Some properties of the Sobolev–Laguerre polynomials
Jet Wimp∗, Harry Kiesel
Department of Mathematics and Computer Science, Drexel University, Philadelphia, PA 19104, USA
Received 5 November 1997; received in revised form 8 June 1998
Abstract
In this paper, we study the case = 0 of the Sobolev–Laguerre polynomials. We determine a generating function for the polynomials and an expansion formula c 1998 Elsevier Science B.V. All rights reserved.
AMS classication:33C49; 33C99
Keywords:Sobolev orthogonal polynomials; Laguerre polynomials; Sobolev–Laguerre polynomials; Bessel polynomials; Reproducing kernels; Generating functions; Laplace transforms
1. Introduction
The so-called Sobolev–Laguerre polynomials have been discussed by a number of dierent writers. The (general) Sobolev–Laguerre polynomials are orthogonal with respect to the inner product
(∗)hf; gi= Z ∞
0
xe−xf(x)g(x) dx+ Z ∞
0
xe−xf′(x)g′(x) dx; ¿0:
The polynomials were exhaustively studied by Brenner in the case = 0 who found a (not very useful) explicit formula for the polynomials [1]. For the more recently investigated case of general
, the article by Marcellan et al. [4] provides an excellent and timely bibliography, as well as a number of original ndings.
In this paper, we study the case = 0. The notation we use for these polynomials is L
n(x). We determine an elegant generating function for the polynomials and some expansions involving the polynomials. These results seem to be new.
∗Corresponding author. E-mail: [email protected].
2. Results
We will employ the notation: ifAis a matrix with elementsai; j, thenAwill denote the determinant whose elements are ai; j, and vice versa.
In our determinants and matrices, i will denote row position, j column position. We can write L
(It is easily veried that (1) provides a legitimate expression for L
n(x) by taking the inner product of the determinant with xi; i= 0;1; : : : ; n−1. The last column of the resulting determinant is equal to another column, so the result is 0.)
With these polynomials are associated two normalization constants, k()
n , the coecient of the highest power of x in L()
n , and
hn()=hLn(); L(n)i: (3)
Taking inner products of the determinant (1) with L()
n shows
a useful formula.
It is convenient to factor (i−1)! out of the ith row, i= 1;2; : : : ; n+ 1, and (j−1)! out of the
jth column, j= 1;2; : : : ; n of L()
n and to consider instead the polynomial
and
We will investigate three (n+ 1)×(n+ 1) determinants, which are auxiliary polynomials in x of degree n: will be used later.
We will also investigate the three following n×n constant (independent of x) determinants:
Cn=
Our proofs of results pertaining to these quantities will utilize the three n×n elementary matrices
En=
Theorem 1. Let
= 4 sinh2: (15)
Proof. A simple computation shows that
EnCnFn=
Next, we have
EnAnFn=
1 0 0 0 : : :
0 + 1 1 1 : : :
0 1 + 2 + 3 : : :
0 1 + 3 2+ 6 : : :
..
. ... ... ... . ..
: (20)
Taking determinants, expanding the determinant on the right by minors of the rst column, and then expanding that determinant by minors of its rst column give
An=An−1+Bn−2: (21)
Applying Sylvester’s theorem (see [7, p. 251]) to An and utilizing the known value of Cn give
AnBn−2=An−1Bn−1−1: (22)
Solving this equation for Bn−2 and substituting it into the previous equation yields
A2n=An+1An−1−: (23)
Induction on this equation gives the An result.
Finally, substituting the known value of An into (21) gives the Bn result.
Theorem 2. Let
D=x2
−(1−x):
Then
(i)
Pn=
(−1)n(x−1)
D +
x(x−1)n
D [(x−1)An+An+1];
(ii)
Qn= (−1)n
D +
(x−1)n+1
D [(+x)An+2−xAn+1];
(iii)
Rn= (−1)n "
An+1−(1−x)An+2−x(1−x)n+1
D
# :
Proof. Considering the determinants of the matrices En+1PnGn+1 and En+1RnGn+1 in much the same way as in the proof of Theorem 1 shows that
Pn= (x−1)Pn−1+x(x−1)Qn−2; (24)
Applying Sylvester’s theorem to Pn gives
PnBn−1=AnxQn−1−Rn−1: (26)
We write (25) as
Rn (x−1)n −
Rn−1 (x−1)n−1=
(−1)ncosh(2n+ 1)
cosh(x−1)n : (27) Expressing the hyperbolic functions as exponentials and iterating this expression, i.e., replacing n by
k and summing from k= 1 to n (note (25) requires interpreting R0= 1), give, after some numbing algebra, (iii).
We now solve (24) for Qn−2, replace n by n+ 1, substitute the result into (26), substitute the just obtained value of Rn into (26), and nally solve forBn−1 from (21) and substitute that into the equation. We obtain the following rst-order dierence equation for Pn:
PnAn+1 (x−1)n −
Pn+1An (x−1)n+1=
D
(−1)n An
(x−1)n + (−1)
n An+1 (x−1)n−1
−xD: (28)
It is easily veried, using the known properties of An, that particular solutions corresponding to the rst and to the second term on the right are, respectively,
P(1)
n =
(−1)n(x
−1)
D ; P
(2)
n =
An+1x(x−1)n
D : (29)
A homogeneous solution is
Pn(3)=An(x−1)n: (30)
We write
Pn=Pn(1)+P
(2)
n +CP
(3)
n : (31)
C is determined by setting n= 1 and using P1=x−1. The result is (i). Finally, using the above value of Pn we solve for Qn from (24). The result is then simplied by expressing An+3 in terms of
An+2 and An+1 and the result is (ii). In what follows, we let
a=−−2 +
√
2+ 4
2 ; n=
n Y
j= 1
j!(j−1)!
−1
: (32)
Theorem 3. For |t|¡min{1;|a|=|x−1|};
h(x; t) =
∞
X
n= 0
Pn(x)(−t)n; (33)
where
h(x; t) =[(1−t)
2+t(x−1) +tx(1−t)]
and
U= (t(x−1) + 1)2+t(x
−1): (35)
Proof. Easy algebra gives
∞
X
n= 0
An(−t)n(x−1)n =
1 +t(x−1)(+ 1)
U ;
∞
X
n= 0
An+1(−t)n(x−1)n =
1 +t(x−1)
U :
(36)
(NoteA0= 1:) Multiplying Theorem 2(i) by (−t)n and summing gives the result. (It requires a rather formidable amount of algebra to show the factor D cancels out of the right-hand side.) The region of convergence of the series is easily determined by examining the location of the poles of the left-hand side. These occur at t= 1; t=a=(x−1); t= 1=[a(x−1)]. Since |a|¡1¡|1=a|, the region of convergence is as indicated.
Theorem 4. For |t|¡|a|;
aext=(t+a)−ext=(t+1=a) (a−1)(1−t) =
∞
X
n= 0
nL(n)(x)(−t)
n: (37)
Proof. We observe that 1
pPn 1
p
=nL{L(n)(x)}; (38)
where L is the Laplace transform operator,
L{f(x)}=
Z ∞
0
e−pxf(x) dx: (39)
Replacing x by 1=p in (33) gives
∞
X
n= 0 1
pPn 1
p
(−t)n= p(1−t)
2+t(1−p) +t(1−t)
(1−t)(p2+ (2 +)tp(1−p) +t2(1−p)2): (40)
The right-hand side as a function of p has two poles which are; generally, simple.
p±=t 2
2t−−2±√4+2 (1−t)2−t
!
: (41)
Note when = 0; (37) yields the standard generating function for the Laguerre polynomials, [3, vol. 2, p. 189, formula (21)]. To see this, observe that
(−1)nnL(n)(x)|x=0=Cn= 1: (42)
The Laguerre polynomials are normalized to have the value 1 at x= 0. Thus,
(−1)n
nL(0)n (x) =Ln(x): (43)
An interesting inequality forL()
n follows from the generating function (37). The singularities of the function on the left are att= 1; t=−a; t=−1=a. Expressed in terms of these aret= 1; t= e−2; t= e2. The smallest of these is e−2. Applying Cauchy’s inequality shows that for any ; 0¡¡e−2
|nL(n)|6 M
(e−2−)n; (44)
where M is some constant independent of x.
Theorem 5.
e−cx=
∞
X
n= 0
nL(n)(x); (45)
where
n=n(1−a) (
c c+ 1
n an−1
(a2n−1−1)+
c
c+ 1
n+1 an
(a2n+1−1)
)
; (46)
and
a=−−2 +
√
2+ 4
2 ; n=
n Y
j= 1
j!(j−1)!
−1
: (47)
The series converges uniformly on compact subsets of the complex x-plane for all ¿0; c¿0; as can be seen from the estimate (44).
Proof. Denote the generating on the left of (37) by g(x; t). We have
he−cx; g(x; t)
i=
∞
X
n= 0
nhe
−cx; L()
n (x)i(−t) n
=− a(ac+a+ 1) (a−1)(ct+ac+a)+
(ac+a+c)
Selecting the coecient of tn from the last expression above gives the value of he−cx; L()
n (x)i. We now need the value of h()
n . However, this is easy to obtain. We have
hg(x; t); g(x; u)i=
∞
X
n= 0
n2hL(n)(x); Ln()(x)i(ut)n=
∞
X
n= 0
2nh(n)(ut)n
=− a
(a−1)2(1−ut=a2)+
a2+ 1
(a−1)2(1−ut)−
a
(a−1)2(1−a2ut): (49) The last expression above is obtained by the evaluation of a Laplace transform integral and partial fractions. Selecting the coecient of (ut)n from this expression gives h()
n ,
h(n)=−(a
2n−1−1)(a2n+1−1)
a2n−1(a−1)22
n
: (50)
We then divide he−cx; L()
n (x)i by this expression and do a little algebra to get the coecients given in formula (46).
Many other expansions of special functions in the polynomials L()
n can be obtained by similarly manipulating the generating function (37), including expansions for the conuent hypergeometric functions and analogous to those expansions in Laguerre polynomials given in [3, vol. 2, p. 215].
Marcellan et al. have shown that the Laguerre–Sobolev polynomials for general satisfy a third-order (four-term) recurrence relation. One starts with the equation [4] (3.6), which relates the general Sobolev–Laguerre polynomials Q()
n (x) (orthogonal with respect to the inner product (∗)) and the monic Laguerre polynomials ˆL()
n (x). In the notation of [4], ˆ
L(n)(x) +nLˆ(n−)1(x) =Q ()
n (x) +dn−1()Q ()
n−1(x): (51)
dn() is a ratio of two Pollaczek polynomials. Now replace n by n−1 then by n−2 and each time use the recurrence for ˆL()
n (x) to express ˆL
()
n−2(x) in terms of ˆL(n)(x) and ˆL
()
n−1(x). Eliminating the latter two quantities from the set of three equations yields the four-term recursion relation given in [4, p. 258]:
Q()
n (x) +
(2n+−2−x) +qn−2n(n+−1)
qn−1
Q(n−)1(x)
+ (n−1)(n+−2)
1 +qn−3(2n+−2−x)
qn−2
Qn(−)2(x)
+qn−4(n−1)(n−2)
qn−3
(n+−2)(n+−3)Qn(−)3(x) = 0: (52)
Using the recurrence in Proposition 3.4 of [4], one nds that qn() may be expressed a linear combination of Pollaczek polynomials
qn() = (+ 1)n
P(1−=2)
n
2 + 1;−=2; =2;
−(+ 1)1 Pn(1−−1=2)
2+ 1;−=2; =2; + 1
;
see also [2, p. 185; 3, vol. 2, p. 220]. There are at least two values of for which the above expression simplies. When = 0 the explicit value of the Pollaczek polynomials above can be found from the formulas [5, (12); 6, (4.4)]. The former provides a formula
3F2
−n; a;
; 1
a+ 1; + 1
= n!a
(−a) 1
(a)n+1 − 1 ()n+1
; (54)
that can be used in the explicit formula for the Pollaczek polynomial given in [6, (4.4)]. We nd that for the special case b=−a,
Pn()(1;a;−a; c) = c (2−1)
(c+ 2
−1)n+1 (c)n+1 −
1
: (55)
Using this formula in (53) shows
qn(0) = (+ 1)n; (56)
and for this case (52) is easily found to be an iteration of the recurrence of the classical monic Laguerre polynomial.
A simplication of qn() also occurs in the special case = 0, since
P(1)
n
2 + 1; 0;0; c
=Un
2 + 1
; (57)
whereUn denotes the Chebyshev polynomial of the second kind. As is well known, these polynomials are elementary functions,
Un
2 + 1
= (−1)na
n+1−a−n−1
a−a−1 ; a=
−−2 +√2+ 4
2 : (58)
We nd that when = 0,
qn() =n!(−1)n an+1
−a−n a−1 ; a=
−−2 +√2+ 4
2 : (59)
Eq. (15) shows
a=−e−2; (60)
so
qn() =n!An+1;
(61)
An=
cosh(2n−1)
cosh =
(p
1 +=4 +√=2)2n−1+ (p
The resulting four-term recurrence relation for the monic ordinary Laguerre–Sobolev polynomials is quite simple:
Q(0)
n (x) +
(2n−2−x) +nAn−1
An
Q(0)n−1(x)
+ (n−1)
(n−2) +(2n−2−x)An−2
An−1
Q(0)n−2(x)
+(n−1)(n−2) 2A
n−3
An−2
Q(0)n−3(x) = 0: (62)
A referee has pointed out that this recurrence can be obtained directly from the generating func-tion (37) without recourse to Pollaczek polynomials. Rewriting that equafunc-tion, we have
(1−t)
∞
X
n=0
nL(n)(x)(−t) n
=ae x t=(t+a)
(a−1) −
ex t=(t+1=a) (a−1)
= a
a−1
1 + t
a ∞
X
n=0
Ln(x)
−at
n
−a 1
−1(1 +at)
∞
X
n=0
Ln(x)(−at)n: (63)
Comparing coecients of (−t)n gives
nL(n)(x) +n−1L ()
n−1(x) = 1
a−1 1
an−1 −a
n
(Ln(x)−Ln−1(x)): (64)
Now, 1
a−1 1
an−1 −a
n
= (−1)nA
n; nLn(x) = An
n!Q (0)
n (x): (65)
When (65) is used and the Laguerre polynomials are eliminated from the relation (64), one recovers the recurrence (62).
Acknowledgements
The authors are greatly indebted to the referees, whose painstaking and attentive readings allowed us to correct many misprints. In addition, they showed us how to correct one false result and how to substantially simplify the recurrence (62).
References
[1] J. Brenner, Uber eine Erweiterung des Orthogonalitatbegries bei Polynomen, Proc. Conf. Constructive Theory of Functions, Akademiai Kiado, Budapest, 1972.
[3] A. Erdelyi et al., Higher Transcendental Functions, vols. 1 and 2, McGraw-Hill, New York, 1953.
[4] F. Marcellan, T.E. Perez, M.A. Pi˜nar, Laguerre–Sobolev orthogonal polynomials, J. Comp. Appl. Math. 71 (1996) 249–265.
[5] J. Wimp, B. Beckermann, Families of some two-point Pade approximants and some4F3(1) identities, SIAM J. Math.
Anal. 26 (1995) 765–773.
[6] J. Wimp, Pollaczek polynomials and Pade approximants: some closed form expressions, J. Comp. Appl. Math. 32 (1990) 301–310.
