Applied Mathematics E-Notes, 14(2014), 221-235 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Orthogonal Polynomials With Respect To A
Nonlinear Form
Mohamed Zaatray
Received 7 May 2014
Abstract
In this paper, we study properties of the form u satisfying
u=
x2
a2
1
v+
0;
where v is a regular symmetric semi-classical form (linear functional). We give a
necessary and su¢ cient condition for the regularity of the form u. The coe¢ cients
of the three-term recurrence relation, satis…ed by the corresponding sequence
of orthogonal polynomials, are given explicitly. A study of the semi-classical
character of the founded families is done. An example related to the Generalized
Gegenbauer form is worked out.
1
Introduction and Preliminaries
The semi-classical forms are a natural generalization of the classical forms (Hermite,
Laguerre, Jacobi, and Bessel). Since the system corresponding to the problem of determining all the semi-classical forms of class s 1 becomes non-linear, the problem
was only solved when s = 1 and for some particular cases [2, 5, 16]. Thus, several
authors use di¤erent processes in order to obtain semi-classical forms of class s
1.
For instance, let v be a regular form and let us de…ne a new form u by the relation
A(x)u = B(x)v, where A(x) and B(x) are non-zero polynomials. When A(x) = 1, v
is positive-de…nite and B(x) is a positive polynomial, Christo¤el [8] has proved that
u is still a positive-de…nite form. This result has been generalized in [9]. The cases
B(x) = 6= 0 and A(x) = x c; x2 ; x3 ; x4 were treated in [15, 17, 18, 22], where it was
shown that under certain regularity conditions the form u is still regular. Moreover, if
v is semi-classical, then u is also semi-classical; see also [1, 4, 6, 11, 23, 24, 25]. When
A(x) = B(x), u is obtained from v by adding …nitely mass points and their derivates
[10, 12, 14] and when A(x) and B(x) have no non-trivial common factor, it was found
a necessary and su¢ cient condition for u to be regular in [13]. When A(x) and B(x)
are of degree equal to one, an extensive study of the form u has been carried in [27].
In this paper, we consider the situation when A(x) and B(x) are of degree equal to
three and one respectively in a particular case. Indeed, we study the form u, ful…lling
x x2
a2 u =
xv; (u)1 = 0; (u)2 =
6= 0;
Mathematics Subject Classi…cations: 42C05, 33C45.
Suprieur de Gestion de Gabs. Avenue Habib Jilani. Gabs 6002, Tunisia
y yInstitut
221
222
Orthogonal Polynomials
where v is a regular symmetric form. The …rst section is devoted to the preliminary
results and notations used in the sequel. In the second section, an explicit necessary
and su¢ cient condition for the regularity of the new form is given. We obtain the coef…cients of the three-term recurrence relation satis…ed by the new family of orthogonal
polynomials. We also analyze some linear relations linking the polynomials orthogonal
with respect to u and v. In the third section, The stability of the semi-classical families
is proved. Finally, we apply our result to Generalized Gegenbauer form.
Let P be the vector space of polynomials with coe¢ cients in C and let P 0 be its
dual. We denote by hv; f i the action of v 2 P 0 on f 2 P. In particular, we denote by
(v)n := hv; xn i ; n 0 , the moments of v. For any form v and any polynomial h let
1
Dv = v 0 , hv, c , and (x c) v be the forms de…ned by:
D
E
1
hv 0 ; f i := hv; f 0 i ; hhv; f i := hv; hf i ; h c ; f i := f (c), (x c) v; f := hv; c f i
where ( c f ) (x) = f (x)x cf (c) , c 2 C, and f 2 P.
Then, it is straightforward to prove that for c ; d 2 C ; c 6= d ; f; g 2 P and v 2 P 0 ,
we have
1
(x c) ((x c) v) = v (v)0 c ;
(1)
(x
c) (x
c)
1
v = v;
(2)
1
( c
(3)
d ):
c d
cf. [21]. Let us de…ne the operator : P ! P by ( f ) (x) = f (x2 ). Then, we de…ne
the even part v of v by h v; f i := hv; f i : Therefore, we have [20]
(x
d)
1
c
=
f (x) ( v) =
f (x2 )v ;
(4)
0
(v ) = 2 ( (xv)) :
(5)
A form v is called regular if there exists a sequence of polynomials fSn gn
such that
hv; Sn Sm i = rn n;m for rn 6= 0 and n 0:
0
(deg Sn
n)
(6)
Then deg Sn = n for n
0 and we can always suppose each Sn is monic. In such a
case, the sequence fSn gn 0 is unique. It is said to be the sequence of monic orthogonal
polynomials with respect to v.
It is a very well known fact that the sequence fSn gn 0 satis…es the recurrence
relation (see, for instance, the monograph by Chihara [7])
Sn+2 (x) = x
n+1 Sn+1 (x)
S1 (x) = x
and
S0 (x) = 1;
0
n+1 Sn (x)
for n
0;
(7)
with n ; n+1 2 C C f0g; n 0 . By convention we set 0 = (v)0 = 1.
(1)
In this case, let fSn gn 0 be the associated sequence of …rst order for the sequence
fSn gn 0 satisfying the recurrence relation
( (1)
(1)
(1)
Sn+2 (x) = x
0;
n+2 Sn+1 (x)
n+2 Sn (x) for n
(8)
(1)
(1)
(1)
S1 (x) = x
1 , S0 (x) = 1; and S 1 (x) = 0:
M. Zaatra
223
(1)
Another important representation of Sn (x) is, (see [7]),
Sn(1) (x) :=
Also, let fSn (:; )gn
isfying
Sn+1 (x)
x
v;
Sn+1 ( )
:
(9)
be the co-recursive polynomials for the sequence fSn gn
0
(1)
Sn (x; ) = Sn (x)
Sn
1 (x)
for n
0
0:
sat(10)
cf. [7].
We recall that a form v is called symmetric if (v)2n+1 = 0 for n 0. The conditions
(v)2n+1 = 0 for n 0 are equivalent to the fact that the corresponding monic orthogonal polynomial sequence fSn gn 0 satis…es the recurrence relation (7) with n = 0 for
n 0: cf. [7].
Throughout this paper, the form v will be supposed normalized, (i.e., (v)0 = 1),
symmetric and regular.
Let us consider the decomposition of fSn gn
S2n (x) = Pn (x2 );
(1)
S2n (x) = Rn (x2 ;
1)
(1)
R1 (x) = x
0:
S2n+1 (x) = xRn (x2 );
(11)
(1)
and S2n+1 (x) = xPn(1) (x2 ):
cf. [7, 20]. The sequences fPn gn 0 and fRn gn
respective to v and x v. We also have
(
R
Rn+2 (x) = x
n+1 Rn+1 (x)
R
0
and fSn gn
0
0
(12)
are respectively orthogonal with
R
n+1 Rn (x)
for n
0;
(13)
and R0 (x) = 1;
with
R
0
=
1
+
R
n+1
2;
=
2n+3
+
2n+4 ;
and
R
n+1
(1)
By virtue of (8), with
n
(1)
1)
2n+2 2n+3
(1)
n+2 Sn (0).
= 0, we get Sn+2 (0) =
S2n (0) = Rn (0;
=
= ( 1)n
n
Y
2
for n
for n
0:
(14)
Consequently,
0:
(15)
=0
PROPOSITION 1 ([7, 21]). v is regular if and only if v and x v are regular.
2
Algebraic Properties
For …xed a 2 C and
relation
2C
0
f0g, we can de…ne a new normalized form u 2 P by the
u=
(x2
a2 )
1
v+
0:
(16)
224
Orthogonal Polynomials
Equivalently, from (1)-(3) we have
x x2
a2 u =
xv; (u)1 = 0; and (u)2 =
:
(17)
The case a = 0 is treated in [1, 18, 26], so henceforth, we assume a 6= 0:
PROPOSITION 2. u is regular if and only if
Rn (a2 ;
1)
6= 0 for n
n
where Rn is de…ned by (13), and for n
0;
(18)
0;
= Rn+1 (a2 ; 1 ) Rn (0; 1 ) + a2 Rn (0)
Rn (a2 ; 1 ) Rn+1 (0; 1 ) + a2 Rn+1 (0) :
n
PROOF. Multiplying (17) by x and applying the operator
tion and using (2), we get
1
x u=
1
1
a2
x
1
1
x v +
(19)
for the obtained equa-
a2 :
(20)
From (20) and (3), we get
u=
1
1x
1
a2
x
1
1
x v + 1+
0
a2
a2
a2 :
(21)
From (16), it is plain that u is a symmetric form. Then, according to Proposition
1, u is regular if and only if x u and u are regular. But
1
x u=
a2
x
1
is regular if and only if 6= 0 and Rn (a2 ;
if and only if Rn (a2 ; 1 ) 6= 0 and
u=
1
1x
a2
x
1)
1
1
1
1
1
1
x v +
6= 0 for n
x v + 1+
a2
0 (see [22]). So u is regular
0
a2
a2
a2
is regular. Or, it was shown in [6] that the form
1x
is regular if and only if
1
x
n
a2
1
6= 0 for n
1
1
x v + 1+
a2
0
a2
a2
0: Then, we deduce the desired result.
REMARK 1. From (11) and (12), we get
Rn (a2 ;
1)
(1)
= S2n (a);
Rn (0;
1)
(1)
0
(0)
= S2n (0); and Rn (0) = S2n+1
M. Zaatra
for n
225
0. Thus, u is regular if and only if
(1)
(1)
S2n (a) S2n+2 (a)
(1)
0
S2n (0) + a2 S2n+1
(0)
(1)
S2n (a)
(22)
(1)
0
S2n+2 (0) + a2 S2n+3
(0)
6= 0 for n
0:
REMARK 2. From (7), we have
0
S10 (0) = 1 and S2n+3
(0) = S2n+2 (0)
0
2n+3 S2n+1 (0)
for n
0:
Therefore, we can easily prove by induction that
0
S2n+1
(0) = ( 1)n
with
n
=1+
n
X1
Y
=0 k=0
2k+1
(1)
n S2n (0)
for n
for n
0 where
2k+2
0;
1
X
(23)
= 0:
(24)
=0
When u is regular, let fZn gn 0 be the corresponding sequence satisfying the recurrence relation
(
Zn+2 (x) = xZn+1 (x)
0;
n+1 Zn (x) for n
(25)
Z1 (x) = x and Z0 (x) = 1:
Let us now consider the quadratic decomposition of the sequence fZn gn
~ n (x2 ) for n
Z2n (x) = P~n (x2 ) and Z2n+1 (x) = xR
0
0:
(26)
From (20) and (21), we can deduce the following results.
~ n gn
PROPOSITION 3 ([22]). The polynomials of the sequence fR
relation
~ n+1 (x) = Rn+1 (x) + an Rn (x) for n 0;
R
where
0
satisfy the
(27)
(1)
an =
S2n+2 (a)
(1)
for n
0:
(28)
S2n (a)
PROPOSITION 4 ([6]). The polynomials of the sequence fP~n gn 0 satisfy the
relation
(
P~n+2 (x) = Rn+2 (x) + cn+1 Rn+1 (x) + bn Rn (x) for n 0;
(29)
P~1 (x) = R1 (x) + c0 ;
226
Orthogonal Polynomials
where
n+1
bn =
for n
0;
(30)
n
and, for n
0;
8
>
>
cn+1 =
>
>
<
>
>
>
>
:
c0 =
n
(1)
1
(1)
0
S2n+2 (0) + a2 S2n+5
(0)
S2n (a)
(1)
(1)
0
S2n (0) + a2 S2n+1
(0)
S2n+4 (a)
(31)
;
2:
1
LEMMA 1.
xZn+3 (x) = Sn+4 (x) + ~bn+2 Sn+2 (x) + a
~n Sn (x) for n
xZ2 (x) = S3 (x) + ~b1 S1 (x);
0;
(32)
xZ1 (x) = S2 (x) + ~b0 ;
with for n
0;
a
~2n = 2n+1 an ; a
~2n+1 = bn ;
~b2n+2 =
~
2n+3 + an ; b2n+3 = cn+1 ;
~b0 =
and ~b1 = c0 :
(33)
1
PROOF. From (26),we have
~ n (x2 ) for n
xZ2n+2 (x) = xP~n+1 (x2 ) and xZ2n+1 (x) = x2 R
0:
Then, from the above equation, (11), (27) and (29), we get (32).
PROPOSITION 5. We may write
1
=
;
n+3
n+2
n+3
=
n+1
= ~bn+2
a
~n+1
;
a
~n
(34)
~bn+3 ;
(35)
~
(36)
and
a
~n+1
for n
a
~n =
~
n+2 bn+2
n+3 bn+1 ;
0:
PROOF. After multiplication of (32) by x, we apply the recurrence relations (7)
and (25), we get
xZn+4 (x) +
n+3 xZn+2 (x)
=
Sn+5 (x) + (
n+4
+(~
an +
~
+ ~bn+2 )Sn+3 (x)
n+2 bn+2 )Sn+1 (x)
+
~n Sn 1 (x)
na
M. Zaatra
227
for n
1. Substituting xZk+3 in the above equation by Sk+4 + ~bk+2 Sk+2 + a
~k Sk
with k = n + 1; n 1, we obtain (34)-(36), after comparing the coe¢ cients of Sk with
k = n + 3; n + 1; n 1:
~ n gn
REMARK 3. From (14), (33) and (34), the sequence fR
rence relation (13) with for n 0;
~
R
0
=
and
3
b0
;
a0
~
R
n+1
~
R
n+1
=
=
2n+2 2n+3
2n+2 2n+3
0
satis…es the recur-
an+1
bn+1
+
;
bn
an+1
an+1
:
an
The Semi-Classical Case
In this section, we compute the exact class of the semi-classical form u.
DEFINITION 1 ([21]). The form v is called semi-classical when it is regular and
satis…es the Riccati equation
(z)S 0 v (z) = C(z)S v (z) + D(z);
(37)
where
monic, C and D are polynomials and S(v)(z) designes the formal Stieltjes
function of the form v de…ned by:
X (v)n
:
z n+1
S(v)(z) =
(38)
n 0
It was shown in [21] that equation (37) is equivalent to
0
( (x)v) + v = 0;
(39)
0
(40)
with
(x) =
(x)
C(x):
We also have the following relation :
D(x) =
(v
0
0
) (x)
(v
0
) (x):
PROPOSITION 6 ([21]). De…ne r = deg( ) and p = deg( ). The semi-classical
form v satisfying (39) is of class s = max (r 2; p 1) if and only if
Y
j 0 (c) + (c)j + hv; 2c + c i 6= 0;
(41)
c2Z
where Z denotes the set of zeros of
.
228
Orthogonal Polynomials
COROLLARY 1 ([19]). The form v satisfying (37) is of class s if and only if
Y
(jC(c)j + jD(c)j) 6= 0:
(42)
c2Z
PROPOSITION 7. If v is a semi-classical form and satis…es (37), then for every
2 C f0g such that Rn (a2 ; 1 ) n 6= 0; n 0, the form u de…ned by (16) is regular
and semi-classical. It satis…es
~ (z)S 0 (u)(z) = C(z)S(u)(z)
~
~
+ D(z);
where
8
2
2
~
a2 (z);
>
< (z) = z z
~
C(z)
= z 2 z 2 a2 C(z)
>
: ~
D(z) = z z 2 a2 C(z)
and u is of class s~ such that s~
(43)
2z 3 (z);
z 2 + a2
(44)
(z)
z 2 D(z);
s + 4.
PROOF. We have [21]
zS(v)(z) = S( v)(z)
(v 0 ( )) (z) = S( v)(z)
1:
Using (17), we get
zS(v)(z) =
1
S( (
a) (
b) u) (z) 1 =
1
(z
a) (z
b) (zS(u)(z)
1) : (45)
Multiplying (37) by z 2 and taking into account (45) we obtain (43)-(44).
From (39) and (43)-(44), the form u satis…es the distributional equation
~ (x)v
0
+ ~ v = 0;
(46)
where ~ is the polynomial de…ned in (44) and
~ (x) =
Then deg( ~ ) = r~
~ 0 (x)
~
C(x)
= x x2
s + 6 and deg( ~ ) = p~
a2 (x (x)
2 (x)) :
s + 5: Thus s~ = max (~
r
2; p~
(47)
1)
s + 4:
PROPOSITION 8. Let u be a semi-classical form satisfying (43). For every zero of
~ di¤erent from 0 and a, the equation (43) is irreducible.
PROOF. Since v is a semi-classical form of class s; S(v)(z) satis…es (37), where the
~ be as in Proposition 7. Let c
polynomials ; C and D are coprime. Let ~ ; C~ and D
~
be a zero of di¤erent from 0 and a, this implies that (c) = 0.
~
We know that jC(c)j+jD(c)j =
6 0, (i) if C(c) 6= 0, then C(c)
6= 0; and (ii) if C(c) = 0
~
~
~
and D(c) 6= 0, then D(c) 6= 0, whence jC(c)j + jD(c)j =
6 0.
Concerning the class of u, we have the following result (see Proposition 9). But
…rst, let us recall this technical lemma.
LEMMA 2. We have the following properties:
M. Zaatra
229
(P1 ) The equation (43)-(44) is irreducible in 0 if and only if
(0) 6= 0:
(P2 ) The equation (43)-(44) is divisible by z but not by z 2 if and only if
(0) = 0 and C(0) +
(P3 ) The equation (43)-(44) is irreducible in a and
0
(0) 6= 0:
a if and only if
j (a)j + jD(a)j =
6 0:
(P4 ) The equation (43)-(44) is divisible by z 2
(a) = D(a) = 0 and jC(a)
a2 but not by (z 2
a
00
a2 )2 if and only if
(a)j + jD00 (a)j =
6 0:
~
~
PROOF. From (44), we have C(0)
= 0 and D(0)
= a2 (0). If (0) 6= 0, then
~
D(0) 6= 0: So, by virtue of (42), we obtain P1 . Now, if (0) = 0, then the equation
(43)-(44) is divisible by z according to (42). Thus S(u)(z) satis…es (43) with
8
2
~
a2 (z);
>
< (z) = z z
~
(48)
C(z)
= z z 2 a2 C(z) 2z 2 (z);
>
: ~
2
2
2
2
D(z) = z
a C(z)
z + a ( 0 ) (z)
zD(z):
~
~
Therefore, C(0)
= 0 and D(0)
= a2 (C(0) + 0 (0)). If C0 (0) + 0 (0) 6= 0, then
the equation (43)-(48) is irreducible in 0. Thus, we deduce P2 . From (44), we get
~
~
C(a)
= 2a3 (a) and D(a)
= a2 ( D(a) + 2 (a)) : Then, we can deduce that
~
~
jC(a)j + jD(a)j =
6 0 if and only if ( (a); D(a)) 6= (0; 0). Thus P3 is proved. If
( (a); D(a)) = (0; 0), then the equation (43)-(44) can be divided by z 2 a2 since
u is symmetric and according to (42). In this case S(u)(z) satis…es (43) with
8
2
~
>
< (z) = z (z);
~
(49)
C(z)
= z 2 C(z) 2z 3 ( a a ) (z);
>
: ~
2
2
2
D(z) = zC(z)
z + a ( a a ) (z)
z ( a a D) (z):
Substituting z by a in (49), we obtain
~
C(a)
= a2 (C(a)
a
00
(a)) ;
Then (43)-(49) is irreducible in a and
Hence P4 .
~
D(z)
= a C(a)
a
00
a if and only if jC(a)
a 00
D (a) :
2
(a)
a
00
(a)j + jD00 (a)j =
6 0.
PROPOSITION 9. Under the conditions of Proposition 7, for the class of u, we
have two di¤erent cases:
230
Orthogonal Polynomials
(A)
(0) 6= 0
(i) s~ = s + 4 if ( (a); D(a)) 6= (0; 0).
(ii) s~ = s + 2 if ( (a); D(a)) = (0; 0) and jC(a)
(B)
(0) = 0 and C(0) +
0
a
00
(a)j + jD00 (a)j =
6 0:
a
00
(a)j + jD00 (a)j =
6 0:
(0) 6= 0
(i) s~ = s + 3 if ( (a); D(a)) 6= (0; 0).
(ii) s~ = s + 1 if ( (a); D(a)) = (0; 0) and jC(a)
PROOF. From Proposition 8, the class of u depends only on the zeros 0 and a. For
the zero 0 we consider the following situation:
(A) (0) 6= 0. In this case the equation (43)-(44) is irreducible in 0 according to
P1 . But what about the zero a? We will analyze the following cases:
(i) ( (a); D(a)) 6= (0; 0), the equation (43)-(44) is irreducible in a and a according
to P3 . Then s~ = s + 4. Thus we proved (A)(i).
(ii) ( (a); D(a)) = (0; 0) and jC(a) a 00 (a)j + jD00 (a)j =
6 0.
2
From P3 and P4 , (43)-(44) is divisible by z 2 a2 but not by z 2 a2 and thus the
order of the class of u decreases in two units. In fact, S(u)(z) satis…es the irreducible
equation (43)-(49) and then s~ = s + 2. Hence (A)(ii).
(B) (0) = 0 and C(0) + 0 (0) 6= 0.
In this condition, (43)-(44) is divisible by z but not by z 2 according to P2 . But
what about the zero a? We have the two following cases:
(i) ( (a); D(a)) 6= (0; 0), the equation (43)-(44) is irreducible in a and a according
to P3 . Therefore S(u)(z) satis…es the irreducible equation (43)-(49) and then s~ = s + 3
and (B)(i) is also proved.
(ii) ( (a); D(a)) = (0; 0) and jC(a) a 00 (a)j + jD00 (a)j =
6 0.
2
2
2
From P3 and P4 , (43)-(48) is divisible by z
a but not by z 2 a2 . Then,
S(u)(z) satis…es the irreducible equation (43) with
8
~
>
< (z) = z (z);
~
(50)
C(z)
= zC(z) 2z 2 ( a a ) (z);
>
: ~
2
2
D(z) = C(z)
z + a ( 0 a a ) (z)
z ( a a D) (z);
and thus s~ = s + 1.
Finally, if we suppose that the form v has the following integral representation:
Z +1
Z +1
hv; f i =
V (x)f (x)dx for f 2 P with (v)0 =
V (x)dx = 1;
1
1
where V is locally integrable function with rapid decay and continuous at a and
Then, from (16) the form u is represented by
hu; f i = f (0) +
2a
P
R +1
V (x)
f (x)dx
1 x+a
P
R +1
V (x)
f (x)dx
1 x a
+ f (a) + f ( a) P
R +1
V (x)
dx
1 x a
a.
(51)
;
M. Zaatra
231
where for c 2 fa;
P
4
Z
+1
ag
Z
V (x)
f (x)dx = lim
"!0
x c
1
c "
1
V (x)
f (x)dx +
x c
Z
+1
c+"
V (x)
f (x)dx :
x c
Application
Proposition 7 shows that we can generate new semi-classical sequences from well known
ones. We apply our results to v := G:G( ; ), where G:G( ; ) is the Generalized
Gegenbauer form. In this case, the form v is symmetric semi-classical of class s = 1.
Thus, we have [7]
8
< 2n+1 = (n+ +1)(n+ + +1)
0;
(2n+ + +1)(2n+ + +2) for n
(52)
(n+1)(n+ +1)
:
0:
2n+2 = (2n+ + +2)(2n+ + +3) for n
The regularity conditions are
6=
n;
6=
n;
+
6=
n; n
1 : We also have
(x) = x x2 1 ;
(x) = 2 ( + + 2) x2 + 2 ( + 1) ;
2
C(x) = (2 + 2 + 1) x
(2 + 1) ; D(x) = 2 ( + + 1) x:
For greater convenience we take a = 1, and
obtain by induction
(1)
S2n (0) = ( 1)n
and
(1)
S2n (1) =
with, for n
n
=
6= 0. From (8) and (52), we can easily
( + + 2) (n + + 1)
for n
( + 1) (2n + + + 2)
+
(2n +
(53)
+1
+ + 2)
n
for n
0;
(54)
0;
(55)
0,
(n +
+ 1) (n +
( + 1)
+
+ 2)
( +
+ 1) (n + 1) (n +
( + 1)
+ 2)
:
From (52), we get
2k+1
2k+2
=
(k + + 1) (k +
(k + 1) (k +
+ + 1) (2k +
+ 1) (2k + +
+ + 3)
:
+ 1)
Then
Y
k=0
2k+1
=
2k+2
=
(2 +
+
+ 3)
( + 1) ( + + 2) ( + + + 2)
( + 1) ( + + 2) ( + 2) ( + + 2)
(2 + + + 3) ( + 1)
( + 1) ( + + 2) ( + 1) ( +
+ 1)
h ;
232
Orthogonal Polynomials
with
hn =
(n + + 2) (n +
(n + 1) (n +
hn+1 =
(n + + 2) (n +
(n + 1) (n +
+ + 2)
; n
+ 1)
0;
+ + 2)
hn ; n
+ 1)
0:
ful…lling
Therefore
( + 1) (2n + + + 3)
hn ; n
(n + 1) (n + + 1)
0;
and consequently, from the above results, we obtain that for n
1;
hn+1
n
X1
Y
2k+1
hn =
( + 1)
( + 2) ( +
=
2k+2
=0 k=0
+ 2)
n
X1
(h
h )
+1
=0
( + 1) (n + + 2) (n + + + 2)
( + 2) ( + + 2) (n + 1) (n + + 1)
=
1:
Finally, (24), (23) and (19) become respectively
n
( + 1) (n + + 2) (n + + + 2)
for n
( + 2) ( + + 2) (n + 1) (n + + 1)
=
(n + + 2) (n + + + 2)
for n
( + 2) (n + 1) (2n + + + 2)
0
S2n+1
(0) =
0;
(56)
0;
(57)
and
n
with for n
=
(2n +
+
+ +1
+ 2) (2n +
+
+ 3)
(
n
+
n)
for n
0
n
= ( 1)n
n
( +
=
+ 2) (n +
( + 1)
+ 1)
(
n+1
+ (n +
+ 1)
+ 2) (n + + + 2)
[(n + 1)
( + 2) (n + 2)
+ (n + + 2) (n + + + 2) n ]:
Thus, u is regular for every
(n +
n+1
6= 0 such that
n(
n
+
n)
6= 0 for n
0:
Using (55) and (58), we obtain for (28) and (30) (for n
0)
n+1
an =
n
(2n +
+
+ 2) (2n +
+
+ 3)
;
n) ;
0;
(58)
M. Zaatra
bn =
233
+
n+1
(
+
n
n ) (2n +
+
+ 2) (2n +
+
n+1
+ 3) (2n +
+
+ 4) (2n +
+
+ 5)
Therefore, we have for (34)
1
2n+2
2n+3
=
;
( n+1 + n+1 )
+ + 4) (2n +
n ) (2n +
n
=
n+2
=
n+1
(
(
+
n
n+1 (
n
+
n ) (n
n+1
+
+
+ 5)
;
+ 1) (n + + 1) (n + + 2) (n + + + 2)
:
+ + 3) (2n + + + 4)
n+1 ) (2n +
Since v is semi-classical, then according to Proposition 7, (40) and (48), the form
u is also semi-classical of class s~ = 4 and ful…ls (43) and (47) with
~ (x) = x2 x2 1 2 ;
~ (x) = x x2 1 (2 + 2 + 5) x2 2
3 ;
2
2
~
C(x) = x x
1 (2 + 2
1) x
2
1 ;
4
~
D(x) = 2 ( + 1) x
2 (( + + 1) ( + 1) + ) x2 + 2 ( + 1) :
The form v has the following integral representation [7], for R > 1, R >
f 2 P,
Z 1
( + + 2)
jxj2 +1 1 x2 f (x)dx:
hv; f i =
( + 1) ( + 1) 1
1,
Then, from (51), we obtain
hu; f i =
"Z
1
jxj2 +1 1 x2
( + + 2)
f (0) +
2 ( + 1) ( + 1)
x+1
1
#
Z 1
jxj2 +1 1 x2
(f (x) f (1)) dx ;
x 1
1
for R > 1 and R > 1. But, if R > 0, we have
Z 1
Z 1
jxj2 +1 1 x2
jxj2 +1 1 x2
dx =
x+1
x 1
1
1
Consequently, if R > 0, R >
(f (x)
( ) ( + 1)
:
( + + 1)
dx =
1, f 2 P,
Z 1
( + + 2)
hu; f i =
jxj2 +1 1 x2
( + 1) ( + 1) 1
+ +1
+f (0)
(f (1) + f ( 1)) :
2
1
f (x)dx
(59)
REMARKS 4. From (59),we have
u=
+
+1
f ( 1)) dx
G:G (
1; ) +
0
+ +1
(
2
1
+
1) :
:
234
Orthogonal Polynomials
For more details see [3]. Using (59), we get
(u)2n+2 =
( + + 2) (n + + 2)
( + 1) (n + + + 2)
+
+1
; n
0:
Acknowledgment. Thanks are due to the referee for his valuable comments and
useful suggestions and for his careful reading of the manuscript.
References
[1] J. Alaya and P. Maroni, Semi-classical and Laguerre-Hahn forms de…ned by
pseudo-functions, Methods Appl. Anal., 3(1996), 12–30.
[2] J. Alaya and P. Maroni, Symmetric Laguerre-Hahn forms of class s = 1, Integral
Transforms Spec. Funct., 4(1996), 301–320.
[3] R.Álvarez-Nodarse, J. Arvesú and F. Marcellán, Modi…cations of quasi-de…nite
linear functionals via addition of delta and derivatives of delta Dirac functions,
Indag. Math., 15(2004), 1–20.
[4] D. Beghdadi and P. Maroni, On the inverse problem of the product of a form by
a polynomial, J. Comput. Appl. Math., 88, 401–417.
[5] B. Bouras and A. Alaya, A large family of semi-classical polynomials of class one,
Integral Transforms Spec. Funct., 18(2007), 913–931.
[6] A. Branquinho and F. Marcellán, Generating new classes of orthogonal polynomials, Int. J. Math. and Math. Sci., 19(1996), 643–656.
[7] T. S. Chihara, An Introduction to Orthogonal Polynomials, Gordon and Breach,
New York, 1978.
• ber die Gaussische Quadratur und eine Verallgemeinerung der[8] E.B. Christo¤el, U
selben, J. f•
ur Reine und Angew. Math., 55(1858), 61–82.
[9] J. Dini and P. Maroni, Sur la multiplication d’une forme semi-classique par un
polynôe, Publ. Sem. Math. Univ. d’Antananarivo, 3(1989), 76–89.
[10] H. Dueñas and F. Marcellán, Perturbations of Laguerre-Hahn functional: modi…cation by the derivative of a Dirac delta , Integral Transforms Spec. Funct.,
20(2009), 59–77.
[11] A. Ghressi and L. Khériji, Orthogonal q-polynomials related to perturbed linear
form. Appl. Math. E-Notes., 7(2007), 111–120.
[12] D. H. Kim, K. H. Kwon, and S. B. Park, Delta perturbation of a moment functional, Appl. Anal., 74(2000), 463–477.
[13] J. H. Lee and K. H. Kwon, Division problem of moment functionals, Rocky Mountain J. Math., 32(2002), 739–758.
M. Zaatra
235
[14] F. Marcellán and P. Maroni, Sur l’adjonction d’une masse de Dirac à une forme
régulière et semi-classique, Ann. Mat. Pura Appl., 162 (IV) (1992), 1–22.
[15] P. Maroni and I. Nicolau, On the inverse problem of the product of a form by
a monomial: the case n=4. Part I, Integral Transforms Spec. Funct., 21(2010),
35–56.
[16] P. Maroni and M. Mejri, Some semiclassical orthogonal polynomials of class one,
Integral Transforms Spect. Funct., 18(2007), 913–931.
[17] P. Maroni and I. Nicolau, On the inverse problem of the product of a form by a
polynomial: The cubic case. Appl. Numer. Math., 45(2003), 419–451.
[18] P. Maroni, On a regular form de…ned by a pseudo-function. Numer. Algorithms,
11(1996), 243–254.
[19] P. Maroni, Variations arround classical orthogonal polynomials. Connected problems, in: 7th Symp. Sobre Polinomios orthogonales y Appliciones, Proc., Granada
(1991), J. Comput. Appl. Math., 48(1993), 133–155.
[20] P. Maroni, Sur la décomposition quadratique d’une suite de polynômes orthogonaux, I, Rivista di Mat. Pura ed Appl., 6(1991), 19–53.
[21] P. Maroni, Une théorie algébrique des polynômes orthogonaux. Application aux
polynômes orthogonaux semi-classiques, in: Orthogonal Polynomials and their
applications, (C. Brezinski et al Editors.) IMACS, Ann. Comput. Appl. Math. 9 (
Baltzer, Basel, 1991), 95–130.
[22] P. Maroni, Sur la suite de polynômes orthogonaux associée à la forme u =
1
(x c) L: Period. Math. Hungar., 21(1990), 223–248.
c
+
[23] M. Sghaier and M. Zaatra, On orthogonal polynomials associated with rational
perturbations of linear functional, Adv. Pure Appl. Math., 4(2013), 139–164.
[24] M. Sghaier, Orthogonal polynomials with respect to the form u = (x a)
Georgian Math. J., 17(2010), 581–596.
1
v + b,
[25] M. Sghaier, Generating semiclassical orthogonal polynomials, Appl. Math. ENotes., 9(2009), 168–176.
[26] M. Sghaier and J.Alaya, Building some symmetric Laguerre-Hahn functionals of
class two at most, through the sum of a symmetric functionals as pseudofunctions
with a Dirac measure at origin, Int. J. Math. Math. Sci., 2006, Art. ID 70835,
1–19.
[27] M. Sghaier and J. Alaya, Orthogonal Polynomials Associated with Some Modi…cations of a Linear Form, Methods Appl. Anal., 11(2004), 267–294.