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Journal of Computational and Applied Mathematics 107 (1999) 133–146
www.elsevier.nl/locate/cam

On the linearization problem involving Pochhammer symbols
and their q-analogues
a;b;∗

R. Alvarez-Nodarse
, N.R. Quinteroc, A. Ronveauxd
a

b

Departamento de Analisis Matematico, Universidad de Sevilla, Apdo, 1160, E-41080, Sevilla, Spain
Instituto CARLOS I de Fsica

Teorica y Computacional Universidad de Granada, E-18071 Granada, Spain
c
Departamento de Matematicas, Universidad Carlos III de Madrid, Ave. Universidad 30, 28911, Leganes,
Madrid, Spain
d

Mathematical Physics, Facultes Universitaires Notre-Dame de la Paix, B-5000 Namur, Belgium
Received 28 July 1998; received in revised form 9 February 1999

Abstract
In this paper we present a simple recurrent algorithm for solving the linearization problem involving some families
c 1999
of q-polynomials in the exponential lattice x(s) = c1 q s + c3 . Some simple examples are worked out in detail.
Elsevier Science B.V. All rights reserved.
MSC: 33C45; 33D45
Keywords: Pochhammer and q-Pochhammer symbols; Linearization problem

1. Introduction
Given three families of polynomials, denoted by Pn (x), Qm (x) and Rj (x), of degree exactly equal
to, respectively, n, m and j, the linearization problem asks to compute the so-called linearization
coecients Lmjn dened by the relation
Qm (x)Rj (x) =

m+j
X

Lmjn Pn (x):

n=0

The study of such a problem has attracted lot of interest in the last few years. Special emphasis
was given to the classical continuous (Hermite, Laguerre, Jacobi and Bessel) [8,9,10,11,15,19–21,24,
∗

Corresponding author. Fax: 341-6249430.

E-mail addresses: renato@gandalf.ugr.es (R. Alvarez-Nodarse),
kinter@math.uc3m.es (N.R. Quintero), andre.ronveaux
@fundp.ac.be (A. Ronveaux)
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ - see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 8 7 - 4

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R. Alvarez-Nodarse
et al. / Journal of Computational and Applied Mathematics 107 (1999) 133–146

26–28] and the discrete cases (Charlier, Meixner, Kravchuk and Hahn) [7,9,12,14,27]. The main aim
of the present paper is to show that the ideas given in [14,26] can be extended in a very easy way
to the q-polynomials on the exponential lattice x(s) = c1 q s + c3 . If fact, if Pn (x), Qm (x) and Rj (x) are
polynomials in x(s) = c1 q s + c3 , then it is possible to nd a recurrence relation for the linearization
coecients Lmjn , which is an alternative approach to the one given in [5]. This approach requires the
knowledge of the so-called structure relation and three-term recurrence relation for the polynomials
Pn (x) as well as the second-order dierence equation which satisfy the other two polynomials Qm (x)
and Rj (x).
In this paper we will present an algorithm for computing the above coecients, showing, as an
example, the case when the polynomials Pn (x), Qm (x) and Rj (x) are the three Pochhammer and
q-Pochhammer symbols. In both cases the solutions are given explicitly. The obtained expression
can be used for solving more “complicated” examples.
The structure of the paper is as follows. In Section 2, a general algorithm for nding a recurrence
relation for the Lmjn , is presented. In Section 3, two special cases, of particular importance, are
worked out, namely, the linearization of a product of two q-Pochhammer or q-Stirling polynomials
in terms of single q-Pochhammer or q-Stirling polynomials, respectively. Finally, some more complicated examples involving the Charlier polynomials and their q-analogues in the lattice x(s) = q s
are presented.

2. General algorithm for solving the linearization problem
In this section we will present a general algorithm to nd a recurrence relation for the linearization
coecients Lmjn in the expansion
Qm (x(s))Rj (x(s)) =

m+j
X

Lmjn Pn (x(s));

x(s) = c1 q s + c3 ;

(2.1)

n=0

where c1 , c3 and q are constants, Qm (x(s)) ≡ Qm (s)q and Rj (x(s)) ≡ Rj (s)q are polynomials which
satisfy a second-order dierence equation of the form
a(s)Qm (s + 1)q + b(s)Qm (s)q + c(s)Qm (s − 1)q = 0

(2.2)

(s)Rj (s + 1)q + (s)Rj (s)q +
(s)Rj (s − 1)q = 0;

(2.3)

and
respectively. A special case of such polynomials are the q-polynomials of hypergeometric type
[3,22,23], which satisfy the dierence equation
3y(s)

y(s)


(s)
+ (s)
+ y(s) = 0;

x(s − 1=2) 3x(s)

x(s)
(2.4)
3f(s) = f(s) − f(s − 1);

f(s) = f(s + 1) − f(s):
Obviously, the Eq. (2.4) is of type (2.2) (y ≡ Qm ), with

(s)
;
a(s) = (s) + (s)
x(s − 21 );
c(s) = (s)
3x(s)
b(s) = 
x(s −

1
)
x(s)
2

− a(s) − c(s):

(2.5)


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135

Notice that when x(s) = c1 q s + c3 , 3x(s),
x(s) and
x(s − 21 ) are polynomials of rst degree in
x(s), then, in this case, the functions a, b, c, ,  and
in (2.2) and (2.3) are polynomials of second
degree at most.
In the following, we will use the operators T and I dened as follows:
T : P → P;

I : P → P;

Tp(s) = p(s + 1);

Ip(s) = p(s):

Using the above operators, we can rewrite the Eqs. (2.2) and (2.3) in the form
a(s + 1)T2 Qm (s)q + b(s + 1)TQm (s)q + c(s + 1)IQm (s)q = 0

(2.6)

(s + 1)T2 Rj (s)q + (s + 1)TRj (s)q +
(s + 1)IRj (s)q = 0:

(2.7)

and

Other important examples of q-polynomials are the so-called q-Pochhammer symbols and the
q-Stirling polynomials (see Section 3 of the present work for more details) which satisfy a rst-order
dierence equation of type (2.6) and (2.7).
2.1. The fourth-order dierence equation for the polynomials Qm (s)q Rj (s)q
Using the two Eqs. (2.6) and (2.7), it is easy to prove that the polynomials u(s)q ≡ Qm (s)q Rj (s)q ,
satisfy a fourth-order dierence equation of the form
L4 u(s) ≡ p4 (s)T4 u(s)q + p3 (s)T3 u(s)q + p2 (s)T2 u(s)q + p1 (s)Tu(s)q + p0 (s)Iu(s)q : (2.8)
To prove this, we will follow the works [13,14]. The idea is the following. Since from (2.6) to
(2.7),
a(s + 1)(s + 1)T2 u(s)
= [b(s + 1)TQm (s)q + c(s + 1)IQm (s)q ][(s + 1)TRj (s)q +
(s + 1)IRj (s)q ];

which can be rewritten as
L2 u(s) ≡ a(s + 1)(s + 1)T2 u(s) − b(s + 1)(s + 1)Tu(s) − c(s + 1)
(s + 1)Iu(s)
= b(s + 1)
(s + 1)[TQm (s)q IRj (s)q ] + c(s + 1)(s + 1)[IQm (s)q TRj (s)q ]
= l1 (s)[TQm (s)q IRj (s)q ] + l2 (s)[IQm (s)q TRj (s)q ]:
Next, we change in the last expression s → s + 1, and substitute in the right-hand side the expression
T2 Qm (s)q and T2 Rj (s)q , using the Eqs. (2.6) and (2.7), respectively. This allows us to rewrite the
resulting expression in the form
M3 u(s) = m1 (s)[TQm (s)q IRj (s)q ] + m2 (s)[IQm (s)q TRj (s)q ];
where M3 is a dierence operator of third order (there is one term proportional to T3 ), m1 and m2
are known functions of s. Repeating the same procedure, but now starting from the above equation

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R. Alvarez-Nodarse
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we obtain

N4 u(s) = n1 (s)[TQm (s)q IRj (s)q ] + n2 (s)[IQm (s)q TRj (s)q ]:
Then


 L2 u(s) l1 (s) l2 (s) 




 M3 u(s) m1 (s) m2 (s)  = 0:




 N u(s) n (s) n (s) 
4
1
2

(2.9)

Expanding the determinant from the rst column, Eq. (2.8) holds.
Remark. The above Eq. (2.9), and its proof, remains true for any lattice function x(s) and not only
for the exponential lattice x(s) = c1 q s + c3 .
2.2. The generalized linearization algorithm
As before, we will suppose that Qm (s)q and Rj (s)q satisfy Eqs. (2.6) and (2.7), respectively, and
that Pn (s)q satises the so-called structure relation in the exponential lattice x(s) = c1 q s + c3 and a
three-term recurrence relation. An example of the latest are, again, the q-hypergeometric orthogonal
polynomials in the aforementioned lattice which satises the structure relation [4] (for the case
x(s) = q s see [6])




1
(s) + (s)
x s −
2




Pn (s)q
= Sn Pn+1 (s)q + Tn Pn (s)q + Rn Pn−1 (s)q ;

x(s)

(2.10)

or, written in its equivalent form
n+2
X

(s)TPn (s)q =



(s) = (s) + (s)
x s −

Ak (n)Pk (s)q ;

k=n−2



(2.11)

n¿0:

(2.12)

1
;
2

as well as the three-term recurrence relation (TTRR)
P−1 (x) ≡ 0;

x(s)Pn (s)q = n Pn+1 (s)q + n Pn (s)q +
n Pn−1 (s)q ;

To obtain (2.11) from (2.10) we use the facts that
x(s) is a polynomial of rst degree in x(s)
(which is not valid in general for any lattice x(s)), (s) + (s)
x(s − 21 ) is a polynomial of degree
two in x(s), and the TTRR (2.12).
From the above expression (2.11), one easily obtains that
2

(s)(s + 1)T Pn (s)q =

n+4
X

Ãk (n)Pk (s)q ;

k=n−4
3

(s)(s + 1)(s + 2)T Pn (s)q =

n+6
X

Âk (n)Pk (s)q ;

k=n−6

(s)(s + 1)(s + 2)(s + 3)T4 Pn (s)q =

n+8
X

k=n−8

Ak (n)Pk (s)q :

(2.13)


R. Alvarez-Nodarse
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137

To obtain a recurrence relation for the linearization coecients we can do the following:
Since (2.8), L4 Qm (s)q Rj (s)q = 0, then applying L4 to both sides of (2.1), we nd
0=

m+j
X

Lmjn (s)(s + 1)(s + 2)(s + 3)L4 Pn (x(s)):

n=0

Taking into account that L4 is a fourth degree operator with polynomial coecients, and using the
structure relation (2.11) as well as (2.13) we nd
0=

m+j
X
n=0

(

Lmjn p4 (s)

n+8
X

Ak (n)Pk (s)q + p3 (s)(s + 3)

k=n−8

+ p2 (s)(s + 2)(s + 3)

n+6
X

Âk (n)Pk (s)q

k=n−6
n+4
X

Ãk (n)Pk (s)q

k=n−4

+ p1 (s)(s + 1)(s + 2)(s + 3)

n+2
X

)

Ak (n)Pk (s)q + p0 (s)Pn (s)q ;

k=n−2

from where, and by taking into account that (s + k), k = 0; 1; 2; 3, is a polynomial of degree two in
x(s) = c1 q s + c3 , as well as the TTRR (2.12) we obtain that the coecients Lmjn satisfy a recurrence
relation of the form
r
X

ck (i; j; n)Lmjn+k = 0:

(2.14)

k=0

In general, the present algorithm may not give the minimal order recurrence for the linearization
coecients. To get the order r minimal it is necessary to use more specic properties of the families
of polynomials involved in (2.1).
Remark. Notice that the present algorithm also works in the case when the product Qm (x(s))Rj (x(s))
satises a kth-linear dierence equation with polynomial coecients (not necessary of order 4 as in
(2.9)), so it can be used for solving more general linearization problems, e.g., linearization problems
involving the product of three or more q-polynomials.
3. Examples
In this section we will work out some examples. For simplicity we will consider the case when
the involved polynomials Qm and Rj satisfy rst-order dierence equations, i.e., equations of form
(2.6) and (2.7) with a(s) = (s) ≡ 0.
3.1. Linearization of a product of two q-Pochhammer symbols
Let us dene the quantities (s)q by
(s)q =

qs − 1
= q s=2−1 [s]q
q−1

(3.1)

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R. Alvarez-Nodarse
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and let [(s)q ]n , the q-Pochhammer symbol, be dened by
[(s)q ]n = (s)q (s + 1)q · · · (s + n − 1)q =

n−1
Y
k=0

q s+k − 1
:
q−1

(3.2)

Notice that [(s)q ]n is a polynomial of degree exactly equal to n in q s . The polynomials [(s)q ]n satisfy
the following dierence equation:
(s)q [(s + 1)q ]n − (s + n)q [(s)q ]n = 0

(3.3)

and a recurrence relation
(s)q [(s)q ]n − q−n [(s)q ]n+1 + q−n (n)q [(s)q ]n = 0:

(3.4)

Notice also that
[(s)q ]n =

(q s ; q)n
;
(1 − q)n

where (a; q)n =

n−1
Y

(1 − aq k ):

(3.5)

k=0

Obviously, in the limit q → 1, the q-Pochhammer symbol [(s)q ]n , becomes into the classical
Pochhammer symbol (s)n = s(s + 1) · · · (s + n − 1).
Since the product [(s)q ]i [(s)q ]j is a polynomial in q s , it can be represented as a linear combination
of the single q-Pochhammer symbols [(sq )]n . In particular,
[(s)q ]i [(s)q ]j =

i+j
X

Lijn (q)[(sq )]n :

(3.6)

n=0

In order to obtain the recurrence relation for the linearization coecients Lijn in (3.6) we apply
the operator
(s)2q T − (s + i)q (s + j)q I

(3.7)

to both sides of (3.6). Using formula (3.3) we obtain the following expression:
0=

i+j
X
n=0

Lijn

"

qs − 1
q−1

2

T[(sq )]n −



q s+i − 1
q−1





#

q s+j − 1
[(sq )]n :
q−1

Taking into account Eq. (3.3) for the q-Pochhammer symbol, we nd
0=







Lijn [(sq )]n

i+j
X

Lijn [(sq )]n [(s)q (s + n)q − (s + i)q (s + j)q ]:

n=0

=



i+j
X

qs − 1
q−1

q s+n − 1
q−1

−

q s+i − 1
q−1



q s+j − 1
q−1



n=0

Next, we will rewrite the expression inside the quadratic brackets using the identity
(s + n)q = q n (s)q + (n)q

(3.8)


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139

to obtain
0=

i+j
X

Lijn [(sq )]n {(s)2q [q n − q i+j ] + (s)q [(n)q − q i (j)q − q j (i)q ] − (i)q (j)q };

n=0

from where, using Eq. (3.4), we arrive at the expression
i+j
X

Lijn {q−2n−1 [q n − q i+j ][(sq )]n+2

n=0

+ [((n)q − q i (j)q − q j (i)q )q−n − (q n − q i+j )(q−2n−1 (n + 1)q + q−2n (n)q )][(sq )]n+1
+ [(q n − q i+j )q−2n (n)2q − ((n)q − q i (j)q − q j (i)q )q−n − (i)q (j)q ][(sq )]n }
=

i+j
X

Lijn {q−2n−1 [q n − q i+j ][(sq )]n+2

n=0

− {q−n−1 (n + 1)q + q i+j−1−2n [1 + q n+j+1 (j)q + q n+i+1 (i)q − 2(n + 1)q ]}[(sq )]n+1
− q i+j−2n [(n)q − q n+j (j)q ][(n)q − q n+i (i)q ][(sq )]n } = 0:
The above equation leads us to the following three-term recurrence relation for the linearization
coecients Lijn :
An Lijn−2 + Bn Lijn−1 + Cn Lijn = 0;

(3.9)

where
An = q−2n+3 [q n−2 − q i+j ];
Bn = −q−n (n)q − q i+j+1−n [q−j (j)q + q−i (i)q − q−n (n)q − q−n+1 (n − 1)q ];

(3.10)

Cn = −q i+j [q−n (n)q − q−j (j)q ][q−n (n)q − q−i (i)q ]
with the initial conditions Lij i+j+1 = 0 and Lij i+j = q−ij .
To solve the above recurrence we apply the algorithm qHyper [1,2,25] which allows us to nd
an equivalent two-term recurrence relation for the linearization coecients. Namely,
Lijn+1 = −

q−k−1 (i + j − n)q
Lijn ;
(i − n − 1)q (j − n − 1)q

(3.11)

so that,
Lijn = (−1)i+j−n q

i(i+1)+j( j+1)−n(n+1)
2

[(−j)q ]i+j−n [(−i)q ]i+j−n
(i + j − n)q !

(3.12)

for n¿max(i; j) and vanishes otherwise.
Notice that, in the limit q → 1, the above recurrence relations (3.9)–(3.11) transform into a
two-term recurrence relation for the standard Pochhammer symbols (s)n of the form
(k − i − j − 1)Lijn−1 − (k 2 − (i + j)k + ij)Lijn = 0;

Lij i+j+1 = 0; Lij i+j = 1;

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which have the solution

i+j+n
(−j)i+j−n (−i)i+j−n

 (−1)
;
(i + j − n)!
Lijn =



n¿max(i; j);

0;

otherwise

that corresponds to (3.12) in the limit q → 1.
3.2. Linearization of a product of two q-Stirling polynomials
Let us dene the q-Stirling polynomials or q-falling factorials (s)[n]
q , by
(s)[n]
q = (s)q (s − 1)q · · · (s − n + 1)q =

n−1
Y
k=0

Also, we will use the notation
(s)[n]
q =

(q s ; q)[n]
;
(1 − q)n

q s−k − 1
:
q−1

(a; q)[n] = (1 − a)(1 − aq−1 ) · · · (1 − aq−n+1 ):

(3.13)

(3.14)

These quantities (sq )[n] are closely related to the q-Stirling numbers S̃ q (n; k); sq∗ (n; k) [29] by
formulas
(s)nq =

n
X

S̃ q (n; k)(sq )[n] ;

(s)[n]
q =

k=0

n
X

sq∗ (n; k)(s)kq

(3.15)

k=0

and satisfy the following dierence equation:
[n]
(s)q (s − 1)[n]
q − (s − n)q (s)q = 0;

(3.16)

as well as the recurrence relation
n
[n+1]
(s)q (s)[n]
− (n)q [(s)q ][n] = 0:
q − q (s)q

(3.17)

[ j]
s
Again, since the product (s)[i]
q (s)q is a polynomial in x(s) = q , it can be represented as a linear
[n]
combination of the q-falling factorials (sq ) . In particular,
[ j]
(s)[i]
q (s)q =

i+j
X

L̃ijn (q)(sq )[n] :

(3.18)

n=0

To obtain the recurrence relation for the linearization coecients L̃ijn in (3.6) we apply the operator
(s)2q T−1 − (s − i)q (s − j)q I

(3.19)

to both sides of (3.18) and do similar calculations as before but now using the equations (3.16) and
(3.17), respectively. This leads us to the expression
i+j
X

L̃ijn {q 2n+1 [q−n − q−i−j ](sq )[n+2]

n=0

+ {(n + 1)q + q n−i−j [(j)q + (i)q − (n)q − (n + 1)q ]}(sq )[n+1]
− q−i−j [(n)q − (j)q ][(n)q − (i)q ](sq )[n] } = 0;


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141

which allows us to obtain the following three-term recurrence relation for the linearization coecients
L̃ijn :
Ãn L̃ijn−2 + B̃n L̃ijn−1 + C̃ n L̃ijn = 0;

(3.20)

where
Ãn = q 2n−3 [q−n+2 − q−i−j ];
B̃n = (n)q + q n−i−j−1 [(j)q + (i)q − (n − 1)q − (n)q ];

(3.21)

C̃ n = −q−i−j [(n)q − (j)q ][(n)q − (i)q ]
with the initial conditions L̃iji+j+1 = 0 and L̃iji+j = q ij .
Again, applying to the above recurrence relation the algorithm qHyper [1,2,25] we nd that the
coecients L̃ijn satisfy an equivalent two-term recurrence relation
L̃ijn+1 =

q(i + j − n)q
L̃ijn ;
(i − n − 1)q (j − n − 1)q

(3.22)

so that,
L̃ijn = q i+j+ij−n

[(−j)q ]i+j−n [(−i)q ]i+j−n
(i + j − n)q !

for n¿max(i; j)

(3.23)

and vanishes otherwise.
We want to point out here that the above recurrence relation (3.20), as well as its solution (3.23),
can be obtained from the previous Eqs. (3.9) and (3.12). In order to do that we use the identity
n −n
(s)[n]
q = (−1) q [(−s)q−1 ]n

(3.24)

and substitute it in (3.6). Comparing the obtained expression with (3.18) we nd
L̃ijn (q) = (−1)i+j−n q n−i−j Lijn (q−1 ):

(3.25)

The above three-term recurrence relation (3.20) in the limit q → 1, transforms into a two-term
recurrence relation for the classical Stirling polynomials (s)[n] = s(s − 1) · · · (s − n + 1), of the form
(j + i + 1 − k)L̃ijn−1 − (k 2 − (i + j)k + ij)L̃ijn = 0;
which have the solution

(−j)i+j−n (−i)i+j−n


;
(i + j − n)!
L̃ijn =



0

n¿max(i; j);

Liji+j+1 = 0;

Liji+j = 1;

(3.26)

(3.27)

otherwise

and that is in agreement with (3.23).
Remark. To conclude this section we want to remark that the method presented here can be used
also for solving the linearization problem
{[(s)q ]i }m =

i+m
X
m=0

Lmin (q)[(sq )]n ;

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or
m
[(s)[i]
q ] =

i+m
X

m

L̃in (q)(sq )[n] ;

m=0

since the m-power of a q-Pochhammer or a q-Stirling polynomial satises a rst-order dierence
equation of the form
(s)mq T{[(s)q ]i }m − (s + i)mq {[(s)q ]i }m = 0;

m
m
[i] m
(s)mq T−1 [(s)[i]
q ] − (s − i)q [(s)q ] ;

respectively. Then, it is easy to see that the corresponding recurrence relations for the coecients
m
Lmin (q) and L̃in (q), are of degree at most m + 1 (for the classical case see [18]).
3.3. Further examples
In this section we will show the usefulness of the obtained relations for solving some linerization
formulas involving some classical hypergeometric polynomials and their q-analogues.
3.3.1. Classical Charlier polynomials
Firstly, we will obtain the linearization of a product of two Stirling polynomials x[m] x[ j] in terms
of the monic Charlier polynomials
m+j
X

x[m] x[ j] =

cm; j; n Cn (x);

(3.28)

n=0

where the Charlier polynomials Cn are dened by
Cn (x)

= (−)

n

2 F0

!
−n; −x  1
n
−
  = x + ··· :
−

(3.29)

Here p Fq denotes the generalized hypergeometric function
p Fq

 !

a1 ; a2 ; : : : ; ap 

x
b1 ; b2 ; : : : ; bq 

=

∞
X
(a1 )k (a2 )k · · · (ap )k xk
k=0

(b1 )k (b2 )k · · · (bq )k k!

:

(3.30)

To obtain the explicit expression for the coecients cm; j; n in (3.28), we will use expression (3.27)
as well as the well-known inversion formula (see, for example [7])
x[k] =

k
X

ckn Cn (x);

ckn =

n=0

 

k
k−n :
n

Then, substituting (3.31) in
x[m] x[ j] =

m+j
X

L̃mjk x[k]

n=0

and interchanging the sums we obtain
m+j−n

cm; j; n =

X
k=0

L̃mjk+n ck+nn :

(3.31)


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143

Substituting in the above expressions Eqs. (3.27) and (3.31) for the coecients L̃mjk+n and ck+nn and
making some straightforward calculations, we nally obtain


cm; j; n =



!
−j; n − m − j  1

 :
−m − j

−m;

m+j
m+j−n 3 F1
n

(3.32)

An analogous expression has been obtained in [7] by others means. Obviously, the same can be
done for other hypergeometric polynomials (e.g. Hahn, Meixner and Kravchuk).
Notice also that, since
Cj
(x) =

j
X

ajk x[k] ;

ajk =

k=0

(−1) j (−j)k
j−k
;
k!

then, multiplying (3.28) by ajk and taking the sum over k, we obtain
x

[m]

Cj
(x)

=

m+j
X

lm; j; n Cn (x);

(3.33)

n=0

where

j m−n

lm; j; n = (−
) 


 
j 
X
m + k (−j)k  k

k!

n

k=0



3 F1

−m;

!
−k; n − m − k  1

 :
−m − k

A similar result was obtained in [7] by a dierent way.
Notice also the niteness of the last sum as well as the termination character of the involved
hypergeometric series 3 F1 .
3.3.2. q-Charlier polynomials
Next, we will obtain some formulas for the linearization coecients involving a q-analogue of
the Charlier polynomials cn (s; q) in the exponential lattice x(s) = q s [3,4].
These monic Charlier q-polynomials cn (s; q) are dened by
cn() (s; q) = n (1

n

− q) q

n2

= n (1 − q)n q n

2

q−n ; q−s

2 ’0

−

n
X
(q−n ; q)k
k=0

(q; q)k k

qs
; q; −
(q − 1)

(s)[k]
q ;

!

1 ¡ q ¡ 1; 0 ¡  ¡ 1;

(3.34)

where (s)[k]
q are the q-Stirling polynomials dened in (3.13), the symbols (a; q)k are given in (3.5),
and the q-basic hypergeometric series is dened by [16]
r ’p

a1 ; a2 ; : : : ; ar
b1 ; b2 ; : : : ; bp

!

; q; z =

∞
X
(a1 ; q)k · · · (ar ; q)k
k=0

zk
[(−1)k q(k=2)(k−1) ]p−r+1 :
(b1 ; q)k · · · (bp ; q)k (q; q)k

(3.35)

Also we need the q-analogue of the inversion formula (3.31) [5]
(s)[k]
q

=

k
X
n=0

q 
ckn
cn (s; q);

q k−(n=2)(n+1)
ckn =
(q − 1)n

 

k
n

q

k−n ;

(3.36)

144

where


R. Alvarez-Nodarse
et al. / Journal of Computational and Applied Mathematics 107 (1999) 133–146

 

k
n

=

q

(q; q)k
:
(q; q)n (q; q)k−n

Notice that the above expressions transform into the classical ones when q → 1. Then, to solve the
linearization problem
[ j]
(s)[m]
q (s)q

=

m+j
X

q

cm;
j; n cn (s; q);

(3.37)

n=0

we apply the same procedure as before, i.e.,
m+j−n
q
cm;
j; n =

X

q
L̃mjk+n (q)ck+nn
;

k=0

q
where L̃mjk+n (q) are given in (3.23) and ck+nn
by (3.36). Again, some straightforward calculations
(in which we use some identities involving the (a; q)n and (a; q)[n] symbols [16,17]) lead us to the
expression
q
cm;
j; n

qm+j+mj−(n=2)(n+1) m+j−n
=
(q − 1)n



j+m
n

bp dened by
Here, we use the function r ’
bp
r’

a1 ; a2 ; : : : ; ar

b1 ; b2 ; : : : ; bp

!

; q; z =



q

b1
3’

q−m ; q−j q n−m−j

∞
X
(a1 ; q)k · · · (ar ; q)k
k=0

q−m−j

zk
:
(b1 ; q)k · · · (bp ; q)k (q; q)k

1
; q;
(1 − q)q n+1 

!

:

(3.38)

Notice also that in the case when q → 1, Eq. (3.37) transforms into the classical one (3.32).
Obviously, the same can be done with others q-polynomials (e.g., q-Hahn, q-Meixner, etc. [3,17]).
Also, analogues of the problem (3.33) can be solved in a similar way.

4. Summary
In the present work we have developed a general recurrent method for solving the linearization
problem for a product of two polynomials in the exponential lattice x(s) = c1 q s + c3 , obtaining
the explicit solution in the case of the q-analogues of the Pochhammer symbol and the Stirling
polynomials. Finally, some linearization problems involving the q-analogues of the classical Charlier
polynomials were solved.

Acknowledgements
This work has been partially supported by the European project INTAS-93-219-ext (R.A.N.)
and by the Spanish Direccion General de Enseñanza Superior (DGES) grant PB-96-0120-C03-01
(R.A.N.).


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et al. / Journal of Computational and Applied Mathematics 107 (1999) 133–146

145

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