INTEGERS 15 (2015)
#A48
SOME APPLICATIONS OF SYMMETRIC FUNCTIONS
Ali Boussayoud 1
Laboratory LMAM, Department of Mathematics, University of Jijel, Jijel, Algeria
aboussayoud@yahoo.fr
Abdelhamid Abderrezzak
Laboratory LITP, University of Paris 7, Paris, France
abderrezzak.abdelhamid@neuf.fr
Mohamed Kerada
Laboratory LMAM, Department of Computer Science, University of Jijel, Jijel,
Algeria
mkerada@yahoo.fr
Received: 11/12/12, Revised: 8/19/15, Accepted: 10/25/15, Published: 11/13/15
Abstract
In this paper, we propose an alternative approach for the determination of the
Fibonacci numbers and some results of Foata, Ramanujan and other results on
Tchebychev polynomials of the first and second kinds. This P
approach is based on
1
the action of the symmetrizing operator Lke1 e2 on the series j=0 aj z j . Obtained
results confirm the e↵ectiveness of the proposed approach.
1. Preliminaries and Notation
Here, we recall some basic definitions and theorems that are needed in the sequel.
Given two sets of indeterminates A and B (called alphabets), we define a symmetric
function Sj (A B) as in [1]:
⇧b2B (1
⇧a2A (1
zb)
za)
=
=
1
X
j=0
1
X
Sj (A
B)z j
Sj (A)z j
j=0
1
X
Sj ( B)z j ,
j=0
where Sj (A B) = 0 for j < 0.
If A = B, then formula (1) can be written as
1=
1
X
j=0
1 Corresponding
author
Sj (A)z j
1
X
j=0
Sj ( A)z j ;
(1)
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INTEGERS: 15 (2015)
therefore,
1
X
j=0
On the other hand, if A =
1
X
Sj (A)z j = P
1
1
.
Sj ( A)z j
j=0
and F = B + D, then formula (1) becomes
Sj ( F )z j =
j=0
1
X
Sj ( B)z j
j=0
1
X
Sj ( D)z j ,
j=0
which can be rewritten as
Sj ( F ) =
j
X
Sj
k(
k=0
B)Sk ( D), for all j 2 N.
Definition 1. [7] Given a function g on Rn , the divided di↵erence operator is
defined as follows:
@xi xi+1 (g) =
g(x1 , · · · , xi , xi+1 , · · · xn ) g (x1 , · · · , xi , xi+1 , · · · xn )
,
xi xi+1
where g is given by
g (x1 , · · · , xi , xi+1 , · · · xn ) = g(x1 , · · · xi
1 , xi+1, xi , xi+2
· · · xn ).
Definition 2. [2] The symmetrizing operator Lkxy is defined by
Lkxy f =
xk+1 f y k+1 f
, for all k 2 N.
x y
(2)
2. The Main Results
Proposition 1. [2] Let E be an alphabet such that E = {e1 , e2 } . The operator
Lke1 e2 is defined as follows:
Lke1 e2 f (e1 ) = Sk
+ e2 )f (e1 ) + ek2 @e1 e2 f (e1 ), for all k 2 N.
P1
j
Theorem 1. Given an alphabet E = {e1 , e2 }, and two series
j=0 aj z and
P1
P
P
1
1
j
j
j
j=0 bj z such that (
j=0 aj z )( j=0 bj z ) = 1, we have
1
X
aj Sk+j
1 (e1
1 (e1
+ e2 )z j
j=0
=
kP1
j=0
bj ej1 ej2 Sk
j 1 (e1
+ e2 )z j
1
P
j=0
ek1 ek2 z k+1
bj ej1 z j
1
P
j=0
1
P
j=0
bj ej2 z j
bj+k+1 Sj (e1 + e2 )z j
.
(3)
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Proof. Let
1
P
1
P
aj z j and
j=0
bj z j be two sequences such that
j=0
1
P
aj z j =
j=0
The left-hand side of the formula (3) can be written as:
0
1
1
1
X
X
Lke1 e2 f (e1 ) = Lke1 e2 @
aj ej1 z j A =
aj Sk+j
j=0
1 (e1
1
P
j=0
+ e2 )z j ,
j=0
while the right-hand side can be expressed as:
Sk
=
1 (e1
+ e2 )f (e1 ) + ek2 @e1 e2 f (e1 )
Sk 1 (e1 + e2 )
1
+ ek2 @e1 e2 P
1
1
P
bj ej1 z j
bj ej1 z j
j=0
=
=
j=0
1
P
Sk 1 (e1 + e2 )
1
P
bj ej1 z j
j=0
1
P
j=0
h
bj ej2 Sk
j=0
1 (e1
1
P
j=0
=
kP1
j=0
h
bj ej2 Sk
1
P
+
j=k+1
h
bj ej2 Sk
1
P
j=0
=
kP1
j=0
bj ej1 ej2 Sk
1
P
j=0
+ e2 )
bj ej1 z j
1
P
bj ej2 z j
1 (e1
bj ej2 z j
1 (e1
bj ej2 z j
+ e2 )
1
P
j=0
ek2 Sj
i
+ e2 ) z j
i
+ e2 ) z j
1 (e1
bj ej2 z j
i
+ e2 ) z j
1
P
ek1 ek2 z k+1
bj+k+1 Sj (e1 + e2 )z j
j=0
!
!
,
1
1
P
P
j j
j j
bj e1 z
bj e2 z
+ e2 )z j
j=0
and the proof is complete.
+ e2 )z j
j=0
ek2 Sj
j=0
bj ej1 z j
j 1 (e1
1
P
ek2 Sj
+ e2 )
1 (e1
1 (e1
bj ej1 z j
bj ej1 z j
1 (e1
j=0
1
P
bj Sj
j=0
1
P
j=0
1
bj z j
.
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3. Applications
P1
3.1. The Case j=0 z j =
1
1
z
Corollary 1. Given an alphabet E = {e1 , e2 } , we have
1
X
Sk+j
1 (e1
+ e2 )z j =
Sk
1 (e1
j=0
If k = 1, then
1
X
+ e2 ) e1 e2 Sk 2 (e1 + e2 )z
, for all k 2 N.
(1 ze1 ) (1 ze2 )
Sj (e1 + e2 )z j =
j=0
(1
1
ze1 ) (1
ze2 )
.
(4)
Replacing e2 by ( e2 ) in (4), we obtain
1
X
Sj (e1 + [ e2 ])z j =
j=0
1
.
ze1 ) (1 + ze2 )
(1
(5)
Choosing e1 and e2 such that
⇢
e1 e2 = 1,
e1 e2 = 1,
and substituting in (5) we end up with
1
X
Sj (e1 + [ e2 ])z j =
j=0
1
1
,
z z2
which represents a generating function for Fibonacci numbers, such that Fj =
Sj (e1 + [ e2 ]).
On the other hand, when replacing e1 and e2 by 2e1 and ( 2e2 ), respectively, in
(5), and under the condition 4e1 e2 = 1, we obtain, for y = e1 e2 ,
1
X
Sj (2e1 + [ 2e2 ])z j =
j=0
1
1
,
2yz + z 2
(6)
which represents a generating function for Tchebychev polynomials of the second
kind, such that Uj (y) = Sj (2e1 + [ 2e2 ]). Moreover, we deduce from (6) that
1
X
j=0
[Sj (2e1 + [ 2e2 ])
ySj
1 (2e1
+ [ 2e2 ])] z j =
1
1 yz
,
2yz + z 2
which represents a generating function for Tchebychev polynomials of the first kind,
such that
Tj (y) = [Sj (2e1 + [ 2e2 ]) ySj 1 (2e1 + [ 2e2 ])] .
(7)
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3.2. The Case
1
P1
zj
j=0 j!
z
=e
Corollary 2. Given an alphabet E = {e1 , e2 } , we have
1
X
j
Sk+j
1 (e1
+ e2 )
j=0
z
j!
=
kP1
j=0
( 1)j ej1 ej2 Sk
⇧ (1
ae1 z) ⇧ (1
a2A
1
P
a2A
j=0
Sj (A)z j =
j
( 1)j+k+1 Sj (e1 + e2 ) zj!
j=0
⇧ (1
P1
j
+ e2 ) zj!
ae2 z)
a2A
ek1 ek2 z k+1
3.3. The Case
j 1 (e1
P1
ae1 z) ⇧ (1
ae2 z)
a2A
j=0
1
Sj (
, for all k 2 N.
A)z j
Corollary 3. Given two alphabets E = {e1 , e2 } and A = {a1 , a2 , . . .} , we have
1
X
Sj (A)Sk+j
1 (e1
+ e2 )z
j
=
j=0
kP1
j=0
Sj ( A)ej1 ej2 Sk
⇧ (1
ae1 z) ⇧ (1
a2A
ek1 ek2 z k+1
j 1 (e1
a2A
1
P
+ e2 )z j
ae2 z)
Sj+k+1 ( A)Sj (e1 + e2 )z j
j=0
⇧ (1
a2A
ae1 z) ⇧ (1
ae2 z)
a1 a2 e1 e2 z 2
ae1 z)⇧a2A (1
ae2 z)
a2A
. (8)
If k = 1 and A = {a1 , a2 } , then
1
X
Sj (a1 + a2 )Sj (e1 + e2 )z j =
j=0
1
⇧a2A (1
.
(9)
Case 1: Substituting e1 = a1 = 1, e2 = x and a2 = y in (9), we obtain the following
identity of Ramanunja [4]:
1
X
Sj (1 + x)Sj (1 + y)z j =
j=0
(1
z) (1
1 xyz 2
zx) (1 zy) (1
zxy)
.
Case 2: Replacing e2 by ( e2 ) and a2 by ( a2 ) in (9) yields
1
X
Sj (a1 + [ a2 ])Sj (e1 + [ e2 ])z j
j=0
=
(1
1 e1 e2 a1 a2 z 2
a1 e1 z) (1 + a2 e1 z) (1 + a1 e2 z) (1
This case consists of three related parts.
a2 e2 z)
.
(10)
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Firstly, the substitutions of
⇢
a1 + a2 = 0,
a1 a2 = 1,
⇢
and
e1 e2 = 1,
e1 e2 = 1,
in (10) give
1
X
Sj (a1 + [ a2 ])Sj (e1 + [ e2 ])z j
=
1
j=0
z
1
X
=
1 z2
4z 2 z 3 + z 4
Fj2 z j ,
j=0
which represents a generating function for Fibonacci numbers of second order (see
[6]), such that Fj2 = Sj (a1 + [ a2 ])Sj (e1 + [ e2 ]).
Secondly, the substitution of
8
< e1 e2 = 1,
e1 e2 = 1,
:
4a1 a2 = 1,
in (10) and set for ease on notations x = a1
a2 , we reach
1
1
X
1 + z2
=
Fj Uj (x)z j ,
2xz + (3 4x2 )z 2 + 2xz 3 + z 4
j=0
which corresponds to a generating function for the combined Fibonacci numbers
and Tchebychev polynomials of the second kind.
In the last case, recall that for y = e1 e2 , the substitution of
⇢
4e1 e2 = 1,
4a1 a2 = 1,
in (10) results in
1
X
Uj (y)Uj (x)z j =
j=0
1
4yxz +
(4x2
1 z2
+ 4y 2 2)z 2
4yxz 3 + z 4
,
which represents a generating function for Tchebychev polynomials of the second
kind.
According to formulas (7), (8), and to the fact that
j
Sj
1 (2a1 + [ 2a2 ]) =
j
(2a1 )
( 2a2 )
,
2a1 + 2a2
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we have
1
X
Uj (y)Tj (x)z j =
j=0
1
1 2yxz + (2x2 1)z 2
,
4yxz + (4x2 + 4y 2 2)z 2 4yxz 3 + z 4
which represents a generating function for the combined Tchebychev polynomials
of the second and first kinds.
Finally, we have
1
X
j=0
Tj (y) Tj (x)z j =
1 3yxz + (2x2 + 2y 2 1)z 2 yxz 3
,
1 4yxz + (4x2 + 4y 2 2)z 2 4yxz 3 + z 4
which corresponds to a generating function for Tchebychev polynomials of the first
kind.
Acknowledgements We would like to thank the anonymous referees for their
valuable comments and suggestions.
References
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(1983), 118–144.
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