Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol106.Issue2.1999:
Journal of Computational and Applied Mathematics 106 (1999) 203–243
www.elsevier.nl/locate/cam
On the positivity of some bilinear functionals in Sobolev spaces
Andre Draux ∗ , Charaf Elhami
LMI - UPRES-A CNRS 6085, INSA de Rouen, Departement de Genie Mathematique, Place Emile Blondel, BP 08,
F-76131 Mont-Saint-Aignan Cedex, France
Received 13 May 1998; received in revised form 30 January 1999
Abstract
The positivity of a bilinear functional a
a(f; g) =
N
X
m c(m) (f(m) ; g(m) )
m=0
is studied as a function of coecients m . The concerned cases are those of Laguerre, Gegenbauer and Jacobi for
c(m) = c(0) ; m = 1; : : : ; N . The domain {m }Nm=1 where a is positive de nite, is given. As a consequence, when N = 1, the
c 1999 Elsevier Science B.V. All rights reserved.
corresponding Markov–Bernstein inequalities are given.
MSC: 33C45; 42C05; 26D05; 26C10
Keywords: Formal orthogonal polynomial; Laguerre; Gegenbauer; Jacobi polynomial; Laguerre–Sobolev; Gegenbauer–
Sobolev; Jacobi–Sobolev polynomial; De nite inner product; Markov–Bernstein inequality; Zeros of polynomial
This paper constitutes the second part of a study devoted to the positivity of bilinear functionals
using positive de nite classical inner products, that is to say those of Hermite, Laguerre and Jacobi
measures. The rst part has solved this problem of positivity for the Hermite measure and for
some other connected problems [5]. That study was the simplest one, since the domain D where
the bilinear functionals are positive de nite, is given by means of explicit equations which de ne
its boundary. This domain D is polyhedric. It remains to solve the more complicated problems of
Laguerre and Jacobi measures. We study them here. In these cases the domain D can not be given
by means of explicit equations of its boundary when N ¿2, because this one is de ned as a limit
of a nappe of an algebraic hypersurface. The question was: does it exist a part of the domain D
∗
Corresponding author.
E-mail addresses: [email protected] (A. Draux), [email protected] (C. Elhami)
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ - see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 6 3 - 1
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
where some coecients m are negative; the answer is the same as in the Hermite case: yes, when
N ¿2. When N = 1; 1 has to be strictly positive in D. But when N = 1, the successive values 1; n ,
obtained for the positivity with polynomials of degree n = 1; 2; : : : ; give the best coecients of the
so-called Markov–Bernstein inequalities (see [1,9]). Except in the Hermite case and Laguerre case
for = 0 (see [9]) these values are not known explicitly, but we have given lower and upper bounds
in the Laguerre and Gegenbauer cases, which gives the behavior of 1; n as a function of n. The help
of Mathematica 3.0 [15] was very useful in that case for some tedious algebraic manipulations.
For replacing this study within the framework of the Sobolev orthogonality, the bibliography,
presented by Marcellan and Ronveaux [8], is essential.
1. Introduction
In order to establish the general framework of our paper, we begin to study the general case where
a bilinear functional a is obtained from two positive de nite inner products. To nd
the domain Dn = { ∈ R | a(p; p) ¿ 0 ∀p ∈ Pn − {0}} is equivalent to solve a generalized eigenvalue problem. Moreover the smallest eigenvalue 1; n of this problem which gives the boundary of
the domain Dn , also gives the best coecient in Markov–Bernstein inequality. Our presentation is
more or less di erent from that given in [9].
P (resp. Pi ) will denote the vector space of polynomials (in one variable) with real coecients
(resp. of degree at most i).
Let c and c(1) be two positive de nite inner products and {Pi }i¿0 the sequence of monic orthogonal
polynomials with respect to c.
Let a be the bilinear functional de ned by
∀p; q ∈ P;
a(p; q) = c(p; q) + c(1) (p′ ; q′ );
′
where is a real parameter and p
(resp. q′ ) is the derivative
of p (resp. q).
Pn
P
p ∈ Pn can be written as p = i=0 yi Pi . Thus p′ = ni=1 yi Pi′ and we get
(1)
a(p; p) = yT Kn; 0 y + y˜ T Kn;−1
y;
˜
(1)
where y (resp. y)
˜ is the vector of Rn+1 (resp. Rn ) of components yi ; i = 0; : : : ; n, (resp. i =
1; : : : ; n); Kn; 0 is the (n + 1) × (n + 1) matrix whose entries Kn; 0 (i; j) are c(Pi ; Pj ) for i; j = 0; : : : ; n
(1)
(1)
(therefore this matrix is diagonal) and Kn;−1
is the n×n matrix whose entries Kn;−1
(i; j) are c(1) (Pi′ ; Pj′ )
for i; j = 1; : : : ; n.
Let Kn;−1 be the n × n matrix deduced from Kn; 0 in cancelling the rst row and the rst column.
Then from (1) we have the following obvious result.
Theorem 1.1. For a xed parameter ∈ R; then; ∀n¿1 and ∀p ∈ Pn − {0}; a(p; p) ¿ 0 if and
(1)
only if Kn;−1 + Kn;−1
is positive de nite.
(1)
Moreover, since Kn;−1
is a positive de nite symmetric matrix, it can be decomposed by means of
Cholesky algorithm as Gn GnT where Gn is a lower triangular matrix. Therefore we have
(1)
Kn;−1 + Kn;−1
= Gn (Gn−1 Kn;−1 (GnT )−1 + I )GnT :
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
205
(1)
Hence Kn;−1 + Kn;−1
is positive de nite if and only if Gn−1 Kn;−1 (GnT )−1 + I is positive de nite.
Let i; n ; i = 1; : : : ; n be the eigenvalues of Gn−1 Kn;−1 (GnT )−1 with 0 ¡ 1; n 62; n 6 · · · 6n; n . Then
we have another obvious results.
(1)
Theorem 1.2. Kn;−1 + Kn;−1
is positive de nite if and only if ¿ − 1; n .
Theorem 1.3. a is positive de nite ∀p ∈ P − {0} if and only if ¿ − limn→∞ 1; n .
Proof. Since Gn−1 Kn;−1 (GnT )−1 is a symmetric matrix, then from an obvious consequence of the
Courant–Fischer theorem (see [14]) we have
0 ¡ 1; n+1 61; n ;
n¿1:
Thus the positive sequence {1; n }n¿1 is decreasing and bounded. Then limn→∞ 1; n exists and
limn→∞ 1; n ¿0. Therefore the result holds.
From Theorem 1.2 we obtain the Markov–Bernstein inequality.
Corollary 1.4.
∀p ∈ Pn ;
c(1) (p′ ; p′ )6
1
c(p; p);
1; n
n¿1:
P
1=1; n is the best constant. An extremal polynomial is p = ni=1 yi Pi where y˜ is the vector deduced
from an eigenvector (GnT )y˜ of Gn−1 Kn;−1 (GnT )−1 corresponding to the eigenvalue 1; n .
In conclusion, in order to de ne the domain Dn where a is positive de nite for any p ∈ Pn − {0},
we have to give the smallest eigenvalue of Gn−1 Kn;−1 (GnT )−1 . Moreover the domain of positivity of
a depends on limn→∞ 1; n . To study the general case can be very dicult. In addition our aim is to
give the domain of positivity of a for any N; N ¿1, de ned by
∀p; q ∈ P;
a(p; q) =
N
X
m c(m) (p(m) ; q(m) );
m=0
where 0 = 1; N 6= 0 and c(m) is a positive de nite inner product, ∀m = 0; : : : ; N , and p(m) is the
derivative of order m of p.
In the cases where the inner products c(m) are linked and the derivatives p(m) are also linked
like for the classical inner products, it is possible to give some basic results about the domain of
positivity as well as the 1; n ’s.
In [5] we have considered the case where c(0) is a classical inner product (Hermite, Laguerre or
Jacobi) and c(m) is the classical inner product deduced from c(0) such that {Pn(m) }n¿m is a sequence
of polynomials orthogonal with respect to c(m) . In the sequel we will study the case where c(m) =
c; m = 0; : : : ; N , is the Laguerre, Jacobi or Gegenbauer inner product.
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
In addition to [5] we will give the Markov–Bernstein inequalities in that di erent cases. We have
kept the notation Cn(1) used in [5]. 1 In the Hermite case Cn(1) was already known (see [9]).
Corollary 1.5 (Markov–Bernstein inequalities). Let c(0) and c(1) be two classical inner products
such that the derivatives of the monic orthogonal polynomials with respect to c(0) are orthogonal
with respect to c(1) .
Let k · k(m) be the norm associated to the inner product c(m) respectively for m =
q0; 1. Then
∀p ∈ Pn − {0}; a(p; p) ¿ 0 if and only if 1 ¿ − 1=Cn(1) . Thus ∀p ∈ Pn ; kp′ k(1) 6 Cn(1) kpk(0)
where
Cn(1) =
2n
in the Hermite case;
in the Laguerre case;
n
n(n + + + 1)
in the Jacobi case:
2. Classical inner products
Let L2 (
; ) be the Hilbert space of square integrable real functions on the open set
⊂ R for
the positive Borel measure supported on
.
On this space we have the classical de nite inner product:
c(f; g) =
Z
f(x)g(x) d(x) ∀f; g ∈ L2 (
; )
R
and the norm c(f; f) = kfk2 =
f2 (x) d(x).
H N (
; ) will denote the Sobolev space
H N (
; ) = {f | f(m) ∈ L2 (
; ); m = 0; 1; : : : ; N };
where the derivatives f(m) of order m are taken in the distributional sense.
The de nite inner product on this Hilbert space is
s(f; g) =
N
X
c(f(m) ; g(m) );
m=0
∀f and g ∈ H N (
; ):
The corresponding norm, denoted by k · ks , is given by
kfk2s
=
N
X
m=0
kf(m) k2
∀f ∈ H N (
; ):
Now we consider the symmetric bilinear functional a
a : H N (
; ) × H N (
; ) → R
We want to indicate a mistake in [5]. n(m) = 2m in the Hermite case instead of 1. Consequently Formula given in the
same paper at the bottom of the page 173 has a version in the Hermite case, di erent from the Laguerre case, and that
1
formula has to be multiplied by 2
N (N +1)
2
.
(1)
Since Cn(m) = (n)m (m)
n ; Cn = 2n in the Hermite case.
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
207
de ned by
a(f; g) =
N
X
m c(f(m) ; g(m) );
m=0
∀f and g ∈ H N (
; );
(2)
where m ; m = 0; : : : ; N; are N + 1 xed real numbers with 0 = 1 and N 6= 0.
We look for the formal orthogonal polynomials with respect to a, that is to say, we look for the
polynomials Sn ; n¿0; such that
deg Sn = n;
a(Sn ; xi ) = 0
for i = 0; : : : ; n − 1:
(3)
These polynomials Sn will be called Sobolev formal orthogonal polynomials.
fn = (a(xj ; xi ))n−1 , we have the following obvious result:
Setting M
i; j=0
Theorem 2.1. The formal orthogonal polynomial Sn exists and is unique up to a normalization for
fn is regular.
the leading coecient is xed; if and only if the matrix M
De nition 2.2. The bilinear functional a is called positive de nite on H N (
; ) × H N (
; ) (resp.
on P × P) if, ∀f ∈ H N (
; ) − {0} (resp. P − {0}); a(f; f) ¿ 0.
Theorem 2.3. The bilinear functional a is positive de nite on P × P if and only if all the formal orthogonal polynomials Sn ; n ∈ N; exist with a positive leading coecient and a(Sn ; Sn ) ¿ 0;
∀n ∈ N.
P
Proof. (i) If a(Sn ; Sn ) ¿ 0; ∀n ∈ N, then ∀p ∈ P (deg p = k) we have p = ki=0 i Si and a(p; p) =
2
i=0 i a(Si ; Si ) ¿ 0.
(0) 2
f
(ii) For p = 1; a(1; 1) ¿ 0. Thus if S0 =
(0)
0 ¿ 0; a(S0 ; S0 ) = (
0 ) a(1; 1) ¿ 0. Moreover M1 is
f1 = a(S0 ; S0 ) ¿ 0. Therefore S1 exists.
regular and det M
fi ¿ 0; i = 1; : : : ; n. Then Sn exists. Let us write Sn as Sn = Pn
(n) x‘
Let us assume that det M
‘
‘=0
fn+1 is
with
n(n) ¿ 0. Let us denote by Xbn the vector of Rn of components
‘(n) ; ‘ = 0; : : : ; n − 1. M
multiplied on the right by the following (n + 1) × (n + 1) regular matrix
Pk
In bxn
0Tn
n(n)
;
where In is the n × n identity matrix and 0n the zero vector of Rn . Thanks to relations (3) we get
fn
M
ATn
0n
a(Sn ; xn )
!
;
where An is the vector of Rn of components a(xj ; xn ); j = 0; : : : ; n − 1. a(Sn ; xn ) = a(Sn ; Sn )=
n(n) ¿ 0.
fn+1 ¿ 0 and the desired conclusion follows.
Therefore det M
In the sequel we will study in which domain D of RN ; of components 1 ; : : : ; N must be located
in order to a be positive de nite on P × P for some particular Borel measures which correspond
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
Table 1
Name
Laguerre
Jacobi
]0; +∞[
] − 1; 1[
−x
kn
x e
n! (n + + 1)
Restrictions
¿−1
Gegenbauer
] − 1; 1[
(1 − x) (1 + x)
2n n! ( n )n (( n()n ))(2n −n)
with n = n + + 1
n = n + + 1
n = n + n − 1
¿ − 1; ¿ − 1
1
(1 − x2 ) − 2
21−2 −2n n! (n+2 )
(n+ +1) (n+ )
¿−
1
2
to the Laguerre, Jacobi and Gegenbauer orthogonal polynomials (respectively denoted by L n ; Pn( ; )
and Gn ). We recall that the Hermite case has been already studied in [5]. We summarize all the
useful informations concerning these three families of monic orthogonal polynomials in Table 1 (see
[2,12]), where kn is the square norm in L2 (
; ) of a monic orthogonal polynomial of degree n.
Moreover we have the following relations for the derivatives of these polynomials (see [2,12]):
d
L (x) = nL +1
n−1 (x);
dx n
(4)
d ( ; )
( +1; +1)
(x);
P (x) = nPn−1
dx n
(5)
d
+1
(x):
G (x) = nGn−1
dx n
(6)
3. Matrix interpretation of some particular relations
Some particular staircase recurrence relations will be of a great interest in order to give a general
expression of a(p; q); ∀p; q ∈ P. These relations are (see [2,12]):
+1
L n (x) = L +1
n (x) + nLn−1 (x);
( +1; )
(x)
Pn( ; ) (x) = Pn( +1; ) (x) + en( = +1; ) Pn−1
2n(n + )
( +1; )
= Pn( +1; ) (x) −
Pn−1
(x);
(2n + + + 1)2
( ; +1)
(x)
Pn( ; ) (x) = Pn( ; +1) (x) + en( ; = +1) Pn−1
2n(n + )
( ; +1)
= Pn( ; +1) (x) +
Pn−1
(x);
(2n + + + 1)2
+1
Gn (x) = Gn +1 (x) + en Gn−2
(x)
n(n − 1) +1
= Gn +1 (x) −
G (x)
4(n + )2 n−2
(7)
(8)
(9)
(10)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
209
and are satis ed ∀n ∈ N. (a)j ; j ∈ N, denotes the shifted factorial: (a)j = a(a − 1) · · · (a − j + 1).
These relations are simple consequences of the orthogonality when the weight function is modi ed
respectively by x; 1 − x; 1 + x; 1 − x2 .
By convention we complete the set of the previous relations (7) – (10) by the following ones:
L −k = L +1
−k = 0;
( ; )
( +1; )
( ; +1)
P−k
= P−k
= P−k
= 0;
+1
G−k
= G−k
= 0;
∀k ¿ 0;
(11)
Moreover the set of relations (4) – (6) is completed by:
d
L−k+1 = L +1
;
−k
dx
d ( ; )
( +1; +1)
(12)
P−k+1 = P−k
; ∀k ¿ 0;
dx
d
+1
;
G−k+1 = G−k
dx
For the sake of simplicity let us denote by Pi ; ∀i ∈ N, the monic orthogonal polynomials with
b −k; n ; ∀k and n ∈ N, the vector of components
respect to any previous inner product c, and by P
b −k; n the vector
Pj ; j = −k; : : : ; n, with the convention that Pj = 0, if j ¡ 0. We will denote by ddx P
whose components are the derivatives of Pj ; j =−k; : : : ; n. When the distinction between the di erent
b ; P
b ( ; ) and
families of orthogonal polynomials will be necessary, then we will use the notation L
−k; n
−k; n
b
b
G−k; n instead of P−k; n .
Now, let us give some matrix expressions deduced from relations (7) – (12).
Property 3.1.
b +1 ;
b = E−k; n L
L
−k; n
−k; n
(13)
( ; ) b ( +1; +1)
;
= E−k;
n P−k; n
(15)
b ( ; ) = E ( = +1; ) P
b ( +1; ) = E ( ; = +1) P
b ( ; +1)
P
−k; n
−k; n
−k; n
−k; n
−k; n
b +1
b = E G
G
−k; n −k; n ;
−k; n
( = +1; )
E−k;
;
n
(14)
(16)
( ; = +1)
E−k;
;
n
( ; )
E−k;
n
where E−k; n ;
and E−k;
n are (n + 1 + k) × (n + 1 + k) nonsingular matrices
whose the entries (i; j) are in row i and column j; i; j = 1; : : : ; n + k + 1:
E−k; n (i; j) =
1
i − k − 1;
0
( = +1; )
E−k;
(i; j)
n
∀i = j;
=
everywhere else;
1
∀i = j;
e( = +1; )
i−k−1
∀i = j + 1; k + 26i6n + k + 1;
0
∀i = j + 1; k + 26i6n + k + 1;
everywhere else;
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
( ; = +1)
E−k;
(i; j)
n
( ; )
E−k;
n (i; j) =
=
1
∀i = j;
e( ; = +1)
i−k−1
∀i = j + 1; k + 26i6n + k + 1;
0 everywhere else;
1 ∀i = j;
2‘( − )
∀i = j + 1;
(2‘ + + )(2‘ + + + 2)
with ‘ = i − k − 1 and 16‘6n;
4‘(‘ − 1)(‘ + )(‘ + )
−
∀i = j + 2;
(2‘
+ + )(2‘ + + + 1)3
with ‘ = i − k − 1 and 26‘6n;
(17)
0 everywhere else;
1
∀i = j;
‘(‘ − 1)
∀i = j + 2; with ‘ = i − k − 1 and 26‘6n;
e‘ = −
E−k;
n (i; j) =
4(‘ + )2
0 everywhere else;
b −k; n = JT
P
−k+1; n+1
d b
P−k+1; n+1 ;
dx
(18)
T
where J−k+1;
n+1 is a (n + 1 + k) × (n + 1 + k) nonsingular matrix obtained by multiplication of
( ; )
b
E−k; n , ( resp. E−k;
n ; E−k; n ) on the right by the diagonal matrix D−k+1; n+1 whose the elements are
b −k+1; n+1 (i; i) =
D
1;
1;
‘
i = 1; : : : ; k;
‘ = 1; : : : ; n + 1 with i = k + ‘:
Proof. Relations (13), (14) and (16) are the matrix version of relations (7) – (10) for the vectors
b −k; n . Eq. (15) is obtained from
P
b ( ; ) = E ( = +1; ) E ( +1; = +1) P
b ( +1; +1) = E ( ; = +1) E ( = +1; +1) P
b ( +1; +1) :
P
−k; n
−k; n
−k; n
−k; n
−k; n
−k; n
−k; n
It is easy to verify that
( = +1; ) ( +1; = +1)
( ; = +1) ( = +1; +1)
E−k;
E−k; n
= E−k;
E−k; n
:
n
n
From Eqs. (4) – (6) it is obvious that a vector Pˆ−k; n can be written as the product of Dˆ −k+1; n+1
and the derivative of a vector Pˆ−k+1; n+1 (but with modi ed parameters ; or and ). Substituting the vector of the right part of any relation (13), (15) or (16) by this last relation, Eq. (18)
is obtained.
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
211
4. The quasi-orthogonal polynomial Ri associated with Pi
Let {Ri }i¿0 be a sequence of monic polynomials which satis es the following conditions:
deg Ri = i; ∀i ∈ N;
di
Ri = i!P0 = i; N P0 ;
d xi
for 06i6N − 1;
dN
Ri = (i)N Pi−N = i; N Pi−N ;
d xN
(19)
for i¿N:
(20)
Let Rˆ n , be the vector of components Ri ; i = 0; : : : ; n, and Dn; N be the (n + 1) × (n + 1) diagonal
matrix (i; N )ni=0 .
The matrix interpretation of relations (19) and (20) is
dN ˆ
Rn = Dn; N Pˆ−N; n−N :
d xN
(21)
Remark that Dn; N contains i; N = i! for i ¡ N and that
Using Eq. (18), Eq. (21) becomes:
dN
d xN
Ri = 0 and P−i = 0 for i ¡ N .
dN ˆ
d ˆ
T
P−N +1; n−N +1
Rn = Dn; N J−N
+1; n−N +1
N
dx
dx
dj ˆ
(22)
P−N +j; n−N +j ∀j = 2; : : : ; N:
d xj
In order to
simplify the writing of products with di erent matrices J, we will Q
use the following
QN −1
N −1
notation: j=i
J−j; n−j = J−i; n−i · · · J−N +1; n−N +1 in this order. If i ¿ N − 1, then j=i
J−j; n−j will
be taken equal to 1. Thus Eq. (22) is
T
T
= Dn; N J−N
+1; n−N +1 · · · J−N +j; n−N +j
T
NY
−1
dN ˆ
dj ˆ
R
P−N +j; n−N +j
=
D
J
n
n; N
−i; n−i
d xN
d xj
i=N −j
∀j = 2; : : : ; N:
(23)
From Eq. (23) written for j = N , we set
Rˆ n = Dn; N
NY
−1
j=0
T
J−j; n−j Pˆ 0; n :
(24)
Therefore, from Eqs. (24) and (18), we also have
T
NY
−1
d N −j ˆ
=
D
J−i; n−i Pˆ−N +j; n−N +j :
R
n
n;
N
d xN −j
i=N −j
(25)
De nition 4.1. An n × n matrix A = (aij ) is a band matrix if aij = 0; | i − j | ¿ h with h xed (h ∈ N)
and h + 1 ¡ n: h is the half width of band.
An n × n matrix A is a lower (resp. upper) triangular band matrix if aij = 0 for i ¡ j (resp. i ¿ j)
and i − j ¿ h (resp. i − j ¡ − h) with h xed (h ∈ N) and h + 1 ¡ n: h is the width of band.
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
Q
N −1
From the de nition of the matrices JT , it is clear that ( j=0
J−j; n−j )T is a lower triangular band
matrix of width N in the Laguerre case, 2N in the Jacobi and Gegenbauer cases. Moreover all the
entries of the last lower diagonal which is the border of the band, are non zero from the second
column (the rst column contains a unique non-zero entry in the rst row). Therefore we have the
following obvious property about the quasi-orthogonality of the Ri ’s (see [2]).
Property 4.2. In the Laguerre case (resp. in the Jacobi and Gegenbauer cases) the polynomials
Ri are quasi-orthogonal of order N (resp. 2N ) with respect to the Laguerre (resp. Jacobi or
Gegenbauer) inner product. They are strictly quasi-orthogonal when i ¿ N (resp. i ¿ 2N ) in the
Laguerre case (resp. in the Jacobi and Gegenbauer cases).
5. Matrix interpretation of a(p; q)
Let p and q be two polynomials of degree n. Since {Ri }i¿0 is a basis of P, we can write p and
q as
p = (Rˆ n )T y and q = (Rˆ n )T z;
where y and z are vectors of Rn+1 which contain the coordinates yi (resp. zi ) of p (resp. q) in the
basis {Ri }ni=0 . Then
d N −j ˆ
d N −j
p
=
Rn
d xN −j
d xN −j
T
y
= (Pˆ−N +j; n−N +j )T
NY
−1
J−i; n−i Dn; N y:
i=N −j
Let us de ne the (t + 1) × (t + 1) matrix Kt; j by Kt; j = (c(Pi ; P‘ ))i;t−j
‘=−j . Thus Kt; j is a diagonal matrix
t−j
(ki )i=−j where ki = c(Pi ; Pi ); i¿0; and ki = 0 if i ¡ 0.
Therefore:
c
Finally
T
NY
−1
d
d
T
p;
q
=
z
D
J
J−i; n−i Dn; N y:
K
n; N
−i; n−i
n; N −j
d xN −j d xN −j
i=N −j
i=N −j
N −j
a(p; q) =
N −j
N
X
NY
−1
i c(p(i) ; q(i) )
i=0
=
N
X
i z T Dn; N
NY
−1
j=i
i=0
J−j; n−j
!T
Kn; i
a(p; q) =
i Fi; n (y; z) = Fb(n); n (y; z)
i=0
where b(n) = min(n; N ).
b(n)
X
i=0
NY
−1
b(n)−1
i
Y
r=i
!
J−j; n−j Dn; N y:
j=i
From Eq. (27) a(p; q) can also be written as
b(n)
X
(26)
!
Fr; n (y; z)
;
Fr+1; n (y; z)
(27)
(28)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
213
Setting Y = Dn; N y and Z = Dn; N z, we have
Q
Q
b(n)−1
b(n)−1
Z T ( j=r
J−j; n−j )T Kn; r ( j=r
J−j; n−j )Y
Fr; n (y; z)
=
:
Q
Q
b(n)−1
b(n)−1
T
T
Fr+1; n (y; z) Z ( j=r+1 J−j; n−j ) Kn; r+1 ( j=r+1 J−j; n−j )Y
Q
b(n)−1
Let u(r; n) (resp. v(r; n) ) be the vector ( j=r
J−j; n−j )Y (resp. (
Its components are numbered from 0 to n.
1
Qb(n)−1
j=r
(29)
J−j; n−j )Z); ∀r = 0; : : : ; b(n).
Let (Kn; r+1 ) 2 be the diagonal matrix whose entries are the square roots of the elements of Kn; r+1 .
1
1
The vector (Kn; r+1 ) 2 u(r+1; n) (resp. (Kn; r+1 ) 2 v(r+1; n) ) has its rst r + 1 components equal to zero. Let
us denote by w(r+1; n) (resp. s(r+1; n) ) the vector of Rn−r which contains the last n − r components of
this vector. From that and the expression of J−r; n−r we get
k0 ur(r+1; n) vr(r+1; n) + (s(r+1; n) )T Jn−r w(r+1; n)
Fr; n (y; z)
=
;
Fr+1; n (y; z)
(s(r+1; n) )T w(r+1; n)
(30)
where Jn−r is a (n − r) × (n − r) matrix. It contains the rst n − r rows and columns of the matrix
Jn which appears in Eq. (30) for r = 0.
Finally we obtain the following expression of a(p; q)
a(p; q) = Fb(n); n (y; z)
b(n)
X
b(n)−1
Y k0 u(r+1; n) v(r+1; n) + (s(r+1; n) )T Jn−r w(r+1; n)
r
r
i
(s(r+1; n) )T w(r+1; n)
r=i
i=0
!
:
(31)
Remark that Jn is independent of N . Indeed, since the n last rows and columns of Kn;1 correspond
to Kn−1;0 ; we have
J0;T n Kn; 0 J0; n
=
k0
0
0
(Kn−1;0 ) 2 Jn (Kn−1;0 ) 2
1
1
!
;
(32)
which can also be written as
1
1
J1;T n Kn−1;−1 J1; n = (Kn−1;0 ) 2 Jn (Kn−1;0 ) 2 :
(33)
Remark that Jn is similar to Gn−1 Kn;−1 (GnT )−1 obtained in Section 1.
Therefore the matrices Jn corresponding to the di erent cases (Laguerre, Jacobi, Gegenbauer) can
be obtained by using Eq. (33).
In the Laguerre case (see also [9]), Jn is an n × n symmetric tridiagonal matrix:
k1
= 1 + for i = j = 1
k0
ki + i2 ki−1
=2+
for i = j = 2; : : : ; n;
2k
i
i
i−1
s
r
Jn (i; j) =
1
ki−1
for i = j + 1;
= 1+
i − 1 ki−2
i−1
for i = j − 1 the matrix is completed by symmetry;
0
everywhere else:
(34)
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
In the Jacobi case, Jn is an n × n symmetric ve diagonal matrix:
k1
for i = j = 1;
k0
4( − )2
k2
+
for i = j = 2;
4k1 (2 + + )2 (4 + + )2
4( − )2
ki−1 ki−2
+
( +1; +1) 2
(2i − 2 + + )2 (2i + + )2
(i − 2)2 (ki−3
)
+
ki
2
i ki−1
for i = j ¿ 2;
2( − )
Jn (i; j) = (2 + + )(4 + + )
s
k1
k0
for (i; j) = (1; 2) and (2; 1);
s
(2 − )
ki−1
ki−2
−
( +1;
+1)
(2i − 2 + + ) ki−2 ki−3
(i − 2)(2i − 4 + + )
1
; for i = j + 1 ¿ 2;
−
(i − 1)(2i + + )
s
k
ki−1
i−2
−
for i = j + 2;
( +1; +1)
2
(i − 2) ki−3
ki−3
for i = j − 2 and i = j − 1 ¿ 1 the matrix is completed by symmetry;
0
(35)
everywhere else:
ki is ki( ; ) in Eq. (35).
In the Gegenbauer case (see also [9]), Jn is an n × n symmetric ve diagonal matrix:
Jn (i; j) =
ki
for i = j = 1; 2;
2k
i
i−1
(ei−1
)2 ki−2
ki
+
for i = j ¿ 2;
i2 ki−1
(i − 2)2 ki−1
ei−1
ki−2
p
for i = j + 2;
2
(i − 2) ki−3 ki−1
for i = j − 2 the matrix is completed by symmetry;
0
(36)
everywhere else:
From Eq. (32) and the de nition of Kn−1;0 and Kn; 0 , we have the following obvious property
Property 5.1. Jn is a positive de nite symmetric matrix.
Property 5.2. If i; n ; i = 1; : : : ; n; are the eigenvalues of Jn ; corresponding to the three previous
cases; with 0 ¡ i; n 6 · · · 6n; n ; then limn→∞ n; n ¡ + ∞.
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
215
Proof. As an obvious consequence of the Courant–Fischer therorem (see [14]) n; n 6n+1; n+1 ; ∀n .
Thus the positive sequence {n; n }n¿1 is increasing.
We use the localization of the eigenvalues given by the Gerschgorin disks. From Eqs. (34) – (36)
it is clear that all the centers of these disks are at a nite distance and all the radii are bounded.
Thus n; n is bounded ∀n and the result holds.
Remark 5.3. Relation (28) can also be written by means of Horner algorithm
d0 = 1;
Fi−1; n (y; z)
di−1 ;
Fi; n (y; z)
a(p; q) = Fb(n); n (y; z)db(n) :
d i = i +
i = 1; : : : ; b(n);
(37)
6. Domain of positivity of a
Let be an element of RN of components m ; m = 1; : : : ; N (N 6= 0).
We want to nd the domain D ⊂ RN such that
D = { ∈ RN | a(p; p) ¿ 0 ∀p ∈ P − {0}};
where a is considered as a function depending on .
∀n¿1; Dn will denote the domain of Rb(n) such that
Dn = { ∈ Rb(n) | a(p; p) ¿ 0 ∀p ∈ Pn − {0}}:
(38)
For any xed p ∈ Pn − {0}; a(p; p) = 0 is a hyperplane of Rb(n) such that the coecients of the
i ’s are positive (see (31)). Thus this hyperplane cuts the axes of Rb(n) in the negative part.
Let us begin to prove a property of N .
Theorem 6.1. Let i; n ; i = 1; : : : ; n; be the eigenvalues of Jn ; (0 ¡ 1; n 6 · · · 6n; n ). n; n is assumed
to be bounded ∀n.
(i) If limn→∞ 1; n = 0 and if = (1 ; : : : ; N ) ∈ D; then N ¿ 0.
(ii) If inf n 1; n ¿ ¿ 0 and if i ¿ 0; i = 1; : : : ; N − 1; then there exists ˆ ¡ 0 depending on the
i ′ s; i = 1; : : : ; N − 1 such that ∀N ¿ ;
ˆ = (1 ; : : : ; N ) ∈ D.
Proof. (i) In FN −1; n (y; y)=FN; n (y; y) we take w(N; n) as the eigenvector corresponding to the smallest
eigenvalue 1; n−N +1 of Jn−N +1 . The vector u(N; n) of components ui(N; n) ; i = 0; : : : ; n, is chosen such
n)
that uN(N;−1
= 0; ui(N; n) ; i = N; : : : ; n, being de ned from the successive components of w(N; n) thanks to
the following relation:
1
(K(n; N ) ) 2 u(N; n) =
0N
w(N; n)
;
where 0N is the zero vector of RN . Then
FN −1; n (y; y) (w(N; n) )T Jn−N +1 w(N; n)
=
= 1; n−N +1 ;
FN; n (y; y)
kw(N; n) k22
where k · k2 is the Euclidean norm of a vector.
(39)
216
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
Thus limn→∞ FN −1; n (y; y)=FN; n (y; y) = 0.
Now we take the vector Y of Rn+1 such that
Yi =
(
ui(N; n)
0
Q
for i = N; : : : ; n;
for i = 0; : : : ; N − 1:
N −1
Since u(r; n) = j=r
J−j; n−j Y , then ui(r; n) = 0 for i = 0; : : : ; r − 1.
Therefore we get
a(p; p) = F(N; n) (y; y)
N
X
i
NY
−1
r=i
i=0
(w(r+1; n) )T Jn−r w(r+1; n)
kw(r+1; n) k22
!
:
(40)
Each factor (w(r+1; n) )T Jn−r w(r+1; n) =kw(r+1; n) k22 which is the Rayleigh quotient Jn−r (w(r+1; n) ) (see [14]),
is bounded by the largest eignevalue n−r; n−r of Jn−r . Since n−r; n−r is also bounded, then Fr−1; n (y; y)=
Fr; n (y; y) which is the coecient of dr−1 in (37), is also bounded, ∀r = 1; : : : ; N − 1. Since the
coecient FN −1; n (y; y)=FN; n (y; y) of dN −1 tends to zero when n tends to in nity, dN remains positive
if and only if N ¿ 0, and the result holds.
(ii) The same technique with the same vectors is used in this second part. Thus we have
FN −1; n (y; y)
¿ inf 1; n ¿ ¿ 0:
n
FN; n (y; y)
Moreover, Jn−r (w(r+1; n) ) is such that
0 ¿ 61; n−r 6Jn−r (w(r+1; n) )6n−r; n−r 6 sup n; n ¡ + ∞:
n
Thus using Remark 5.3
N
−1
X
i N −1−i 6dN −1 6
i=0
N
−1
X
i sup n; n
i=0
n
N −1−i
:
Hence if i ¿ 0; i =1; : : : ; N −1; dN which is such that dN ¿N −dN −1 is positive for any N ¿ ˆ =
−dN −1 . Thus the result holds.
6.1. Case N = 1
In the particular case where N = 1, the domains Dn and D are characterized by means of 1; n and
limn→∞ 1; n . We give lower and upper bounds of 1; n in the Laguerre case, a lower bound of 1; n
in the Gegenbauer case, and the behavior of an upper bound in the Gegenbauer and Jacobi cases.
Theorem 6.2.
Dn = {1 ∈ R; 1 ¿ − 1; n };
where 1; n is the smallest eigenvalue of Jn .
Proof. From (31) we have
a(p; p) = kw(1; n) k22
(w(1; n) )T Jn w(1; n)
k0 y02
+
1 +
kw(1; n) k22
kw(1; n) k22
!
:
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
217
The case where w(1; n) = 0 is without interest, since p = y0 R0 in this case, and of course a(p; p) =
c(p; p) = k0 y02 ¿ 0. Therefore we have to study
1 +
k0 y02
(w(1; n) )T Jn w(1; n)
+
¿ 0 ∀p ∈ Pn − {0};
kw(1; n) k22
kw1; n) k22
that is to say, to nd 1 depending on n such that
1 ¿ −
min
w(1; n) ∈Rn −{0}
(w(1; n) )T Jn w(1; n)
k0 y02
+
2
kw(1; n) k2
kw1; n) k22
!
(41)
:
(w(1; n) )T Jn w(1; n) =kw(1; n) k22 is the Rayleigh quotient Jn (w(1; n) ). It is independent of y0 . Therefore the
minimum of the right part of Eq. (41) is obtained for y0 = 0 and w(1; n) being the eigenvector
corresponding to the smallest eigenvalue 1; n of Jn ; this minimum is equal to 1; n .
Now we give some properties about the smallest eigenvalue 1; n . A upper bound will be obtained
from the value of the Rayleigh quotient which corresponds to a particular vector. A lower bound
will be given in the case where the eigenvalues are the zeros of orthogonal polynomials. It is given
by the rst step of the Newton method with 0 as starting point.
In the Laguerre case the eigenvalues of Jn are the one’s of a tridiagonal matrix which corresponds to a Jacobi matrix of sequence of orthogonal polynomials. These polynomials correspond to
co-recursive polynomials of the generalized Pollaczek polynomials (this fact was presented in the
doctoral dissertation of Perez [10] and in the paper of Marcellan et al. [7]). See also [9].
Theorem 6.3 (Laguerre case). The eigenvalues i; n ; i = 1; : : : ; n; are the zeros of the orthogonal
polynomials An (x) de ned from the following three-term recurrence relation:
An (x) = x − 2 −
An−1 (x) − 1 +
An−2 (x);
n
n−1
(42)
n¿2;
with A0 (x) = 1 and A1 (x) = x − − 1.
These zeros are real; positive; distinct and
√
3( + 2) − ( + 2)( + 10)
1;1 = + 1; 1; 2 =
;
4
2( + 1)
3
;
¡ 1; n 6(n + n )
n(n + 1)
n (22n + 1)
∀n¿3;
(43)
where n = [ n+1
] ([:] denotes the integer part).
2
!
2
2n − 2 +
2 2 − 2
( − 1) 1 − +
+
1−
ln
;
+
n
2
2 2+
2
2
2+
n =
!
2
2
2n +
−
1)
1
−
+
+
1
−
ln
; if ¿2;
(
n
2
2
2
2+
if 62;
218
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
!
2
2(2n − 2 + )
4n +
2
1
−
+
ln
+
4
(
−
1)
+
(
−
2))
ln
−
;
(8
n
n
n
2
4n − 2 +
4
2n +
if 60;
!
2
(2n − 1)(2n + )
2
n 1 − +
+ 4 n ln
− (8n ( − 1) + ( − 2))
2
(
−
1)(4
+
)
4
n
n
n =
4n − 2 +
; if 06 6 ˆn ;
ln
2
n −2+
!
(2n − 1)(2n + )
2
2
n 1 − +
+ 4 n ln
− (8n ( − 1)
2
(
−
1)(4
+
)
4
n
n
4n +
+ ( − 2)) ln
; if ¿ ˆn ;
2n +
p
where ˆn = −4n + 1 + 162n + 1; (1 ¡ ˆn ¡ 2); is the largest zero of 8n ( − 1) + ( − 2)
considered as a polynomial in one variable .
Proof. The rst zeros 1;1 and 1; 2 are obtained in an obvious way from (42). The rst part is
obtained from a classical property for tridiagonal matrices:
An (x) = det(xI − Jn )
= x−2−
det(xI − Jn−1 ) − 1 +
det(xI − Jn−2 ):
n
n−1
Since the coecient 1 + =(n − 1) of An−2 is strictly positive, we can deduce that the eigenvalues
of the positive de nite symmetric matrix Jn which are real and positive, are, besides, distinct.
The zeros of An being distinct real, those of A′n ; A′′n , and so on, also are real distinct and are located
between 1; n and n; n . Thus A′′n has a constant sign for x ¡ 1; n . Therefore ’n = −An (0)=A′n (0) ¡ 1; n .
Using Eq. (42) it is easy to verify by recurrence on n that
An (0) = (−1)n
( + n)n
:
n!
(44)
Since
A′n (0) = − 2 +
A′n−1 (0) + An−1 (0) − 1 +
A′n−2 (0);
n
n−1
we get by recurrence on n:
A′n (0) = (−1)n−1
(n + 1)( + n)n−1
:
2(n − 1)!
From Eqs. (44) and (45), the lower bound of Eq. (43) is satis ed.
(45)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
219
From Eq. (34) we get for any vector y ∈ Rn
T
y Jn y =
n−1
X
1
i
i=1
=
s
r
n−1
X
i=1
ki
ki−1
!2
+
1 kn 2
y
n2 kn−1 n
!2
+
n+ 2
yn :
n
yi + yi+1
i+
yi + yi+1
i
Let us choose the vector y such that
(−1)i i
n+1
if i6n =
;
2
yi =
(−1)i (2n − i) if n 6i62n :
(46)
Remark that yn+1 = 0 if n is odd.
For the vector y given by (46) we have
n
X
yi2
=
i=1
n
X
n −1
2
i +
i=1
X
i2 =
i=1
n (22n + 1)
:
3
The corresponding numerator of the Rayleigh quotient Jn can be written as
X p
2
n−1
i=1
(i + )i − i − 1 +
n
X
j=1
s
2n − j +
j−j+1
2n − j
!2
:
Let us denote by i (resp. q
j ) an element in the rst (resp. second) sum.
√
2
2
Since (i + )i = (i + 2 ) 1 − (2i+ )
2 ¿(i + 2 )(1 − (2i+ )2 ), we get
p
i = i(i + ) + (i + 1)2 − 2(i + 1) i(i + )61 − +
2 2 (2 − )
+
:
2
2(2i + )
In the same way we have
j =
p
1
((2n − j + )j 2 + (2n − j)( j − 1)2 − 2j( j − 1) (2n − j)(2n − j + ))
2n − j
1
6
2n − j
2
+j( j − 1)
4n − 2j +
= 1+
(2n − j + )j + (2n − j)( j − 1) − 2j( j − 1) 2n − j +
2
2
!
2
( − 2)
42n
82n
(8n ( − 1) + ( − 2))
+
−
−
:
2
2n − j 4n − 2j +
2(4n − 2j + )
220
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
Now we will use the following inequalities. Let a be a xed real. If a + x ¿ 0 for x ∈ [‘ − 1; k +
1]; ‘; k ∈ N, then
ln
a+k +1
=
a+‘
Z
k+1
‘
k
X
1
dx
6
6
a + x i=‘ a + i
Z
‘−1
Pn −1
Therefore, in using Eq. (47), we obtain
Finally
3(n + n )
:
1; n = min
Jn (y)6
n
y∈R −{0}
n (22n + 1)
j=1
k
a+k
dx
= ln
:
a+x
a+‘−1
i 6n and
Pn
j=1
(47)
j 6 n .
Remark that n = O(n ) and n = O(n ). Therefore 1; n = O(1=n2 ).
When = 0, Turan [13] (see also [9]) gave the precise expression of 1; n which is
:
1; n = 4 sin2
2(2n + 1)
For any , Dor
er [4] gave an upper bound of 1=1; n in using the Frobenius norm of a certain
matrix. But the entries of that matrix are wrong. Indeed the coecient c which appears in his relations
depends on k : c(k) = (1 + =k)−1=2 . He wrote all the relations with a coecient c independent of
k. Therefore his upper bound of 1=1; n is false.
As a straightforward consequence of Theorems 6.2 and 6.3 we get the Markov–Bernstein inequality
in the Laguerre case.
Corollary 6.4. In the Laguerre case the following Markov–Bernstein inequality is satis ed.
∀p ∈ Pn − {0};
1
kpk ¡
kp k6 √
1; n
′
√
1= 1; n is the best constant.
An extremal polynomial is
p=
s
n(n + 1)
kpk:
2( + 1)
n
X
w(1; n)
pi
Ri ;
i=1
i ki−1
where w(1; n) = (w1(1; n) ; : : : ; wn(1; n) )T is the eigenvector of Jn corresponding to the eigenvalue 1; n .
Proof. The form of an extremal polynomial is a consequence of relations (41) and (39) with
N = 1.
Corollary 6.5. D = R+ .
Remark that this result could be obtained from the region of R which is a region of density of
the set of all zeros of all An ; ∀n¿1 (see Chihara [2,3]). Since limn→∞ (2 + =n) = 2 and limn→∞ (1 +
=(n − 1)) = 1, then [0,4] is that region. Therefore a is positive de nite for 1 ¿ 0.
Corollary 6.5 also is a consequence of Theorem 6.1. The assumptions of this theorem are satis ed,
since limn→∞ 1; n =0 and n; n bounded ∀n (Property 5.2). That implies, in the Laguerre case, if ∈ D,
then N ¿ 0.
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
221
In the Gegenbauer case, the matrix Jn has ve diagonals, but the two diagonals which are here
and there of the main diagonal are zero. Therefore Jn can be transformed in a similar matrix which
has a structure of 2 × 2 diagonal block matrix. Moreover the two diagonal blocks are tridiagonal.
The matrices used in the transformation are permutation matrices Vij (i 6= j) which are such that
Vij (‘; ‘) = 1
for ‘ = 1; : : : ; n; ‘ 6= i and ‘ 6= j;
V (i; j) = Vij ( j; i) = 1;
ij
0 everywhere else:
Of course Vij Vij = I . Thus Jn is multiplied on the right and on the left by such matrices having
the same indices. They permute rows and columns having the same couple of numbers. Finally we
get a [(n + 1)=2] × [(n + 1)=2] rst tridiagonal block, denoted by Jˆ[ n+1
; and a [n=2] × [n=2] second
2 ]
˜
tridiagonal block, denoted by J [ 2n ] ; where [:] denotes the integer part. We have
Jˆ[ n+1
(i; j) =
2 ]
k1
for i = j = 1;
k0
(e2i−2
)2 k2i−3
1
k2i−1
+
(2i − 1)2 k
(2i − 3)2 k
2i−2
for i = j ¿ 1;
2i−2
e2i
k2i−1
p
for i = j − 1;
2
(2i − 1) k2i−2 k2i
for i = j + 1 the matrix is completed by symmetry;
0
everywhere else:
Using the expression of ki ; we get the following sequence of monic orthogonal polynomials for this
Jacobi matrix
Aˆ i+1 (x) = (x − i )Aˆ i (x) − i Aˆ i−1 (x);
(48)
i¿1;
with Aˆ 0 = 1; Aˆ 1 (x) = x − 1=2( + 1); and
1
i =
2
i =
i+
i
+
;
(2i + 1)(2i + + 1)2 (2i + )2 (2i + 2 − 1)
i(i + − 1)
;
4(2i − 1)(2i + 2 − 1)(2i + )3 (2i + − 1)
J˜ [ n2 ] (i; j) =
k2
for i = j = 1;
4k1
(e2i−1
)2 k2i−2
1 k2i
+
(2i)2 k2i−1
(2i − 2)2 k2i−1
for i = j ¿ 1;
e2i+1
k2i
p
for i = j − 1;
2
(2i) k2i+1 k2i−1
for i = j + 1 the matrix is completed by symmetry;
0
everywhere else:
(49)
(50)
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
We get another sequence of monic orthogonal polynomials for this Jacobi matrix
A˜ i+1 (x) = (x − i+ 1 )A˜ i (x) − i+ 1 A˜ i−1 (x); i¿1;
2
(51)
2
with A˜ 0 = 1; A˜ 1 (x) = x − (2 + 1)=(8( + 2)2 ); i+ 1 and i+ 1 are obtained from (49) and (50) in
2
2
replacing i by i + 1=2.
Remark that these polynomials should be those found by Perez in her doctoral dissertation [10],
but she gave a wrong expression for i . See also [9].
Theorem 6.6 (Gegenbauer case). The eigenvalues ˆi; n ; i = 1; : : : ; n; of Jˆn which are the zeros of the
orthogonal polynomials Aˆ n given by (48); are real; positive; distinct. Moreover
(1 + 2 )
1
¡ ˆ1; n 6ˆ n = O 4 ;
2
2n(2n + 2n − 1)(n + )
n
n¿1:
Proof. It is still obvious that the ˆi; n ’s are real, positive, distinct.
(i) For nding the lower bound, we will use the same technique as in Theorem 6.3.
Using Eq. (48) it is easy to verify by recurrence on n that
n!
Aˆ n (0) = (−1)n
:
(2n)!(2n − 1 + )n
(52)
Indeed Eq. (52) is satis ed for Aˆ 1 (0) = −1=(2(1 + )). If Eq. (52) is assumed to be satis ed up to
the degree n, we have
2n + 2
(−1)n n!
2n
+
(2n + 1)(2n + + 1)2 (2n + )2 (2n + 2 − 1) (2n)!(2n − 1 + )n
2n(2n + 2 − 2)(−1)n−1 (n − 1)!
−
16(2n − 1)(2n + 2 − 1)(2n + )3 (2n + − 1)(2n − 2)!(2n + − 3)n−1
2(n + )2(n + 1)n!
= (−1)n+1
4(2n + 1)(2n + 2)(2n + + 1)2 (2n)!(2n + − 1)n
2n(−1)n n!
−
4(2n + )2 (2n + 2 − 1)(2n)!(2n + − 1)n
2n(2n + 2 − 2)(−1)n−1 (n − 1)!
−
16(2n − 1)!(2n + )n+2 (2n + 2 − 1)(2n + − 1)
(−1)n+1 (n + 1)!
:
=
(2n + 2)!(2n + 1 + )n+1
1
Aˆ n+1 (0) = −
4
On the other hand, we have
′
Aˆ n (0) = (−1)n−1
n!(2n2 + 2n − 1)
;
(2n − 1)!(1 + 2 )(2n − 1 + )n−1
(53)
n¿1:
′
Aˆ 1 (0) = 1 satis es Eq. (53). If Eq. (53) is assumed to be satis ed up to the rank n; we have
′
(−1)
Aˆ n+1 (0) =
2
n
n+
n
+
(2n + 1)(2n + + 1)2 (2n + )2 (2n + 2 − 1)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
223
n!(2n2 + 2n − 1)
n!
+ (−1)n
(2n − 1)!(1 + 2 )(2n − 1 + )n−1
(2n)!(2n + − 1)n
n(n
+
−
1)
+(−1)n−1
4(2n − 1)(2n + 2 − 1)(2n + )3 (2n + − 1)
(n − 1)! 2(n − 1)2 + 2(n − 1) − 1
×
(2n − 3)! (1 + 2 )(2n − 3 + )n−2
(n + )(2n2 + 2n − 1)
n!
= (−1)n
2(2n + 1)(2n − 1)! (2 + 1)(2n + + 1)n+1
1
n!
((1 + 2 )(2n + ) + n(2n − 1))
+(−1)n
(2n)! (1 + 2 )(2n + )n+1
(n + )(2n2 + 2n − 1)
n!
= (−1)n
2(2n + 1)(2n − 1)! (1 + 2 )(2n + + 1)n+1
n! (n + )(2n + 2 + 1)
+(−1)n
(2n)! (1 + 2 )(2n + )n+1
(n
+ 1)! 2(n + 1)2 + 2(n + 1) − 1
:
= (−1)n
(2n + 1)! (1 + 2 )(2n + + 1)n
×
′
The ratio −Aˆ n (0)= Aˆ n (0) gives the lower bound.
(ii) In order to obtain the upper bound, we will use the Rayleigh quotient Jˆ of Jˆn . Its numerator
is
n−1
X
i=1
=
1
2i − 1
s
s
n−1
X
i=1
k2i−1
e2i
yi +
k2i−2
2i − 1
s
k2i−1
yi+1
k2i
i−1+
yi −
2(2i − 1)(2i + − 1)2
!2
s
n−1+
y2 :
2(2n − 1)(2n + − 1)2 n
Let us choose the vector y such that
n+1
2
;
i
for i6n =
yi =
2
2
(2n − i) for n 6i62n :
+
1
k2n−1 2
y
2
(2n − 1) k2n−1 n
i
yi+1
2(2i + )2 (2i + 2 − 1)
!2
+
(54)
Since yn+1 = 0 if n is odd, we can write this numerator as
n −1
X
i=1
s
i+ −1
i2 −
2(2i − 1)(2i + − 1)2
2n −1
+
X
i=n
s
−
s
s
i
(i + 1)2
2(2i + )2 (2i + 2 − 1)
i+ −1
(2n − i)2
2(2i − 1)(2i + − 1)2
i
(2n − 1 − i)2
2(2i + )2 (2i + 2 − 1)
!2
!2
224
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
n −1
=
X
i=1
2n −1
ˆi +
X
ˆi ;
i=n
i(i + 1)4
(i + − 1)i4
+
2(2i − 1)(2i + − 1)2 2(2i + )2 (2i + 2 − 1)
√
i2 (i + 1)2 2i(2i + )(2i − 1)(2i + 2 − 1)(2i + − 2)(2i + 2 − 2)
−
2(2i − 1)(2i + )3 (2i + 2 − 1)
4
i (i + − 1)
i2 (i + 1)2 $i
i(i + 1)4
6
−
;
+
2(2i − 1)(2i + − 1)2 2(2i + )2 (2i + 2 − 1) (2i − 1)(2i + )3 (2i + 2 − 1)
ˆi =
where
1
2
$i =
4i + −
8
4i +
!
2
2i − 1 + −
2i − 1 +
!
2
4i − 4 + 3 −
4i − 4 + 3
!
:
In the same way we get
(2n − i)2 (2n − 1 − i)2 $i
(i + − 1)(2n − i)4
i(2n − 1 − i)4
ˆi 6
−
:
+
2(2i − 1)(2i + − 1)2 2(2i + )2 (2i + 2 − 1) (2i − 1)(2i + )3 (2i + 2 − 1)
The computation of a precise upper bound is so tedious that, rstly this work was made with the
help of Mathematica 3.0 [15], secondly the expression of this upper bound using the same technique
as in Theorem 6.3 is so long that it is unusable and we have preferred to give an equivalent as a
O(:).
32 − 32 + 11 2
qˆi
ˆi 6
+
;
128
128(2i − 1)(2i + − 1)(2i + )3 (2i + 2 − 1)(4i + )(4i + 3 − 4)
where qˆi is a polynomial in the variable i of degree 7 whose leading coecient is −211 ( − 2)(16 −
Pn −1 ˆ
i can be represented as
24 + 11 2 ). Then the sum i=1
2
32 − 32 + 11
(n − 1) + O(1n n ):
(55)
128
32 − 32 + 11 2
rˆi
+
;
128
128(2i − 1)(2i + − 1)(2i + )3 (2i + 2 − 1)(4i + )(4i + 3 − 4)
where rˆi is a polynomial in the variable i of degree 7 whose coecients also are expressions which
contain n . Considered as a polynomial in the two variables i and n ; rˆi is of degree 8. Since
n 6i62n − 1; the behavior of rˆi for large n is that of the set of terms ij n8−j ; j = 0; : : : ; 7; which is
ˆi 6
213 (−(16 − 24 + 11 2 )n i7 + (16 − 48 + 33 2 )2n i6 − 4(−8 + 11 2 )3n i5 + 22 2 4n i4 ):
Therefore the behavior of ˆi is
32−32 +11 2
128
+ O(1) and the one of
P2n −1
i=n
ˆi is
32 − 32 + 11 2
n + O(n ):
128
The denominator of the Rayleigh quotient is
1
n (124n + 453n + 702n − 45n − 22):
60
Therefore, gathering (55) – (57), the result concerning the upper bound holds.
(56)
(57)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
225
Theorem 6.7 (Gegenbauer case). The eigenvalues ˜ i; n ; i = 1; : : : ; n; of J˜ n which are the zeros of the
orthogonal polynomials A˜ n given by (51); are positive; real; distinct. Moreover
1 + 2
1
¡ ˜ 1; n 6˜ n = O 4 :
4n(n + 1)( + n)( + n + 1)
n
Proof. The nature of the ˜ i; n ’s is still obvious.
(i) From (51) it is very easy to verify by recurrence on n that
A˜ n (0) = (−1)n
(2n + 2 )2n
:
+ )2n (n + )n
24n n!(2n
On the other hand, we have
′
A˜ n (0) = (−1)n−1
(n + 1)(2n + 2 )2n−1 ( + n + 1)
;
(n − 1)!24n−2 (2n + )2n ( + n − 1)n−1
(58)
n¿1:
′
A˜ 1 (0) satis es Eq. (58) which can be assumed to be satis ed up to the rank n. Then we have
2n + 1 + 2
2n + 1
+
2(n + 1)(2n + + 2)2 (2n + + 1)2 (2n + 2 )
(2n + 2 )2n−1 (n + 1)( + n + 1)
(−1)n−1
(n − 1)!24n−2 (2n + )2n ( + n − 1)n−1
(2n + 2 )2n
+ (−1)n−1
+ (−1)n 4n
2 n!(2n + )2n (n + )n
(2n + 1)(2n + 2 − 1)(2n + 2 − 2)2n−3 (n + )
(2n + 2 )(2n + + 1)3 (2n + )(n − 2)!24n−1 (2n + − 2)2n−2 ( + n − 2)n−2
(2n + 2 + 2)2n+1
= (−1)n 4n+2
2
(n − 1)!(2n + + 2)2n+2 (n + − 1)n−1
(2n + 1)(n + 1)(2n + 2 )2n−1 ( + n + 1)
+ (−1)n
(n − 1)!24n+1 (2n + + 1)2n+1 ( + n)n (2n + )
(2n + 2 )2n
+ (−1)n 4n
2 n!(2n + )2n (n + )n
(2n + 1)(2n + 2 )2n−1
+ (−1)n−1
4n+1
(n − 2)!2
(2n + + 1)2n+1 ( + n − 2)n−2 (n + )(2n + )
(2n
+
2 + 2)2n+1
= (−1)n 4n+2
2
(n − 1)!(2n + + 2)2n+2 ( + n − 1)n−1
(2n + 2 + 2)2n+1
+ (−1)n 4n+1
2
n!(2n + + 1)2n+1 ( + n)n
(2n + 2 + 2)2n+1 (n + 2)( + n + 2)
= (−1)n
:
24n+2 n!(2n + + 2)2n+2 ( + n)n
′
1
A˜ n+1 (0) = −
4
′
The ratio −A˜ n (0)= A˜ n (0) gives the lower bound.
226
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
(ii) The numerator of the Rayleigh quotient J˜ of J˜ n is
n−1
X
1
2i
i=1
=
s
e
yi + 2i+1
k2i−1
2i
k2i
n−1
X
1
2i
i=1
+
s
s
k2i
yi+1
k2i+1
!2
+
1 k2n 2
y
(2n)2 k2n−1 n
2i + 1
i(2i − 1 + 2 )
1
yi −
2(2i + )2
4 (2i + + 1)2
s
2(2i + )(2i + + 1)
yi+1
(2i + 1)(i + )
!2
1 2n − 1 + 2 2
y:
8n (2n + )2 n
We do the same choice (see (54)) for the vector y as in the previous theorem and the numerator
becomes
s
n −1
X 1
4
i=1
2i + 2 − 1 2
i −
2i(2i + )2
s
2n −1
+
X 1
i=n
4
=
i=1
4
2i + 1
(i + 1)2
2(2i + + 1)2 (i + )
2i + 2 − 1
(2n − i)2 −
2i(2i + )2
s
!2
2i + 1
(2n − 1 − i)2
2(2i + + 1)2 (i + )
!2
2n −1
n −1
X ˜i
s
+
X ˜ i
i=n
4
;
(2i + 2 − 1)i3
(2i + 1)(i + 1)4
˜i =
+
2(2i + )2
2(2i + + 1)2 (i + )
√
2
i(i + 1) i(i + )(2i + 1)(2i + + 1)(2i + − 1)(2i + 2 − 1)
−
(i + )(2i + + 1)3
3
i(i + 1)2 &i
(2i + 1)(i + 1)4
(2i + 2 − 1)i
−
;
+
6
2(2i + )2
2(2i + + 1)2 (i + ) (2i + + 1)3 (i + )
where
2
www.elsevier.nl/locate/cam
On the positivity of some bilinear functionals in Sobolev spaces
Andre Draux ∗ , Charaf Elhami
LMI - UPRES-A CNRS 6085, INSA de Rouen, Departement de Genie Mathematique, Place Emile Blondel, BP 08,
F-76131 Mont-Saint-Aignan Cedex, France
Received 13 May 1998; received in revised form 30 January 1999
Abstract
The positivity of a bilinear functional a
a(f; g) =
N
X
m c(m) (f(m) ; g(m) )
m=0
is studied as a function of coecients m . The concerned cases are those of Laguerre, Gegenbauer and Jacobi for
c(m) = c(0) ; m = 1; : : : ; N . The domain {m }Nm=1 where a is positive de nite, is given. As a consequence, when N = 1, the
c 1999 Elsevier Science B.V. All rights reserved.
corresponding Markov–Bernstein inequalities are given.
MSC: 33C45; 42C05; 26D05; 26C10
Keywords: Formal orthogonal polynomial; Laguerre; Gegenbauer; Jacobi polynomial; Laguerre–Sobolev; Gegenbauer–
Sobolev; Jacobi–Sobolev polynomial; De nite inner product; Markov–Bernstein inequality; Zeros of polynomial
This paper constitutes the second part of a study devoted to the positivity of bilinear functionals
using positive de nite classical inner products, that is to say those of Hermite, Laguerre and Jacobi
measures. The rst part has solved this problem of positivity for the Hermite measure and for
some other connected problems [5]. That study was the simplest one, since the domain D where
the bilinear functionals are positive de nite, is given by means of explicit equations which de ne
its boundary. This domain D is polyhedric. It remains to solve the more complicated problems of
Laguerre and Jacobi measures. We study them here. In these cases the domain D can not be given
by means of explicit equations of its boundary when N ¿2, because this one is de ned as a limit
of a nappe of an algebraic hypersurface. The question was: does it exist a part of the domain D
∗
Corresponding author.
E-mail addresses: [email protected] (A. Draux), [email protected] (C. Elhami)
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ - see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 6 3 - 1
204
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
where some coecients m are negative; the answer is the same as in the Hermite case: yes, when
N ¿2. When N = 1; 1 has to be strictly positive in D. But when N = 1, the successive values 1; n ,
obtained for the positivity with polynomials of degree n = 1; 2; : : : ; give the best coecients of the
so-called Markov–Bernstein inequalities (see [1,9]). Except in the Hermite case and Laguerre case
for = 0 (see [9]) these values are not known explicitly, but we have given lower and upper bounds
in the Laguerre and Gegenbauer cases, which gives the behavior of 1; n as a function of n. The help
of Mathematica 3.0 [15] was very useful in that case for some tedious algebraic manipulations.
For replacing this study within the framework of the Sobolev orthogonality, the bibliography,
presented by Marcellan and Ronveaux [8], is essential.
1. Introduction
In order to establish the general framework of our paper, we begin to study the general case where
a bilinear functional a is obtained from two positive de nite inner products. To nd
the domain Dn = { ∈ R | a(p; p) ¿ 0 ∀p ∈ Pn − {0}} is equivalent to solve a generalized eigenvalue problem. Moreover the smallest eigenvalue 1; n of this problem which gives the boundary of
the domain Dn , also gives the best coecient in Markov–Bernstein inequality. Our presentation is
more or less di erent from that given in [9].
P (resp. Pi ) will denote the vector space of polynomials (in one variable) with real coecients
(resp. of degree at most i).
Let c and c(1) be two positive de nite inner products and {Pi }i¿0 the sequence of monic orthogonal
polynomials with respect to c.
Let a be the bilinear functional de ned by
∀p; q ∈ P;
a(p; q) = c(p; q) + c(1) (p′ ; q′ );
′
where is a real parameter and p
(resp. q′ ) is the derivative
of p (resp. q).
Pn
P
p ∈ Pn can be written as p = i=0 yi Pi . Thus p′ = ni=1 yi Pi′ and we get
(1)
a(p; p) = yT Kn; 0 y + y˜ T Kn;−1
y;
˜
(1)
where y (resp. y)
˜ is the vector of Rn+1 (resp. Rn ) of components yi ; i = 0; : : : ; n, (resp. i =
1; : : : ; n); Kn; 0 is the (n + 1) × (n + 1) matrix whose entries Kn; 0 (i; j) are c(Pi ; Pj ) for i; j = 0; : : : ; n
(1)
(1)
(therefore this matrix is diagonal) and Kn;−1
is the n×n matrix whose entries Kn;−1
(i; j) are c(1) (Pi′ ; Pj′ )
for i; j = 1; : : : ; n.
Let Kn;−1 be the n × n matrix deduced from Kn; 0 in cancelling the rst row and the rst column.
Then from (1) we have the following obvious result.
Theorem 1.1. For a xed parameter ∈ R; then; ∀n¿1 and ∀p ∈ Pn − {0}; a(p; p) ¿ 0 if and
(1)
only if Kn;−1 + Kn;−1
is positive de nite.
(1)
Moreover, since Kn;−1
is a positive de nite symmetric matrix, it can be decomposed by means of
Cholesky algorithm as Gn GnT where Gn is a lower triangular matrix. Therefore we have
(1)
Kn;−1 + Kn;−1
= Gn (Gn−1 Kn;−1 (GnT )−1 + I )GnT :
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
205
(1)
Hence Kn;−1 + Kn;−1
is positive de nite if and only if Gn−1 Kn;−1 (GnT )−1 + I is positive de nite.
Let i; n ; i = 1; : : : ; n be the eigenvalues of Gn−1 Kn;−1 (GnT )−1 with 0 ¡ 1; n 62; n 6 · · · 6n; n . Then
we have another obvious results.
(1)
Theorem 1.2. Kn;−1 + Kn;−1
is positive de nite if and only if ¿ − 1; n .
Theorem 1.3. a is positive de nite ∀p ∈ P − {0} if and only if ¿ − limn→∞ 1; n .
Proof. Since Gn−1 Kn;−1 (GnT )−1 is a symmetric matrix, then from an obvious consequence of the
Courant–Fischer theorem (see [14]) we have
0 ¡ 1; n+1 61; n ;
n¿1:
Thus the positive sequence {1; n }n¿1 is decreasing and bounded. Then limn→∞ 1; n exists and
limn→∞ 1; n ¿0. Therefore the result holds.
From Theorem 1.2 we obtain the Markov–Bernstein inequality.
Corollary 1.4.
∀p ∈ Pn ;
c(1) (p′ ; p′ )6
1
c(p; p);
1; n
n¿1:
P
1=1; n is the best constant. An extremal polynomial is p = ni=1 yi Pi where y˜ is the vector deduced
from an eigenvector (GnT )y˜ of Gn−1 Kn;−1 (GnT )−1 corresponding to the eigenvalue 1; n .
In conclusion, in order to de ne the domain Dn where a is positive de nite for any p ∈ Pn − {0},
we have to give the smallest eigenvalue of Gn−1 Kn;−1 (GnT )−1 . Moreover the domain of positivity of
a depends on limn→∞ 1; n . To study the general case can be very dicult. In addition our aim is to
give the domain of positivity of a for any N; N ¿1, de ned by
∀p; q ∈ P;
a(p; q) =
N
X
m c(m) (p(m) ; q(m) );
m=0
where 0 = 1; N 6= 0 and c(m) is a positive de nite inner product, ∀m = 0; : : : ; N , and p(m) is the
derivative of order m of p.
In the cases where the inner products c(m) are linked and the derivatives p(m) are also linked
like for the classical inner products, it is possible to give some basic results about the domain of
positivity as well as the 1; n ’s.
In [5] we have considered the case where c(0) is a classical inner product (Hermite, Laguerre or
Jacobi) and c(m) is the classical inner product deduced from c(0) such that {Pn(m) }n¿m is a sequence
of polynomials orthogonal with respect to c(m) . In the sequel we will study the case where c(m) =
c; m = 0; : : : ; N , is the Laguerre, Jacobi or Gegenbauer inner product.
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
In addition to [5] we will give the Markov–Bernstein inequalities in that di erent cases. We have
kept the notation Cn(1) used in [5]. 1 In the Hermite case Cn(1) was already known (see [9]).
Corollary 1.5 (Markov–Bernstein inequalities). Let c(0) and c(1) be two classical inner products
such that the derivatives of the monic orthogonal polynomials with respect to c(0) are orthogonal
with respect to c(1) .
Let k · k(m) be the norm associated to the inner product c(m) respectively for m =
q0; 1. Then
∀p ∈ Pn − {0}; a(p; p) ¿ 0 if and only if 1 ¿ − 1=Cn(1) . Thus ∀p ∈ Pn ; kp′ k(1) 6 Cn(1) kpk(0)
where
Cn(1) =
2n
in the Hermite case;
in the Laguerre case;
n
n(n + + + 1)
in the Jacobi case:
2. Classical inner products
Let L2 (
; ) be the Hilbert space of square integrable real functions on the open set
⊂ R for
the positive Borel measure supported on
.
On this space we have the classical de nite inner product:
c(f; g) =
Z
f(x)g(x) d(x) ∀f; g ∈ L2 (
; )
R
and the norm c(f; f) = kfk2 =
f2 (x) d(x).
H N (
; ) will denote the Sobolev space
H N (
; ) = {f | f(m) ∈ L2 (
; ); m = 0; 1; : : : ; N };
where the derivatives f(m) of order m are taken in the distributional sense.
The de nite inner product on this Hilbert space is
s(f; g) =
N
X
c(f(m) ; g(m) );
m=0
∀f and g ∈ H N (
; ):
The corresponding norm, denoted by k · ks , is given by
kfk2s
=
N
X
m=0
kf(m) k2
∀f ∈ H N (
; ):
Now we consider the symmetric bilinear functional a
a : H N (
; ) × H N (
; ) → R
We want to indicate a mistake in [5]. n(m) = 2m in the Hermite case instead of 1. Consequently Formula given in the
same paper at the bottom of the page 173 has a version in the Hermite case, di erent from the Laguerre case, and that
1
formula has to be multiplied by 2
N (N +1)
2
.
(1)
Since Cn(m) = (n)m (m)
n ; Cn = 2n in the Hermite case.
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
207
de ned by
a(f; g) =
N
X
m c(f(m) ; g(m) );
m=0
∀f and g ∈ H N (
; );
(2)
where m ; m = 0; : : : ; N; are N + 1 xed real numbers with 0 = 1 and N 6= 0.
We look for the formal orthogonal polynomials with respect to a, that is to say, we look for the
polynomials Sn ; n¿0; such that
deg Sn = n;
a(Sn ; xi ) = 0
for i = 0; : : : ; n − 1:
(3)
These polynomials Sn will be called Sobolev formal orthogonal polynomials.
fn = (a(xj ; xi ))n−1 , we have the following obvious result:
Setting M
i; j=0
Theorem 2.1. The formal orthogonal polynomial Sn exists and is unique up to a normalization for
fn is regular.
the leading coecient is xed; if and only if the matrix M
De nition 2.2. The bilinear functional a is called positive de nite on H N (
; ) × H N (
; ) (resp.
on P × P) if, ∀f ∈ H N (
; ) − {0} (resp. P − {0}); a(f; f) ¿ 0.
Theorem 2.3. The bilinear functional a is positive de nite on P × P if and only if all the formal orthogonal polynomials Sn ; n ∈ N; exist with a positive leading coecient and a(Sn ; Sn ) ¿ 0;
∀n ∈ N.
P
Proof. (i) If a(Sn ; Sn ) ¿ 0; ∀n ∈ N, then ∀p ∈ P (deg p = k) we have p = ki=0 i Si and a(p; p) =
2
i=0 i a(Si ; Si ) ¿ 0.
(0) 2
f
(ii) For p = 1; a(1; 1) ¿ 0. Thus if S0 =
(0)
0 ¿ 0; a(S0 ; S0 ) = (
0 ) a(1; 1) ¿ 0. Moreover M1 is
f1 = a(S0 ; S0 ) ¿ 0. Therefore S1 exists.
regular and det M
fi ¿ 0; i = 1; : : : ; n. Then Sn exists. Let us write Sn as Sn = Pn
(n) x‘
Let us assume that det M
‘
‘=0
fn+1 is
with
n(n) ¿ 0. Let us denote by Xbn the vector of Rn of components
‘(n) ; ‘ = 0; : : : ; n − 1. M
multiplied on the right by the following (n + 1) × (n + 1) regular matrix
Pk
In bxn
0Tn
n(n)
;
where In is the n × n identity matrix and 0n the zero vector of Rn . Thanks to relations (3) we get
fn
M
ATn
0n
a(Sn ; xn )
!
;
where An is the vector of Rn of components a(xj ; xn ); j = 0; : : : ; n − 1. a(Sn ; xn ) = a(Sn ; Sn )=
n(n) ¿ 0.
fn+1 ¿ 0 and the desired conclusion follows.
Therefore det M
In the sequel we will study in which domain D of RN ; of components 1 ; : : : ; N must be located
in order to a be positive de nite on P × P for some particular Borel measures which correspond
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
Table 1
Name
Laguerre
Jacobi
]0; +∞[
] − 1; 1[
−x
kn
x e
n! (n + + 1)
Restrictions
¿−1
Gegenbauer
] − 1; 1[
(1 − x) (1 + x)
2n n! ( n )n (( n()n ))(2n −n)
with n = n + + 1
n = n + + 1
n = n + n − 1
¿ − 1; ¿ − 1
1
(1 − x2 ) − 2
21−2 −2n n! (n+2 )
(n+ +1) (n+ )
¿−
1
2
to the Laguerre, Jacobi and Gegenbauer orthogonal polynomials (respectively denoted by L n ; Pn( ; )
and Gn ). We recall that the Hermite case has been already studied in [5]. We summarize all the
useful informations concerning these three families of monic orthogonal polynomials in Table 1 (see
[2,12]), where kn is the square norm in L2 (
; ) of a monic orthogonal polynomial of degree n.
Moreover we have the following relations for the derivatives of these polynomials (see [2,12]):
d
L (x) = nL +1
n−1 (x);
dx n
(4)
d ( ; )
( +1; +1)
(x);
P (x) = nPn−1
dx n
(5)
d
+1
(x):
G (x) = nGn−1
dx n
(6)
3. Matrix interpretation of some particular relations
Some particular staircase recurrence relations will be of a great interest in order to give a general
expression of a(p; q); ∀p; q ∈ P. These relations are (see [2,12]):
+1
L n (x) = L +1
n (x) + nLn−1 (x);
( +1; )
(x)
Pn( ; ) (x) = Pn( +1; ) (x) + en( = +1; ) Pn−1
2n(n + )
( +1; )
= Pn( +1; ) (x) −
Pn−1
(x);
(2n + + + 1)2
( ; +1)
(x)
Pn( ; ) (x) = Pn( ; +1) (x) + en( ; = +1) Pn−1
2n(n + )
( ; +1)
= Pn( ; +1) (x) +
Pn−1
(x);
(2n + + + 1)2
+1
Gn (x) = Gn +1 (x) + en Gn−2
(x)
n(n − 1) +1
= Gn +1 (x) −
G (x)
4(n + )2 n−2
(7)
(8)
(9)
(10)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
209
and are satis ed ∀n ∈ N. (a)j ; j ∈ N, denotes the shifted factorial: (a)j = a(a − 1) · · · (a − j + 1).
These relations are simple consequences of the orthogonality when the weight function is modi ed
respectively by x; 1 − x; 1 + x; 1 − x2 .
By convention we complete the set of the previous relations (7) – (10) by the following ones:
L −k = L +1
−k = 0;
( ; )
( +1; )
( ; +1)
P−k
= P−k
= P−k
= 0;
+1
G−k
= G−k
= 0;
∀k ¿ 0;
(11)
Moreover the set of relations (4) – (6) is completed by:
d
L−k+1 = L +1
;
−k
dx
d ( ; )
( +1; +1)
(12)
P−k+1 = P−k
; ∀k ¿ 0;
dx
d
+1
;
G−k+1 = G−k
dx
For the sake of simplicity let us denote by Pi ; ∀i ∈ N, the monic orthogonal polynomials with
b −k; n ; ∀k and n ∈ N, the vector of components
respect to any previous inner product c, and by P
b −k; n the vector
Pj ; j = −k; : : : ; n, with the convention that Pj = 0, if j ¡ 0. We will denote by ddx P
whose components are the derivatives of Pj ; j =−k; : : : ; n. When the distinction between the di erent
b ; P
b ( ; ) and
families of orthogonal polynomials will be necessary, then we will use the notation L
−k; n
−k; n
b
b
G−k; n instead of P−k; n .
Now, let us give some matrix expressions deduced from relations (7) – (12).
Property 3.1.
b +1 ;
b = E−k; n L
L
−k; n
−k; n
(13)
( ; ) b ( +1; +1)
;
= E−k;
n P−k; n
(15)
b ( ; ) = E ( = +1; ) P
b ( +1; ) = E ( ; = +1) P
b ( ; +1)
P
−k; n
−k; n
−k; n
−k; n
−k; n
b +1
b = E G
G
−k; n −k; n ;
−k; n
( = +1; )
E−k;
;
n
(14)
(16)
( ; = +1)
E−k;
;
n
( ; )
E−k;
n
where E−k; n ;
and E−k;
n are (n + 1 + k) × (n + 1 + k) nonsingular matrices
whose the entries (i; j) are in row i and column j; i; j = 1; : : : ; n + k + 1:
E−k; n (i; j) =
1
i − k − 1;
0
( = +1; )
E−k;
(i; j)
n
∀i = j;
=
everywhere else;
1
∀i = j;
e( = +1; )
i−k−1
∀i = j + 1; k + 26i6n + k + 1;
0
∀i = j + 1; k + 26i6n + k + 1;
everywhere else;
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
( ; = +1)
E−k;
(i; j)
n
( ; )
E−k;
n (i; j) =
=
1
∀i = j;
e( ; = +1)
i−k−1
∀i = j + 1; k + 26i6n + k + 1;
0 everywhere else;
1 ∀i = j;
2‘( − )
∀i = j + 1;
(2‘ + + )(2‘ + + + 2)
with ‘ = i − k − 1 and 16‘6n;
4‘(‘ − 1)(‘ + )(‘ + )
−
∀i = j + 2;
(2‘
+ + )(2‘ + + + 1)3
with ‘ = i − k − 1 and 26‘6n;
(17)
0 everywhere else;
1
∀i = j;
‘(‘ − 1)
∀i = j + 2; with ‘ = i − k − 1 and 26‘6n;
e‘ = −
E−k;
n (i; j) =
4(‘ + )2
0 everywhere else;
b −k; n = JT
P
−k+1; n+1
d b
P−k+1; n+1 ;
dx
(18)
T
where J−k+1;
n+1 is a (n + 1 + k) × (n + 1 + k) nonsingular matrix obtained by multiplication of
( ; )
b
E−k; n , ( resp. E−k;
n ; E−k; n ) on the right by the diagonal matrix D−k+1; n+1 whose the elements are
b −k+1; n+1 (i; i) =
D
1;
1;
‘
i = 1; : : : ; k;
‘ = 1; : : : ; n + 1 with i = k + ‘:
Proof. Relations (13), (14) and (16) are the matrix version of relations (7) – (10) for the vectors
b −k; n . Eq. (15) is obtained from
P
b ( ; ) = E ( = +1; ) E ( +1; = +1) P
b ( +1; +1) = E ( ; = +1) E ( = +1; +1) P
b ( +1; +1) :
P
−k; n
−k; n
−k; n
−k; n
−k; n
−k; n
−k; n
It is easy to verify that
( = +1; ) ( +1; = +1)
( ; = +1) ( = +1; +1)
E−k;
E−k; n
= E−k;
E−k; n
:
n
n
From Eqs. (4) – (6) it is obvious that a vector Pˆ−k; n can be written as the product of Dˆ −k+1; n+1
and the derivative of a vector Pˆ−k+1; n+1 (but with modi ed parameters ; or and ). Substituting the vector of the right part of any relation (13), (15) or (16) by this last relation, Eq. (18)
is obtained.
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
211
4. The quasi-orthogonal polynomial Ri associated with Pi
Let {Ri }i¿0 be a sequence of monic polynomials which satis es the following conditions:
deg Ri = i; ∀i ∈ N;
di
Ri = i!P0 = i; N P0 ;
d xi
for 06i6N − 1;
dN
Ri = (i)N Pi−N = i; N Pi−N ;
d xN
(19)
for i¿N:
(20)
Let Rˆ n , be the vector of components Ri ; i = 0; : : : ; n, and Dn; N be the (n + 1) × (n + 1) diagonal
matrix (i; N )ni=0 .
The matrix interpretation of relations (19) and (20) is
dN ˆ
Rn = Dn; N Pˆ−N; n−N :
d xN
(21)
Remark that Dn; N contains i; N = i! for i ¡ N and that
Using Eq. (18), Eq. (21) becomes:
dN
d xN
Ri = 0 and P−i = 0 for i ¡ N .
dN ˆ
d ˆ
T
P−N +1; n−N +1
Rn = Dn; N J−N
+1; n−N +1
N
dx
dx
dj ˆ
(22)
P−N +j; n−N +j ∀j = 2; : : : ; N:
d xj
In order to
simplify the writing of products with di erent matrices J, we will Q
use the following
QN −1
N −1
notation: j=i
J−j; n−j = J−i; n−i · · · J−N +1; n−N +1 in this order. If i ¿ N − 1, then j=i
J−j; n−j will
be taken equal to 1. Thus Eq. (22) is
T
T
= Dn; N J−N
+1; n−N +1 · · · J−N +j; n−N +j
T
NY
−1
dN ˆ
dj ˆ
R
P−N +j; n−N +j
=
D
J
n
n; N
−i; n−i
d xN
d xj
i=N −j
∀j = 2; : : : ; N:
(23)
From Eq. (23) written for j = N , we set
Rˆ n = Dn; N
NY
−1
j=0
T
J−j; n−j Pˆ 0; n :
(24)
Therefore, from Eqs. (24) and (18), we also have
T
NY
−1
d N −j ˆ
=
D
J−i; n−i Pˆ−N +j; n−N +j :
R
n
n;
N
d xN −j
i=N −j
(25)
De nition 4.1. An n × n matrix A = (aij ) is a band matrix if aij = 0; | i − j | ¿ h with h xed (h ∈ N)
and h + 1 ¡ n: h is the half width of band.
An n × n matrix A is a lower (resp. upper) triangular band matrix if aij = 0 for i ¡ j (resp. i ¿ j)
and i − j ¿ h (resp. i − j ¡ − h) with h xed (h ∈ N) and h + 1 ¡ n: h is the width of band.
212
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
Q
N −1
From the de nition of the matrices JT , it is clear that ( j=0
J−j; n−j )T is a lower triangular band
matrix of width N in the Laguerre case, 2N in the Jacobi and Gegenbauer cases. Moreover all the
entries of the last lower diagonal which is the border of the band, are non zero from the second
column (the rst column contains a unique non-zero entry in the rst row). Therefore we have the
following obvious property about the quasi-orthogonality of the Ri ’s (see [2]).
Property 4.2. In the Laguerre case (resp. in the Jacobi and Gegenbauer cases) the polynomials
Ri are quasi-orthogonal of order N (resp. 2N ) with respect to the Laguerre (resp. Jacobi or
Gegenbauer) inner product. They are strictly quasi-orthogonal when i ¿ N (resp. i ¿ 2N ) in the
Laguerre case (resp. in the Jacobi and Gegenbauer cases).
5. Matrix interpretation of a(p; q)
Let p and q be two polynomials of degree n. Since {Ri }i¿0 is a basis of P, we can write p and
q as
p = (Rˆ n )T y and q = (Rˆ n )T z;
where y and z are vectors of Rn+1 which contain the coordinates yi (resp. zi ) of p (resp. q) in the
basis {Ri }ni=0 . Then
d N −j ˆ
d N −j
p
=
Rn
d xN −j
d xN −j
T
y
= (Pˆ−N +j; n−N +j )T
NY
−1
J−i; n−i Dn; N y:
i=N −j
Let us de ne the (t + 1) × (t + 1) matrix Kt; j by Kt; j = (c(Pi ; P‘ ))i;t−j
‘=−j . Thus Kt; j is a diagonal matrix
t−j
(ki )i=−j where ki = c(Pi ; Pi ); i¿0; and ki = 0 if i ¡ 0.
Therefore:
c
Finally
T
NY
−1
d
d
T
p;
q
=
z
D
J
J−i; n−i Dn; N y:
K
n; N
−i; n−i
n; N −j
d xN −j d xN −j
i=N −j
i=N −j
N −j
a(p; q) =
N −j
N
X
NY
−1
i c(p(i) ; q(i) )
i=0
=
N
X
i z T Dn; N
NY
−1
j=i
i=0
J−j; n−j
!T
Kn; i
a(p; q) =
i Fi; n (y; z) = Fb(n); n (y; z)
i=0
where b(n) = min(n; N ).
b(n)
X
i=0
NY
−1
b(n)−1
i
Y
r=i
!
J−j; n−j Dn; N y:
j=i
From Eq. (27) a(p; q) can also be written as
b(n)
X
(26)
!
Fr; n (y; z)
;
Fr+1; n (y; z)
(27)
(28)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
213
Setting Y = Dn; N y and Z = Dn; N z, we have
Q
Q
b(n)−1
b(n)−1
Z T ( j=r
J−j; n−j )T Kn; r ( j=r
J−j; n−j )Y
Fr; n (y; z)
=
:
Q
Q
b(n)−1
b(n)−1
T
T
Fr+1; n (y; z) Z ( j=r+1 J−j; n−j ) Kn; r+1 ( j=r+1 J−j; n−j )Y
Q
b(n)−1
Let u(r; n) (resp. v(r; n) ) be the vector ( j=r
J−j; n−j )Y (resp. (
Its components are numbered from 0 to n.
1
Qb(n)−1
j=r
(29)
J−j; n−j )Z); ∀r = 0; : : : ; b(n).
Let (Kn; r+1 ) 2 be the diagonal matrix whose entries are the square roots of the elements of Kn; r+1 .
1
1
The vector (Kn; r+1 ) 2 u(r+1; n) (resp. (Kn; r+1 ) 2 v(r+1; n) ) has its rst r + 1 components equal to zero. Let
us denote by w(r+1; n) (resp. s(r+1; n) ) the vector of Rn−r which contains the last n − r components of
this vector. From that and the expression of J−r; n−r we get
k0 ur(r+1; n) vr(r+1; n) + (s(r+1; n) )T Jn−r w(r+1; n)
Fr; n (y; z)
=
;
Fr+1; n (y; z)
(s(r+1; n) )T w(r+1; n)
(30)
where Jn−r is a (n − r) × (n − r) matrix. It contains the rst n − r rows and columns of the matrix
Jn which appears in Eq. (30) for r = 0.
Finally we obtain the following expression of a(p; q)
a(p; q) = Fb(n); n (y; z)
b(n)
X
b(n)−1
Y k0 u(r+1; n) v(r+1; n) + (s(r+1; n) )T Jn−r w(r+1; n)
r
r
i
(s(r+1; n) )T w(r+1; n)
r=i
i=0
!
:
(31)
Remark that Jn is independent of N . Indeed, since the n last rows and columns of Kn;1 correspond
to Kn−1;0 ; we have
J0;T n Kn; 0 J0; n
=
k0
0
0
(Kn−1;0 ) 2 Jn (Kn−1;0 ) 2
1
1
!
;
(32)
which can also be written as
1
1
J1;T n Kn−1;−1 J1; n = (Kn−1;0 ) 2 Jn (Kn−1;0 ) 2 :
(33)
Remark that Jn is similar to Gn−1 Kn;−1 (GnT )−1 obtained in Section 1.
Therefore the matrices Jn corresponding to the di erent cases (Laguerre, Jacobi, Gegenbauer) can
be obtained by using Eq. (33).
In the Laguerre case (see also [9]), Jn is an n × n symmetric tridiagonal matrix:
k1
= 1 + for i = j = 1
k0
ki + i2 ki−1
=2+
for i = j = 2; : : : ; n;
2k
i
i
i−1
s
r
Jn (i; j) =
1
ki−1
for i = j + 1;
= 1+
i − 1 ki−2
i−1
for i = j − 1 the matrix is completed by symmetry;
0
everywhere else:
(34)
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A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
In the Jacobi case, Jn is an n × n symmetric ve diagonal matrix:
k1
for i = j = 1;
k0
4( − )2
k2
+
for i = j = 2;
4k1 (2 + + )2 (4 + + )2
4( − )2
ki−1 ki−2
+
( +1; +1) 2
(2i − 2 + + )2 (2i + + )2
(i − 2)2 (ki−3
)
+
ki
2
i ki−1
for i = j ¿ 2;
2( − )
Jn (i; j) = (2 + + )(4 + + )
s
k1
k0
for (i; j) = (1; 2) and (2; 1);
s
(2 − )
ki−1
ki−2
−
( +1;
+1)
(2i − 2 + + ) ki−2 ki−3
(i − 2)(2i − 4 + + )
1
; for i = j + 1 ¿ 2;
−
(i − 1)(2i + + )
s
k
ki−1
i−2
−
for i = j + 2;
( +1; +1)
2
(i − 2) ki−3
ki−3
for i = j − 2 and i = j − 1 ¿ 1 the matrix is completed by symmetry;
0
(35)
everywhere else:
ki is ki( ; ) in Eq. (35).
In the Gegenbauer case (see also [9]), Jn is an n × n symmetric ve diagonal matrix:
Jn (i; j) =
ki
for i = j = 1; 2;
2k
i
i−1
(ei−1
)2 ki−2
ki
+
for i = j ¿ 2;
i2 ki−1
(i − 2)2 ki−1
ei−1
ki−2
p
for i = j + 2;
2
(i − 2) ki−3 ki−1
for i = j − 2 the matrix is completed by symmetry;
0
(36)
everywhere else:
From Eq. (32) and the de nition of Kn−1;0 and Kn; 0 , we have the following obvious property
Property 5.1. Jn is a positive de nite symmetric matrix.
Property 5.2. If i; n ; i = 1; : : : ; n; are the eigenvalues of Jn ; corresponding to the three previous
cases; with 0 ¡ i; n 6 · · · 6n; n ; then limn→∞ n; n ¡ + ∞.
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
215
Proof. As an obvious consequence of the Courant–Fischer therorem (see [14]) n; n 6n+1; n+1 ; ∀n .
Thus the positive sequence {n; n }n¿1 is increasing.
We use the localization of the eigenvalues given by the Gerschgorin disks. From Eqs. (34) – (36)
it is clear that all the centers of these disks are at a nite distance and all the radii are bounded.
Thus n; n is bounded ∀n and the result holds.
Remark 5.3. Relation (28) can also be written by means of Horner algorithm
d0 = 1;
Fi−1; n (y; z)
di−1 ;
Fi; n (y; z)
a(p; q) = Fb(n); n (y; z)db(n) :
d i = i +
i = 1; : : : ; b(n);
(37)
6. Domain of positivity of a
Let be an element of RN of components m ; m = 1; : : : ; N (N 6= 0).
We want to nd the domain D ⊂ RN such that
D = { ∈ RN | a(p; p) ¿ 0 ∀p ∈ P − {0}};
where a is considered as a function depending on .
∀n¿1; Dn will denote the domain of Rb(n) such that
Dn = { ∈ Rb(n) | a(p; p) ¿ 0 ∀p ∈ Pn − {0}}:
(38)
For any xed p ∈ Pn − {0}; a(p; p) = 0 is a hyperplane of Rb(n) such that the coecients of the
i ’s are positive (see (31)). Thus this hyperplane cuts the axes of Rb(n) in the negative part.
Let us begin to prove a property of N .
Theorem 6.1. Let i; n ; i = 1; : : : ; n; be the eigenvalues of Jn ; (0 ¡ 1; n 6 · · · 6n; n ). n; n is assumed
to be bounded ∀n.
(i) If limn→∞ 1; n = 0 and if = (1 ; : : : ; N ) ∈ D; then N ¿ 0.
(ii) If inf n 1; n ¿ ¿ 0 and if i ¿ 0; i = 1; : : : ; N − 1; then there exists ˆ ¡ 0 depending on the
i ′ s; i = 1; : : : ; N − 1 such that ∀N ¿ ;
ˆ = (1 ; : : : ; N ) ∈ D.
Proof. (i) In FN −1; n (y; y)=FN; n (y; y) we take w(N; n) as the eigenvector corresponding to the smallest
eigenvalue 1; n−N +1 of Jn−N +1 . The vector u(N; n) of components ui(N; n) ; i = 0; : : : ; n, is chosen such
n)
that uN(N;−1
= 0; ui(N; n) ; i = N; : : : ; n, being de ned from the successive components of w(N; n) thanks to
the following relation:
1
(K(n; N ) ) 2 u(N; n) =
0N
w(N; n)
;
where 0N is the zero vector of RN . Then
FN −1; n (y; y) (w(N; n) )T Jn−N +1 w(N; n)
=
= 1; n−N +1 ;
FN; n (y; y)
kw(N; n) k22
where k · k2 is the Euclidean norm of a vector.
(39)
216
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
Thus limn→∞ FN −1; n (y; y)=FN; n (y; y) = 0.
Now we take the vector Y of Rn+1 such that
Yi =
(
ui(N; n)
0
Q
for i = N; : : : ; n;
for i = 0; : : : ; N − 1:
N −1
Since u(r; n) = j=r
J−j; n−j Y , then ui(r; n) = 0 for i = 0; : : : ; r − 1.
Therefore we get
a(p; p) = F(N; n) (y; y)
N
X
i
NY
−1
r=i
i=0
(w(r+1; n) )T Jn−r w(r+1; n)
kw(r+1; n) k22
!
:
(40)
Each factor (w(r+1; n) )T Jn−r w(r+1; n) =kw(r+1; n) k22 which is the Rayleigh quotient Jn−r (w(r+1; n) ) (see [14]),
is bounded by the largest eignevalue n−r; n−r of Jn−r . Since n−r; n−r is also bounded, then Fr−1; n (y; y)=
Fr; n (y; y) which is the coecient of dr−1 in (37), is also bounded, ∀r = 1; : : : ; N − 1. Since the
coecient FN −1; n (y; y)=FN; n (y; y) of dN −1 tends to zero when n tends to in nity, dN remains positive
if and only if N ¿ 0, and the result holds.
(ii) The same technique with the same vectors is used in this second part. Thus we have
FN −1; n (y; y)
¿ inf 1; n ¿ ¿ 0:
n
FN; n (y; y)
Moreover, Jn−r (w(r+1; n) ) is such that
0 ¿ 61; n−r 6Jn−r (w(r+1; n) )6n−r; n−r 6 sup n; n ¡ + ∞:
n
Thus using Remark 5.3
N
−1
X
i N −1−i 6dN −1 6
i=0
N
−1
X
i sup n; n
i=0
n
N −1−i
:
Hence if i ¿ 0; i =1; : : : ; N −1; dN which is such that dN ¿N −dN −1 is positive for any N ¿ ˆ =
−dN −1 . Thus the result holds.
6.1. Case N = 1
In the particular case where N = 1, the domains Dn and D are characterized by means of 1; n and
limn→∞ 1; n . We give lower and upper bounds of 1; n in the Laguerre case, a lower bound of 1; n
in the Gegenbauer case, and the behavior of an upper bound in the Gegenbauer and Jacobi cases.
Theorem 6.2.
Dn = {1 ∈ R; 1 ¿ − 1; n };
where 1; n is the smallest eigenvalue of Jn .
Proof. From (31) we have
a(p; p) = kw(1; n) k22
(w(1; n) )T Jn w(1; n)
k0 y02
+
1 +
kw(1; n) k22
kw(1; n) k22
!
:
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
217
The case where w(1; n) = 0 is without interest, since p = y0 R0 in this case, and of course a(p; p) =
c(p; p) = k0 y02 ¿ 0. Therefore we have to study
1 +
k0 y02
(w(1; n) )T Jn w(1; n)
+
¿ 0 ∀p ∈ Pn − {0};
kw(1; n) k22
kw1; n) k22
that is to say, to nd 1 depending on n such that
1 ¿ −
min
w(1; n) ∈Rn −{0}
(w(1; n) )T Jn w(1; n)
k0 y02
+
2
kw(1; n) k2
kw1; n) k22
!
(41)
:
(w(1; n) )T Jn w(1; n) =kw(1; n) k22 is the Rayleigh quotient Jn (w(1; n) ). It is independent of y0 . Therefore the
minimum of the right part of Eq. (41) is obtained for y0 = 0 and w(1; n) being the eigenvector
corresponding to the smallest eigenvalue 1; n of Jn ; this minimum is equal to 1; n .
Now we give some properties about the smallest eigenvalue 1; n . A upper bound will be obtained
from the value of the Rayleigh quotient which corresponds to a particular vector. A lower bound
will be given in the case where the eigenvalues are the zeros of orthogonal polynomials. It is given
by the rst step of the Newton method with 0 as starting point.
In the Laguerre case the eigenvalues of Jn are the one’s of a tridiagonal matrix which corresponds to a Jacobi matrix of sequence of orthogonal polynomials. These polynomials correspond to
co-recursive polynomials of the generalized Pollaczek polynomials (this fact was presented in the
doctoral dissertation of Perez [10] and in the paper of Marcellan et al. [7]). See also [9].
Theorem 6.3 (Laguerre case). The eigenvalues i; n ; i = 1; : : : ; n; are the zeros of the orthogonal
polynomials An (x) de ned from the following three-term recurrence relation:
An (x) = x − 2 −
An−1 (x) − 1 +
An−2 (x);
n
n−1
(42)
n¿2;
with A0 (x) = 1 and A1 (x) = x − − 1.
These zeros are real; positive; distinct and
√
3( + 2) − ( + 2)( + 10)
1;1 = + 1; 1; 2 =
;
4
2( + 1)
3
;
¡ 1; n 6(n + n )
n(n + 1)
n (22n + 1)
∀n¿3;
(43)
where n = [ n+1
] ([:] denotes the integer part).
2
!
2
2n − 2 +
2 2 − 2
( − 1) 1 − +
+
1−
ln
;
+
n
2
2 2+
2
2
2+
n =
!
2
2
2n +
−
1)
1
−
+
+
1
−
ln
; if ¿2;
(
n
2
2
2
2+
if 62;
218
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
!
2
2(2n − 2 + )
4n +
2
1
−
+
ln
+
4
(
−
1)
+
(
−
2))
ln
−
;
(8
n
n
n
2
4n − 2 +
4
2n +
if 60;
!
2
(2n − 1)(2n + )
2
n 1 − +
+ 4 n ln
− (8n ( − 1) + ( − 2))
2
(
−
1)(4
+
)
4
n
n
n =
4n − 2 +
; if 06 6 ˆn ;
ln
2
n −2+
!
(2n − 1)(2n + )
2
2
n 1 − +
+ 4 n ln
− (8n ( − 1)
2
(
−
1)(4
+
)
4
n
n
4n +
+ ( − 2)) ln
; if ¿ ˆn ;
2n +
p
where ˆn = −4n + 1 + 162n + 1; (1 ¡ ˆn ¡ 2); is the largest zero of 8n ( − 1) + ( − 2)
considered as a polynomial in one variable .
Proof. The rst zeros 1;1 and 1; 2 are obtained in an obvious way from (42). The rst part is
obtained from a classical property for tridiagonal matrices:
An (x) = det(xI − Jn )
= x−2−
det(xI − Jn−1 ) − 1 +
det(xI − Jn−2 ):
n
n−1
Since the coecient 1 + =(n − 1) of An−2 is strictly positive, we can deduce that the eigenvalues
of the positive de nite symmetric matrix Jn which are real and positive, are, besides, distinct.
The zeros of An being distinct real, those of A′n ; A′′n , and so on, also are real distinct and are located
between 1; n and n; n . Thus A′′n has a constant sign for x ¡ 1; n . Therefore ’n = −An (0)=A′n (0) ¡ 1; n .
Using Eq. (42) it is easy to verify by recurrence on n that
An (0) = (−1)n
( + n)n
:
n!
(44)
Since
A′n (0) = − 2 +
A′n−1 (0) + An−1 (0) − 1 +
A′n−2 (0);
n
n−1
we get by recurrence on n:
A′n (0) = (−1)n−1
(n + 1)( + n)n−1
:
2(n − 1)!
From Eqs. (44) and (45), the lower bound of Eq. (43) is satis ed.
(45)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
219
From Eq. (34) we get for any vector y ∈ Rn
T
y Jn y =
n−1
X
1
i
i=1
=
s
r
n−1
X
i=1
ki
ki−1
!2
+
1 kn 2
y
n2 kn−1 n
!2
+
n+ 2
yn :
n
yi + yi+1
i+
yi + yi+1
i
Let us choose the vector y such that
(−1)i i
n+1
if i6n =
;
2
yi =
(−1)i (2n − i) if n 6i62n :
(46)
Remark that yn+1 = 0 if n is odd.
For the vector y given by (46) we have
n
X
yi2
=
i=1
n
X
n −1
2
i +
i=1
X
i2 =
i=1
n (22n + 1)
:
3
The corresponding numerator of the Rayleigh quotient Jn can be written as
X p
2
n−1
i=1
(i + )i − i − 1 +
n
X
j=1
s
2n − j +
j−j+1
2n − j
!2
:
Let us denote by i (resp. q
j ) an element in the rst (resp. second) sum.
√
2
2
Since (i + )i = (i + 2 ) 1 − (2i+ )
2 ¿(i + 2 )(1 − (2i+ )2 ), we get
p
i = i(i + ) + (i + 1)2 − 2(i + 1) i(i + )61 − +
2 2 (2 − )
+
:
2
2(2i + )
In the same way we have
j =
p
1
((2n − j + )j 2 + (2n − j)( j − 1)2 − 2j( j − 1) (2n − j)(2n − j + ))
2n − j
1
6
2n − j
2
+j( j − 1)
4n − 2j +
= 1+
(2n − j + )j + (2n − j)( j − 1) − 2j( j − 1) 2n − j +
2
2
!
2
( − 2)
42n
82n
(8n ( − 1) + ( − 2))
+
−
−
:
2
2n − j 4n − 2j +
2(4n − 2j + )
220
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
Now we will use the following inequalities. Let a be a xed real. If a + x ¿ 0 for x ∈ [‘ − 1; k +
1]; ‘; k ∈ N, then
ln
a+k +1
=
a+‘
Z
k+1
‘
k
X
1
dx
6
6
a + x i=‘ a + i
Z
‘−1
Pn −1
Therefore, in using Eq. (47), we obtain
Finally
3(n + n )
:
1; n = min
Jn (y)6
n
y∈R −{0}
n (22n + 1)
j=1
k
a+k
dx
= ln
:
a+x
a+‘−1
i 6n and
Pn
j=1
(47)
j 6 n .
Remark that n = O(n ) and n = O(n ). Therefore 1; n = O(1=n2 ).
When = 0, Turan [13] (see also [9]) gave the precise expression of 1; n which is
:
1; n = 4 sin2
2(2n + 1)
For any , Dor
er [4] gave an upper bound of 1=1; n in using the Frobenius norm of a certain
matrix. But the entries of that matrix are wrong. Indeed the coecient c which appears in his relations
depends on k : c(k) = (1 + =k)−1=2 . He wrote all the relations with a coecient c independent of
k. Therefore his upper bound of 1=1; n is false.
As a straightforward consequence of Theorems 6.2 and 6.3 we get the Markov–Bernstein inequality
in the Laguerre case.
Corollary 6.4. In the Laguerre case the following Markov–Bernstein inequality is satis ed.
∀p ∈ Pn − {0};
1
kpk ¡
kp k6 √
1; n
′
√
1= 1; n is the best constant.
An extremal polynomial is
p=
s
n(n + 1)
kpk:
2( + 1)
n
X
w(1; n)
pi
Ri ;
i=1
i ki−1
where w(1; n) = (w1(1; n) ; : : : ; wn(1; n) )T is the eigenvector of Jn corresponding to the eigenvalue 1; n .
Proof. The form of an extremal polynomial is a consequence of relations (41) and (39) with
N = 1.
Corollary 6.5. D = R+ .
Remark that this result could be obtained from the region of R which is a region of density of
the set of all zeros of all An ; ∀n¿1 (see Chihara [2,3]). Since limn→∞ (2 + =n) = 2 and limn→∞ (1 +
=(n − 1)) = 1, then [0,4] is that region. Therefore a is positive de nite for 1 ¿ 0.
Corollary 6.5 also is a consequence of Theorem 6.1. The assumptions of this theorem are satis ed,
since limn→∞ 1; n =0 and n; n bounded ∀n (Property 5.2). That implies, in the Laguerre case, if ∈ D,
then N ¿ 0.
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
221
In the Gegenbauer case, the matrix Jn has ve diagonals, but the two diagonals which are here
and there of the main diagonal are zero. Therefore Jn can be transformed in a similar matrix which
has a structure of 2 × 2 diagonal block matrix. Moreover the two diagonal blocks are tridiagonal.
The matrices used in the transformation are permutation matrices Vij (i 6= j) which are such that
Vij (‘; ‘) = 1
for ‘ = 1; : : : ; n; ‘ 6= i and ‘ 6= j;
V (i; j) = Vij ( j; i) = 1;
ij
0 everywhere else:
Of course Vij Vij = I . Thus Jn is multiplied on the right and on the left by such matrices having
the same indices. They permute rows and columns having the same couple of numbers. Finally we
get a [(n + 1)=2] × [(n + 1)=2] rst tridiagonal block, denoted by Jˆ[ n+1
; and a [n=2] × [n=2] second
2 ]
˜
tridiagonal block, denoted by J [ 2n ] ; where [:] denotes the integer part. We have
Jˆ[ n+1
(i; j) =
2 ]
k1
for i = j = 1;
k0
(e2i−2
)2 k2i−3
1
k2i−1
+
(2i − 1)2 k
(2i − 3)2 k
2i−2
for i = j ¿ 1;
2i−2
e2i
k2i−1
p
for i = j − 1;
2
(2i − 1) k2i−2 k2i
for i = j + 1 the matrix is completed by symmetry;
0
everywhere else:
Using the expression of ki ; we get the following sequence of monic orthogonal polynomials for this
Jacobi matrix
Aˆ i+1 (x) = (x − i )Aˆ i (x) − i Aˆ i−1 (x);
(48)
i¿1;
with Aˆ 0 = 1; Aˆ 1 (x) = x − 1=2( + 1); and
1
i =
2
i =
i+
i
+
;
(2i + 1)(2i + + 1)2 (2i + )2 (2i + 2 − 1)
i(i + − 1)
;
4(2i − 1)(2i + 2 − 1)(2i + )3 (2i + − 1)
J˜ [ n2 ] (i; j) =
k2
for i = j = 1;
4k1
(e2i−1
)2 k2i−2
1 k2i
+
(2i)2 k2i−1
(2i − 2)2 k2i−1
for i = j ¿ 1;
e2i+1
k2i
p
for i = j − 1;
2
(2i) k2i+1 k2i−1
for i = j + 1 the matrix is completed by symmetry;
0
everywhere else:
(49)
(50)
222
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
We get another sequence of monic orthogonal polynomials for this Jacobi matrix
A˜ i+1 (x) = (x − i+ 1 )A˜ i (x) − i+ 1 A˜ i−1 (x); i¿1;
2
(51)
2
with A˜ 0 = 1; A˜ 1 (x) = x − (2 + 1)=(8( + 2)2 ); i+ 1 and i+ 1 are obtained from (49) and (50) in
2
2
replacing i by i + 1=2.
Remark that these polynomials should be those found by Perez in her doctoral dissertation [10],
but she gave a wrong expression for i . See also [9].
Theorem 6.6 (Gegenbauer case). The eigenvalues ˆi; n ; i = 1; : : : ; n; of Jˆn which are the zeros of the
orthogonal polynomials Aˆ n given by (48); are real; positive; distinct. Moreover
(1 + 2 )
1
¡ ˆ1; n 6ˆ n = O 4 ;
2
2n(2n + 2n − 1)(n + )
n
n¿1:
Proof. It is still obvious that the ˆi; n ’s are real, positive, distinct.
(i) For nding the lower bound, we will use the same technique as in Theorem 6.3.
Using Eq. (48) it is easy to verify by recurrence on n that
n!
Aˆ n (0) = (−1)n
:
(2n)!(2n − 1 + )n
(52)
Indeed Eq. (52) is satis ed for Aˆ 1 (0) = −1=(2(1 + )). If Eq. (52) is assumed to be satis ed up to
the degree n, we have
2n + 2
(−1)n n!
2n
+
(2n + 1)(2n + + 1)2 (2n + )2 (2n + 2 − 1) (2n)!(2n − 1 + )n
2n(2n + 2 − 2)(−1)n−1 (n − 1)!
−
16(2n − 1)(2n + 2 − 1)(2n + )3 (2n + − 1)(2n − 2)!(2n + − 3)n−1
2(n + )2(n + 1)n!
= (−1)n+1
4(2n + 1)(2n + 2)(2n + + 1)2 (2n)!(2n + − 1)n
2n(−1)n n!
−
4(2n + )2 (2n + 2 − 1)(2n)!(2n + − 1)n
2n(2n + 2 − 2)(−1)n−1 (n − 1)!
−
16(2n − 1)!(2n + )n+2 (2n + 2 − 1)(2n + − 1)
(−1)n+1 (n + 1)!
:
=
(2n + 2)!(2n + 1 + )n+1
1
Aˆ n+1 (0) = −
4
On the other hand, we have
′
Aˆ n (0) = (−1)n−1
n!(2n2 + 2n − 1)
;
(2n − 1)!(1 + 2 )(2n − 1 + )n−1
(53)
n¿1:
′
Aˆ 1 (0) = 1 satis es Eq. (53). If Eq. (53) is assumed to be satis ed up to the rank n; we have
′
(−1)
Aˆ n+1 (0) =
2
n
n+
n
+
(2n + 1)(2n + + 1)2 (2n + )2 (2n + 2 − 1)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
223
n!(2n2 + 2n − 1)
n!
+ (−1)n
(2n − 1)!(1 + 2 )(2n − 1 + )n−1
(2n)!(2n + − 1)n
n(n
+
−
1)
+(−1)n−1
4(2n − 1)(2n + 2 − 1)(2n + )3 (2n + − 1)
(n − 1)! 2(n − 1)2 + 2(n − 1) − 1
×
(2n − 3)! (1 + 2 )(2n − 3 + )n−2
(n + )(2n2 + 2n − 1)
n!
= (−1)n
2(2n + 1)(2n − 1)! (2 + 1)(2n + + 1)n+1
1
n!
((1 + 2 )(2n + ) + n(2n − 1))
+(−1)n
(2n)! (1 + 2 )(2n + )n+1
(n + )(2n2 + 2n − 1)
n!
= (−1)n
2(2n + 1)(2n − 1)! (1 + 2 )(2n + + 1)n+1
n! (n + )(2n + 2 + 1)
+(−1)n
(2n)! (1 + 2 )(2n + )n+1
(n
+ 1)! 2(n + 1)2 + 2(n + 1) − 1
:
= (−1)n
(2n + 1)! (1 + 2 )(2n + + 1)n
×
′
The ratio −Aˆ n (0)= Aˆ n (0) gives the lower bound.
(ii) In order to obtain the upper bound, we will use the Rayleigh quotient Jˆ of Jˆn . Its numerator
is
n−1
X
i=1
=
1
2i − 1
s
s
n−1
X
i=1
k2i−1
e2i
yi +
k2i−2
2i − 1
s
k2i−1
yi+1
k2i
i−1+
yi −
2(2i − 1)(2i + − 1)2
!2
s
n−1+
y2 :
2(2n − 1)(2n + − 1)2 n
Let us choose the vector y such that
n+1
2
;
i
for i6n =
yi =
2
2
(2n − i) for n 6i62n :
+
1
k2n−1 2
y
2
(2n − 1) k2n−1 n
i
yi+1
2(2i + )2 (2i + 2 − 1)
!2
+
(54)
Since yn+1 = 0 if n is odd, we can write this numerator as
n −1
X
i=1
s
i+ −1
i2 −
2(2i − 1)(2i + − 1)2
2n −1
+
X
i=n
s
−
s
s
i
(i + 1)2
2(2i + )2 (2i + 2 − 1)
i+ −1
(2n − i)2
2(2i − 1)(2i + − 1)2
i
(2n − 1 − i)2
2(2i + )2 (2i + 2 − 1)
!2
!2
224
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
n −1
=
X
i=1
2n −1
ˆi +
X
ˆi ;
i=n
i(i + 1)4
(i + − 1)i4
+
2(2i − 1)(2i + − 1)2 2(2i + )2 (2i + 2 − 1)
√
i2 (i + 1)2 2i(2i + )(2i − 1)(2i + 2 − 1)(2i + − 2)(2i + 2 − 2)
−
2(2i − 1)(2i + )3 (2i + 2 − 1)
4
i (i + − 1)
i2 (i + 1)2 $i
i(i + 1)4
6
−
;
+
2(2i − 1)(2i + − 1)2 2(2i + )2 (2i + 2 − 1) (2i − 1)(2i + )3 (2i + 2 − 1)
ˆi =
where
1
2
$i =
4i + −
8
4i +
!
2
2i − 1 + −
2i − 1 +
!
2
4i − 4 + 3 −
4i − 4 + 3
!
:
In the same way we get
(2n − i)2 (2n − 1 − i)2 $i
(i + − 1)(2n − i)4
i(2n − 1 − i)4
ˆi 6
−
:
+
2(2i − 1)(2i + − 1)2 2(2i + )2 (2i + 2 − 1) (2i − 1)(2i + )3 (2i + 2 − 1)
The computation of a precise upper bound is so tedious that, rstly this work was made with the
help of Mathematica 3.0 [15], secondly the expression of this upper bound using the same technique
as in Theorem 6.3 is so long that it is unusable and we have preferred to give an equivalent as a
O(:).
32 − 32 + 11 2
qˆi
ˆi 6
+
;
128
128(2i − 1)(2i + − 1)(2i + )3 (2i + 2 − 1)(4i + )(4i + 3 − 4)
where qˆi is a polynomial in the variable i of degree 7 whose leading coecient is −211 ( − 2)(16 −
Pn −1 ˆ
i can be represented as
24 + 11 2 ). Then the sum i=1
2
32 − 32 + 11
(n − 1) + O(1n n ):
(55)
128
32 − 32 + 11 2
rˆi
+
;
128
128(2i − 1)(2i + − 1)(2i + )3 (2i + 2 − 1)(4i + )(4i + 3 − 4)
where rˆi is a polynomial in the variable i of degree 7 whose coecients also are expressions which
contain n . Considered as a polynomial in the two variables i and n ; rˆi is of degree 8. Since
n 6i62n − 1; the behavior of rˆi for large n is that of the set of terms ij n8−j ; j = 0; : : : ; 7; which is
ˆi 6
213 (−(16 − 24 + 11 2 )n i7 + (16 − 48 + 33 2 )2n i6 − 4(−8 + 11 2 )3n i5 + 22 2 4n i4 ):
Therefore the behavior of ˆi is
32−32 +11 2
128
+ O(1) and the one of
P2n −1
i=n
ˆi is
32 − 32 + 11 2
n + O(n ):
128
The denominator of the Rayleigh quotient is
1
n (124n + 453n + 702n − 45n − 22):
60
Therefore, gathering (55) – (57), the result concerning the upper bound holds.
(56)
(57)
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
225
Theorem 6.7 (Gegenbauer case). The eigenvalues ˜ i; n ; i = 1; : : : ; n; of J˜ n which are the zeros of the
orthogonal polynomials A˜ n given by (51); are positive; real; distinct. Moreover
1 + 2
1
¡ ˜ 1; n 6˜ n = O 4 :
4n(n + 1)( + n)( + n + 1)
n
Proof. The nature of the ˜ i; n ’s is still obvious.
(i) From (51) it is very easy to verify by recurrence on n that
A˜ n (0) = (−1)n
(2n + 2 )2n
:
+ )2n (n + )n
24n n!(2n
On the other hand, we have
′
A˜ n (0) = (−1)n−1
(n + 1)(2n + 2 )2n−1 ( + n + 1)
;
(n − 1)!24n−2 (2n + )2n ( + n − 1)n−1
(58)
n¿1:
′
A˜ 1 (0) satis es Eq. (58) which can be assumed to be satis ed up to the rank n. Then we have
2n + 1 + 2
2n + 1
+
2(n + 1)(2n + + 2)2 (2n + + 1)2 (2n + 2 )
(2n + 2 )2n−1 (n + 1)( + n + 1)
(−1)n−1
(n − 1)!24n−2 (2n + )2n ( + n − 1)n−1
(2n + 2 )2n
+ (−1)n−1
+ (−1)n 4n
2 n!(2n + )2n (n + )n
(2n + 1)(2n + 2 − 1)(2n + 2 − 2)2n−3 (n + )
(2n + 2 )(2n + + 1)3 (2n + )(n − 2)!24n−1 (2n + − 2)2n−2 ( + n − 2)n−2
(2n + 2 + 2)2n+1
= (−1)n 4n+2
2
(n − 1)!(2n + + 2)2n+2 (n + − 1)n−1
(2n + 1)(n + 1)(2n + 2 )2n−1 ( + n + 1)
+ (−1)n
(n − 1)!24n+1 (2n + + 1)2n+1 ( + n)n (2n + )
(2n + 2 )2n
+ (−1)n 4n
2 n!(2n + )2n (n + )n
(2n + 1)(2n + 2 )2n−1
+ (−1)n−1
4n+1
(n − 2)!2
(2n + + 1)2n+1 ( + n − 2)n−2 (n + )(2n + )
(2n
+
2 + 2)2n+1
= (−1)n 4n+2
2
(n − 1)!(2n + + 2)2n+2 ( + n − 1)n−1
(2n + 2 + 2)2n+1
+ (−1)n 4n+1
2
n!(2n + + 1)2n+1 ( + n)n
(2n + 2 + 2)2n+1 (n + 2)( + n + 2)
= (−1)n
:
24n+2 n!(2n + + 2)2n+2 ( + n)n
′
1
A˜ n+1 (0) = −
4
′
The ratio −A˜ n (0)= A˜ n (0) gives the lower bound.
226
A. Draux, C. Elhami / Journal of Computational and Applied Mathematics 106 (1999) 203–243
(ii) The numerator of the Rayleigh quotient J˜ of J˜ n is
n−1
X
1
2i
i=1
=
s
e
yi + 2i+1
k2i−1
2i
k2i
n−1
X
1
2i
i=1
+
s
s
k2i
yi+1
k2i+1
!2
+
1 k2n 2
y
(2n)2 k2n−1 n
2i + 1
i(2i − 1 + 2 )
1
yi −
2(2i + )2
4 (2i + + 1)2
s
2(2i + )(2i + + 1)
yi+1
(2i + 1)(i + )
!2
1 2n − 1 + 2 2
y:
8n (2n + )2 n
We do the same choice (see (54)) for the vector y as in the previous theorem and the numerator
becomes
s
n −1
X 1
4
i=1
2i + 2 − 1 2
i −
2i(2i + )2
s
2n −1
+
X 1
i=n
4
=
i=1
4
2i + 1
(i + 1)2
2(2i + + 1)2 (i + )
2i + 2 − 1
(2n − i)2 −
2i(2i + )2
s
!2
2i + 1
(2n − 1 − i)2
2(2i + + 1)2 (i + )
!2
2n −1
n −1
X ˜i
s
+
X ˜ i
i=n
4
;
(2i + 2 − 1)i3
(2i + 1)(i + 1)4
˜i =
+
2(2i + )2
2(2i + + 1)2 (i + )
√
2
i(i + 1) i(i + )(2i + 1)(2i + + 1)(2i + − 1)(2i + 2 − 1)
−
(i + )(2i + + 1)3
3
i(i + 1)2 &i
(2i + 1)(i + 1)4
(2i + 2 − 1)i
−
;
+
6
2(2i + )2
2(2i + + 1)2 (i + ) (2i + + 1)3 (i + )
where
2