof is completed by taking 1
i and
applying the limit from i(k) to k on u ( )
[1,3]
k successively for i=4,5,,n. The following theorem gives a numerical solution of the equation (7).
Theorem 3.5 Consider the assumptions of Theorem 3.4. Then, for i n
i k
1 =
,
)) ~ ( ( u =
)
( 1
1] [1, ] , [ 1 =
k k
v m
m n m n
m
(17)
is the numerical solution of the difference equation (7).
Proof. From equation (6), we have
(say) ), ( = )) ( ( u ) ( u = ) ~ ( = | )
( 1 1
[1,1] [1,1]
[1,1] ) ( 1 1
1u k u k k k z k
k
k
(18)
is a numerical solution of the equation (7) for n=1. Again taking 1
2
on z1(k) and applying equation (6), we get
(say) ), ( = ) ( = | )
( 1 2
[2,2] ) ( 2 1 1
2z k z k z k
k
k
(19)
which is a numerical solution of the equation (7) for n=2. Replacing k by kr22 in (18), we obtain
)). (
( u ) (
u = )
( 1 2 2
[1,1] 2 2 [1,1] 2
2
1
r k r
k r
k
z (20)
Substituting (20) in (19), we find that
)) ~ ( ( u )
~ ( u =
)
( 1
[1,1] [2,2] [1,1]
[2,2]
2 k k k
z
(21)
which is the same as
)). ~ ( ( u ))
( ( u ) ( u = )
( 1
[1,1] [2,2] 2
[1,2] [1,2]
2 k k k k
z
(22)
Applying the numerical solution ( )= (~)
[1,2]
2 k u k
z
on (22), we get
)), ~ ( ( u )
~ ( = | ) (
u 1
[1,1] [2,2] [1,2]
) ( 2 [1,2]
k k
u
k k k
(23)where the values u ( 1( 2 2))
[1,1]
r
k can be evaluated by replacing k by 1(kr22) in the
closed form solution k
k k ( )
1 [1,1]
| ) (
u
given in Theorem 3.4 for n=1.
Taking 1
3
on z2(k) and applying equation (7) yield
(say). ) ( = ) ( =
| )
( 2 3
[3,3] ) ( 3 2 1
3z k z k z k
k
k
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Page 103 Replacing k by kr33 in (22), we havez2(kr33) = u ( ) u ( ( )) u ( 1( 3 3 2 2)).
[1,1] =1 2 3 3 2 [1,2] 3 3 [1,2]
2 3 3
r r k r
k r
k
r k
r
(25)
Substituting (25) in (24), we obtain
)) ~ ( ( u ))
~ ( ( u )
~ ( u = )
( 1
[1,1] [2,3] 2
[1,2] [3,3] [1,2]
[3,3]
3 k k k k
z
which is the same as
)). ~ ( ( u ))
~ ( ( u ))
( ( u ) ( u = )
( 1
[1,1] [2,3] 2
[1,2] [3,3] 3
[1,3] [1,3]
3 k k k k k
z
Since z3(k) is a solution of the equation (7), taking the numerical solution for it, then for n=3, we find that
)) ~ ( ( u ))
~ ( ( u )
~ ( = | ) (
u 2
[1,2] [3,3] 1
[1,1] [2,3] [1,3]
) ( 3 [1,3]
k k
k u
k k k
(26)where the values u ( 1( 3 3 2 2))
[1,1]
r r
k and u ( 2( 3 3))
[1,2]
r
k can be evaluated by replacing
k by 1(kr33r22) and 2(kr33) in the closed form solutions kn k n
k ( )
] [1,
| ) (
u
given in
Theorem 3.4 for n=1,2.
The proof is completed by taking 1
i
on z3(k) and applying the numerical solution mentioned
in (7) successively for i=4,5,,n.
The following theorem is the n_multi-series of u(k).
Theorem 3.6 The numerical-complete relation of the difference equation (7) is given by
)) ~ ( (
u 1
1] [1, ] , [ 1 =
k
m m n m n
m
= t
n J t t s s m n
t n
k
u( ) ( 1)
) ( =1 } { =1 1
] [1,
. ))
( ( ))
(
( 1 (0)
] , [1 (0) 1 1
] 1 , [1 1 = 1 1
] 1 [1,
k k
k u
n t m i
m i m i m t
i m m
(27)
Proof. The proof follows by equating the numerical solution given in Theorem 3.5 and the complete solution given in Theorem 3.4.
Theorem 3.7 If n and 1 n are odd positive integers, then 2
1 ) ( ) ( 0 = 2
1 2
0 = 2 2
1 1
0 = 1 1 2 1
2 1 1 2
1 1
] [1,
1) ( 2
1) ( = cos
sin n t s
n L t A n
t n
s n
s n n n
n
n n
n
qk
pk
. cos 1
) ( sin
cos 1
) ( sin ! !
=1 =1
2 ) 2 ( 2 1
) 1 ( 1
in
i i n
i s s
P A k P
P A k P s
n s n
(28)
Proof. Applying 1
1
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Page 104 !! 2
1) ( =
cos sin
2 ) 2 ( 2 1
) 1 ( 1 2
1
1 ) ( 2
1 1 2
1 2
0 = 2 2
1 1
0 = 1 2 1
1
1 s
n s n qk
pk
s s
n n
s t n n n
s n
s n
n
1 cos
.) ( sin ) ( sin cos
1
) ( sin ) ( sin
1 1 1
1
P
k P k
P P
k P k
P
(29)
Similarly we find n pk n qk
cos
sin 1 2
1 [1,2]
,
. cos 1
) ( sin cos
1
) ( sin ! ! 2
1) (
2
=1 2
=1 2
) 2 ( 2 1
) 1 ( 1 1 2 1
1 ) (2 2
1 1
) 2 ( 2
0 = 2
1 2
0 = 2 2
1 1
0 =
1
i i
i i
s s
n n
s t n
L t A t n
s n
s
P A k P
P A k P s
n s n
Proceeding like this, we get (28).
Remark 3.8 n pk n
sin 1 1
] [1,
and n pk
n
cos 2 1
] [1,
can be obtained by putting n2 =0 and n1=0 i n
(28) respectively.
Theorem 3.9 If n is an odd and 1 n is an even positive integers, then 2
! 1)
( !
2 1) ( = cos sin
2 ) 2 ( 2 2
1 2
0 = 2 1 ) ( ) ( 0 = 1
) 1 ( 1 2
1 1
0 = 1 1 2 1
2 1 1 2
1 1
]
[1, s
n s
n qk
pk
s n
s s t n
n L t A n
t s n
s n n n
n
n n
n
.
2 cos 1
) ( 2 sin
! 2 cos
1
) ( sin
cos 1
) ( sin
=1 2
2 2
2
=1
=1
i n
i n
i n
i i n
i
P P
A k P P
n n
P A k P
P A k P
(30)
Theorem 3.10 If n is an even and 1 n is an odd positive integers, then 2
( ) 11 ) 1 ( 1 2
1 1
0 = 1 ) ( 0 = 2
) 2 ( 2 2
1 2
0 = 2 1 2 1
2 1 2
1 1
] [1,
1) ( ! !
2 1) ( = cos
sin n t s
s n
s n L t A n
t s n
s n n n
n
n n
n s
n s
n qk
pk
.
2 cos 1
) ( 2 cos
! 2 cos
1
) ( cos
cos 1
) ( cos
=1 1
2 1
1
=1
=1
i n
i n
i n
i i n
i
P P
A k P P
n n
P A k P
P A k P
(31)
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Page 105Theorem 3.11 If n and 1 n are even positive integers, then 2
( ) 11 ) 1 ( 1 2
2 1
0 = 1 ) ( 0 = 2
) 2 ( 2 2
2 2
0 = 2 1 2 1
2 1 2
1 1
] [1,
1) ( ! !
2 1) ( = cos
sin n t s
s n
s n L t A n
t s n
s n n n
n
n n
n s
n s
n qk
pk
i n
i n
i n
i i n
i
P P
A k P P
n n
P A k P
P A k P
2 cos 1
) ( 2 cos
! 2 cos
1
) ( cos cos
1
) ( cos
=1 2
2 2
2
=1 =1
. ) (1
) (1 !
2 ! 2 2
2 cos 1
) ( 2 cos
! 2
1 1
) 1 (1
1 1 _ 1 1
_ 1 1
=1 1
2 = 2
2 2
2
1 2 1
1 1
=1 1
2 1
1
n nn p
n
r r
r r p
r r p r p
r p n
r n n
n
i n
i n
p k
p S
n n n n
P P
A k P P
n n
(32)
Note 3.12 The n_multiseries to sinn1 pkcosn2qk for the four cases can be obtained by applying (28), (30), (31) and (32) to equation (27) respectively.
Remark 3.13 If we take 1=2 =3 ==n =, we get 1 ( )= ( ).
] [1,
k u k
u m
n
and all
n_multi-series becomes m-series denoted in [1].
The following example illustrates a 2_multi-series for n pk n qk
cos
sin 1 2
Example 3.14 Consider the case n=2 in equation (27), n1 =4, n2 =4, k =10.3, 1 =3.1,
4.2 =
2
, p=7,q=3, then 1(k)=1, 2(k)=1.9. Let P=
7(22s1)3(22s2)
and
7(2 2 ) 3(2 2 )
= s1 s2
P .
)) ~ ( (
u 1
1] [1, ,2] [ 2
1 =
k
m m m m
= t
J t t s s m t k
u( ) ( 1)
) 2 ( =1 } { 2
=1 1
[1,2]
. ))
( ( ))
~ (
( 1 (0)
] , [1 (0) 1 1
] 1 , [1 1 = 1 1
] 1 [1,
k k
k u
n t m i
m i m i m t
i m m
(33)
LHS of equation (33) is the sum of the terms 1
=
m ; (~)
[1,2]
k u
.
2 =
m ; u ( (~))= u (3) u (1.9).
[1,1] [1,1]
1 [1,1]
[2,2]
kISSN: 2231-5373
http://www.ijmttjournal.org
Page 106
(2 ) 11 ) 1 ( 2 2 4 0 = 1 ) 2 ( 2 0 = 2 ) 2 ( 2 2 4 0 = 2 9 2 4 4 4 1
2 ! ( 1)
4 ! 4 2 1) ( = cos sin
(i) t s
s s L t A t s
s s s
qk pk
! 2 4 4 cos 1 ) (10.3 cos cos 1 ) (10.3 cos 2 4 2 =1 2 =1
i i i i P A P P A P
1 2(2) 2 2 4 2 4 2 =1 2 4 2 =1 2 10.3 ! 2 4 4 ! 2 4 4 2 2 cos 1 ) (10.3 2 cos ! 2 4 4 2 cos 1 ) (10.3 2 cos i i i i P P A P P P P A P P
( the terms for 1(J4)=
1, 2
)
(1 ) 11 ) 1 ( 2 2 4 0 = 1 ) 2 ( 1 0 = 2 ) 2 ( 2 2 4 0 = 2 8 2 4 2 (1) 2 1 4 1 4 1
1 ! ( 1)
4 ! 4 2 1) ( = 10.3 cos sin
(ii) t s
s s L t A t s
s s s
q p
i ii P P
A P P P A P P A P 2 cos 1 ) (1 2 cos ! 2 4 4 cos 1 ) (1 cos cos 1 ) (1 cos 2 4 , 10.3 1 ! 2 4 4 ! 2 4 4 ) 2 cos 1 ) (1 2 cos ! 2 4 4 2 (1) 2 1 2 4 2 4 2 4 i P P A P P
(4 ) 11 ) 1 ( 2 2 4 0 = 1 ) 2 ( 4 0 = 2 ) 2 ( 2 2 4 0 = 2 9 2 4 2 4 2 4 1
2 ! ( 1)
4 ! 4 2 1) ( = cos sin
(iii) t s
s s L t A t s
s s s
q p
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Page 107 (the terms for 2(J4)=
1,2 )
! 4 !
4 2
1) ( = 1.9 cos
sin (iv)
1 ) 1 ( 2
2 4
0 = 1 ) 2 ( 1
0 = 2
) 2 ( 2
2 4
0 = 2 8
2 4
2 (1)
2 1 4 1 4 1
1 p q s s
s
s L t A t s
s
i i
i s
t
P P
A P
P
P A P P
A P
2 cos 1
) (1 2 cos
! 2 4 4 cos
1
) (1 cos cos
1
) (1 cos 1)
(
2 4
1 ) (1
1 1.9 .
! 2 4 4
! 2 4 4 )
2 cos 1
) (1 2 cos
! 2 4 4
2 (1)
2
1 2 4 2 4 2
4
i
P P
A P
P
References
[1] G.Britto Antony Xavier, S.Sathya, S. U. Vasanthakumar, m-Series of the Generalized Difference Equation to Circular Functions,International Journal of Mathematical Archive-4(7), (2013), 200-209.
[2] Kwakernak, H., Sivan, R., Strubos, R.C.W., Modern Signals and Systems, New Jersey: Prentice Hall, 1991.
[3] M.Maria Susai Manuel, G.Britto Antony Xavier and E.Thandapani, Theory of Generalized Difference Operator and its Applications, Far East Journal of Mathematical Sciences, 20(2) (2006), 163 - 171.
[4] M.Maria Susai Manuel, G.Britto Antony Xavier and E.Thandapani, Generalized Bernoulli Polynomials Through Weighted Pochhammer Symbols, Far East Journal of Applied Mathematics, 26(3) (2007), 321-333.
[5] M.Maria Susai Manuel, V.Chandrasekar and G.Britto Antony Xavier, Some Applications of the Generalized Difference Operator of the nth Kind, Far East Journal of Applied Mathematics, 66(2) (2012), 107 - 126.
[6] Mitra, S.K. Digital Signal Processing: A Computer Based Approach, Fourth Edition., New York: McGraw-Hill, 2006. [7] Smekal, Z., Sysel, P. Digital Signal Processors and their Applications, Prague: Sdelovací technika s.r.o, 2006 (In Czech). [8] Zdenek Smekal, Difference Equations with Forward and Backward Differences and their Usage in Digital Signal Processor Algorithms , Radio Engineering, Vol. 15, No. 2, June 2006.
[9] A.A.Kilbas, H.M.Srivastava and J.J.Trujillo, Theory and application of fractional differential equations, Elsevier, Amsterdam, 2006.
[10] K.S.Miller,B.Ross , An introduction to the fractional calculus and fractional differential equations, Wiley, New york, 1993. [11] S.G.Samko, A.A.Kibas and O.I.Marichev, Fractional integrals and derivatives, Translated from the 1987 Russian original, Gordon and Breach, Yverdon, 1993.
