On The Connection Problem Between Two Classical Orthogonal Polynomial Sequences

On The Connection Problem Between Two Classical Orthogonal Polynomial Sequences

Imed Ben Salah

Received 16 December 2014

Abstract

In this paper, we solve the following connection problem

(x)Qn(x) =

n+deg

X

k=0

n;kPk(x)forn 0;

wherefPng^{n 0} and fQng^{n 0} are two MOPS and is a monic polynomial. We establish a method for computing the coe¢ cient n;kstep by step. As application, we apply this process for some continuous, discrete and quantum classical MOPS with the choicedeg 2and some new relationships are obtained. In particular, some well known formulas such as duplication, addition are derived.

1 Introduction and Preliminaries

Given two MPS fPng^{n 0} andfQng^{n 0} and a monic polynomial , the so-called con- nection problem between them, i.e. the computation of coe¢ cients n;kin the following expression

(x)Qn(x) =

n+degX

k=0

n;kPk(x)forn 0: (1)

plays an important role in many problems in pure and applied mathematics (see for instance [6] for adequate references). The literature on this topic is extremely vast and a wide variety of methods, based on speci…c properties of the involved polynomi- als, have been developed using several techniques for (x) = 1 [1, 2, 6, 7, 8, 9]. In the context of the connection problem (1), we are dealing in this contribution with a numerical method to compute the coe¢ cient n;k step by step. Some illustrative examples from the classical continuous, discrete andq-discrete case (Hermite, Meixner and Little q-Laguerre) are highlighted for some monic polynomials with deg 2.

As consequence, some new connections are obtained and some well known formulas such as duplication, addition are recovered.

LetP be the vector space of polynomials with coe¢ cients in C and letP⁰ be its dual. We denote by hu; fi the e¤ect of u 2 P⁰ on f 2 P. In particular, we denote

Mathematics Sub ject Classi…cations: 33C45, 42C05.

yFaculté des Sciences de Monastir, Département de Mathématiques,5019, Monastir, Tunisia.

63

by (u)_n := hu; xⁿi; n 0, the moments of the form u (linear functional). Let us introduce some useful operations in P⁰. For any form u, any polynomial g, and any (A; B)2C f0g C, letgu,hAu, and Bube the forms de…ned by duality

hgu; fi:=hu; gfi; hhAu; fi:=hu; hAfi; h ^Bu; fi=hu; Bfi; for allf 2 P where(hAf)(x) =f(Ax)and( Bf)(x) :=f(x+B)[3, 5].

LetfPng^{n 0} be a sequence of monic polynomials with degPn =n; n 0 (MPS) and let fu_ngn 0 be its dual sequence, u_n 2 P⁰ de…ned byhu_n; P_mi:= _n;m; n; m 0:

The sequence fP_ngn 0 is called orthogonal (MOPS) if we can associate with it a form u(with(u)₀= 1) and a sequence of numbersfr_ngn 0 (r_n 6= 0; n 0) such that [3, 5]

hu; P_mP_ni=r_{n n;m} forn; m 0:

The form u is then said to be regular. The MOPS fPng^{n 0} ful…ls the three-term recurrence relation [3, 5]

( P0(x) = 1,P1(x) =x ₀;

P_n+2(x) = (x _n+1)P_n+1(x) _n+1P_n(x)forn 0; (2) where

n= hu; xP_n²i rn

and _n+1= r_n+1

rn 6= 0forn 0:

The regular formuis positive de…nite if and only if _n2Rand n+1>0forn 0. cf.

[3, 5].

If we consider the shifted monic polynomialsPe_n(x) =A ⁿP_n(Ax+B) forn 0, thenfPe_ngn 0 is also a MOPS and its recurrence coe¢ cients are [3, 5]

e_n = ⁿ B

A anden+1= ⁿ⁺¹

A² forn 0: (3)

A formuis said to be symmetric if and only if(u)_2n+1= 0forn 0. A MPSfP_ngn 0

is symmetric if and only ifP_n( x) = ( 1)ⁿP_n(x)forn 0. cf. [3, 5]. LetfP_ngn 0 be a MOPS with respect tou, then

uis symmetric() fPng^{n 0} is symmetric() n= 0forn 0:

cf. [3, 5].

In the sequel, letfPng^{n 0} be a MOPS with respect tou0 and satisfying (2) and fQng^{n 0} be a MOPS ful…lling

( Q0(x) = 1; Q1(x) =x ₀;

Q_n+2(x) = (x _n+1)Q_n+1(x) _n+1Q_n(x)forn 0: (4)

2 The Method

The scope of this section is to give recurrence relations in order to be able to calculate by induction the coe¢ cients n;k between fPng^{n 0} and fQng^{n 0} with respect to (t= deg 0) given by the expansion of Qn in terms of thePn basis. We may write (1) in the following way

(x)Q_n(x) =X

k2Z

n;kP_k(x)forn 0: (5)

By virtue of (5), (4) and (2), we get the following formula

n;n+t j = 0;t j+

nX1 k= j

( _{k+t+1 j k}) k;k+t+1 j+ k+t+2 j k;k+t+2 j

k k 1;k+t+1 j (6)

for n max(1; j+ 1); where j = max(0; j t 1); 0 j n+t; and the initial conditions are reached in the following values( 0;k)0 k t. Moreover,

n;k= 0for either k 1ork n+t+ 1; n 0: (7) We are going to detail the process (6)–(7). For j= 0, we have

n;n+t= 1forn 0: (8)

Forj= 1;we have 1= 0:Taking (6)–(7) into account, we get

n;n+t 1= 0;t 1+

n 1

X

k=0

( _{k+t k})forn 1: (9)

Forj= 2 in (6)-(7), two cases arise:

(i) If t 1, then 2= 0. Therefore, forn 2,

n;n+t 2= _{0;t 2}+

nX1 k=0

( _{k+t 1 k}) _{k;k+t 1}+ _t+

n 1

X

k=1

( _{k+t k}) (10) and, forn= 1;

1;t 1= 0;t 2+ ( _{t 1 0}) 0;t 1+ t: (11) (ii) If t= 0, then ₂= 1. Therefore, forn 2;

n;n 2=

n 1

X

k=1

n

( _{k 1 k}) k;k 1+ ( k k)o

: (12)

If we suppose that for an integerj satisfying0 j+ 1 n+t;all the coe¢ cients

k;k+t (j 1)and k;k+t j;0 k n 1have been calculated, then using (6)-(7) with the changej j+ 1 yields

n;n+t (j+1) = 0;t j 1+

nX1 k= j+1

( _{k+t j k}) k;k+t j

+ _{k+t+1 j k;k+t (j 1) k k 1;k 1+t (j 1)} : (13) Hence, it is possible to determine _{n;n+t (j+1)}forn max(1; j+1+ 1):

REMARK 1. On account of (4), we obtain

(x+c)Qn(x) =Qn+1(x) + (c+ _n)Qn(x) + _nQn 1(x)forn 0; c2C: (14)

REMARK 2. When (x) =x²+cx+d; c; d2Cand using the previous relation, the coe¢ cientsf n;kgn;k 0 betweenfQ_ngn 0 andfQ_ngn 0 by respect to are given by

8>

>>

<

>>

>:

n;n+2= 1; _n;n+1=c+ _n+ _n+1 forn 0;

n;n=d+c _n+ ²_n+ _n+ _n+1 forn 0;

n;n 1= _n(c+ _n+ _{n 1}); n 1; n;n 2= _{n n 1} forn 2;

n;k = 0; 0 k n 3 forn 3:

(15)

PROPOSITION 1. Let consider the following connection problems Q_n(x) =

Xn k=0

n;kP_k(x)and (x)Q_n(x) =

n+tX

k=0

n;kP_k(x)forn 0:

Then the following two statements hold.

(i) If (x) =x+c;then n;k= _n+1;k+ ( _n+c) _n;k+ _{n n 1;k} forn; k 0:

(ii) If (x) =x²+cx+d;then

n;k = _n+2;k+ _{n;n+1 n+1;k}+ _{n;n n;k}+ _{n;n 1 n 1;k}+ _{n;n 2 n 2;k} forn; k 0where n;k is given in (15).

PROOF. (i)(respectively (ii)) is an immediate consequence of (14)(respectively (15)).

3 Applications

3.1 The Continuous Classical Hermite MOPS f H

g

Let fHng^{n 0} be the Hermite MOPS satisfying (2) with _n = 0and n+1 = ¹₂(n+ 1) forn 0 [3]. Let consider the two shifted MOPSfHeng^{n 0}andfHbng^{n 0}de…ned by

He_n(x) = ( _yH_n)(x) =H_n(x+y)fory2C and

Hb_n(x) =a ⁿH_n(ax)fora2Cn f0g: Accordingly to (3), we obtain

e_n= y anden+1=1

2(n+ 1)forn 0; (16)

and

b_n= 0and bn+1= 1

2a²(n+ 1)forn 0: (17) 3.1.1 The Connection Problem Hbn(x) =Pn

k=0 n;kHk(x)

Choosing Q_n(x) =Hb_n(x),P_n=H_n and (x) = 1(t= 0) in (1), (6)-(7) and by virtue of (17), then (9) and (12) lead to

n;n 1= 0forn 1and n;n 2=1

2 1 1

a²

n

n 2 forn 2:

By induction and (13), we get n;n (2j+1)= 0:Suppose that

k;k 2j = Qj

=1(2 1)

2^j 1 1

a²

j k

2j for0<2j n 2and2j k n:

On account of (13) an other time, we obtain

n;n (2j+2)=

nX1 k=2j+1

f ^{k 2j k;k 2j} k k 1;k 1 2jg:

It’s easy to verify that

k 2j k;k 2j = Qj

=0(2 + 1)

2^j+1 1 1

a²

j k 2j+1

and

k k 1;k 1 2j= Qj

=0(2 + 1)

2^j+1 1 1

a²

j 1 a²

k 2j+1 :

Then

n;n (2j+2) = Qj

=0(2 + 1)

2^j+1 1 1

a²

j+1 nX1 k=2j+1

k 2j+1

= Qj

k=0(2k+ 1)

2^j+1 1 1

a²

j+1 n 2j+2 :

Lastly, we obtain 8>

<

>:

n;n j= 0 forj = 2k+ 1 andk [ⁿ₂¹];

n;n j=

Qj

k=0(2k+1) 2^k 1 _a¹2

k n

2k forj = 2kand1 k [ⁿ₂];

n;n= 1 forn 0:

Hence,

a ⁿHn(ax) =

[ⁿ₂]

X

k=0

(2k)!

2^2kk! 1 1 a²

k n

2k Hn 2k(x)forn 0:

Consequently, we recover again the so-called duplication formula for the Hermite poly- nomials [6].

3.1.2 The Connection Problem He_n(x) =Pn

k=0 n;kH_k(x)

On account of (9), (12) with (16), whereP_n=H_n, Q_n =He_n andt= 0, we get

n;n 1= _nⁿ₁ y; n 1 and _{n;n 2}= _nⁿ₂ y² forn 2:

Suppose that

k;k i= _{k i}^k yⁱ fori j n 1andj k n:

By virtue of (13), we obtain

n;n (j+1)=

n 1

X

k=j

y _{k;k j}+ k+1 j k;k (j 1) k k 1;k 1 (j 1) :

But k+1 j k;k (j 1)= k k 1;k 1 (j 1):Hence,

n;n (j+1)=y^j+1

nX1 k=j

k

j =y^j+1 _j+1ⁿ =y^j+1 _n ⁿ_(j+1) : Consequently, we recover again the well known addition formula [6]

Hn(x+y) = Xn k=0

n

k y^{n k}Hk(x); n 0:

3.1.3 The Connection Problem (x²+cx+d)He_n(x) =Pn+2

k=0 n;kH_k(x); c; d2C Using the connection problem in 3.1.2 and applying Proposition 1., we get that, for n 0 andk n,

n;n+1=c+ny, n;n+2= 1;

n;k = y^{n 2 k n}_k y⁴ (n+ 2)(n+ 1)

(n+ 2 k)(n+ 1 k)+ (c 2y)y³ n+ 1 n+ 1 k +y²(n+1

2 +d+y² cy) +y(c 2y)(n k)

2 +(n k)(n k 1)

4 :

3.2 The Discrete Classical Meixner MOPS f M

g

Let us consider the Meixner MOPS fMn^{( ;a)}gn 0 of parameters ; a. It satis…es (2) with [3]

n= a +n(1 +a)

1 a forn 0 and n=an( +n 1)

(1 a)² forn 1: (18) 3.2.1 The Connection Problem Mn^{( ;a)}(x) =Pn

k=0 n;kM_k^{( ;a)}(x) Choosing P_n=Mn^{( ;a)}andQ_n=Mn^{( ;a)} in (1). On account of (18), (9) gives

n;n 1=

nX1 k=0

( _{k k}) =n a

1 a ( ) = a

1 a

n

n 1 ( )₁; where

(x)n:=

nY1 k=0

(x+k) = (x+n)

(x) forn 1;

being the well-known Pochhammer’s symbol and be the Gamma function [3]. Then

n;n 1= _nⁿ₁ a a 1

n (n 1)

( )_{n (n 1)}: Likewise, taking (12) into account and by virtue of (18), we have

n;n 2 =

n 1

X

k=1

( _{k 1 k}) k;k 1+

nX1 k=1

( k k)

=

n 1

X

k=1

a( ) (1 +a) 1 a

a

1 a( )k +

n 1

X

k=1

ak( )

(1 a)²

= a

1 a

2

( )( 1)(n 1)n

2

= _nⁿ₂ a a 1

n (n 2)

( )_{n (n 2)}:

Suppose that

k;k i= ⁿ_i a a 1

i

( )ifori j andk n:

Using (13) and by virtue of (18) an other time, we obtain

n;n (j+1)

=

nX1 k=j

( _{k j k}) _{k;k j}+ k+1 j k;k (j 1) k k 1;k 1 (j 1)

= a^j

(a 1)^j+1( )_j 8<

:[a( +j) +j]

nX1 k=j

k j

nX1 k=j

(k j+ 1) _j^k₁ 9=

;

= a

a 1

j+1

( )_j+1

n 1

X

k=j k j

= _j+1ⁿ a a 1

j+1

( )j+1:

Hence,

n;n j= ⁿ_j a a 1

j

( )j for0 j nandn 1:

3.2.2 The Connection Problem (x+c)Mn^{( ;a)}(x) =Pn

k=0 n;kM_k^{( ;a)}(x); c2C Taking into account Proposition 1. and the connection problem 3.2.1, the coe¢ cients betweenfMn^{( ;a)}gn 0andfMn^{( ;b)}gn 0by respect to (x) =x+care given by _n;n+1= 1 and

n;k = ⁿ

k

a a 1

n+1 k

( )_{n k} (n+ 1)( +n k) n+ 1 k (1 a)c+a +n(1 +a)

a + ( +n 1)(n k)

a( +n k 1)

forn 0 and0 k n:

3.3 The Quantum Classical Little q-Laguerre MOPS f L

(:; a j q) g

Let us consider the Little q-Laguerre MOPS fLn(:;ajq)g^{n 0} of parameters a6= 0. It satis…es (2) with [4]

(

n=f1 +a a(1 +q)qⁿgqⁿ forn 0,

n=a(1 qⁿ)(1 aqⁿ)q^{2n 1} forn 1: (19)

On account of (3), its shifted MOPSfq ⁿh_qL_n(:;ajq)gn 0 satis…es (4) with (

n=f1 +a a(1 +q)qⁿgq^{n 1} forn 0,

n =a(1 qⁿ)(1 aqⁿ)q^{2n 3} forn 1: (20)

3.3.1 The Connection Problem q ⁿLn(qx;ajq) =Pn

k=0 n;kLk(x;ajq)

Choosing Pn(x) =Ln(x;ajq)andQn(x) =q ⁿLn(qx;ajq)in (1). On account of (19)- (20), (9) gives

n;n 1= q ¹(1 qⁿ)(1 aqⁿ)forn 1:

Moreover, after some calculations taking into account (19)-(20) and (12)-(13) we get

n;n 2= 0forn 2and _{n;n 3}= 0forn 3:

Consequently, formula (13) an other time yields _{n;n k} = 0for0 k n 2:Therefore, q ⁿLn(qx;ajq) =Ln(x;ajq) q ¹(1 qⁿ)(1 aqⁿ)Ln 1(x;ajq)forn 0 (21) withL 1(x;ajq) := 0.

3.3.2 The Connection Problem(x+c)q ⁿLn(qx;ajq) =Pn+1

k=0 n;kLk(x;ajq); c2 C

Taking into account Proposition 1. and the relationship (21) in the connection problem 3.3.1, the coe¢ cients betweenfq ⁿh_qL_n(:;ajq)gn 0 andfL_n(:;ajq)gn 0 by respect to

(x) =x+c; c2Cand

(x+c)q ⁿL_n(qx;ajq) =

n+1X

k=n 2

n;kL_k(x;ajq)

are given by

n;n+1= 1, _n;n =c q ¹+ (1 +a)(1 +q) aqⁿ(1 +q+q²) q^{n 1};

n;n 1= q ¹(1 qⁿ)(1 aqⁿ) c+q^{n 1} aq^{n 1} 1 q^{n 1}(1 +q+q²) ; and

n;n 2= a(1 q^{n 1})(1 qⁿ)(1 aq^{n 1})(1 aqⁿ)q^{2n 4}:

References

[1] Y. Ben Cheikh and H. Chaggara, Connection coe¢ cients via lowering operators, J.

Comput. Appl. Math., 178(2005), 45–61.

[2] H. Chaggara and W. Koepf, Duplication coe¢ cients via generating functions, Com- plex Var. Elliptic Equ., 52(2007), 537–549.

[3] T. S. Chihara, An Introduction to Orthogonal Polynomials, Gordon and Breach, New york, 1978.

[4] A. Ghressi and L. Khériji, Orthogonal q-polynomials related to perturbed form, Appl. Math. E-Notes, 7(2007), 111–120.

[5] A. Ghressi and L. Khériji, A Survey OnD-Semiclassical Orthogonal Polynomials, Appl. Math. E-Notes, 10(2010), 210–234.

[6] E. Godoy, A. Ronveaux, A. Zarzo, I. Area, Minimal recurrence relations for con- nection coe¢ cients between classical orthogonal polynomials: Continuous case, J.

Comput. Appl. Math., 84(1997), 257–275.

[7] E. Godoy, A. Ronveaux, A. Zarzo, I. Area, Minimal recurrence relations for connec- tion coe¢ cients between classical orthogonal polynomials: Discrete case, J. Com- put. Appl. Math., 89(1998), 309–325.

[8] P. Maroni and Z. da Rocha, Connection coe¢ cients between orthogonal polynomials and the canonical sequence: an approach based on symbolic computation, Numer.

Algor., 47(2008), 291–314.

[9] P. Maroni and Z. da Rocha, Connection coe¢ cients for orthogonal polynomials:

symbolic computations, veri…cations and demonstrations in the Mathematica lan- guage, Numer. Algor., 63(2013), 507–520.