A unified construction of all the hypergeometric and basic hypergeometric families of orthogonal polynomial sequences
aa r X i v : . [ m a t h . C A ] F e b A unified construction of all the hypergeometric and basichypergeometric families of orthogonal polynomial sequences
Luis Verde-Star
Department of Mathematics, Universidad Aut´onoma Metropolitana, Iztapalapa, Apartado 55-534,Mexico City 09340, Mexico
Abstract
We construct a set H of orthogonal polynomial sequences that contains all the familiesin the Askey scheme and the q -Askey scheme. The polynomial sequences in H aresolutions of a generalized first-order difference equation which is determined by threelinearly recurrent sequences of numbers. Two of these sequences are solutions of thedifference equation s k +3 = z ( s k +2 − s k +1 ) + s k , where z is a complex parameter, and theother sequence satisfies a related difference equation of order five.We obtain explicit expressions for the coefficients of the orthogonal polynomials andfor the generalized moments with respect to a basis of Newton type of the space ofpolynomials. We also obtain explicit formulas for the coefficients of the three-termrecurrence relation satisfied by the polynomial sequences in H .The set H contains all the 15 families in the Askey scheme of hypergeometric orthog-onal polynomials [5, p. 183] and all the 29 families of basic hypergeometric orthogonalpolynomial sequences in the q -Askey scheme [5, p. 413]. Each of these families is ob-tained by direct substitution of appropriate values for the parameters in our generalformulas. The only cases that require some limits are the Hermite and continuous q -Hermite polynomials. We present the values of the parameters for some of the families. AMS classification:
Keywords: Orthogonal polynomials, recurrence coefficients, generalized difference op-erators, generalized moments, infinite matrices.
Email address: [email protected] (Luis Verde-Star)
Preprint submitted to Elsevier February 20, 2020 . Introduction
Among the families of orthogonal polynomial sequences the hypergeometric and ba-sic hypergeometric families are certainly some of the most important and have beenextensively studied for a long time. See [4] and [5]. In the present paper, we present aconstruction of a class H of orthogonal polynomial sequences that includes all the hy-pergeometric and basic hypergeometric families. We find first the polynomial solutionsof certain generalized difference equation of first order, which is determined by threesequences of numbers, and then we show that when the three sequences satisfy certaindifference equations the polynomial solutions are orthogonal.We obtain explicit expressions for the coefficients of the orthogonal polynomials andfor the generalized moments with respect to a basis of Newton type of the space ofpolynomials. We also obtain explicit formulas for the coefficients of the three-termrecurrence relation satisfied by the polynomial sequences in H . We have verified that H contains all the 15 families in the Askey scheme and all the 29 families in the q -Askey scheme. The coefficients of the normalized recurrence relation satisfied by eachfamily can be obtained by substitution of suitable values of the parameters in our generalformulas for coefficients of the recurrence relations satisfied by the elements of H .We present some examples of cases in which the generalized difference equation offirst order becomes the usual second order differential or difference equations used tocharacterize some families of orthogonal polynomial sequences.This article is organized as follows. In Section 2 we present preliminary material andsome definitions and notation. In Section 3 we introduce the generalized difference equa-tion of first order and find explicit expressions for its polynomial solutions. In Section4 we find the matrix L that represents the operator of multiplication by the indepen-dent variable with respect to a basis that is a sequence of solutions of the generalizeddifference equation. In Section 5 we consider a special class of generalized differenceequations determined by certain linearly recurrent sequences and show that in that casethe entries of the matrix L satisfy some recurrence relations. Then we show that L becomes tridiagonal if the initial values of the linearly recurrent sequences are relatedin an appropriate way. If L is tridiagonal then the associated polynomial sequence isorthogonal. In Section 6 we define the class H of orthogonal polynomial sequences,describe its main properties and divide it in three sets. In section 7 we study the set H q , which contains all the families in the q -Askey scheme, and we give the values of ourparameters that yield some of the families. In Section 8 we consider the set H , whichcorresponds to q = 1 and contains all the families in the Askey scheme, and we giveexamples of the parameters that produce some of the families. In Section 9 we look atthe set H − , which corresponds to q = −
2. Preliminary material
We present in this section some preliminary material and introduce notation that willbe used in the paper. A more detailed account of the matrix approach to polynomialsequences can be found in [14] and [15]. See also [1] where some properties of doubly-infinite matrices are obtained.A polynomial sequence is a sequence of polynomials p ( t ) , p ( t ) , p ( t ) , . . . with com-plex coefficients such that p n ( t ) has degree n for n ≥
0. Every polynomial sequence is abasis for the complex vector space P of all polynomials in one variable.If { u n } and { v n } are two polynomial sequences then there exists a unique matrix A = [ a n,k ], where ( n, k ) ∈ N × N , such that u n ( t ) = n X k =0 a n,k v k ( t ) , n ≥ . (2 . A is lower triangular and invertible. If all the polynomials u n and v n are monic then a n,n = 1 for n ≥
0. If we consider a fixed polynomial sequence v n thenevery lower triangular invertible matrix A determines another polynomial sequence by(2.1). The n -th row of the matrix A is the vector of coefficients of u n with respect tothe basis { v n } . Equation (2.1) is equivalent to the matrix equation[ u , u , u , . . . ] T = A [ v , v , v , . . . ] T . (2 . A is the matrix of the sequence of polynomials { u k ( t ) : k ∈ N } with respectto the basis { v k ( t ) : k ∈ N } .Let x , x , x , . . . be a sequence of complex numbers and define the polynomials v ( t ) = 1, and v k ( t ) = ( t − x )( t − x )( t − x ) · · · ( t − x k − ) , k ≥ . (2 . { v n } is a polynomial sequence and therefore it is a basis for the space ofpolynomials. It is called the Newton basis associated with the sequence x n . Let V bethe infinite matrix that satisfies[ v ( t ) , v ( t ) , v ( t ) , . . . ] T = V [1 , t, t , . . . ] T . (2 . n -row of V are the coefficients of v n ( t ) with respect to the basis ofmonomials and therefore they are elementary symmetric functions of x , x , x , . . . , x n − ,with the appropriate signs. Therefore the entries in V − are complete homogeneoussymmetric functions of the x k .The dual basis of the Newton basis { v n } is the sequence of divided difference function-als ∆[ x , x , x n − ], which give us the coefficients in the representation of any polynomialin terms of the Newton basis. The basic theory of divided differences can be found in[2] and [13].Let τ be a linear operator on the space of polynomials and let T be its matrixrepresentation with respect to the basis { v k ( t ) : k ∈ N } . If A is the matrix of thesequence of polynomials { u k ( t ) : k ∈ N } with respect to the basis { v k ( t ) : k ∈ N } then AT is the matrix of the sequence { τ u k ( t ) : k ∈ N } with respect to the same basis.Note that the multiplication by T is on the right-hand side, because the polynomial u k corresponds to the k -th row of A .We introduce next some infinite matrices that will be used in the rest of the paper.Let S = . . . . . . . . . . . . ... ... ... . . . . . . , S T = . . . . . . . . . . (2 . S is called the left shift and S T is the right shift.Any sequence of numbers g , g , g , . . . can be used to construct an infinite matrix G defined by G k,k = g k for k ≥ G j,k = 0 if j = k . We say that G is thediagonal matrix associated with the sequence g k . A matrix of the form GS , where G is diagonal and G k,k = 0 for k ≥
1, is called generalized difference matrix of firstorder. If the linear operator represented by GS with respect to the basis { v k ( t ) } isdenoted by γ then γv k ( t ) = g k v k − ( t ), for k ≥
1, and γv ( t ) = 0. A diagonal matrix G , acting by multiplication on the right-hand side, represents an operator of the form φv k ( t ) = g k v k ( t ), which is a rescaling of the basis and can be considered as a generalizeddifference operator of order zero. 4efine the matrix D = . . . . . . . . . . . . ... ... ... . . . . . . . (2 . D is the matrix representation of theusual differential operator, and S T is the representation of multiplication by the variable t of the polynomials.If { u k } k > is a sequence of monic orthogonal polynomials we write the correspondingthree-term recurrence relation in the form α k u k − ( t ) + β k u k ( t ) + u k +1 ( t ) = tu k ( t ) , k > , (2 . σ k = β + β + · · · + β k , for k > . Using the σ k is convenient becausetheir explicit formulas are simpler than those for the β k .
3. The generalized difference equation
We denote by P the complex vector space of all polynomials in one variable. Let v k ( t )be a monic polynomial sequence in P . It is clear that { v k : k ≥ } is a basis for the space P . A generalized difference operator of order one with respect to the basis { v k : k ≥ } is a linear operator γ defined by γv k = g k v k − for k ≥
0, where the sequence of complexnumbers g k satisfies g = 0 and g k = 0 for k ≥
1. A linear operator φ defined by φv k = h k v k , where the h k are complex numbers, is called generalized difference operatorof order zero.Let γ and φ be generalized difference operators with respect to the basis { v k } , of or-ders one and zero, respectively, and let { u n ( t ) : n ≥ } be a monic polynomial sequence.Let us consider the generalized difference equation γu n ( t ) + φu n ( t ) = h n u n ( t ) , n ≥ . (3 . Theorem 3.1.
If the sequence h k satisfies h k = h j , for k = j , then the solution of (3.1)is the polynomial sequence u n ( t ) = n X k =0 c n,k v k ( t ) , n ≥ , (3 . here the coefficients c n,k are given by c n,k = n − Y j = k g j +1 h n − h j , ≤ k ≤ n − , (3 . and c n,n = 1 for n ≥ .Proof: Since { u n ( t ) : n ≥ } is a monic polynomial sequence it is clear that u n ( t )can be written as in equation (3.2) for some matrix of coefficients c n,k with c n,n = 1 for n ≥
0. Then, by the definition of γ we have γu n ( t ) = n X k =1 c n,k g k v k − ( t ) = n − X k =0 c n,k +1 g k +1 v k ( t ) , and the definition of φ gives φu n ( t ) = n X k =0 c n,k h k v k ( t ) . Therefore the difference equation (3.1) becomes n − X k =0 ( c n,k +1 g k +1 + c n,k h k ) v k ( t ) + h n v n ( t ) = n − X k =0 c n,k h n v k ( t ) + h n v n ( t ) , and this gives n − X k =0 ( c n,k +1 g k +1 + c n,k ( h k − h n )) v k ( t ) = 0 . By the linear independence of the polynomial sequence { v k } we obtain c n,k +1 g k +1 = c n,k ( h n − h k ) , ≤ k ≤ n − , and since the numbers h j are pairwise distinct we can write the previous equation in theform c n,k = g k +1 h n − h k c n,k +1 , ≤ k ≤ n − . This recurrence relation clearly gives us (3.3). (cid:3)
Let C be the matrix of coefficients c n,k . It is an invertible lower triangular infinitematrix that satisfies C [ v ( t ) , v ( t ) , v ( t ) , . . . ] T = [ u ( t ) , u ( t ) , u ( t ) , . . . ] T , n ≥ . (3 . w k ( t ) as follows, w ( t ) = 1 and w k ( t ) = ( t − h )( t − h )( t − h ) · · · ( t − h k − ) , k ≥ . (3 . h k . We will use the notation w n,k ( t ) = w n ( t ) w k ( t ) = n − Y j = k ( t − h j ) , ≤ k ≤ n − . (3 . Theorem 3.2.
Let C − = [ˆ c n,k ] . Then ˆ c n,k = Q nj = k +1 g j w ′ n +1 ,k ( h k ) = n Y j = k +1 g j h k − h j , ≤ k ≤ n − , (3 . and ˆ c n,n = 1 for n ≥ . Proof:
Let us note that the denominator in (3.3) is equal to w ′ n +1 ,k ( h n ) if k < n . Let k < n and let ˆ c n,j be defined by (3.7). Then n X j = k ˆ c n,j c j,k = n X j = k (cid:16)Q ni = j +1 g i (cid:17) (cid:16)Q ji = k +1 g i (cid:17) w ′ n +1 ,j ( h j ) w ′ j +1 ,k ( h j ) = n Y i = k +1 g i ! n X j = k w ′ n +1 ,k ( h j ) = 0 , (3 . h k , h k +1 , . . . , h n , whichare at least two nodes, since k < n . That sum is also the sum of the residues of1 /w n +1 ,k ( t ). The basic properties of divided differences can be found in [2] or [13].For k = n we get ˆ c n,n c n,n = 1 for n ≥
0. Therefore the matrix product [ˆ c n,k ][ c n,k ] isequal to the infinite identity matrix and this completes the proof. (cid:3)
4. The operator of multiplication by the variable t We let now { v k } be the Newton basis associated with a sequence x , x , x , . . . . Thatis, v ( t ) = 1 and v k ( t ) = ( t − x )( t − x ) · · · ( t − x k − ) , k ≥ . (4 . V be the matrix whose ( n, k ) entry is the Taylor coefficient of t k in the polynomial v n ( t ). The entries of the inverse matrix V − are the complete homogeneous symmetricpolynomials of the nodes x j , that is,( V − ) n,k = X x i x i · · · x i n − k , (4 . i , i , . . . , i n − k ) with entries in { x , x , . . . , x k } .Therefore V and V − are change of bases matrices that satisfy V [1 , t, t , . . . ] T = [ v ( t ) , v ( t ) , v ( t ) , . . . ] T , and V − [ v ( t ) , v ( t ) , v ( t ) , . . . ] T = [1 , t, t , . . . ] T . Let us note that CV is the matrix of coefficients of the polynomials u k ( t ) with respectto the basis of monomials { t k : k ≥ } .Since tv k ( t ) = (( t − x k ) + x k ) v k ( t ) = v k +1 ( t ) + x k v k ( t ) , k ≥ , the matrix representation with respect to the basis { v k ( t ) : k ≥ } of the map p ( t ) → tp ( t ) on the space P is S T + F , where F is the diagonal matrix whose ( k, k ) entry is x k ,for k ≥ L be the matrix representation with respect to the basis { u k ( t ) : k ≥ } of theoperator of multiplication by t . Using equation (3.4) we obtain L = C ( S T + F ) C − . (4 . C and C − give us L n,k = n Y j = k +1 g j ! n +1 X j = k ( h n − h j − ) x j + g j w j +1 ,k +1 ( h k ) w n,j − ( h n ) , ≤ k ≤ n, (4 . L n,n +1 = 1 for n ≥ L n,n = x n + g n +1 h n − h n +1 − g n h n − − h n , (4 . L n,n − = g n h n − − h n (cid:18) g n − h n − − h n − g n h n − − h n + g n +1 h n − − h n +1 + x n − x n − (cid:19) . (4 . h k , x k , and g k are certainlinearly recurrent sequences the matrix L is tridiagonal and therefore the polynomialsequence { u k ( t ) : k ≥ } is orthogonal, and the entries of L are the coefficients of thethree-term recurrence relation. 8 . The family of orthogonal polynomial sequences In this section we study the polynomial sequences obtained when the sequences h k and x k satisfy a particular type of linear difference equation of third order and g k satisfiesa linear difference equation of fifth order related with the equation satisfied by h k and x k . We will show that choosing appropriate initial values for the sequence g k the matrix L becomes tridiagonal and the polynomial sequence { u k ( t ) : k ≥ } becomes orthogonal.Let z be a complex number and consider the difference equation s k +3 = z ( s k +2 − s k +1 ) + s k , k ≥ . (5 . t − zt + zt −
1. The sum of its rootsequals z and the product of the roots is equal to 1. Since 1 is a root we see that theroots can be expressed as 1, q , and q − for some nonzero complex number q . If z isreal and − ≤ z ≤ z = 1 + 2 cos( θ ) and then it is easy to see that q = cos( θ ) + i sin( θ ) and hence q and q − have modulus one. If z = 3 then 1 is a rootof the characteristic polynomial with multiplicity three. If z = − − h k and x k are solutions of the differenceequation (5.1) and that h k = h n for k = n . It is easy to verify that the pointwise (orHadamard) product of two solutions of (5.1) satisfies the difference equation of orderfive s k +5 = ( z − z − s k +4 − s k +1 ) − ( z − z − z − s k +3 − s k +2 ) + s k , k ≥ . (5 . t − ( z − z − t + 1)( t − zt + zt − , q, q − , q , q − .From now on we suppose that the sequence g k satisfies (5.2) and also g = 0 and g k = 0 for k ≥ . Therefore the matrices C , C − , and L are completely determined by z and the initial values h , h , h , x , x , x , and g , g , g , g .Let us note that the entries of the matrices C , C − , and L are functions of thedifferences h k − h n .For fixed m > δh ) m,k = h k − h k + m satisfies the difference equation s k +2 = ( z − s k +1 − s k , k ≥ , (5 . δh ) m, = h − h m and ( δh ) m, = ( h − h m ). This is a simple three-term recurrence relation which is related with the recurrence satisfied by the Chebyshevfamilies of orthogonal polynomials. 9et p k , r k , and y k be the solutions of (5.3) determined by the initial values p = 1and p = z − r = 1 and r = z ; y = 2 and y = z −
1. The sequence p k satisfies p k = 2 k ˆ U k (cid:18) z − (cid:19) , k ≥ , (5 . U k ( t ) are the monic Chebyshev polynomials of the second kind. The sequence r k satisfies a similar equation with modified initial conditions, and also satisfies r k +2 = zp k +1 − p k for k ≥
0. The sequence y k satisfies y k = 2 k ˆ T k (cid:18) z − (cid:19) , k ≥ , (5 . T k ( t ) are the monic Chebyshev polynomials of the first kind. The polynomials y k also satisfy y k +2 = ( z − p k +1 − p k for k ≥ p k and the shifted sequence p k − form a basis for the space ofsolutions of (5.3) it is easy to see that( δh ) m,k +2 = ( δh ) m, p k +1 − ( δh ) m, p k , k ≥ , m ≥ , (5 . δh ) m,k = ( δh ) ,k − m p m − , (5 . δh ) m +1 ,k = ( δh ) ,k + m r m . (5 . h k − h n = 0 for k = n is satisfied if z is not a root of any ofthe orthogonal polynomials p k and r k for k ≥ δh ) ,k = 0 and ( δh ) ,k = 0 for k ≥
0. Since the roots of p k +1 and p k are interlaced, by (5.6), (5.7), and (5.8) we seethat it is enough to have h , h , h pairwise distinct and z ( h − h ) = h − h . Let usnote that because of their relation with the Chebyshev polynomials the polynomials p k and r k have their roots in the interval [ − , L depend on the recurrentsequences h k , x k , g k . Therefore it seems reasonable to expect that such entries satisfysome kind of recurrence relation. We find first a recurrence relation for three consecutiveentries on a diagonal.Define τ ( n, k ) = n Y j = k +1 g j ! w n +2 ,k +1 ( h k ) w n,k +2 ( h n ) , ≤ k ≤ n − . (5 . L n,k is a divisor of τ ( n, k ).10 straightforward computation gives us the recurrence relation τ ( n + 2 , k + 2) L n +2 ,k +2 − y n − k τ ( n + 1 , k + 1) L n +1 ,k +1 + τ ( n, k ) L n,k = 0 , ≤ k ≤ n − , (5 . y j is the sequence of orthogonal polynomials in z related with the monic Chebyshevpolynomials of the first kind that satisfies (5.3) and (5.5).We will find next a recurrence relation on L similar to the recurrences satisfied bythe entries of the matrix of Stirling numbers of the second kind and other matrices ofgeneralized binomial coefficients.Define the functions ǫ ( n, k ) = w n +2 ,k +1 ( h k ) w n,k +2 ( h n ) ,ǫ ( n, k ) = g n +1 w k +1 ,k ( h k +1 ) w n +3 ,k +1 ( h k ) w n +1 ,k +2 ( h n +1 ) w n +1 ,n − ( h n +1 ) ,ǫ ( n, k ) = g k w n,n − ( h n ) w n +2 ,k +2 ( h k − ) w n,k +1 ( h n ) . (5 . L satisfy the recurrence relation p n − k − ǫ ( n, k ) L n,k − ǫ ( n − , k ) L n − ,k + ǫ ( n, k +1) L n,k +1 = 0 , ≤ k ≤ n − , (5 . p j is the sequence of orthogonal polynomials of z that satisfies (5.3) and (5.4).From the recurrence relations (5.10) and (5.12) we can see that if L , = 0 and L , = 0 then L m + k,k = 0 for m ≥ k ≥
0. That is, L is tridiagonal. Using theexplicit formula (4.4) for the entries of L we can solve L , = 0 for g and then solve L , = 0 for g . We obtain g = z ( x ( h − h ) + x ( h − h ) + x ( h − h ) + g − g ) , (5 . g = ( x − x )( h − h ) z − ( x − x )( h − h ) z + ( g − g ) z +( x ( h − h ) + x ( h − h ) + x ( h − h )) z − g z + g . (5 .
6. The three main classes of orthogonal polynomial sequences
Let us summarize some of the results in the previous sections. Let z be a complexnumber and let h k and x k be sequences that satisfy the recurrence relation (5.1) s k +3 = z ( s k +2 − s k +1 ) + s k , k ≥ , g k be a sequence that satisfies the recurrence relation (5.2) s k +5 = ( z − z − s k +4 − s k +1 ) − ( z − z − z − s k +3 − s k +2 ) + s k , k ≥ . Suppose that z and the initial values h , h , h are such that h k − h n = 0 if k = n and that g = 0 and g and g are given by (5.13) and (5.14) respectively. Define thepolynomial sequence u n ( t ) = n X k =0 c n,k v k ( t ) , n ≥ , where the v k are the Newton polynomials associated with the sequence x k and thecoefficients c n,k are c n,k = n − Y j = k g j +1 h n − h j , ≤ k = n − , and c n,n = 1 for n ≥
0. Then the monic polynomial sequence u n ( t ) is orthogonal andsatisfies the three-term recurrence relation u n +1 ( t ) = ( t − β n ) u n ( t ) − α n u n − , n ≥ , (6 . α n = L n,n − , given by (4.6), and β n = L n,n , given by (4.5).The sequence u n ( t ) also satisfies the generalized difference equation of first order γu n ( t ) + φu n ( t ) = h n u n ( t ) , n ≥ , (6 . γ and φ were defined in Section 3.The generalized moments with respect to the basis { v k ( t ) : k ≥ } are m n = ˆ c n, = n Y k =1 g k h − h k , n ≥ , (6 . m = 1. Note that the generalized moments satisfy a recurrence relation of orderone.The standard moments µ n , with respect to the basis of monomials t n , are given by[ µ , µ , µ . . . ] T = V − [ m , m , m , . . . ] T , (6 . V − is defined in (4.2).Let H denote the set of all the orthogonal polynomial sequences u k ( t ) obtainedby the procedure described above. The sequences in H are determined by z and the12nitial values h , h , h , x , x , x , g , g . The parameters z, h , h , h are restricted by thecondition h k = h n if k = n .If the nodes x k are all distinct, that is, x j = x k for j = k , then, if some additionalconditions are satisfied, the orthogonality of the polynomials u k ( t ) can be expressed asa discrete orthogonality of the form h u k , u n i = ∞ X j =0 r j u k ( x j ) u n ( x j ) = δ k,n , (6 . r j is a weight associated with the node x j for j ≥
0. The weights r j are determinedby the equations m k = ∞ X j =0 r j v k ( x j ) , k ≥ , (6 . m , m , m , . . . ] = [ r , r , r , . . . ] P, (6 . P is the lower triangular matrix defined by P j,k = v k ( x j ), for j ≥ k ≥
0. Ifthe nodes x j are distinct then P is invertible, and using some basic properties of divideddifferences we obtain P − j,k = ( v ′ j +1 ( x k )) − . Therefore (6.7) yields[ r , r , r , . . . ] = [ m , m , m , . . . ] P − , (6 . r k = ∞ X j = k m j v ′ j +1 ( x k ) , k ≥ . (6 . P and P − are scaled generalized lower triangular Pascalmatrices.The roots of the characteristic polynomial of the difference equation (5.1) are 1 , q, q − where q is a nonzero complex number. If we consider the possible multiplicities of theroots in order to classify the elements of H we find three cases:1. The 3 roots are distinct, that is q = 1 and q = − q = 1, that is, 1 is a root with multiplicity 3.3. q = −
1, that is, − z = 1 + q + q − , in case 2 we have z = 3 and in case 3 we have z = −
1. Let H q , H , and H − denote the corresponding subsets of H for each one of the 3 cases.13 . The class H q In this case the roots of the characteristic polynomial t − zt + zt − , q, q − ,with q = 1 and q = −
1. Therefore the roots are distinct and the general solution of thedifference equation (5.1) can be expressed as s k = λ + λ q k + λ q − k , k ≥ , (7 . λ j are complex numbers. Since the sequences h k and x k aresolutions of (5.1) they can be expressed as h k = a + a q k + a q − k , k ≥ , (7 . a = ( q + 1) h − q ( h + h )( q − ,a = − q ( h − h ) + h − h ( q + 1)( q − ,a = q ( q ( h − h ) + h − h )( q + 1)( q − , (7 . x k = b + b q k + b q − k , k ≥ , (7 . b = ( q + 1) x − q ( x + x )( q − ,b = − q ( x − x ) + x − x ( q + 1)( q − ,b = q ( q ( x − x ) + x − x )( q + 1)( q − . (7 . b corresponds to a translation of the Newton polynomials v k ( t ) and therefore to a translation of the polynomials u k ( t ).The characteristic polynomial of the difference equation (5.2) has roots 1 , q, q − , q , q − and therefore the general solution of (5.2) has the form s k = λ + λ q k + λ q − k + λ q k + λ q − k , k ≥ , (7 . λ j are complex numbers. Since the sequence g k is a solution of(5.2) it can be expressed as g k = d + d q k + d q − k + d q k + d q − k , k ≥ . (7 . g = 0 and g and g are given by (5.13) and (5.14) respectively, we obtain d = − ( a b q + d + d + a b q − ) , d = a b q − , d = a b q, (7 . d and d are arbitrary parameters. Therefore we have g k = ( q k − d + ( q − k − d + q − ( q k − a b + q ( q − k − a b , k ≥ . (7 . a , a , a , b , b , b , d , d by usingequations (7.2), (7.4), and (7.7). The entries α k = L k,k − , for k ≥
1, are given by α k = ( q k − a + a )( q k a b − a b ) + q k ( q k − a d − a d )( q k a − a )( q k − a − a ) ( q k − a − a ) × ( q k − q k − a − a ) (( q k + 1)( q k − a b − a b ) + q k − ( q k d − d )) . (7 . β k = L k,k and let σ k = β + β + · · · + β k , for k ≥
0. Then we obtain σ k = ( k + 1) b + (cid:18) q k +1 − q − (cid:19) ( q k a + a )( qb − b ) − q k +1 d + d q k +1 a − a , (7 . α k = 0, for k ≥
1, the parameter q mustnot be a root of 1. Let us note that if q k a − a = 0 for k ≥ α k and β k are welldefined. Note also that α k is independent of a and b , and β k is independent of a .Equation (7.2) gives us h k − h j = ( q k − q j ) a + ( q − k − q − j ) a = ( q − j − q − k )( q k + j a − a ) , k = j, (7 . q n a = a for n ≥
1. With equations (7.12) and (7.9)we can express the entries of the matrices C and C − in terms of the parameters a , a , a , b , b , b , d , d . For example, the quotients g k h − h k = d − q k d − q k − (1 + q k ) a b + q (1 + q − k ) a b q k a − a , k ≥ , (7 . m n = ˆ c n, , given by equation (6.3).15n the generalized difference equation (3.1) the operators γ and φ are defined withrespect to the basis { v k ( t ) : k ≥ } . Using the matrices V and V − we can transformequation (3.1) and obtain an equation with respect to the basis of monomials { t k : k ≥ } . If h = 0, b = 0, and b = 0 we obtain the q -difference equation of second order (cid:18) ( q − q f ( t ) D q D /q + q − q f ( t ) D q (cid:19) u k ( t ) = h k u k ( t ) , k ≥ , (7 . f ( t ) = ( qa − a ) t + qd + a b + a b − d , f ( t ) = a t + ( d − a b ) t − b d ,h k is given by (7 .
2) and D q and D /q are the usual q -difference and q − -difference oper-ators. There is a similar equation in the case with h = 0, b = 0, and b = 0.The class H q contains all the families of basic hypergeometric orthogonal polynomialsequences in the q -Askey scheme [5]. The coefficients of the normalized three-termrecurrence relation for each of the families listed in Chapter 14 of [5] are obtained bydirect substitution of appropriate values of the parameters a , a , b , b , b , d , d , withouttaking limits. The only case that requires a limit is the family of continuous q -Hermitepolynomials. We give next some examples.The Askey-Wilson polynomials are obtained with a = abcdq − a , b = 0 , b = a, b = a − ,d = − a ( abcd + q ( bc + bd + cd )) q − a , d = − (( b + c + d ) + qa − ) a , (7 . a is an arbitrary nonzero number and a, b, c, d are the parameters used in [5, eq.14.1.5].The q -Racah polynomials are obtained with a = αβqa , b = 0 , b = γδq, b = 1 ,d = − q ( αβγδ + αβδ + βγδ + αγ ) a , d = − q ( βδ + α + γ + 1) a , (7 . a is an arbitrary nonzero number and α, β, γ, δ are the parameters used in [5, eq.14.2.4].The continuous dual q -Hahn polynomials are obtained with a = 0 , b = 0 , b = a , b = 12 a ,d = − abca q , d = − ( ab + ac + q ) a a , (7 . a is an arbitrary nonzero number and a, b, c are the parameters used in [5, eq.14.3.5].The Al-Salam-Chihara polynomials are obtained with a = 0 , b = 0 , b = a, b = a − ,d = 0 , d = − a ( b + qa − ) , (7 . a is any nonzero number and a, b are the parameters in [5, eq. 14.8.5].For the big q -Jacobi polynomials we have a = abqa , b = 0 , b = aq, b = 0 ,d = − ( ab + b + c ) aqa , d = − cqa , (7 . a is any nonzero number and a, b, c are the parameters in [5, eq. 14.5.4].For the q -Meixner polynomials we have a = 0 , b = 0 , b = bq, b = 0 ,d = a ( bc − b − , d = ca , (7 . a is an arbitrary nonzero number and b, c are the parameters in [5, eq. 14.13.4].
8. The class H In this section we consider the family of orthogonal polynomial sequences obtainedwhen we take z = 3. In this case the characteristic polynomial of the difference equation(5.1) is t − t + 3 t − t − and therefore the general solution of (5.1) has theform s k = λ + λ k + λ k ( k − , k ≥ , (8 . λ j are arbitrary complex numbers.Since the sequences h k and x k are solutions of (5.1) we can write them as follows. h k = a + a k + a k ( k − , k ≥ , (8 . a = h , a = h − h , and a = ( h − h + h ) /
2, and x k = b + b k + b k ( k − , k ≥ , (8 . b = x , b = x − x , and b = ( x − x + x ) / t − and therefore the sequence g k , which is a solution of (5.2), with g = 0 and g and g given by (5.13) and (5.14) respectively, can be expressed as g k = d k + d k ( k −
1) + d k ( k − k −
2) + d k ( k − k − k − , k ≥ , (8 . d and d are arbitrary numbers and d = a b + a b + 2 a b , d = a b . (8 . h m − h n = ( m − n )( a + ( m − n ) a ) , m ≥ , n ≥ , and therefore the condition h m = h n , for m = n , is satisfied if a + ka = 0 for k ≥ α k = g k ( a + ( k − a )( a + (2 k − a )( a + (2 k − a ) ( a + (2 k − a ) × (( a + ( k − a )( a b k ( k −
1) + ( a b − a b ) k − a b − a b + d ) − d a ) , (8 . g k is given by (8.4) and σ k = ( k + 1) (cid:18) b − d + e k + e k ( k −
1) + e k ( k − k − a + 2 ka (cid:19) , k ≥ , (8 . e = − (1 / a b + 2 a b − d ) , e = (2 / b ( a + a ) , e = (1 / a b . Recallthat β = σ and β k = σ k − σ k − for k ≥ g k h − h k = − d + d ( k −
1) + 2( a b + a b + 2 a b ) (cid:0) k − (cid:1) + 6 a b (cid:0) k − (cid:1) a + ( k − a , k ≥ , (8 . m n = ˆ c n, given by (6.3). The entries of thematrices C and C − are products of terms similar to (8.8).If b , b , and b are equal to zero then the basis { v k ( t ) } is the standard basis ofmonomials { t k } and in such case the generalized difference equation (3.1) can be writtenin the form (( d t + a t ) D + ( d + a t ) D + a ) u k ( t ) = h k u k ( t ) , k ≥ , (8 . D denotes differentiation with respect to t and h k is given by (8.2).If we take b = 0 and b = 0 then the basis { v k ( t ) } becomes the basis of translatedpowers { ( t − b ) k : k ≥ } and the matrices V and V − are generalized Pascal matrices.Transforming the generalized difference equation (3.1) to the basis of monomials weobtain the equation(( b ( a b − d ) + ( d − a b ) t + a t ) D + ( d − a b + a t ) D + a ) u k ( t ) = h k u k ( t ) , (8 . D denotes differentiation with respect to t . If we put b = 0 in (8.10) we obtain(8.9).If b = 0 and b = 0 then the basis { v k ( t ) } is the Newton basis associated with thesequence b + b k , for k ≥
0. In this case the generalized difference equation (3.1) canbe expressed in matrix form as C ( a b D ( S T ) + D ( d S T + a ( S T ) ) + D ( d I + a S T ) + a I ) = HC. (8 . D is the matrix representation of the difference operator ∆ defined by∆ v k ( t ) = v k ( t + b ) − v k ( t ) b = kv k − ( t ) , k ≥ , (8 . S T is the matrix representation of the forward shift operator E defined by Ev k ( t ) = v k +1 ( t ) , k ≥ , (8 . H is the diagonal matrix with entries h , h , h , . . . in the main diagonal. Observethat (8.11) becomes a difference equation of order 2 when a = 0.The class H contains all the families of hypergeometric orthogonal polynomial se-quences in the Askey scheme. The coefficients of the normalized three-term recurrencerelation for each family listed in Chapter 9 of [5] are obtained by giving appropriatevalues to our parametrs a , a , b , b , b , d , d , without taking limits. The only case thatrequires limits is the family of Hermite polynomials. We give next some examples.The Wilson polynomials are obtained with a = ( a + b + c + d ) a , b = − a , b = − a − , b = − ,d = − a ( a + b )( a + c )( a + d ) , d = − a ((2 a +1)( a + b + c + d +1)+ a + bc + bd + cd ) , (8 . a is any nonzero number and a, b, c, d are the parameters in [5, eq. 9.1.5].The Racah polynomials, in the case α + 1 = − N , are obtained when a = (1 + β − N ) a , b = 0 , b = 2 + γ + δ, b = 1 , = − N (1 + γ )(1 + β + δ ) a , d = − a ((3 + β + γ + δ ) N − (2 + γ )(2 + δ + γ )) , (8 . a is any nonzero number and αβ, γ, δ, N are the parameters in [5, eq. 9.2.4].For the continuous dual Hahn polynomials we have a = 0 , b = − a , b = − a − , b = − ,d = − a ( a + c )( a + b ) , d = − a (2 a + b + c + 1) , (8 . a is any nonzero number and a, b, c are the parameters in [5, eq. 9.3.5].For the continuous Hahn polynomials we have a = a ( a + b + c + d ) , b = ib, b = i, b = 0 ,d = ia ( b + bc + bd + cd ) , d = ia (2 b + c + d + 1) , (8 . a is any nonzero number and a, b, c, d are the parameters in [5, eq. 9.4.4].The Meixner-Pollaczek polynomials are obtained with a = 0 , b = − iλ, b = − i, b = 0 ,d = a λ (cid:18) cos( φ )sin( φ ) − i (cid:19) , d = a (cid:18) cos( φ )sin( φ ) − i (cid:19) , (8 . a is any nonzero number and λ, φ are the parameters in [5, eq. 9.7.4].For the Jacobi polynomials we have a = a (2 + α + β ) , b = 1 , b = 0 , b = 0 ,d = 2 a ( α + 1) , d = 2 a , (8 . a is any nonzero number and α, β are the parameters in [5, eq.9.8.5].The Bessel polynomials are obtained with a = ( a + 2) a , b = 0 , b = 0 , b = 0 , d = 2 a , d = 0 , (8 . a is any nonzero number and a is the parameter in [5, eq. 9.13.4].20 . The class H − We consider now the class of orthogonal polynomial sequences obtained when wetake z = −
1. In this case the characteristic polynomial of the difference equation (5.1)is t + t − t − t − t + 1) and its roots are 1 , − , −
1. Therefore the generalsolution of (5.1) has the form s k = λ + λ ( − k + λ k ( − k , k ≥ , (9 . λ j are arbitrary numbers. We can write the sequences h k and x k , which aresolutions of (5.1), as h k = a + a ( − k + 2 a k ( − k , k ≥ , (9 . x k = b + b ( − k + 2 b k ( − k , k ≥ . (9 . a and b in order to simplify the notation.When z = − , , , − , − g k = d + d ( − k + 2 d k ( − k + 2 d k + 2 d k ( k − , k ≥ . (9 . g = 0 and g and g must satisfy (5.13) and (5.14) respectively, we obtain d = − d , d = − a b − a b , d = − a b . (9 . d and d are arbitrary numbers.Equation (9.2) gives us h n − h m = (( − n − ( − m ) a + 2(( − n n − ( − m m ) a , and this shows that the condition h n − h m = 0 for n = m is satisfied if a is not zeroand a − na = 0 for n ≥ h k are a + a , a − a − a , a + a + 4 a , a − a − a , . . . . This shows that the terms with even indices and the terms with odd indicesbehave differently. The sequences x k and g k have a similar property.The coefficients of the three-term recurrence relation also have different expressionsfor even and odd indices. For n even we have α n = − n ( a + ( n − a )( a b + a b − d + 2( n − a b )(2 na b + a b − a b + d ) a ( a + (2 n − a ) . (9 . n odd we have α n = − d − n ( a b + a b + d ) − n ( n − a b a ( a + (2 n − a ) × (( a ( a b − a b − d )+ a d + a ((3 n − a b − ( n − a b + d )+2 n ( n − a b ))) . (9 . α n is a rational function of n with numerator of degree 4 anddenominator of degree 2.For the coefficients σ n = β + β + · · · + β n we have, if n is even σ n = ( n + 1) b + n ( a b − a b − d ) + a b − a b − d − d a + (2 n + 1) a , (9 . n is odd then σ n = ( n + 1) (cid:18) b + a b − a b − d ) a + (2 n + 1) a (cid:19) . (9 . g k h − h k = 4 k a b − ( − k ( d + 2 kd ) − k (2 a b − a b − a b ) + d ( − k ( a + 2 ka ) − a , k ≥ . (9 . b = 0 and b = 0 then the matrix representation B of the difference operator in(3.1) with respect to the basis of monomials is a sum of Kronecker products of infinitematrices and 2 × B = I ⊗ B + D ⊗ B + ( DS T ) ⊗ B , (9 . ⊗ denotes the Kronecker product and B = (cid:20) a + a a b − d − d a − a − a (cid:21) , B = (cid:20) − b d − a b + 4 d b d (cid:21) , and B = (cid:20) a a b − d − a (cid:21) . If A = CV is the matrix of coefficients of the sequence { u k ( t ) } with respect to themonomial basis then the difference equation (3.1) is in this case equivalent to the matrixequation AB = HA . 22 eferences [1] M. I. Arenas-Herrera and L. Verde-Star, Representation of doubly infinite matricesas non-commutative Laurent series, Spec. Matrices 5 (2017) 250–257.[2] S. D. Conte and C. de Boor, Elementary numerical analysis, McGraw-Hill, NewYork, 1980.[3] R. S. Costas-Santos and F. Marcell´an, q -classical orthogonal polynomials: a generaldifference calculus approach. Acta Appl. Math. 111 (2010), no. 1, 107–128.[4] M. E. H. Ismail, Classical and quantum orthogonal polynomials in one variable,Cambridge Univ. Press, Cambridge, 2005.[5] R. Koekoek, P. A. Lesky, and R. F. Swarttouw, Hypergeometric orthogonal poly-nomials and their q -analogues, Springer-Verlag, Berlin, Heidelberg, 2010.[6] W. Koepf and D. Schmersau, Recurrence equations and their classical orthogonalpolynomial solutions, Orthogonal systems and applications, Appl. Math. Comput.128 (2002) 303–327.[7] T. H. Koornwinder, The Askey scheme as a four-manifold with corners. RamanujanJ., 20 (2009) 409–439.[8] F. Marcell´an, A. Branquinho, and J. Petronilho, Classical orthogonal polynomials:a functional approach, Acta Appl. Math. 34 (1994) 283–303.[9] M. Masjed-Jamei, F. Soleyman, I. Area, and J. J. Nieto, On ( p, q )-classical orthog-onal polynomials and their characterization theorems, Adv. Difference Equ. (2017)Paper No. 186, 17 pp.[10] M. Njinkeu Sandjon, A. Branquinho, M. Foupouagnigni, and I. Area, Characteriza-tion of classical orthogonal polynomials on quadratic lattices, J. Differ. Equ. Appl.23 (2017) 983–1002.[11] P. Njionou Sadjang, W. Koepf, and M. Foupouagnigni, On moments of classicalorthogonal polynomials. J. Math. Analysis Appl. 424 (2015) 122–151.[12] D. D. Tcheutia, Recurrence equations and their classical orthogonal polynomialsolutions on a quadratic or a q-quadratic lattice. J. Difference Equ. Appl. 25 (2019),no. 7, 969–993. 2313] L. Verde-Star, Divided differences and combinatorial identities, Stud. Appl. Math.85 (1991) 215–242.[14] L. Verde-Star, Characterization and construction of classical orthogonal polynomi-als using a matrix approach, Linear Algebra Appl. 438 (2013) 3635–3648.[15] L. Verde-Star, Recurrence coefficients and difference equations of classical discreteand qq