c
Applied Mathematics E-Notes, 9(2009), 34-39 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
Some Discrete Representations Of q-Classical Linear
Forms∗
Olfa Fériel Kamech†, Manoubi Mejri‡
Received 25 September 2007
Abstract
We give a discrete measure for some Hq -classical forms and some consequent
summation formulas.
1
Introduction and Preliminaries
In [4], Hq -classical orthogonal polynomials are exhaustively described and integral or
discrete representations of corresponding regular forms are given, except in some cases
where the problem remains open (see also [3] for the Hq -semiclassical case). So, the aim
of this contribution is to establish discrete representations of two canonical situations
in [4] which are the q-analogous of Hermite (for 0 < q < 1, q > 1) and the q-analogous
of Laguerre (for q > 1).
Let P be the vector space of polynomials with coefficients in C and let P 0 be its
dual. We denote by hu, fi the action of u ∈ P 0 on f ∈ P. In particular, for any f ∈ P,
we let fu , be the form defined by duality hfu, pi := hu, fpi, p ∈ P.
Let hδc , pi = p(c), c ∈ C, p ∈ P.
The form u is called regular if we can associate with it a sequence {Pn }n≥0 of monic
polynomials, deg Pn = n , n ≥ 0 such that
hu, PmPn i = rn δn,m , n, m ≥ 0 ; rn 6= 0 , n ≥ 0.
The sequence {Pn }n≥0 is orthogonal with respect to u and fulfils the standard recurrence relation:
(
P0 (x) = 1 , P1 (x) = x − β0 ,
(1)
Pn+2 (x) = (x − βn+1 )Pn+1 (x) − γn+1 Pn (x), n ≥ 0
∗ Mathematics
Subject Classifications: 42C05, 35C45.
of Mathematics, Instiut Preparatoire aux Etudes d’Ingenieurs EL Manar 2090 EL
Manar, B.P 244 Tunis Tunisia
‡ Department of Mathematics, Institut Superieur Des Sciences Appliquees et de Technologie Rue
Omar Ibn EL Khattab Gabes 6072. Tunisia. e-mail: mejri− manoubi@yahoo.fr
† Department
34
35
O. F. Kamech and M. Mejri
with γn+1 6= 0, n ≥ 0.
The form u is said to be normalized if (u)0 = 1 where in general (u)n = hu, xn i, n ≥
0, are the moments of u. In this paper we suppose that any form will be normalized.
Let us introduce the Hahn’s operator
(Hq f)(x) :=
e := C− {0} ∪
where C
By duality we have
f(qx) − f(x)
e
, f ∈ P , q ∈ C,
(q − 1)x
∪ {z ∈ C, z n = 1}
n≥0
.
hHq u, fi = −hu, Hq fi, u ∈ P 0 , f ∈ P.
DEFINITION. A form u is called Hq - classical when it is regular and there exists
two polynomials φ (monic) and ψ with deg(φ) ≤ 2, deg(ψ) = 1 such that
Hq (φu) + ψu = 0.
(2)
The corresponding orthogonal sequence {Pn }n≥0 is called Hq -classical.
We are going to use the following notations and results [1,2,5]

n = 0,

 1,
n−1
(a; q)n =
Q

(1 − aq k ), n ≥ 1.

(3)
k=0
(a; q)n = (−1)n an q
n(n−1)
2
(a; q)∞ =
(a; q)n =
+∞
Y
(a−1 ; q −1 )n , n ≥ 0, a, q 6= 0.
(1 − aq k ), | q |< 1.


(aq −1 q n ; q −1 )∞


, | q |> 1.
(aq −1 ; q −1 )∞
(z; q)∞ =
k=0
(6)
k(k−1)
(−1)k q 2
(q; q)k
z k , | q |< 1.
+∞
X 1
1
=
z k , | q |< 1 , | z |< 1.
(z, q)∞
(q; q)k
k=0
(5)
k=0

(a; q)∞



 (aq n ; q)∞ , | q |< 1,
+∞
X
(4)
(7)
(8)
36
Discrete Representation of Classical Functional
2
Discrete measure for some Hq -classical forms
2.1
Consider the symmetric Hq -classical linear form u which is the q-analog of Hermite
functional. We have [4]


 βn = 0, n ≥ 0,


1 − q n+1 n
(9)
γ
=
q , n ≥ 0,
n+1

2(1 − q)



Hq (u) + 2xu = 0.
 √
Z

(q −2 ; q −2 )∞ +∞
f(x)
2


dx, f ∈ P, q > 1,
(q − 1)1/2 −1 −2


(q ; q )∞ −∞ −2(q − 1)x2 ; q −2 ∞
 π
hu, fi =
Z + √1

q 2(1−q)


2q 2 (1 − q)x2 ; q 2 ∞ f(x)dx, f ∈ P, 0 < q < 1,
 K1

1

− √
q
2(1−q)
(10)
with
K1 =
1
2
Z
1
2(1−q)
+ √
q
0
(u)2n =
2q 2 (1 − q)x2 ; q 2
dx
∞
−1
.
1 (q; q 2 )n
, (u)2n+1 = 0, n ≥ 0.
2n (1 − q)n
(11)
(12)
PROPOSITION 1. We have the following discrete representations:
For f ∈ P, q > 1
2
+∞
X
1
iq k
(−1)k q −k −iq k p
p
+
f
.
hu, fi =
f
2(q −1 ; q −2 )∞
(q −2 ; q −2 )k
2(q − 1)
2(q − 1)
(13)
k=0
For f ∈ P, 0 < q < 1
hu, fi = 2−1 (q; q 2 )∞
+∞
X
k=0
o
qk n
−q k
qk
p
p
+
f
.
f
(q 2 ; q 2 )k
2(1 − q)
2(1 − q)
PROOF. Let q > 1 by (6), equation (12) becomes
(u)2n =
2n (1
1
(q 2n−1 ; q −2 )∞
, n ≥ 0.
n
− q) (q −1 ; q −2 )∞
On account of (7), we get
(u)2n =
2
+∞
2n
X
1
(−1)k q −k iq k
p
, n ≥ 0.
(q −1 ; q −2 )∞
(q −2 ; q −2 )k
2(q − 1)
k=0
(14)
37
O. F. Kamech and M. Mejri
Therefore
(u)2n =
D
2
+∞
E
X
1
(−1)k q −k
2n
δ
k
,
x
, n ≥ 0.
iq
(q −1 ; q −2 )∞
(q −2 ; q −2 )k √2(q−1)
k=0
But (u)2n+1 = 0, n ≥ 0, yields to
(u)n = hu, xni =
D
2
+∞
o
E
X
1
(−1)k q −k n
n
+
δ
k
,
x
, n ≥ 0.
δ
k
iq
−iq
√
2(q −1 ; q −2 )∞
(q −2 ; q −2 )k √2(q−1)
2(q−1)
k=0
Consequently
u=
2
+∞
o
X
1
(−1)k q −k n
δ √−iq k + δ √ iq k
.
−1
−2
−2
−2
2(q ; q )∞
(q ; q )k
2(q−1)
2(q−1)
k=0
Then we get the desired result (13).
When 0 < q < 1, by virtue of (6), equation (12) becomes
(u)2n =
(q; q 2 )∞
, n ≥ 0,
2n (1 − q)n (q 2n+1 ; q 2 )∞
on account of (8), it follows that
2
(u)2n = (q; q )∞
2n
qk
qk
p
, n ≥ 0.
(q 2 ; q 2 )k
2(1 − q)
+∞
X
k=0
Then
+∞
D
X
(u)n = 2−1 (q; q 2 )∞
k=0
qk
(q 2 ; q 2 )k
Consequently, we are lead to
u = 2−1 (q; q 2 )∞
+∞
X
k=0
n
δ √ −q k
2(1−q)
+ δ√
qk
2(1−q)
o
E
, xn , n ≥ 0.
n
o
qk
δ
+
δ
.
−q k
qk
√
(q 2 ; q 2 )k √2(1−q)
2(1−q)
(15)
Hence (14).
2.2
Consider the q-analogous of Laguerre linear form u given in [4,pp 68] .We have

β = {1 − (1 + q)q n }q n−1 , n ≥ 0,


 n
γn+1 = (q n+1 − 1)q 3n , n ≥ 0,



Hq (xu) − (q − 1)−1 (x + 1)u = 0.
(16)
38
Discrete Representation of Classical Functional
For q > 1, we have the following representations [4]:

Z 0
ln2 | x | 
−3/2
−1/2 −1/8

(2π
ln
q)
q
f(x)dx, f ∈ P,
|
x
|
exp
−


2 ln q

−∞
hu, fi =
+∞
−k 2

X

s(k)

k q

(−1)
f(−q k ), f ∈ P,

−1
(q ; q −1 )k
(17)
k=0
where
s(k) =
+∞
X
m=0
1
q −( 2 m(m+1)+km)
(u)φm+k , k ≥ 0,
(q −1 ; q −1 )m
(18)
and (u)φ2n = (q − 1)n , (u)φ2n+1 = 0, n ≥ 0.
The moments of u are given by the following formulas:
1
(u)n = (−1)n q 2 n(n−1), n ≥ 0.
(19)
PROPOSITION 2. The form u possesses the following discrete representation:
For f ∈ P, q > 1
(−1; q −1 )∞ (−q −1 ; q −1 )∞ hu, fi =
+∞
X
q−
k(k−1)
2
k
X
µ=0
k=0
2
q −µ +(k−1)µ
f −q 2µ−k ,
−1
−1
−1
−1
(q ; q )µ (q ; q )k−µ
(20)
PROOF. From (4), for (19) we obtain
(u)n = (−1)n
(−1; q)n
, n ≥ 0.
(−1; q −1 )n
(21)
Let q > 1, taking (6) into account, equation (21) can be written in the following way
(u)n =
(−1)n
(−1; q −1 )∞ (−q −1 ; q −1 )∞
(−q n−1 ; q −1 )∞ (−q −n ; q −1 )∞ , n ≥ 0.
In accordance of (7), we get
k(k−1)
k(k−1)
+∞
+∞
X
X
(−1)n
q− 2
q− 2
k(n−1)
q
q −kn , n ≥ 0.
(u)n =
(−1; q −1 )∞ (−q −1 ; q −1 )∞
(q −1 ; q −1 )k
(q −1 ; q −1 )k
k=0
k=0
Using the Cauchy product, the last expression becomes (for n ≥ 0)
(u)n =
2
+∞
k
X
n
k(k−1) X
1
q −µ +(k−1)µ
−
2
q
−q 2µ−k .
−1
−1
−1
−1
−1
−1
−1
(−1; q )∞ (−q ; q )∞
(q ; q )µ (q ; q )k−µ
µ=0
k=0
Then, the discrete measure in (20) is deduced.
Acknowledgment. We would like to thank the referee for his valuable review.
O. F. Kamech and M. Mejri
39
References
[1] T. S. Chihara, An introduction to Orthogonal Polynomials, Gordon and Breach,
New York, 1978.
[2] G. Gasper, M. Rahman, Basic Hypergeometric Series, Cambridge University Press,
Cambridge, 1990.
[3] A. Ghressi and L. Khériji, Orthogonal q-polynomials related to perturbed linear
form, Appl. Math. E-Notes, 7 (2007) 111-120.
[4] L. Khériji and P. Maroni, The Hq -classical orthogonal polynomials, Acta Appl.
Math., 71 (2002) 49-115.
[5] R. Koekoek and R. F Swarttow, The ASkey-scheme of hypergeometric orthogonal
polynomials and its q-analogue, Report 98-17, TU Delft, 1998.