Table Of ContentFebruary1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page1
Chapter 1
6
The orthogonality of Al-Salam-Carlitz polynomials for
1
0 complex parameters
2
n
a Howard S. Cohl∗, Roberto S. Costas-Santos† and Wenqing Xu‡
J
∗Applied and Computational Mathematics Division,
9
2 National Institute of Standards and Technology,
Gaithersburg, MD 20899-8910, USA
] [email protected]
A
C †Departamento de F´ısica y Matem´aticas, Facultad de Ciencias,
. Universidad de Alcal´a, 28871 Alcal´a de Henares, Madrid, Spain
h
[email protected]
t
a
m ‡Department of Mathematics and Statistics, California Institute of
Technology, CA 91125, USA
[
[email protected]
1
v
In this contribution, we study the orthogonality conditions satisfied by
78 Al-Salam-Carlitz polynomials Un(a)(x;q) when the parameters a and q
are not necessarily real nor ‘classical’, i.e., the linear functional u with
1
respect to such polynomial sequence is quasi-definite and not positive
8
0 definite. We establish orthogonality on a simple contour in the com-
. plex plane which dependson theparameters. In all cases we show that
1
the orthogonality conditions characterize the Al-Salam-Carlitz polyno-
60 mials Un(a)(x;q) of degree n up to a constant factor. We also obtain a
generalization of theuniquegenerating function for these polynomials.
1
:
v Keywords: q-orthogonal polynomials; q-difference operator; q-integral rep-
i resentation; discrete measure.
X
MSC classification: 33C45; 42C05
r
a
1. Introduction
The Al-Salam-CarlitzpolynomialsU(a)(x;q) wereintroducedbyW.A.Al-
n
Salam and L. Carlitz in1 as follows:
U(a)(x;q):=(−a)nq(n2) n (q−n;q)k(x−1;q)k qkxk. (1)
n (q;q) ak
k
k=0
X
1
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page2
2 H. S. Cohl, R. S. Costas-Santos and W. Xu
In fact, these polynomials have a Rodrigues-type formula [2, (3.24.10)]
anq(n2)(1−q)n
U(a)(x;q)= Dn w(x;a;q) ,
n qnw(x;a;q) q−1
(cid:0) (cid:1)
where
w(x;a;q):=(qx;q) (qx/a;q) ,
∞ ∞
the q-Pochhammer symbol (q-shifted factorial) is defined as
n−1
(z;q) :=1, (z;q) := (1−zqk),
0 n
k=0
Y
∞
(z;q) := (1−zqk), |z|<1,
∞
k=0
Y
and the q-derivative operator is defined by
f(qz)−f(z)
if q 6=1 and z 6=0,
D f(z):= (q−1)z
q
f′(z) if q =1 or z =0.
Remark 1. Observe that by the definition of the q-derivative
Dq−1f(z)=Dqf(qz), and Dqn−1f(z):=Dqn−−11 Dq−1f(z) , n=2,3,...
The expression (1) shows us that U(a)(x;q) is an(cid:0)analytic fu(cid:1)nction for any
n
complex value parameters a and q, and thus can be considered for general
a,q ∈C\{0}.
The classical Al-Salam-Carlitz polynomials correspond to parameters
a < 0 and 0 < q < 1. For these parameters, the Al-Salam-Carlitz polyno-
mials are orthogonalon [a,1] with respect to the weight function w. More
specifically, for a<0 and 0<q <1 [2, (14.24.2)],
1
U(a)(x;q)U(a)(x;q)(qx,qx/a;q) d x=d2δ ,
n m ∞ q n n,m
Za
where
d2n :=(−a)n(1−q)(q;q)n(q;q)∞(a;q)∞(q/a;q)∞q(n2),
and the q-Jacksonintegral [2, (1.15.7)] is defined as
b b a
f(x)d x:= f(x)d x− f(x)d x,
q q q
Za Z0 Z0
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page3
The orthogonality of Al-Salam-Carlitz polynomials for complex parameters 3
where
a ∞
f(x)d x:=a(1−q) f(aqn)qn.
q
Z0 n=0
X
Takingintoaccountthepreviousorthogonalityrelation,itisadirectresult
that if a and q are classical, i.e., a, q ∈ R, with a 6= 1, 0 < q < 1 all
the zeros of U(a)(x;q) are simple and belong to the interval [a,1], but this
n
is no longer valid for general a and q complex. In this paper we show
that for general a, q complex numbers, but excluding some special cases,
the Al-Salam-Carlitz polynomials U(a)(x;q) may still be characterized by
n
orthogonality relations. The case a < 0 and 0 < q < 1 or 0 < aq < 1
and q > 1 are classical, i.e., the linear functional u with respect to such
polynomial sequence is orthogonal is positive definite and in such a case
there exists a weight function ω(x) so that
1
hu,pi= p(x)ω(x)dx, p∈P[x].
Za
Note that this is the key for the study of many properties of Al-Salam-
Carlitz polynomials I and II. Thus, our goal is to establish orthogonality
conditions for most of the remaining cases for which the linear form u is
quasi-definite, i.e., for all n,m∈N
0
hu,p p i=k δ , k 6=0.
n m n n,m n
We believe that these new orthogonality conditions can be useful in the
study of the zeros of Al-Salam-Carlitz polynomials. For general a,q ∈
C\{0}, the zeros are not confined to a real interval, but they distribute
themselves in the complex plane as we can see in Figure 1. Throughout
this paper denote p:=q−1.
2. Orthogonality in the complex plane
Theorem 1. Let a,q ∈ C, a 6= 0,1, 0 < |q| < 1, the Al-Salam-Carlitz
polynomials are the unique polynomials (up to a multiplicative constant)
satisfying the property of orthogonality
1
U(a)(x;q)U(a)(x;q)w(x;a;q)d x=d2δ . (2)
n m q n n,m
Za
Remark 2. I if 0<|q|<1, the lattice {qk :k ∈N }∪{aqk :k ∈N } is a
0 0
set of points which are located inside on a single contour that goes from 1
to 0, and then from 0 to a, through the spirals
S :z(t)=|q|texp(itargq), S :z(t)=|a||q|texp(itargq+iarga),
1 2
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page4
4 H. S. Cohl, R. S. Costas-Santos and W. Xu
y(t)
a
1
0.8
0.6
0.4
0.2
x(t)
−0.5 0.5 1
−0.2
Fig.1. ZerosofU(1+i)(cid:0)x;4exp(πi/6)(cid:1)
30 5
where 0 < |q| < 1, t ∈ [0,∞), which we can see in Figure 2. Taking into
account(2),weneedtoavoidthea=1case. Forthea=0case,wecannot
apply Favard’s result,3 because in such a case this polynomial sequence
fulfills the recurrence relation2
U(0) (x;q)=(x−qn)U(0)(x;q), U(0)(x;q)=1.
n+1 n 0
Proof. Let 0 < |q| < 1, and a ∈ C, a 6= 0,1. We are going to express the
q-Jackson integral (2) as the difference of the two infinite sums and apply
the identity
M f(qM)g(qM)−f(q−1)g(q−1)
f(qk)Dq−1g(qk)qk = q−1−1
k=0
X
M
− g(qk−1)Dq−1f(qk)qk. (3)
k=0
X
Let n ≥ m. Then, for one side since w(q−1;a;q) = 0, and using the
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page5
The orthogonality of Al-Salam-Carlitz polynomials for complex parameters 5
1.0
0.5
-0.5 0.5 1.0
Fig.2. Thelattice{qk :k∈N0}∪{(1+i)qk :k∈N0}withq=4/5exp(πi/6).
identities [2, (14.24.7), (14.24.9)], one has
∞
U(a)(qk;q)U(a)(qk;q)w(qk;a;q)qk
m n
k=0
X
M
a(1−q)
= q2−n Mli→m∞ Dq−1[w(qk;a;q)Un(a−)1(qk;q)]Um(a)(qk;q)qk
k=0
X
=aqn−1 lim U(a)(qM;q)U(a) (qM;q)w(qM;a;q)
m n−1
M→∞
M−1
+aqn−1(qm−1) lim w(qk;a;q)U(a) (qk;q)U(a) (qk;q)qk.
n−1 m−1
M→∞
k=0
X
Following an analogous process as before, and since w(aq−1;a;q) = 0, we
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page6
6 H. S. Cohl, R. S. Costas-Santos and W. Xu
have
∞
U(a)(aqk;q)U(a)(aqk;q)w(aqk;a;q)aqk
m n
k=0
X
=aqn−1 lim U(a)(aqM;q)U(a) (aqM;q)w(aqM;a;q)
m n−1
M→∞
M−1
+aqn−1(qm−1) lim w(aqk;a;q)U(a) (aqk;q)U(a) (aqk;q)aqk.
n−1 m−1
M→∞
k=0
X
Therefore, if m<n, and since m is finite one can first repeat the previous
process m+1 times obtaining
∞
U(a)(qk;q)U(a)(qk;q)w(qk;a;q)qk
m n
k=0
X
m+1
= lim (−aqn)νq−ν(ν+1)/2(q−m+ν−1;q)
ν
M→∞
ν=1
X
×U(a) (qM;q)U(a) (qM;q)w(qM;a;q),
m−ν+1 n−ν
and
∞
U(a)(aqk;q)U(a)(aqk;q)w(aqk;a;q)aqk
m n
k=0
X
m+1
= lim (−aqn)νq−ν(ν+1)/2(q−m+ν−1;q)
ν
M→∞
ν=1
X
×U(a) (aqM;q)U(a) (aqM;q)w(aqM;a;q).
m−ν+1 n−ν
Hence since the difference of both limits, term by term, goes to 0 since
|q|<1, then
1
U(a)(x;q)U(a)(x;q)(qx,qx/a;q) d x=0.
n m ∞ q
Za
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page7
The orthogonality of Al-Salam-Carlitz polynomials for complex parameters 7
For n=m, following the same idea, we have
1
U(a)(x;q)U(a)(x;q)w(x;a;q)d x
n n q
Za
a(qn−1) ∞ 2
= w(qk;a;q) U(a) (qk;q) qk
q1−n n−1
Xk=0 (cid:16) (cid:17)
2
−aw(aqk;a;q) U(a) (aqk;q) qk
n−1
!
∞ (cid:16) (cid:17)
=(−a)n(q;q) q(n2) w(qk;a;q)qk−a w(aqk;a;q)qk
n
k=0
=(−a)n(q;q)n(q;q)X∞q(cid:0)(n2) ∞ (qk+1/a;q)∞−a(aqk+1(cid:1);q)∞ qk ,
(q;q)
k
k=0
X(cid:0) (cid:1)
since it is known that in this case [2, (14.24.2)]
1
U(a)(x;q)U(a)(x;q)w(x;a;q)d x
n n q
Za
=(−a)n(q;q)n(q;q)∞(a;q)∞(q/a;q)∞q(n2).
Due to the normality of this polynomial sequence, i.e., degU(a)(x;q) = n
n
foralln∈N ,theuniquenessisstraightforward,hencetheresultholds.
0
From this result, and taking into account that the squared norm for the
Al-Salam-Carlitz polynomials is known, we got the following consequence
for which we could not find any reference.
Corollary 1. Let a,q∈C\{0}, |q|<1. Then
∞ qk
(qk+1/a;q) −a(aqk+1;q) =(a;q) (q/a;q) .
∞ ∞ ∞ ∞
(q;q)
k
k=0
X(cid:0) (cid:1)
The following case, which is just the Al-Salam-Carlitz polynomials for the
|q|>1 case, is commonly called the Al-Salam-Carlitz II polynomials.
Theorem 2. Let a,q ∈ C, a 6= 0,1, |q| > 1. Then, the Al-Salam-Carlitz
polynomials areunique(uptoamultiplicative constant)satisfying theprop-
erty of orthogonality given by
1
Z Un(a)(x;q−1)Um(a)(x;q−1)(q−1x;q−1)∞(q−1x/a;q−1)∞dq−1x
a
=(−a)n(1−q−1)(q−1;q−1)n(q−1;q−1)∞(a;q−1)∞(q−1/a;q−1)∞q−(n2)δm,n.(4)
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page8
8 H. S. Cohl, R. S. Costas-Santos and W. Xu
Proof. Let us denote q−1 by p, then 0 < |p| < 1. For a ∈ C, a 6= 0,1.
Then, by using the identity (3) replacing q 7→ p, and taking into account
that w(aq;a;p)=w(q;a;p)=0 and [2, (14.24.9)], for m<n one has
∞
aw(apk;a;p)U(a)(apk;p)U(a)(apk;p)pk
m n
k=0
X=apn−1 lim U(a)(apM;p)U(a) (apM;p)w(apM;a;p)
m n−1
M→∞
M−1
+apn−1(1−pm) lim aw(apk;a;p)U(a) (apk;p)U(a) (apk;p)pk.
n−1 m−1
M→∞
k=0
X
Following the same idea from the previous result, we have
∞
w(pk;a;p)U(a)(pk;p)U(a)(pk;p)pk
m n
k=0
X=apn−1 lim U(a)(pM;p)U(a) (pM;p)w(pM;a;p)
m n−1
M→∞
M−1
+apn−1(1−pm) lim w(pk;a;p)U(a) (pk;p)U(a) (pk;p)pk.
n−1 m−1
M→∞
k=0
X
Therefore, the property of orthogonality holds for m<n. Next, if n=m,
we have
1
U(a)(x;p)U(a)(x;p)w(x;a;p)d x
n n p
Za a(pn−1)∞ 2
= aw(apk;a;p) U(a) (apk;p) pk
p1−n n−1
Xk=0 (cid:16) (cid:17)
2
−w(pk;a;p) U(a) (pk;p) pk
n−1
!
(cid:16) (cid:17)
∞
=(−a)n(p;p) p(n2) aw(apk;a;p)pk−w(pk;a;p)pk
n
!
k=0
=(−a)n (q−1;q−1)n(Xp;p)∞p(n2) ∞ qk a(pk+1a;p)∞−(pk+1/a;p)∞
(p;p)
k=0 (cid:0) k (cid:1)
=(−a)n(q−1;q−1)n(p;p)∞(a;p)∞X(p/a;p)∞p(n2).
Using the same argument as in Theorem 1, the uniqueness holds, so the
claim follows.
Remark 3. Observethat inthe previoustheorems if a=qm, with m∈Z,
a 6= 0, after some logical cancellations, the set of points where we need to
calculate the q-integralis easy to compute. For example, if 0<aq <1 and
0<q <1, one obtains the sum [2, p. 537, (14.25.2)].
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page9
The orthogonality of Al-Salam-Carlitz polynomials for complex parameters 9
Remark 4. The a = 1 case is special because it is not considered in the
literature. In fact, the linear form associated with the Al-Salam-Carlitz
polynomials u is quasi-definite and fulfills the Pearson-type distributional
equations
x−2 x−2
Dq[(x−1)2u]= 1−qu and Dq−1[q−1u]= 1−qu.
Moreover, the Al-Salam-Carlitz polynomials fulfill the three-term recur-
rence relation [2, (14.24.3)]
xU(a)(x;q)=U(a) (x;q)+(a+1)qnU(a)(x;q)−aqn−1(1−qn)U(a) (x;q),
n n+1 n n−1
(5)
where n = 0,1,..., with initial conditions U(a)(x;q) = 1, U(a)(x;q) =
0 1
x−a−1.
Therefore,we believe that it will be interesting to study such a case for
its peculiarity because the coefficientqn−1(1−qn)6=0 for all n, so one can
apply Favard’s result.
2.1. The |q| = 1 case.
In this section we only consider the case where q is a root of unity. Let N
be a positive integer such that qN =1 then, due to the recurrence relation
(5) and following the same idea that the authors did in [4, Section 4.2], we
apply the following process:
(1) The sequence (U(a)(x;q))N−1 is orthogonal with respect to the Gaus-
n n=0
sian quadrature
N
p(x )
hv,pi:= γ(a)...γ(a) s ,
1 N−1 2
Xs=1 UN(a−)1(xs)
where {x ,x ,...,x } are the zeros of U(cid:16)(a)(x;q) for(cid:17)such value of q.
1 2 N N
(2) Since hv,U(a)(x;q)U(a)(x;q)i = 0, we need to modify such a linear
n n
form.
Next, we can prove that the sequence (U(a)(x;q))2N−1 is orthogonal
n n=0
with respect to the bilinear form
hp,ri =hv,pqi+hv,DNpDNri,
2 q q
since D U(a)(x;q)=(qn−1)/(q−1)U(a) (x;q).
q n n−1
(3) Since hU(a)(x;q),U(a)(x;q)i =0 and taking into accountwhat we did
2N 2N 2
before, we consider the linear form
hp,ri =hv,pqi+hv,DNpDNri+hv,D2NpD2Nri.
3 q q q q
(4) Therefore one can obtain a sequence of bilinear forms such that the
Al-Salam-Carlitz polynomials are orthogonal with respect to them.
February1,2016 1:28 ws-rv9x6 BookTitle AlSalamCarlitz page10
10 H. S. Cohl, R. S. Costas-Santos and W. Xu
3. A generalized generating function for Al-Salam-Carlitz
polynomials
For this section, we are going to assume |q|>1, or 0<|p|<1. Indeed, by
startingwith the generatingfunctions forAl-Salam-Carlitzpolynomials[2,
(14.25.11-12)], we derive generalizations using the connection relation for
these polynomials.
Theorem 3. Let a,b,p∈C\{0}, |p|<1, a,b6=1. Then
U(a)(x;p)=(−1)n(p;p) p−(n2) n (−1)kan−k(b/a;p)n−kp(k2)U(b)(x;p).
n n (p;p) (p;p) k
n−k k
k=0
X
(6)
Proof. If we consider the generating function for Al-Salam-Carlitz poly-
nomials [2, (14.25.11)]
(xt;p)∞ = ∞ (−1)np(n2)U(a)(x;p)tn,
(t,at;p) (p;p) n
∞ n
n=0
X
and multiply both sides by (bt;p) /(bt;p) , obtaining
∞ ∞
∞ (−1)np(n2)U(a)(x;p)tn = (bt;p)∞ ∞ (−1)np(n2)U(b)(x;p)tn. (7)
(p;p) n (at;p) (p;p) n
n ∞ n
n=0 n=0
X X
If we now apply the q-binomial theorem [2, (1.11.1)]
∞
(az;p) (ap;p)
∞ = nzn, 0<|p|<1, |z|<1,
(z;p) (p;p)
∞ n
k=0
X
to (7), and then collect powers of t, we obtain
∞ tk k (−1)mak−m(b/a;p)k−mp(m2)U(b)(x;p)
(p;p) (p;p) m
k−m m
k=0 m=0
X X
∞ (−1)np(n2)
= U(a)(x;p)tn.
(p;p) n
n
n=0
X
Taking into account this expression, the result follows.
Theorem 4. Let a,b,p∈C\{0}, |p|<1, a,b6=1, t∈C, |at|<1. Then
x ∞ pk(k−1) b/a
(at;p) φ ;p,t = φ ;p,atpk U(b)(x;p)tk, (8)
∞1 1 at (p;p) 1 1 0 k
(cid:18) (cid:19) k=0 k (cid:18) (cid:19)
X
