February1,2016 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