Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol99.Issue1-2.1998:

Journal of Computational and Applied Mathematics 99 (1998) 275–286

Recurrence relations for the connection coecients
of orthogonal polynomials of a discrete variable on the lattice
x(s) = q2s
Stanis law Lewanowicz ∗
Institute of Computer Science, University of Wroclaw, 51-151 Wroclaw, Poland
Received 30 October 1997; received in revised form 30 March 1998

Abstract
We give explicitly recurrence relations satised by the connection coecients between two families of the classical
orthogonal polynomials of a discrete variable on a non-uniform lattice x(s) = q 2s (i. e., the q-analogues of Charlier, Meixner,
Krawtchouk and Hahn polynomials), in terms of the coecients  and  of the Pearson equation satised by the weight
function %, and the coecients of the three-term recurrence relation and of two structure relations obeyed by these
c 1998 Elsevier Science B.V. All rights reserved.
polynomials.
AMS classications: primary 33C45, 33E30
Keywords: Classical orthogonal polynomials of a discrete variable; q-Meixner polynomials; q-Krawtchouk polynomials;
Connection coecients; Recurrence relations

1. Introduction

Let {Pk (x)} be any system of the classical orthogonal polynomials of a discrete variable, orthogonal
on the exponential lattice x = x(s) := q2s (s ∈ {0; 1; : : : ; B − 1}) with q = e! ,
B−1
X

Pk (x(s))Pl (x(s))%(s)
x(s − 1=2) = kl d2k

(k; l = 0; 1; : : :);

s=0

where %(s)
x(s−1=2)¿0 (s = 0; 1; : : : ; B−1), i.e., q-Charlier polynomials Ck() (x; q); q-Meixner polynomials Mk(
; ) (x; q); q-Krawtchouk polynomials Kk(p) (x; N; q), or q-Hahn polynomials Qk(; ) (x; N; q).
Here B equals +∞; +∞; N + 1 and N , respectively.
∗

E-mail: stanislaw.lewanowicz@ii.uni.wroc.pl.

c 1998 Elsevier Science B.V. All rights reserved.
0377-0427/98/$ – see front matter

PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 6 2 - 9

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S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

We are looking for a formula of the type
Pn =

n
X

(1.1)

cn; k Pk ;

k=0

where {Pk } and {Pk } are any two families of classical orthogonal polynomials.
The coecients cn; k in (1.1) are called the connection coecients between the polynomials {Pk }

and {Pk } (see [4], Lecture 7).
In a recent paper [3], an algorithmic way has been proposed of obtaining a recurrence relation
(in k) of the form
Lcn; k ≡

r
X

Ai (k)cn; k+i = 0:

(1.2)

i=0

Now, the coecients cn; k can be found by use of this recurrence relation in the backward direction
(see [11], Section 7.2).
In the present paper we propose an alternative technique of derivation of the recurrence relation (1.2), based on an generalization of an idea introduced in [8] (see also [5]). The dierence
operator L is given in terms of the coecients  and  of the dierence Pearson equation for the
weight %, and the coecients of the three-term recurrence relation and of two structure relations
obeyed by {Pk } (see Theorems 3.1 and 3.6). Also, it should be stressed that the order r of the

obtained recurrence relation is signicantly lower than in [3]. Applications of the result to some
pairs of the classical discrete orthogonal polynomials are given.
2. Properties of the classical orthogonal polynomials
2.1. Basics of classical orthogonal polynomials of a discrete variable
For the sake of compactness, the following notation will be used in the sequel:
D :=



;

x(s)

D̂ :=



;

x̂(s)

N :=

B

;
Bx(s)

(2.1)
x̂(s) := x(s − 1=2);

(2.2)
(2.3)

+ (s) := (s) + (s)
x̂(s);

− (s) := (s);

(2.4)

U := q−1 − N + I ;

(2.5)

V := q+ D + I :

(2.6)
m

m

Here
:= E − I ; B := I − E ; E (m ∈ Z) is the mth shift operator, E f(s) = f(s + m); I is
the identity operator, If(s) = f(s). The meaning of  and  is given below. (By convention, all the
bold letter operators act on the variable s).
−1

S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

277

In the sequel, we make use of certain properties enjoyed by all classical families of orthogonal polynomials on the lattice x(s) = q2s ([9, Chapter II]; [10]; [1–3, 6]). Besides the three-term
recurrence relation
x(s)Pk (x(s)) = 0 (k)Pk−1 (x(s)) + 1 (k)Pk (x(s)) + 2 (k)Pk+1 (x(s))
(k = 0; 1; : : : ; P−1 (x(s)) ≡ 0; P0 (x(s)) ≡ 1)

(2.7)

we need ve other properties.
First, the weight function % satises a dierence equation of the type
D̂[(s)%(s)] = (s)%(s);

(2.8)

where (s) := (x(s));
˜
(s) := (x(s)),
˜
and where ;
˜ ˜ are polynomials in x; deg ˜ 6 2; deg ˜ = 1.
Second, for arbitrary n, the polynomial Pn obeys the second order dierence equation
Ln Pn (x(s)) ≡ {(s)D̂N + (s)D + n I }Pn (x(s)) = 0:

(2.9)

Here n is a constant given by

n := −[n]q { 12 [n − 1]q ˜ ′′+ + cosh(n − 1)! · ˜′ }

(n ∈ N);

(2.10)

where ˜ + (x) := (x)
˜
+ 12 (q − q−1 )x˜(x) (notice that
x̂(s) = (q − q−1 )x(s)), and we use the notation
[n]q :=

qn − q−n sinh(!n)
;
=
q − q−1
sinh(!)

q = e! :

Notice that

Ln = UD + n I = VN + n I :

(2.11)

Third, we have the dierence analogue of the Rodrigues formula:
k
Y
B
B
Bk
B
Pk (x(s)) =
%(s)
(k + i) ;
···
%(s) Bx1 (s) Bx2 (s)
Bxk (s)
i=1

"

#

(2.12)

where xi (s) := x(s + i=2) (i = 1; 2 : : :). One consequence of this formula is the following explicit
expression for the leading coecient k in the expansion Pk (x(s)) = k x k (s) + · · ·:
k = Bk

k−1
Y
l=0

cosh(n + l − 1)! · ˜′ +

sinh(n + l − 1)! ′′
· ˜ +
2 sinh(!)



(k = 0; 1; : : :):

(2.13)

Fourth, we have a pair of the so-called structure relations [3],
+ (s)DPk (x(s)) = 0 (k)Pk−1 (x(s)) + 1 (k)Pk (x(s)) + 2 (k)Pk+1 (x(s));

(2.14)

− (s)NPk (x(s)) = 0 (k)Pk−1 (x(s)) + 1 (k)Pk (x(s)) + 2 (k)Pk+1 (x(s)):

(2.15)

and

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S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

Here
i (k) :=

k −k
(q i (k) − ’i (k));
[k]q

k k
(q i (k) − ’i (k))
i (k) :=
[k]q

(i = 0; 1; 2);

(2.16)

and
’0 (k) := 0;

’1 (k) :=

k (0)
;
′k

’2 (k) :=

Bk
:
′
k Bk+1

(2.17)

We use the notation
k (s) := [+ (s + k) − − (s)]=
xk−1 (s):
Fifth,
k

(s)%(s)x (s)x

l



1
s−
2

 s=B

 =0


(k; l = 0; 1; : : :):

(2.18)

s=0

2.2. Identities involving the discrete Fourier coecients
We shall need certain properties of the Fourier coecients of an arbitrary polynomial P, deg P¡B,
dened by
ak [P] := d−2
k bk [P]

(k = 0; 1; : : : ; B − 1);

(2.19)

where
bk [P] :=

B−1
X

Pk (x(s))P(x(s))%(s)
x̂(s)

(2.20)

s=0

P
i.e., the coecients in the expansion P = deg
k=0 ak [P]Pk .
Let X; D and N be the dierence operators (acting on k) dened by

P

X := 0 (k)E−1 + 1 (k)I + 2 (k)E;

(2.21)

D := 0 (k)E−1 + 1 (k)I + 2 (k)E;

(2.22)

N := 0 (k)E−1 + 1 (k)I + 2 (k)E= D + 2k sinh ! · X

(2.23)

(cf. (2.7), (2.14) and (2.15), respectively) where I is the identity operator, and Em – the mth shift
operator: Ibk [f] = bk [f]; Em bk [f] = bk+m [f] (m ∈ Z). For the sake of simplicity, we write E in
place of E1 . (We adopt the convention that all the script letter operators act on the variable k).
We prove the following lemma.

S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

279

Lemma 2.1. The coecients (2.20) obey the identities:
bk [xP] = Xbk [P];

(2.24)

Nbk [N P] = qk bk [P];

(2.25)

Dbk [DP] = q−1 k bk [P];

(2.26)

bk [U P] = −qDbk [P];

(2.27)

bk [V P] = −q−1 Nbk [P];

(2.28)

bk [Ln P] = (n − k )bk [P]:

(2.29)

Here P stands for P(x(s)).
Proof. We shall use the notation
p(s) := P(x(s));

pk (s) := Pk (x(s)):

In view of (2.7) and (2.21), identity (2.24) is obviously true.
We will prove the identity (2.25). Using (2.14), summing by parts, and then using (2.18) and the
equation
D̂[(s)%(s)Npk (s)] = − k %(s)pk (s);
(cf. (2.8) and (2.9)), we get
Nbk [NP] =

B−1
X

Npk (s)E −1 Dp(s)%(s)
x̂(s) = q

s=0

B−1
X

(s)Npk (s)
p(s − 1)%(s)

s=0

s=B
B−1

X

= q(s)Npk (s)p(s − 1)%(s) − q
p(s)D̂[%(s)(s)Npk (s)]
x̂(s)

s=0

s=0

= qk

B−1
X

%(s)pk (s)p(s)
x̂(s) = qk bk [P]:

s=0

The proof of (2.26) goes as follows:
Dbk [DP] =

B−1
X

Dpk Dp(s)%(s)
x̂(s) = q−1

s=0

=q

−1

B−1
X

%(s)+ (s)Dpk (s)
p(s)

s=0

B−1
X

%(s + 1)(s + 1)ENpk (s)
p(s) = q

−1

s=0

= q−1

B−1
X

t=1

%(s)(s)Npk (s)
p(s − 1) = − q−1

s=0

=

B−1
X

B
X

B−1
X

%(t)(t)

Bpk (t)

p(t − 1)
Bx(t)

p(s)D̂[(s)%(s)Npk (s)]
x̂(s)

s=0

%(s)pk (s)p(s)
x̂(s) = q−1 k bk [P]:

s=0

Here we used, a.o., the equation (s + 1)%(s + 1) = + (s)%(s) (cf. (2.8)).

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S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

Similarly, we obtain
bk [− NP] =

B−1
X

%(s)(s)pk (s)Np(s)
x̂(s) = q

s=0

= −q

B−1
X

%(s)(s)pk (s)
p(s − 1)

s=0

B−1
X

p(s)D̂[(s)%(s)pk (s)]
x̂(s)

B−1
X

p(s){D̂[(s)%(s)]pk (s) + (s + 1)%(s + 1)D̂pk (s)}
x̂(s)

s=0

= −q

s=0

= −q

B−1
X

(s)%(s)pk (s)p(s)
x̂(s) − q

s=0

= −q

B−1
X

B−1
X

+ (s)%(s)D̂pk (s)p(s)
x̂(s)

s=0

(s)%(s)pk (s)p(s)
x̂(s) − q2

s=0

B−1
X

%(s)Dpk p(s)
x̂(s)

s=0

= −qbk [P] − q2 Dbk [P]:
Hence follows the identity (2.27).
Identity (2.28) may be proved in an analogous way.
Using (2.11), (2.27), and (2.26), we have
bk [Ln P] = bk [UDP] + n bk [P] = − qDbk [DP] + n bk [P] = (n − k )bk [P]:
This proves the validity of (2.29).
Remark 1. In the proof of Lemma 2.1 we use the fact that identity (2.24) can be easily generalized
to the form
bk [#P] = #(X)bk [P];

(2.30)

where # is any polynomial in x.
3. Main result
Let {Pk } and {Pk } be any two families of the classical discrete orthogonal polynomials. We shall
give a recurrence relation (in k) of the form
Lcn; k ≡

r
X

Ai (k)cn; k+i = 0;

(3.1)

i=0

obeyed by the connection coecients cn; k in
Pn =

n
X
k=0

cn; k Pk :

(3.2)

S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

281

Obviously, cn; k are the Fourier coecients ak [Pn ]. Let us write
bn; k := bk [Pn ] = d2k cn; k :

(3.3)

Let Pn satises Eq. (2.9), and let
L n Pn (x(s)) ≡ {(s)
 D̂N + (s)D +  n I }Pn (x(s)) = 0:

(3.4)

Here (s)
 := ∗ (x(s)); (s) := ∗ (x(s)); ∗ and  are polynomials in x; deg  6 2; deg  = 1. The
constant  n is given by  n := − [n]q { 12 [n − 1]q ∗+′′ + ∗ ′ cosh(n − 1)!}, where ∗+ (x) := ∗ (x) + 12 (q −
q−1 )x∗ (x). We shall use the notation
˜
’(x) := ∗ (x) − (x);

(3.5)

(x) := ∗ (x) − ˜(x):

(3.6)

We can write
L n = Ln + ’(x(s))D̂N + (x(s))D + ( n − n )I :

(3.7)

3.1. Connection between q-Charlier, q-Meixner and q-Krawtchouk families
Now we consider the case where none of the families {Pk }, {P k } is a Hahn family. We will prove
the following:
Theorem 3.1. Let {Pk }; {Pk } be (independently chosen) families of q-Charlier; or q-Meixner; or
q-Krawtchouk polynomials. The coecients (3.3) satisfy the recurrence relation
L̃bn; k = 0;

(3.8)

where the dierence operator L̃ is given by
L̃ := D(k I) + q−1 k (q−2 X)

(3.9)

with
k :=  n − k −

′ −2

(3.10)

q :

The order of the recurrence relation (3.8) is not greater than 2.
Proof. Under the assumptions of the theorem, we have ∗ = ˜ (cf., e.g., [9, Table 3.3], or [1]), so
that Eq. (3.7) simplies to
L n P n (x(s)) = Ln P n (x(s)) + (x(s))DP n (x(s)) + ( n − n )P n (x(s));
which can be rewritten in the form
L n P n (x(s)) = Ln P n (x(s)) + D[ (x(s − 1))P n (x(s))] + – P n (x(s))
with – :=  n − n − D (x(s − 1)) =  n − n −
bk [L n P n ] = 0;

′ −2

q . Using this result in the equation
(3.11)

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S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

we obtain
bk [Ln P n ] + bk [D{ (q−2 x)P n }] + bk [– P n ] = 0:
Applying the operator D to both sides of the above equation, and making a repeated use of
Lemma 2.1, we arrive at the recurrence relation (3.8) with the operator L̃ given in (3.9).
Corollary 3.2. The connection coecients in (1.1) satisfy
Lcn; k = 0

(3.12)

2
L := d−2
k L̃(dk I);

(3.13)

with

where L̃ is the dierence operator given in (3.9).
Remark 2. Notice that in (3.13) we need not the explicit form for d2k , but only for the quotients
d2k+h =d2k . Using the equation d2k+1 =d2k = 0 (k + 1)=2 (k) (see [9, p. 106]), we obtain
h
d2k+h Y
0 (k + h)
=
2
 (k + h − 1)
dk
m=1 2

(h ¿ 0):

In particular, for the monic case we have 2 ≡ 1. Making use of the forms of the operators D
(see (2.22)) and X (see (2.21)), we arrive at the following:
Corollary 3.3. Scalar form of Eq. (3.12) is
A0 (k)cn; k−1 + A1 (k)cn; k + A2 (k)cn; k+1 = 0;

(3.14)

where


A0 (k) := 0 (k)( n −




′

− k−1 ) + q−3 k

A (k) :=  (k){ (k)( −

′

′

0 (k);

−  ) + q−1 k [ ′ q−2 1 (k) + (0)]};

1
0
1
n
k



 A (k) :=  (k) (k + 1){ (k)( −
2
0
0
2
n

′

− k+1 ) + q−3 k

′

(3.15)

2 (k)}:

Example 3.4. Let us consider the following formula, connecting two q-Meixner families:
Mn
;  (x; q) =

n
X

cn; k Mk
;  (x; q):

(3.16)

k=0

The specic expressions for ; ; k as well as the forms for the coecients of the operators X
(see (2.21)) and D (see (2.22)) for the monic q-Meixner polynomials are given in Table 1 (see
Appendix A).

S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

283

The coecients cn; k in (3.16) satisfy Eq. (3.14) with Ai ’s given in (3.15), where, in particular,
 n := − [n]q q$ [n + $ − 1]q ;
′

:= q$ [$]
 q − q$ [$]q ;


(0) := q+1 [ + 1]q − q+1 [ + 1]q ;


$ :=
 +  + 1:

q := ;


Example 3.5. Let us consider the following formula, connecting two q-Charlier families:
n
X


Cn()
(x; q) =

cn; k Ck() (x; q):

(3.17)

k=0

The specic expressions for ; ; k as well as the forms for the coecients of the operators X
(see (2.21)) and D (see (2.22)), and N (see (2.23)) for the q-Charlier polynomials are given in
Table 2 (see Appendix A). The coecients cn; k in (3.17) satisfy Eq. (3.14) with Ai ’s given in (3.15),
where, in particular,
 n = n := q1−n [n]q =–q ;

′

:= 0;

(0) := ( − )q3 :

Here –q := q − q−1 . Noticing that the coecient 2 of the operator D vanishes we see that A2 ≡ 0,
so that we obtain the rst-order recurrence
A0 (k)cn; k−1 + A1 (k)cn; k = 0;
where
A0 (k) = ( n − k−1 )0 (k);
A1 (k) = 0 (k){( n − k )1 (k) + q2 ( − )k }:
Hence the formula
n−k

cn; k = (−–q )

 n Y
n




j=k+1

q 2j−1

( n − j )1 (j) + q2 ( − )j
 n − j−1

(0 6 k 6 n):

3.2. Case when a q-Hahn family is involved
Observe that for arbitrary pair of classical orthogonal sequences we have
(x)
˜
= (x)(x − 1);

∗ (x) = (x)(x − 1);

(3.18)

where  and  are rst-degree polynomials in x. If none of the sequences {Pk }; {Pk } belong to the
Hahn family we have the case  =  discussed in the preceding subsection. However, equation  = 
 N; q), with
holds also for the case of two Hahn sequences Pk = Qk (·; ; ; N; q), and Pk = Qk (·; ; ;
the same  and N parameters. It is easy to observe that Theorem 3.1 remains valid in this case, as
we have ∗ = ,
˜ which is the only assumption used explicitly in the proof of the theorem.

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S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

The case  6=  is discussed in the following theorem.
Theorem 3.6. Let {Pk } and {Pk } be such two sequences of classical orthogonal polynomials that
(i) exactly one of the sequences belongs to the Hahn family; or (ii) {Pk }; {Pk } are both in the
 N ; q); with  6= ;
 ;
 or N 6= N .
Hahn family; Pk = Qk (·; ; ; N; q); P k = Qk (·; ;
2

The coecients bn; k := bk [Pn ] = dk cn; k ; where cn; k ’s are the connection coecients in
P n =

n
X

(3.19)

cn; k Pk ;

k=0

satisfy the fourth-order recurrence relation
L̃bn; k = 0;

(3.20)

where the dierence operator L̃ is given by
L̃ := D(X)(n − k )I + q−1 k #(q−2 X) + D(X);

(3.21)

with
# := ∗ − ˜;

 :=  n − n ;

(x) := (x) − D#(q−2 x):

(3.22)

Proof. Notice that the assumptions of the theorem imply that  6=  (cf., e.g., [9, Table 3.3], or [1]).
Multiplying both sides of (3.4) by , and making use of ∗ = ,
˜ we write
L n P n (x(s)) = Ln P n (x(s)) + #(x(s))DP n (x(s)) + P n (x(s))
= Ln P n (x(s)) + D[#(x(s − 1))P n (x(s))] + P n (x(s));
notation used being that of (3.22).
Now, in view of (3.4) we have
0 = bk [Ln P n ] + bk [D{#(q−2 x)P n }] + bk [P n ]:
Applying the operator D to both sides of the above equation, and making a repeated use of
Lemma 2.1, we arrive at the recurrence relation (3.20) with the operator L̃ given in (3.21).

Acknowledgements
The author wishes to thank the referees for their valuable comments.

Appendix A
Data of the q-Meixner and q-Charlier polynomials are presented in Tables 1 and 2, respectively.

S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286
Table 1
Data for the monic q-Meixner polynomials [1, 2, 6]
Mk
;  (x; q); x = q2s ;  = q2
(
¿0; 0¡; q¡1)
(x)
˜

x(x − 1)

˜(x)

q$ [$]q x − q+1 [ + 1]q

k

−q$ [k]q [k + $ − 1]q

0 (k)

q2−k−3
[k]q [k +
− 1]q [k + ]q [k + $ − 2]q
[2k + $ − 1]q [2k + $ − 2]2q [2k + $ − 3]q

1 (k)

q−

2 (k)

1

0 (k)

−q−2
+3

1 (k)

−q k++1 [k]q [k + $ − 1]q

2 (k)

−q$ (q k [k + $ − 1]q + [2k + $ − 1]q )

[k + 1]q [k +  + 1]q
[k]q [k + ]q
− q−
[2k + $]q
[2k + $ − 2]q
[k]q [k +
− 1]q [k + ]q [k + $ − 1]q [k + $ − 2]q
[2k + $ − 1]q [2k + $ − 2]2q [2k + $ − 3]q



q

[k +  + 1]q
[k + ]q
−
[2k + $]q
[2k + $ − 2]q



Note: $ :=
+  + 1.

Table 2
Data for the q-Charlier polynomials [1, 2]
Ck() (x; q); x = q2s ,
(¿0; 0¡(1 − q2 )¡1)
(x)
˜

x(x − 1)

(x)
˜

q3 + (1 − x)=–q

k

q1−k [k]q =–q

0 (k)

−q2k+2 –q k

1 (k)

1 + –q q4k+3 { + q−3k−3 k }

2 (k)

−–q q3k

k

q−3k(k−1)=2 (−–q )−k

0 (k)

−q2 k

1 (k)

q−k k + q2k+4 (–q − 1)=–q + q(1 − q−2k –q )=–q2

2 (k)

0

0 (k)

−q2k+2 k

1 (k)

q k k + q 2k+4 (q2k –q − 1)=–q − q(–q − 1)=–q2

2 (k)

q k(k+1) (1 − q 2k )

Note: –q := q − q−1 ; k := [k]q (1 − (1 − q2 )q2k ).

285

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S. Lewanowicz / Journal of Computational and Applied Mathematics 99 (1998) 275–286

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