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Classical multiple orthogonal polynomials of Angelesco system
Applied Numerical Mathematics 60 (2010) 1342–1351
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Applied Numerical Mathematics www.elsevier.com/locate/apnum
Classical multiple orthogonal polynomials of Angelesco system D.W. Lee 1 Department of Mathematics, Teachers College, Kyungpook National University, Daegu 702-701, South Korea
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 18 November 2009 Received in revised form 28 April 2010 Accepted 5 May 2010 Available online 12 May 2010
In this paper, we analyze more carefully the Rodrigues formula of classical multiple orthogonal polynomials of Angelesco system and consider the sequence of them for the diagonal index case, and then find many properties of such a sequence of orthogonal polynomials. More precisely, we consider the diagonal index sequence { P n,n (x)}n∞=0 of classical multiple orthogonal polynomials { P n1 ,n2 (x)}n∞,n =0 of Angelesco system. For such a
Keywords: Multiple orthogonal polynomial Angelesco system Differential equation Recurrence relation
sequence we give recurrence relations, differential-difference equations, and then we finally find a third order differential equation having { P n,n (x)}n∞=0 as solutions. © 2010 IMACS. Published by Elsevier B.V. All rights reserved.
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1. Introduction A sequence { P n1 ,n2 (x)}n∞,n =0 of polynomials is called a multiple orthogonal polynomial system (multiple OPS) if 1 2 (i) deg( P n1 ,n2 ) = n1 + n2 ; (ii) there exist two positive weights w 1 and w 2 such that
∞ xk P n1 ,n2 (x) w 1 (x) dx = 0
for k = 0, 1, 2, . . . , n1 − 1
xk P n1 ,n2 (x) w 2 (x) dx = 0
for k = 0, 1, 2, . . . , n2 − 1.
−∞
and
∞ −∞
In this case, w 1 and w 2 are called the orthogonalizing weights for { P n1 ,n2 (x)}n∞,n =0 . 1 2 The multiple OPS appeared in [1] to obtain simultaneous Hermite–Padé approximants. See also [4,5,10,12]. Recently it attracts a renewed interest as a natural extension of orthogonal polynomials, in particular, of classical orthogonal polynomials [2,3,11,15]. There are two kinds of systems in multiple OPS’s that are now extensively studied. They are AT system and Angelesco system.
E-mail address: [email protected]. The author thanks to Claude Brezinski for his hospitality, and also thanks to the referee for his/her valuable remarks and corrections after careful reading of the manuscript that improved this paper very much. This work was supported by Korea Research Foundation Grant funded by the Korean Government (KRF-2008-313-C00027). 1
0168-9274/$30.00 © 2010 IMACS. Published by Elsevier B.V. All rights reserved. doi:10.1016/j.apnum.2010.05.004
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
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The AT system is a class of multiple OPS’s that the orthogonalizing weights have the same support of intervals. Five families of AT system are well known whose weights consist of the same type of classical weights. They are Jacobi–Piñeiro, multiple Bessel, multiple Laguerre I, multiple Laguerre II, and multiple Hermite polynomials. Of course, they are named after their orthogonalizing weights that come from the classical weight function. Many properties for classical multiple OPS’s of AT system were found by many authors [3,6,9,15], such as recurrence relations, differential-difference equations, raising operators, lowering operators, differential equations, and so on. These properties are fundamentally induced from the Rodrigues formula obtained in [15]. On the other hand, the Angelesco system were not investigated as much extensively as AT system. The Angelesco system is a class of multiple OPS’s that the orthogonalizing weights have pairwisely disjoint support allowing the touch of their boundary points. Three families of Angelesco system are known as an extension of classical orthogonal polynomials, which we call the classical multiple OPS of Angelesco system. They are Jacobi–Angelesco, Jacobi–Laguerre, and Laguerre–Hermite polynomials, named after the orthogonalizing weights and the interval of orthogonality [7,8,13–15]. It is remarkable that the Rodrigues formula was also obtained in [15] for these classical multiple OPS’s of Angelesco system. In this paper, we analyze more carefully the Rodrigues formula (in [15]) of classical multiple OPS of Angelesco system and consider the sequence of them for the diagonal index case, and then find many properties of such a sequence of OPS’s. More precisely, we consider the diagonal index sequence { P n,n (x)}n∞=0 of classical multiple OPS { P n1 ,n2 (x)}n∞,n =0 of 1 2 Angelesco system. For such a sequence we give recurrence relations, differential-difference equations, and then we finally find a third order differential equation having { P n,n (x)}n∞=0 as solutions. 2. Laguerre–Hermite polynomials The Laguerre–Hermite polynomial { H n1 ,n2 (x)}n∞,n =0 (β > −1) is a multiple OPS of Angelesco system which was first 1 2 2 2 considered by Sorokin [14]. The orthogonalizing weights are w 1 (x) = |x|β e −x on (−∞, 0] and w 2 (x) = xβ e −x on [0, ∞). Hence, (β)
0
xk H n1 ,n2 (x)|x|β e −x dx = 0 for k = 0, 1, 2, . . . , n1 − 1 2
(β)
−∞
and
∞
xk H n1 ,n2 (x)xβ e −x dx = 0 2
(β)
for k = 0, 1, 2, . . . , n2 − 1.
0
Iterations of integration by parts with orthogonality give a Rodrigues formula
dn β+n −x2 (β+n) 2 (β) x e H k,0 (x) = (−2)n xβ e −x H n+k,n (x). dxn (β)
(β)
(β)
If we let H n (x) := H n,n (x) with normalization H 0,0 (x) = 1, the Rodrigues formula for the diagonal index case can be written by
2 2 (n) (β) (−2)n xβ e −x H n (x) = xβ+n e −x . (β)
Note that H n (x) defined as above is a polynomial of degree 2n even when β −1, in which case the orthogonalities
of { H n (x)}n∞=0 are broken. In this section, by
{ H n (x)}n∞=0 we mean the sequence of Laguerre–Hermite polynomials of diagonal index defined by the Rodrigues formula with real β . (β) In the following, theorems for { H n (x)}n∞=0 hold for all real β since we use only Rodrigues’ formula in proving them. (β)
(β)
Theorem 2.1. The Laguerre–Hermite polynomials { H n (x)}n∞=0 satisfy the following recurrence relations. (β)
(β+1) (x) + n2 H n−1 (x), n 1. (β+ 2 ) (β) (b) 2H n (x) = 2x2 H n−1 (x) − (β + n) H n−1 (x), (β)
(β+1)
(a) H n (x) = H n (β)
n 1.
Proof. From the Rodrigues formula and the identity
xβ+1+n e −x
2
(n)
2 (n) 2 (n−1) = x xβ+n e −x + n xβ+n e −x ,
we have
(−2)n xβ+1 e −x H n 2
(β+1)
(x) = x(−2)n xβ e −x H n (x) + n(−2)n−1 xβ+1 e −x H n−1 (x) 2
(β)
2
(β+1)
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D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
which implies (a). On the other hand, from the identity
xβ+n e −x
2
(n)
2 (n−1) 2 (n−1) = (β + n) xβ+n−1 e −x − 2 xβ+n+1 e −x
and the Rodrigues formula, (b) follows.
2
Theorem 2.2. Let { H n (x)}n∞=0 be the Laguerre–Hermite polynomials. Then for n 0, (β)
(β)
(β+2)
(β)
H n (x) = 2xH n (x) − 2xH n
(x). (β) (β) (β−1) (b) x H n (x) = 2x2 − β H n (x) − 2H n+1 (x). (β) (β) (β) (c) x H n (x) = 2x2 − β − n − 1 H n (x) − 2H n+1 (x). (a)
(d)
(β)
(β)
(β)
H n+2 (x) = 2(n + 2)xH n+1 (x) − (n + 2) H n+1 (x) +
(n + 1)(n + 2) 2
(β)
xH n (x) −
(n + 1)(n + 2) 4
(β)
H n (x) .
Proof. The relation (a) follows from
2 (β) 2 2 (β) (β) (−2)n xβ e −x H n (x) = (−2)n xβ e −x H n (x) − (−2)n β − 2x2 xβ−1 e −x H n (x) β 2 (n+1) 2 (n) 2 (β) = xβ+n e −x − xβ+n e −x − (−2)n+1 xβ+1 e −x H n (x)
x 1 2 (n) 2 (β) 2 β+n−1 −x2 (n) − (−2)n+1 xβ+1 e −x H n (x) = x β + n − 2x x e − β xβ+n e −x x 1 2 (n) 2 (n−1) 2 (n) β + n − 2x2 xβ+n e −x = − n β + n − 2x2 xβ+n−1 e −x − β xβ+n e −x x
− (−2)n+1 xβ+1 e −x H n (x) 2 2 (n) 2 (β) − (−2)n+1 xβ+1 e −x H n (x) = − xβ+n+2 e −x 2
(β)
x
= (−2)n+1 xβ+1 e −x H n 2
(β+2)
(x) − (−2)n+1 xβ+1 e −x H n (x). 2
(β)
From the Rodrigues formula, we have
2 (β) 2 2 (n+1) (β) (−2)n xβ e −x H n (x) + (−2)n β − 2x2 xβ−1 e −x H n (x) = xβ+n e −x = (−2)n+1 xβ−1 e −x H n+1 (x), 2
(β−1)
which is (b). We obtain (c) combining (b) and Theorem 2.1(a). Combining (a) with Theorem 2.1(a), we have (β+1)
(β)
(β+1)
2xH n+2 (x) = 2xH n+2 (x) + (n + 2)xH n+1 (x)
(n + 2)(n + 1)x (β+2) (β+2) (β+2) = 2xH n+2 (x) + 2(n + 2)xH n+1 (x) + (x) Hn 2 (β) (β) (β) (β) = 2xH n+2 (x) − H n+2 (x) + (n + 2) 2xH n+1 (x) − H n+1 (x) (β) (n + 2)(n + 1) (β) 2xH n (x) − H n (x) , + 4
from which (d) follows.
2
From Theorem 2.2, we can find a third order differential equation for { H n (x)}n∞=0 . More precisely, we have: (β)
Theorem 2.3. The Laguerre–Hermite polynomial { H n (x)}n∞=0 satisfies a differential equation (β)
x2 H n (x) + 2x β + 1 − 2x2 H n (x) +
+ 2nx n + 3 + 2β − 4x2 H n (x) = 0, (β)
where H n (x) = H n (x).
2x2 − β 2x2 − β − 1 + 4(n − 1)x2 H n (x)
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
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Proof. By (c) and (d) in Theorem 2.2, we have for n 0,
4H n +2 (x) = 8(n + 2)xH n+1 (x) − 4(n + 2) H n +1 (x) + 2(n + 1)(n + 2)xH n (x) − (n + 1)(n + 2) H n (x)
= 2(n + 2)xH n (x) − (n + 2) 8x2 − 2β − n − 3 H n (x) + 2(n + 2)x 4x2 − 2β − n − 5 H n (x).
(2.1)
On the other hand, applying (c) in Theorem 2.2 iteratively, we have for n 0,
4H n +2 (x) = 8xH n+1 (x) + 2 2x2 − β − n − 3 H n +1 (x) − 2xH n+1 (x)
= x2 H n (x) − 2x 2x2 − β − n − 3 H n (x) + 2x2 − β − n − 3 2x2 − β − n − 2 − 12x2 H n (x) + 4 4x2 − 2β − 2n − 5 xH n (x).
(2.2)
2
Equating (2.1) to (2.2) and then simplifying it, we obtain the result. 3. Jacobi–Laguerre polynomials (α ,β)
The Jacobi–Laguerre polynomial { L n1 ,n2 (x; a)}n∞,n =0 (α , β > −1, a < 0) is a multiple OPS of Angelesco system which was 1 2 first considered by Sorokin [13]. The orthogonalizing weights are w 1 (x) = |(x − a)α xβ |e −x on [a, 0] and w 2 (x) = (x − a)α xβ e −x on [0, ∞). Hence,
0
xk L n1 ,n2 (x; a) (x − a)α xβ e −x dx = 0 for k = 0, 1, 2, . . . , n1 − 1 (α ,β)
a
and
∞
xk L n1 ,n2 (x; a)(x − a)α xβ e −x dx = 0 for k = 0, 1, 2, . . . , n2 − 1. (α ,β)
0
Iterations of integration by parts with orthogonality give a Rodrigues formula (see [15])
dn dxn
(α +n,β+n)
(x − a)α +n xβ+n e −x Lk,0
(α ,β)
(α ,β) (x; a) = (−1)n (x − a)α xβ e −x Ln+k,n (x; a).
(α ,β)
(α ,β)
If we let L n (x) := Ln,n (x; a) with normalization L 0,0 (x; a) = 1, the Rodrigues formula for the diagonal index case can be written by
(−1)n (x − a)α xβ e −x Ln
(α ,β)
(α ,β)
Note that L n
(n) (x) = (x − a)α +n xβ+n e −x .
(x) defined as above is a polynomial of degree 2n even when α −1 or β −1, in which case the (α ,β)
(α ,β)
orthogonalities of { L n (x)}n∞=0 are broken. In this section, by { Ln (x)}n∞=0 we mean the sequence of Jacobi–Laguerre polynomials of diagonal index defined by the Rodrigues formula with real α and β . (α ,β) In the following, theorems for { L n (x)}n∞=0 hold for all real α and β since we use only Rodrigues’ formula in proving them. (α ,β)
Theorem 3.1. The Jacobi–Laguerre polynomials { L n
(x)}n∞=0 satisfy the following recurrence relations.
(α ,β+1) (α +1,β) (α ,β) (x) = (x − a) Ln (x) + aLn (x), (a) xL n
(b)
(α ,β)
(x) =
(α ,β)
(x) = x(x − a) Ln−1
Ln (c)
(α +1,β+1) Ln (x) + nxLn−1 (x), n 1; (α ,β+1) (α +1,β+1) Ln (x) + n(x − a) Ln−1 (x), n 1.
(x) =
Ln
Ln
n 0.
(α +1,β)
(α ,β)
(α +1,β+1)
(α ,β) (x) − (α + n)x + (β + n)(x − a) Ln−1 (x) (α +1,β+1)
(x), n 2. + (n − 1)(α + β + 2n)x(x − a) Ln−2 (α ,β) (α ,β) (x) = − (α + n)x + (β + n)(x − a) − x(x − a) Ln−1 (x) (d) L n (α +1,β+1)
+ (n − 1)(α + β + 2n + a − 2x)x(x − a) Ln−2 (α +2,β+2)
+ (n − 1)(n − 2)x2 (x − a)2 Ln−3
(x),
n 3.
(x)
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(e)
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
(α ,β)
Ln
(α +1,β+1)
(x) = Ln
(α +1,β+1)
(x) + n(2x − a) Ln−1
(α +2,β+2)
(x) − n(n − 1)x(x − a) Ln−2
(x),
n 2.
Proof. (a) follows from the Rodrigues formula and the identity
(x − a)α +n xβ+1+n e −x
(n)
(n) (n) = (x − a)α +n+1 xβ+n e −x + a (x − a)α +n xβ+n e −x .
The first equation of (b) can be obtained from the identity
(x − a)α +n+1 xβ+n e −x
(n)
(n) (n−1) = (x − a) (x − a)α +n xβ+n e −x + n (x − a)α +n xβ+n e −x
and the second equation can be obtained by the same way. (c) It follows from
(x − a)α +n xβ+n e −x
(n)
(n−1) (x − a)α +n xβ+n e −x (n−1) = (α + n)x + (β + n)(x − a) (x − a)α +n−1 xβ+n−1 e −x (n−1) (n−2) − (x − a)α +n xβ+n e −x + (n − 1)(α + β + 2n) (x − a)α +n−1 xβ+n−1 e −x . =
(d) By Leibniz rule, we have
(x − a)α +n xβ+n e −x
(n)
(n−1) = h(α + n, β + n; x) (x − a)α +n−1 xβ+n−1 e −x (n−2) + (n − 1)(α + β + 2n + a − 2x) (x − a)α +n−1 xβ+n−1 e −x (n−3) − (n − 1)(n − 2) (x − a)α +n−1 xβ+n−1 e −x ,
where h(α , β; x) = α x + β(x − a) − x(x − a), from which (d) follows by the Rodrigues formula. (e) Comparing (c) with (d) on both sides we have (α +1,β+1)
x(x − a) L n−1
(α +1,β+1)
(α ,β)
(x) = x(x − a) Ln−1 (x) + (n − 1)(a − 2x)x(x − a) Ln−2
(x)
(α +2,β+2) + (n − 1)(n − 2)x2 (x − a)2 Ln−3 (x),
2
which implies (e).
(α ,β)
Theorem 3.2. Let { L n
(x)}n∞=0 be the Jacobi–Laguerre polynomials. Then for n 1,
(α +1,β+1) (α +1,β+1) (x) = −x(x − a) Ln−1 (x) − (α + 1)x + (β + 1)(x − a) − x(x − a) Ln−1 (x). (α ,β+1) (α ,β+1) (α ,β) (x) = −x(x − a) Ln−1 (x) − (α + n)x + (β + 1)(x − a) − x(x − a) Ln−1 (x); (b) L n (α +1,β) (α +1,β) (α ,β) Ln (x) = −x(x − a) Ln−1 (x) − (α + 1)x + (β + n)(x − a) − x(x − a) Ln−1 (x). (a)
(α ,β)
Ln
Proof. From the Rodrigues formula and the identity
(x − a)α +n xβ+n e −x
(n)
=
(n−1) (α +1,β+1) (x − a)α +n xβ+n e −x = (−1)n−1 (x − a)α +1 xβ+1 e −x Ln−1 (x) ,
the relation (a) follows. Combining (a) and Theorem 3.1(b), we have (α ,β)
Ln
(α ,β+1)
(x) + nxLn−1
from which (b) follows.
(α −1,β)
(x) = Ln
(α ,β+1) (α ,β+1) (x) = −x(x − a) Ln−1 (x) − α x + (β + 1)(x − a) − x(x − a) Ln−1 (x),
2 (α ,β)
Now we are ready to find a third order differential equation for Jacobi–Laguerre polynomials { L n (α ,β)
Theorem 3.3. The Jacobi–Laguerre polynomial { L n
(x)}n∞=0 .
(x)}n∞=0 satisfies a differential equation
x2 (x − a)2 L n (x) + 2x(x − a)h(α + 1, β + 1; x) L n (x)
+ h(α , β; x)h(α + 1, β + 1; x) − (n − 1)(α + β + n + 2 + a − 2x)x(x − a) Ln (x) + n (n + 3)x(x − a) − (α + β + n + 1 + a − 2x)h(α , β; x) Ln (x) = 0, (α ,β)
where L n (x) = L n
(x) and h(α , β; x) = α x + β(x − a) − x(x − a).
(3.1)
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
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Proof. Applying Theorem 3.2 (a) into (d) in Theorem 3.1 for n + 1 and n + 2, iteratively, we have for n 0, (α −2,β−2)
L n +3
(x) = −φ 2 (x)h(α + n + 1, β + n + 1; x) Ln (x) − φ(x) 2h(α , β; x)h(α + n + 1, β + n + 1; x) + (n + 2)(α + β + 2n + 2 + a − 2x)φ(x) Ln (x) − φ(x)(α + β + a − 2x)h(α + n + 1, β + n + 1; x) + (n + 2)(α + β + 2n + 2 + a − 2x)φ(x)h(α , β; x)
+ h(α , β; x)h(α − 1, β − 1; x)h(α + n + 1, β + n + 1; x) − (n + 1)(n + 2)φ 2 (x) Ln (x),
(3.2)
where φ(x) = x(x − a). On the other hand, by Theorem 3.2 (a) and (b), we have for n 0, (α −2,β−2)
L n +3
(α −1,β−1) (α −1,β−1) (x) = −φ(x) Ln+2 (x) − h(α − 1, β − 1; x) Ln+2 (x) (α −1,β) (α −1,β) 2 = φ (x) Ln+1 (x) + φ(x) 2h(α , β; x) + (n + 1)x Ln+1 (x) (α −1,β) + φ(x)[α + β + n + 1 + a − 2x] + h(α − 1, β − 1; x)h(α + n + 1, β; x) Ln+1 (x) = −φ 3 (x) Ln (x) − φ 2 (x) 3h(α , β; x) + (n + 3)(2x − a) Ln (x) − φ(x) [3α + 3β + 3n + 5 + 3a − 6x]φ(x) + h(α − 1, β − 1; x)h(α + n + 1, β; x) + h(α + 1, β + n + 2; x) 2h(α , β; x) + (n + 1)x Ln (x) − h(α − 1, β − 1; x)h(α , β + n + 1; x)h(α + n + 1, β; x) + φ(x)(α + β + n + 1 + a − 2x) 3h(α , β; x) + (n + 1)(2x − a) − 2φ 2 (x) Ln (x).
Equating (3.2) to (3.3) and then simplifying it, we obtain the result.
(3.3)
2
4. Jacobi–Angelesco polynomials (α ,β,γ )
(x; a)}n∞1 ,n2 =0 (α , β, γ > −1, a < 0) is a multiple OPS of Angelesco system which was first investigated in [7,8]. The orthogonalizing weights are w 1 (x) = |(x − a)α xβ (1 − x)γ | on [a, 0] and w 2 (x) = (x − a)α xβ (1 − x)γ on [0, 1]. Hence, The Jacobi–Angelesco polynomial { P n1 ,n2
0
(α ,β,γ )
xk P n1 ,n2
(x; a) (x − a)α xβ (1 − x)γ dx = 0 for k = 0, 1, 2, . . . , n1 − 1
a
and
1
(α ,β,γ )
xk P n1 ,n2
(x; a)(x − a)α xβ (1 − x)γ dx = 0 for k = 0, 1, 2, . . . , n2 − 1.
0
Iterations of integration by parts with orthogonality give a Rodrigues formula (see [15])
dn dxn
(α +n,β+n,γ +n)
(x − a)α +n xβ+n (1 − x)γ +n P k,0
(x; a) (α ,β,γ )
= (−1)n (α + β + γ + k + 2n + 1)n (x − a)α xβ (1 − x)γ P n+k,n (x; a), (α ,β,γ )
where (a)n = a(a + 1)(a + 2) · · · (a + n − 1) is the Pochhammer symbol. If we let P n
(α ,β,γ )
(x) := P n,n
(x; a) with normal-
(α ,β,γ ) (x; a) = 1, the Rodrigues formula for the diagonal index case can be written by ization P 0,0
(α ,β,γ )
(−1)n (α + β + γ + 2n + 1)n (x − a)α xβ (1 − x)γ P n (α ,β,γ )
Note that P n
(n) (x) = (x − a)α +n xβ+n (1 − x)γ +n .
(x) defined as above is a polynomial of degree 2n if α + n, β + n, γ + n > −1 for all non-negative (α ,β,γ )
(x)}n∞=0 we mean the sequence of Jacobi–Angelesco polynomials of diagonal index integer n. In this section, by { P n defined by the Rodrigues formula for such α , β , and γ . (α ,β,γ )
(x)}n∞=0 hold for all such In the following, theorems for { P n proving them. For convenience, we will use the notations
φ(x) = (x − a)x(1 − x)
α , β , and γ , since we use only Rodrigues’ formula in
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D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
and
h(α , β, γ ; x) = α x(1 − x) + β(x − a)(1 − x) − γ x(x − a) throughout this section. (α ,β,γ )
Theorem 4.1. Let { P n
(x)}n∞=0 be the Jacobi–Angelesco polynomials. Then for α , β, γ > −1 and n 0, we have
a(α + β + γ + 2n + 1) (α ,β,γ ) (a) (x − a) P n(α +1,β,γ ) (x) = xP n(α ,β+1,γ ) (x) − (x); Pn (α ,β+1,γ )
xP n
(α ,β,γ +1)
(x) = (x − 1) P n
α + β + γ + 3n + 1 α + β + γ + 2n + 1 (α ,β,γ ) (x) + (x). P α + β + γ + 3n + 1 n
(n + 1)x(1 − x) α + β + γ + 2n + 2 (α −1,β,γ ) (α ,β+1,γ +1) (x) − (x); P n +1 P α + β + γ + 3n + 3 α + β + γ + 3n + 3 n (n + 1)(x − a)(1 − x) (α +1,β,γ +1) α + β + γ + 2n + 2 (α ,β−1,γ ) (α ,β,γ ) P n+1 (x) = (x) − (x); P P α + β + γ + 3n + 3 n+1 α + β + γ + 3n + 3 n α + β + γ + 2n + 2 (α ,β,γ −1) (n + 1)(x − a)x (α ,β,γ ) (α +1,β+1,γ ) P P P n+1 (x) = (x) + (x). α + β + γ + 3n + 3 n+1 α + β + γ + 3n + 3 n (α + β + γ + 2n − 1)n+2 (α −2,β−2,γ −2) (α −2,β−2,γ −2) P n +3 (x) = − h(α + n + 1, β + n + 1, γ + n + 1; x) P n+2 (x) (α + β + γ + 2n + 1)n+3 (n + 2)(α + β + γ + 2n)n+1 (α −1,β−1,γ −1) + φ(x)h (α + n + 1, β + n + 1, γ + n + 1; x) P n+1 (x) (α + β + γ + 2n + 1)n+3 (n + 2)(n + 1)(α + β + γ + 2n + 1)n 2 (α ,β,γ ) − φ (x)h (α + n + 1, β + n + 1, γ + n + 1; x) P n (x). 2(α + β + γ + 2n + 1)n+3
(b) P (α ,β,γ ) (x) = n +1
(c)
Proof. (a) follows from the Rodrigues formula and the identity
(x − a)α +1+n xβ+n (1 − x)γ +n
(n)
(n) (n) = (x − a)α +n xβ+n+1 (1 − x)γ +n − a (x − a)α +n xβ+n (1 − x)γ +n
(n)
(n) (n) = − (x − a)α +n xβ+n (1 − x)γ +n+1 + (x − a)α +n xβ+n (1 − x)γ +n .
and
(x − a)α +n xβ+1+n (1 − x)γ +n
(b) The first equation follows from
(x − a)α +n+1 xβ+n+1 (1 − x)γ +n+1
(n+1)
(n+1) = (x − a) (x − a)α +n xβ+n+1 (1 − x)γ +n+1 (n+1) = (x − a) (x − a)α +n xβ+n+1 (1 − x)γ +n+1 (n) + (n + 1) (x − a)α +n xβ+n+1 (1 − x)γ +n+1 ,
and the other equations can be proved by the same way. (c) From the Leibniz rule, we have
(x − a)α +n+1 xβ+n+1 (1 − x)γ +n+1
(n+3)
(n+2) (x − a)α +n+1 xβ+n+1 (1 − x)γ +n+1 (n+2) = h(x)(x − a)α +n xβ+n (1 − x)γ +n (n+2) = h(x) (x − a)α +n xβ+n (1 − x)γ +n (n+1) + (n + 2)h (x) (x − a)α +n xβ+n (1 − x)γ +n (n) (n + 2)(n + 1) + , h (x) (x − a)α +n xβ+n (1 − x)γ +n =
2
where h(x) = h(α + n + 1, β + n + 1, γ + n + 1; x). If α , β, γ > −1 and n 0, then the Rodrigues formula generates polynomials of degree 2n, that is the Jacobi–Angelesco polynomials, so that the recurrence relation follows. 2 (α ,β,γ )
Theorem 4.2. Let { P n (a)
(α ,β,γ )
Pn
(x) = −
(x)}n∞=0 be the Jacobi–Angelesco polynomials. Then for n 1,
(α +1,β+1,γ +1) (α + β + γ + 2n + 2)n−1 (α +1,β+1,γ +1) φ(x) P n−1 (x) + h(α + 1, β + 1, γ + 1; x) P n−1 (x) . (α + β + γ + 2n + 1)n
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
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(α ,β+1,γ +1) (α + β + γ + 2n + 1)n−1 (α ,β+1,γ +1) φ(x) P n−1 (x) + h(α + n, β + 1, γ + 1; x) P n−1 (x) ; (α + β + γ + 2n + 1)n (α +1,β,γ +1) (α + β + γ + 2n + 1)n−1 (α ,β,γ ) (α +1,β,γ +1) Pn (x) = − (x) + h(α + 1, β + n, γ + 1; x) P n−1 (x) ; φ(x) P n−1 (α + β + γ + 2n + 1)n (α +1,β+1,γ ) (α + β + γ + 2n + 1)n−1 (α ,β,γ ) (α +1,β+1,γ ) Pn (x) = − (x) + h(α + 1, β + 1, γ + n; x) P n−1 (x) . φ(x) P n−1 (α + β + γ + 2n + 1)n (α ,β,γ )
(b)
Pn
(x) = −
Proof. (a) It follows from (α ,β,γ )
(−1)n (α + β + γ + 2n + 1)n (x − a)α xβ (1 − x)γ P n (x) α +n β+n γ +n (n−1) = (x − a) x (1 − x) (α +1,β+1,γ +1) n −1 = (−1) (α + β + γ + 2n + 2)n−1 (x − a)α +1 xβ+1 (1 − x)γ +1 P n−1 (x) . (b) From (a), we have (α −1,β,γ )
Pn
(x) = −
(α ,β+1,γ +1) (α + β + γ + 2n + 1)n−1 (α ,β+1,γ +1) φ(x) P n−1 (x) + h(α , β + 1, γ + 1; x) P n−1 (x) . (α + β + γ + 2n)n (4.1)
Substituting (4.1) into the first equation of Theorem 4.1(b), we obtain the first equation of (b).
2 (α ,β,γ )
Now we are ready to find a third order differential equation for Jacobi–Angelesco polynomials { P n (α ,β,γ )
Theorem 4.3. For α , β, γ > −1 and n 0, the Jacobi–Angelesco polynomial P n
(x)}n∞=0 .
(x) satisfies a differential equation
φ 2 (x) P n (x) + 2φ(x)h(α + 1, β + 1, γ + 1; x) P n (x) + h(α , β, γ ; x)h(α + 1, β + 1, γ + 1; x) + (n − 1)φ(x) (2α + 2β + 2γ + 3n + 6)x − (α + β + n + 2) − a(β + γ + n + 2) P n (x) + n (n + 3)(α + β + γ ) + 2(n + 1)(n + 2) φ(x) + h(α , β, γ ; x) (2α + 2β + 2γ + 3n + 3)x − (α + β + n + 1) − a(β + γ + n + 1) P n (x) = 0, (α ,β,γ )
where P n (x) = P n
(4.2)
(x).
Proof. Applying Theorem 4.2(a), we have for n 0, (α −1,β−1,γ −1)
(x) = −
(α −2,β−2,γ −2)
(x) = −
P n +1
(α + β + γ + 2n + 1)n φ(x) P n (x) + h(α , β, γ ; x) P n (x) (α + β + γ + 2n)n+1
(4.3)
and
P n +2
(α + β + γ + 2n)n+1 (α + β + γ + 2n − 1)n+2 (α −1,β−1,γ −1) (α −1,β−1,γ −1) × φ(x) P n+1 (x) + h(α − 1, β − 1, γ − 1; x) P n+1 (x) .
(4.4)
Substituting Eqs. (4.3) and (4.4) into Theorem 4.1(c), we have for n 0, (α −2,β−2,γ −2)
P n +3
(α + β + γ + 2n + 1)n (α + β + γ + 2n + 1)n+3 × φ 2 (x)h(α + n + 1, β + n + 1, γ + n + 1; x) P n + A P n + B P n ,
(x) = −
(4.5)
where
A = 2φ(x)h(α , β, γ ; x)h(α + n + 1, β + n + 1, γ + n + 1; x)
+ (n + 2)φ 2 (x)h (α + n + 1, β + n + 1, γ + n + 1; x) and
B = h(α + n + 1, β + n + 1, γ + n + 1; x) φ(x)h (α , β, γ ; x) + h(α − 1, β − 1, γ − 1; x)h(α , β, γ ; x)
+ (n + 2)φ(x)h (α + n + 1, β + n + 1, γ + n + 1; x)h(α , β, γ ; x) − (n + 1)(n + 2)(α + β + γ + 3n + 3)φ 2 (x).
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On the other hand, by applying Theorem 4.2(b) iteratively, we have for n 0, (α −2,β−2,γ −2)
P n +3
(α −2,β−1,γ −1) (α + β + γ + 2n + 1)n+2 φ(x) P n+2 (x) (α + β + γ + 2n + 1)n+3 (α −2,β−1,γ −1) + h(α + n + 1, β − 1, γ − 1; x) P n+2 (x) (α −1,β−1,γ ) (α + β + γ + 2n + 1)n+1 = φ(x) φ(x) P n+1 (x) (α + β + γ + 2n + 1)n+3 (α −1,β−1,γ ) + h(α − 1, β + n + 1, γ ; x) P n+1 (x) (α −1,β−1,γ ) + h(α + n + 1, β − 1, γ − 1; x) φ(x) P n+1 (x) (α −1,β−1,γ ) + h(α − 1, β + n + 1, γ ; x) P n+1 (x) (α + β + γ + 2n + 1)n 3 φ (x) P n (x) =− (α + β + γ + 2n + 1)n+3
(x) = −
+ φ 2 (x)h(3α + n + 3, 3β + n + 3, 3γ + n + 3; x) P n (x) + C P n (x) + D P n (x) ,
(4.6)
where
C = φ(x) φ(x)h (3α , 3β + n + 2, 3γ + 2n + 3; x) + φ (x)h(3α + n + 1, 3β + n + 1, 3γ + n + 1; x)
+ h(2α + n, 2β + n, 2γ − 1; x)h(α , β, γ + n + 1; x)
+ h(α + n + 1, β − 1, γ − 1; x)h(α − 1, β + n + 1, γ ; x) and
D = h(α + n + 1, β − 1, γ − 1; x)h(α − 1, β + n + 1, γ ; x)h(α , β, γ + n + 1; x)
+ φ(x)h(2α + n + 1, 2β + n + 1, 2γ ; x)h (α , β, γ + n + 1; x) + φ(x)h (α − 1, β + n + 1, γ ; x)h(α , β, γ + n + 1; x) − 2(α + β + γ + n + 1)φ 2 (x). Equating (4.5) to (4.6), we have a differential equation
φ 3 (x) P n (x) + φ 2 (x)h(2α + 2, 2β + 2, 2γ + 2; x) P n (x) + (C − A ) P n (x) + ( D − B ) P n (x) = 0, which is (4.2) by tedious calculations.
2
It is worth to remark that the Jacobi–Laguerre polynomials and the Laguerre–Hermite polynomials can be regarded as limiting cases of Jacobi–Angelesco polynomials. More precisely, they satisfy the relations (α ,β)
Ln
(α ,β,γ )
(x) = lim γ 2n P n γ →∞
(x/γ ; a/γ )
and (β)
H n (x) = lim
α →∞
√
αn P n(α ,β,α ) (x/ α ; −1).
Hence, the results in section two and section three can be obtained by taking a limit process on the results in section four. For example, if we put x by x/γ , a by a/γ , the corresponding derivatives, and then taking γ → ∞ on the differential (α ,β)
equation (4.2), we obtain the differential equation (3.1) for L n
(x).
References [1] [2] [3] [4] [5]
A. Angelesco, Sur l’approximatión simultaneé de plusieurs integrales definies, C. R. Paris 167 (1918) 629–631. A.I. Aptekarev, Multiple orthogonal polynomials, J. Comput. Appl. Math. 99 (1998) 423–447. A.I. Aptekarev, A. Branquinho, W. Van Assche, Multiple orthogonal polynomials for classical weights, Trans. Amer. Math. Soc. 355 (2003) 3887–3914. M.G. de Bruin, Simultaneous Padé approximation and orthogonality, in: Lecture Notes in Mathematics, vol. 1171, Springer, 1985, pp. 74–83. M.G. de Bruin, Some aspects of simultaneous rational approximation, in: Numerical Analysis and Mathematical Modeling, in: Banach Center Publications, vol. 24, PWN Polish Scientific Publishers, Warsaw, 1990, pp. 51–84. [6] J. Coussement, W. Van Assche, Differential equations for multiple orthogonal polynomials with respect to classical weights, J. Phys. A: Math. Gen. 39 (2006) 3311–3318. [7] V.A. Kalyagin, On a class of polynomials defined by two orthogonality relations, Mat. Sb. 110 (1979) 609–627, translated in Math. USSR Sb. 38 (1981) 563–580. [8] V.A. Kalyagin, A. Ronveaux, On a system of classical polynomials of simultaneous orthogonality, J. Comput. Appl. Math. 67 (1996) 207–217.
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[9] D.W. Lee, Properties of multiple Hermite and multiple Laguerre polynomials by the generating function, Integral Transforms Spec. Funct. 18 (2007) 855–869. [10] K. Mahler, Perfect systems, Compositio Math. 19 (1968) 95–166. [11] P. Maroni, L’orthogonalité et les récurrences de polynômes d’ordre supérieur à deux, Ann. Fac. Sci. Toulouse 10 (1989) 105–139. [12] E.M. Nikishin, V.N. Sorokin, Rational Approximations and Orthogonality, Translations of Mathematical Monographs, vol. 92, Amer. Math. Soc., Providence, RI, 1991. [13] V.N. Sorokin, Simultaneous Padé approximants for finite and infinite intervals, Izv. Vyssh. Uchebn. Zaved. Mat. 8 (267) (1984) 45–52, translated in J. Soviet Math. 28 (1984) 56–64. [14] V.N. Sorokin, A generalization of Laguerre polynomials and convergence of simultaneous Padé approximants, Uspekhi Mat. Nauk 41 (1986) 207–208. [15] W. Van Assche, E. Coussement, Some classical multiple orthogonal polynomials, J. Comput. Appl. Math. 127 (2001) 317–347.
Contents lists available at ScienceDirect
Applied Numerical Mathematics www.elsevier.com/locate/apnum
Classical multiple orthogonal polynomials of Angelesco system D.W. Lee 1 Department of Mathematics, Teachers College, Kyungpook National University, Daegu 702-701, South Korea
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 18 November 2009 Received in revised form 28 April 2010 Accepted 5 May 2010 Available online 12 May 2010
In this paper, we analyze more carefully the Rodrigues formula of classical multiple orthogonal polynomials of Angelesco system and consider the sequence of them for the diagonal index case, and then find many properties of such a sequence of orthogonal polynomials. More precisely, we consider the diagonal index sequence { P n,n (x)}n∞=0 of classical multiple orthogonal polynomials { P n1 ,n2 (x)}n∞,n =0 of Angelesco system. For such a
Keywords: Multiple orthogonal polynomial Angelesco system Differential equation Recurrence relation
sequence we give recurrence relations, differential-difference equations, and then we finally find a third order differential equation having { P n,n (x)}n∞=0 as solutions. © 2010 IMACS. Published by Elsevier B.V. All rights reserved.
1
2
1. Introduction A sequence { P n1 ,n2 (x)}n∞,n =0 of polynomials is called a multiple orthogonal polynomial system (multiple OPS) if 1 2 (i) deg( P n1 ,n2 ) = n1 + n2 ; (ii) there exist two positive weights w 1 and w 2 such that
∞ xk P n1 ,n2 (x) w 1 (x) dx = 0
for k = 0, 1, 2, . . . , n1 − 1
xk P n1 ,n2 (x) w 2 (x) dx = 0
for k = 0, 1, 2, . . . , n2 − 1.
−∞
and
∞ −∞
In this case, w 1 and w 2 are called the orthogonalizing weights for { P n1 ,n2 (x)}n∞,n =0 . 1 2 The multiple OPS appeared in [1] to obtain simultaneous Hermite–Padé approximants. See also [4,5,10,12]. Recently it attracts a renewed interest as a natural extension of orthogonal polynomials, in particular, of classical orthogonal polynomials [2,3,11,15]. There are two kinds of systems in multiple OPS’s that are now extensively studied. They are AT system and Angelesco system.
E-mail address: [email protected]. The author thanks to Claude Brezinski for his hospitality, and also thanks to the referee for his/her valuable remarks and corrections after careful reading of the manuscript that improved this paper very much. This work was supported by Korea Research Foundation Grant funded by the Korean Government (KRF-2008-313-C00027). 1
0168-9274/$30.00 © 2010 IMACS. Published by Elsevier B.V. All rights reserved. doi:10.1016/j.apnum.2010.05.004
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
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The AT system is a class of multiple OPS’s that the orthogonalizing weights have the same support of intervals. Five families of AT system are well known whose weights consist of the same type of classical weights. They are Jacobi–Piñeiro, multiple Bessel, multiple Laguerre I, multiple Laguerre II, and multiple Hermite polynomials. Of course, they are named after their orthogonalizing weights that come from the classical weight function. Many properties for classical multiple OPS’s of AT system were found by many authors [3,6,9,15], such as recurrence relations, differential-difference equations, raising operators, lowering operators, differential equations, and so on. These properties are fundamentally induced from the Rodrigues formula obtained in [15]. On the other hand, the Angelesco system were not investigated as much extensively as AT system. The Angelesco system is a class of multiple OPS’s that the orthogonalizing weights have pairwisely disjoint support allowing the touch of their boundary points. Three families of Angelesco system are known as an extension of classical orthogonal polynomials, which we call the classical multiple OPS of Angelesco system. They are Jacobi–Angelesco, Jacobi–Laguerre, and Laguerre–Hermite polynomials, named after the orthogonalizing weights and the interval of orthogonality [7,8,13–15]. It is remarkable that the Rodrigues formula was also obtained in [15] for these classical multiple OPS’s of Angelesco system. In this paper, we analyze more carefully the Rodrigues formula (in [15]) of classical multiple OPS of Angelesco system and consider the sequence of them for the diagonal index case, and then find many properties of such a sequence of OPS’s. More precisely, we consider the diagonal index sequence { P n,n (x)}n∞=0 of classical multiple OPS { P n1 ,n2 (x)}n∞,n =0 of 1 2 Angelesco system. For such a sequence we give recurrence relations, differential-difference equations, and then we finally find a third order differential equation having { P n,n (x)}n∞=0 as solutions. 2. Laguerre–Hermite polynomials The Laguerre–Hermite polynomial { H n1 ,n2 (x)}n∞,n =0 (β > −1) is a multiple OPS of Angelesco system which was first 1 2 2 2 considered by Sorokin [14]. The orthogonalizing weights are w 1 (x) = |x|β e −x on (−∞, 0] and w 2 (x) = xβ e −x on [0, ∞). Hence, (β)
0
xk H n1 ,n2 (x)|x|β e −x dx = 0 for k = 0, 1, 2, . . . , n1 − 1 2
(β)
−∞
and
∞
xk H n1 ,n2 (x)xβ e −x dx = 0 2
(β)
for k = 0, 1, 2, . . . , n2 − 1.
0
Iterations of integration by parts with orthogonality give a Rodrigues formula
dn β+n −x2 (β+n) 2 (β) x e H k,0 (x) = (−2)n xβ e −x H n+k,n (x). dxn (β)
(β)
(β)
If we let H n (x) := H n,n (x) with normalization H 0,0 (x) = 1, the Rodrigues formula for the diagonal index case can be written by
2 2 (n) (β) (−2)n xβ e −x H n (x) = xβ+n e −x . (β)
Note that H n (x) defined as above is a polynomial of degree 2n even when β −1, in which case the orthogonalities
of { H n (x)}n∞=0 are broken. In this section, by
{ H n (x)}n∞=0 we mean the sequence of Laguerre–Hermite polynomials of diagonal index defined by the Rodrigues formula with real β . (β) In the following, theorems for { H n (x)}n∞=0 hold for all real β since we use only Rodrigues’ formula in proving them. (β)
(β)
Theorem 2.1. The Laguerre–Hermite polynomials { H n (x)}n∞=0 satisfy the following recurrence relations. (β)
(β+1) (x) + n2 H n−1 (x), n 1. (β+ 2 ) (β) (b) 2H n (x) = 2x2 H n−1 (x) − (β + n) H n−1 (x), (β)
(β+1)
(a) H n (x) = H n (β)
n 1.
Proof. From the Rodrigues formula and the identity
xβ+1+n e −x
2
(n)
2 (n) 2 (n−1) = x xβ+n e −x + n xβ+n e −x ,
we have
(−2)n xβ+1 e −x H n 2
(β+1)
(x) = x(−2)n xβ e −x H n (x) + n(−2)n−1 xβ+1 e −x H n−1 (x) 2
(β)
2
(β+1)
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D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
which implies (a). On the other hand, from the identity
xβ+n e −x
2
(n)
2 (n−1) 2 (n−1) = (β + n) xβ+n−1 e −x − 2 xβ+n+1 e −x
and the Rodrigues formula, (b) follows.
2
Theorem 2.2. Let { H n (x)}n∞=0 be the Laguerre–Hermite polynomials. Then for n 0, (β)
(β)
(β+2)
(β)
H n (x) = 2xH n (x) − 2xH n
(x). (β) (β) (β−1) (b) x H n (x) = 2x2 − β H n (x) − 2H n+1 (x). (β) (β) (β) (c) x H n (x) = 2x2 − β − n − 1 H n (x) − 2H n+1 (x). (a)
(d)
(β)
(β)
(β)
H n+2 (x) = 2(n + 2)xH n+1 (x) − (n + 2) H n+1 (x) +
(n + 1)(n + 2) 2
(β)
xH n (x) −
(n + 1)(n + 2) 4
(β)
H n (x) .
Proof. The relation (a) follows from
2 (β) 2 2 (β) (β) (−2)n xβ e −x H n (x) = (−2)n xβ e −x H n (x) − (−2)n β − 2x2 xβ−1 e −x H n (x) β 2 (n+1) 2 (n) 2 (β) = xβ+n e −x − xβ+n e −x − (−2)n+1 xβ+1 e −x H n (x)
x 1 2 (n) 2 (β) 2 β+n−1 −x2 (n) − (−2)n+1 xβ+1 e −x H n (x) = x β + n − 2x x e − β xβ+n e −x x 1 2 (n) 2 (n−1) 2 (n) β + n − 2x2 xβ+n e −x = − n β + n − 2x2 xβ+n−1 e −x − β xβ+n e −x x
− (−2)n+1 xβ+1 e −x H n (x) 2 2 (n) 2 (β) − (−2)n+1 xβ+1 e −x H n (x) = − xβ+n+2 e −x 2
(β)
x
= (−2)n+1 xβ+1 e −x H n 2
(β+2)
(x) − (−2)n+1 xβ+1 e −x H n (x). 2
(β)
From the Rodrigues formula, we have
2 (β) 2 2 (n+1) (β) (−2)n xβ e −x H n (x) + (−2)n β − 2x2 xβ−1 e −x H n (x) = xβ+n e −x = (−2)n+1 xβ−1 e −x H n+1 (x), 2
(β−1)
which is (b). We obtain (c) combining (b) and Theorem 2.1(a). Combining (a) with Theorem 2.1(a), we have (β+1)
(β)
(β+1)
2xH n+2 (x) = 2xH n+2 (x) + (n + 2)xH n+1 (x)
(n + 2)(n + 1)x (β+2) (β+2) (β+2) = 2xH n+2 (x) + 2(n + 2)xH n+1 (x) + (x) Hn 2 (β) (β) (β) (β) = 2xH n+2 (x) − H n+2 (x) + (n + 2) 2xH n+1 (x) − H n+1 (x) (β) (n + 2)(n + 1) (β) 2xH n (x) − H n (x) , + 4
from which (d) follows.
2
From Theorem 2.2, we can find a third order differential equation for { H n (x)}n∞=0 . More precisely, we have: (β)
Theorem 2.3. The Laguerre–Hermite polynomial { H n (x)}n∞=0 satisfies a differential equation (β)
x2 H n (x) + 2x β + 1 − 2x2 H n (x) +
+ 2nx n + 3 + 2β − 4x2 H n (x) = 0, (β)
where H n (x) = H n (x).
2x2 − β 2x2 − β − 1 + 4(n − 1)x2 H n (x)
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Proof. By (c) and (d) in Theorem 2.2, we have for n 0,
4H n +2 (x) = 8(n + 2)xH n+1 (x) − 4(n + 2) H n +1 (x) + 2(n + 1)(n + 2)xH n (x) − (n + 1)(n + 2) H n (x)
= 2(n + 2)xH n (x) − (n + 2) 8x2 − 2β − n − 3 H n (x) + 2(n + 2)x 4x2 − 2β − n − 5 H n (x).
(2.1)
On the other hand, applying (c) in Theorem 2.2 iteratively, we have for n 0,
4H n +2 (x) = 8xH n+1 (x) + 2 2x2 − β − n − 3 H n +1 (x) − 2xH n+1 (x)
= x2 H n (x) − 2x 2x2 − β − n − 3 H n (x) + 2x2 − β − n − 3 2x2 − β − n − 2 − 12x2 H n (x) + 4 4x2 − 2β − 2n − 5 xH n (x).
(2.2)
2
Equating (2.1) to (2.2) and then simplifying it, we obtain the result. 3. Jacobi–Laguerre polynomials (α ,β)
The Jacobi–Laguerre polynomial { L n1 ,n2 (x; a)}n∞,n =0 (α , β > −1, a < 0) is a multiple OPS of Angelesco system which was 1 2 first considered by Sorokin [13]. The orthogonalizing weights are w 1 (x) = |(x − a)α xβ |e −x on [a, 0] and w 2 (x) = (x − a)α xβ e −x on [0, ∞). Hence,
0
xk L n1 ,n2 (x; a) (x − a)α xβ e −x dx = 0 for k = 0, 1, 2, . . . , n1 − 1 (α ,β)
a
and
∞
xk L n1 ,n2 (x; a)(x − a)α xβ e −x dx = 0 for k = 0, 1, 2, . . . , n2 − 1. (α ,β)
0
Iterations of integration by parts with orthogonality give a Rodrigues formula (see [15])
dn dxn
(α +n,β+n)
(x − a)α +n xβ+n e −x Lk,0
(α ,β)
(α ,β) (x; a) = (−1)n (x − a)α xβ e −x Ln+k,n (x; a).
(α ,β)
(α ,β)
If we let L n (x) := Ln,n (x; a) with normalization L 0,0 (x; a) = 1, the Rodrigues formula for the diagonal index case can be written by
(−1)n (x − a)α xβ e −x Ln
(α ,β)
(α ,β)
Note that L n
(n) (x) = (x − a)α +n xβ+n e −x .
(x) defined as above is a polynomial of degree 2n even when α −1 or β −1, in which case the (α ,β)
(α ,β)
orthogonalities of { L n (x)}n∞=0 are broken. In this section, by { Ln (x)}n∞=0 we mean the sequence of Jacobi–Laguerre polynomials of diagonal index defined by the Rodrigues formula with real α and β . (α ,β) In the following, theorems for { L n (x)}n∞=0 hold for all real α and β since we use only Rodrigues’ formula in proving them. (α ,β)
Theorem 3.1. The Jacobi–Laguerre polynomials { L n
(x)}n∞=0 satisfy the following recurrence relations.
(α ,β+1) (α +1,β) (α ,β) (x) = (x − a) Ln (x) + aLn (x), (a) xL n
(b)
(α ,β)
(x) =
(α ,β)
(x) = x(x − a) Ln−1
Ln (c)
(α +1,β+1) Ln (x) + nxLn−1 (x), n 1; (α ,β+1) (α +1,β+1) Ln (x) + n(x − a) Ln−1 (x), n 1.
(x) =
Ln
Ln
n 0.
(α +1,β)
(α ,β)
(α +1,β+1)
(α ,β) (x) − (α + n)x + (β + n)(x − a) Ln−1 (x) (α +1,β+1)
(x), n 2. + (n − 1)(α + β + 2n)x(x − a) Ln−2 (α ,β) (α ,β) (x) = − (α + n)x + (β + n)(x − a) − x(x − a) Ln−1 (x) (d) L n (α +1,β+1)
+ (n − 1)(α + β + 2n + a − 2x)x(x − a) Ln−2 (α +2,β+2)
+ (n − 1)(n − 2)x2 (x − a)2 Ln−3
(x),
n 3.
(x)
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(e)
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
(α ,β)
Ln
(α +1,β+1)
(x) = Ln
(α +1,β+1)
(x) + n(2x − a) Ln−1
(α +2,β+2)
(x) − n(n − 1)x(x − a) Ln−2
(x),
n 2.
Proof. (a) follows from the Rodrigues formula and the identity
(x − a)α +n xβ+1+n e −x
(n)
(n) (n) = (x − a)α +n+1 xβ+n e −x + a (x − a)α +n xβ+n e −x .
The first equation of (b) can be obtained from the identity
(x − a)α +n+1 xβ+n e −x
(n)
(n) (n−1) = (x − a) (x − a)α +n xβ+n e −x + n (x − a)α +n xβ+n e −x
and the second equation can be obtained by the same way. (c) It follows from
(x − a)α +n xβ+n e −x
(n)
(n−1) (x − a)α +n xβ+n e −x (n−1) = (α + n)x + (β + n)(x − a) (x − a)α +n−1 xβ+n−1 e −x (n−1) (n−2) − (x − a)α +n xβ+n e −x + (n − 1)(α + β + 2n) (x − a)α +n−1 xβ+n−1 e −x . =
(d) By Leibniz rule, we have
(x − a)α +n xβ+n e −x
(n)
(n−1) = h(α + n, β + n; x) (x − a)α +n−1 xβ+n−1 e −x (n−2) + (n − 1)(α + β + 2n + a − 2x) (x − a)α +n−1 xβ+n−1 e −x (n−3) − (n − 1)(n − 2) (x − a)α +n−1 xβ+n−1 e −x ,
where h(α , β; x) = α x + β(x − a) − x(x − a), from which (d) follows by the Rodrigues formula. (e) Comparing (c) with (d) on both sides we have (α +1,β+1)
x(x − a) L n−1
(α +1,β+1)
(α ,β)
(x) = x(x − a) Ln−1 (x) + (n − 1)(a − 2x)x(x − a) Ln−2
(x)
(α +2,β+2) + (n − 1)(n − 2)x2 (x − a)2 Ln−3 (x),
2
which implies (e).
(α ,β)
Theorem 3.2. Let { L n
(x)}n∞=0 be the Jacobi–Laguerre polynomials. Then for n 1,
(α +1,β+1) (α +1,β+1) (x) = −x(x − a) Ln−1 (x) − (α + 1)x + (β + 1)(x − a) − x(x − a) Ln−1 (x). (α ,β+1) (α ,β+1) (α ,β) (x) = −x(x − a) Ln−1 (x) − (α + n)x + (β + 1)(x − a) − x(x − a) Ln−1 (x); (b) L n (α +1,β) (α +1,β) (α ,β) Ln (x) = −x(x − a) Ln−1 (x) − (α + 1)x + (β + n)(x − a) − x(x − a) Ln−1 (x). (a)
(α ,β)
Ln
Proof. From the Rodrigues formula and the identity
(x − a)α +n xβ+n e −x
(n)
=
(n−1) (α +1,β+1) (x − a)α +n xβ+n e −x = (−1)n−1 (x − a)α +1 xβ+1 e −x Ln−1 (x) ,
the relation (a) follows. Combining (a) and Theorem 3.1(b), we have (α ,β)
Ln
(α ,β+1)
(x) + nxLn−1
from which (b) follows.
(α −1,β)
(x) = Ln
(α ,β+1) (α ,β+1) (x) = −x(x − a) Ln−1 (x) − α x + (β + 1)(x − a) − x(x − a) Ln−1 (x),
2 (α ,β)
Now we are ready to find a third order differential equation for Jacobi–Laguerre polynomials { L n (α ,β)
Theorem 3.3. The Jacobi–Laguerre polynomial { L n
(x)}n∞=0 .
(x)}n∞=0 satisfies a differential equation
x2 (x − a)2 L n (x) + 2x(x − a)h(α + 1, β + 1; x) L n (x)
+ h(α , β; x)h(α + 1, β + 1; x) − (n − 1)(α + β + n + 2 + a − 2x)x(x − a) Ln (x) + n (n + 3)x(x − a) − (α + β + n + 1 + a − 2x)h(α , β; x) Ln (x) = 0, (α ,β)
where L n (x) = L n
(x) and h(α , β; x) = α x + β(x − a) − x(x − a).
(3.1)
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
1347
Proof. Applying Theorem 3.2 (a) into (d) in Theorem 3.1 for n + 1 and n + 2, iteratively, we have for n 0, (α −2,β−2)
L n +3
(x) = −φ 2 (x)h(α + n + 1, β + n + 1; x) Ln (x) − φ(x) 2h(α , β; x)h(α + n + 1, β + n + 1; x) + (n + 2)(α + β + 2n + 2 + a − 2x)φ(x) Ln (x) − φ(x)(α + β + a − 2x)h(α + n + 1, β + n + 1; x) + (n + 2)(α + β + 2n + 2 + a − 2x)φ(x)h(α , β; x)
+ h(α , β; x)h(α − 1, β − 1; x)h(α + n + 1, β + n + 1; x) − (n + 1)(n + 2)φ 2 (x) Ln (x),
(3.2)
where φ(x) = x(x − a). On the other hand, by Theorem 3.2 (a) and (b), we have for n 0, (α −2,β−2)
L n +3
(α −1,β−1) (α −1,β−1) (x) = −φ(x) Ln+2 (x) − h(α − 1, β − 1; x) Ln+2 (x) (α −1,β) (α −1,β) 2 = φ (x) Ln+1 (x) + φ(x) 2h(α , β; x) + (n + 1)x Ln+1 (x) (α −1,β) + φ(x)[α + β + n + 1 + a − 2x] + h(α − 1, β − 1; x)h(α + n + 1, β; x) Ln+1 (x) = −φ 3 (x) Ln (x) − φ 2 (x) 3h(α , β; x) + (n + 3)(2x − a) Ln (x) − φ(x) [3α + 3β + 3n + 5 + 3a − 6x]φ(x) + h(α − 1, β − 1; x)h(α + n + 1, β; x) + h(α + 1, β + n + 2; x) 2h(α , β; x) + (n + 1)x Ln (x) − h(α − 1, β − 1; x)h(α , β + n + 1; x)h(α + n + 1, β; x) + φ(x)(α + β + n + 1 + a − 2x) 3h(α , β; x) + (n + 1)(2x − a) − 2φ 2 (x) Ln (x).
Equating (3.2) to (3.3) and then simplifying it, we obtain the result.
(3.3)
2
4. Jacobi–Angelesco polynomials (α ,β,γ )
(x; a)}n∞1 ,n2 =0 (α , β, γ > −1, a < 0) is a multiple OPS of Angelesco system which was first investigated in [7,8]. The orthogonalizing weights are w 1 (x) = |(x − a)α xβ (1 − x)γ | on [a, 0] and w 2 (x) = (x − a)α xβ (1 − x)γ on [0, 1]. Hence, The Jacobi–Angelesco polynomial { P n1 ,n2
0
(α ,β,γ )
xk P n1 ,n2
(x; a) (x − a)α xβ (1 − x)γ dx = 0 for k = 0, 1, 2, . . . , n1 − 1
a
and
1
(α ,β,γ )
xk P n1 ,n2
(x; a)(x − a)α xβ (1 − x)γ dx = 0 for k = 0, 1, 2, . . . , n2 − 1.
0
Iterations of integration by parts with orthogonality give a Rodrigues formula (see [15])
dn dxn
(α +n,β+n,γ +n)
(x − a)α +n xβ+n (1 − x)γ +n P k,0
(x; a) (α ,β,γ )
= (−1)n (α + β + γ + k + 2n + 1)n (x − a)α xβ (1 − x)γ P n+k,n (x; a), (α ,β,γ )
where (a)n = a(a + 1)(a + 2) · · · (a + n − 1) is the Pochhammer symbol. If we let P n
(α ,β,γ )
(x) := P n,n
(x; a) with normal-
(α ,β,γ ) (x; a) = 1, the Rodrigues formula for the diagonal index case can be written by ization P 0,0
(α ,β,γ )
(−1)n (α + β + γ + 2n + 1)n (x − a)α xβ (1 − x)γ P n (α ,β,γ )
Note that P n
(n) (x) = (x − a)α +n xβ+n (1 − x)γ +n .
(x) defined as above is a polynomial of degree 2n if α + n, β + n, γ + n > −1 for all non-negative (α ,β,γ )
(x)}n∞=0 we mean the sequence of Jacobi–Angelesco polynomials of diagonal index integer n. In this section, by { P n defined by the Rodrigues formula for such α , β , and γ . (α ,β,γ )
(x)}n∞=0 hold for all such In the following, theorems for { P n proving them. For convenience, we will use the notations
φ(x) = (x − a)x(1 − x)
α , β , and γ , since we use only Rodrigues’ formula in
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D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
and
h(α , β, γ ; x) = α x(1 − x) + β(x − a)(1 − x) − γ x(x − a) throughout this section. (α ,β,γ )
Theorem 4.1. Let { P n
(x)}n∞=0 be the Jacobi–Angelesco polynomials. Then for α , β, γ > −1 and n 0, we have
a(α + β + γ + 2n + 1) (α ,β,γ ) (a) (x − a) P n(α +1,β,γ ) (x) = xP n(α ,β+1,γ ) (x) − (x); Pn (α ,β+1,γ )
xP n
(α ,β,γ +1)
(x) = (x − 1) P n
α + β + γ + 3n + 1 α + β + γ + 2n + 1 (α ,β,γ ) (x) + (x). P α + β + γ + 3n + 1 n
(n + 1)x(1 − x) α + β + γ + 2n + 2 (α −1,β,γ ) (α ,β+1,γ +1) (x) − (x); P n +1 P α + β + γ + 3n + 3 α + β + γ + 3n + 3 n (n + 1)(x − a)(1 − x) (α +1,β,γ +1) α + β + γ + 2n + 2 (α ,β−1,γ ) (α ,β,γ ) P n+1 (x) = (x) − (x); P P α + β + γ + 3n + 3 n+1 α + β + γ + 3n + 3 n α + β + γ + 2n + 2 (α ,β,γ −1) (n + 1)(x − a)x (α ,β,γ ) (α +1,β+1,γ ) P P P n+1 (x) = (x) + (x). α + β + γ + 3n + 3 n+1 α + β + γ + 3n + 3 n (α + β + γ + 2n − 1)n+2 (α −2,β−2,γ −2) (α −2,β−2,γ −2) P n +3 (x) = − h(α + n + 1, β + n + 1, γ + n + 1; x) P n+2 (x) (α + β + γ + 2n + 1)n+3 (n + 2)(α + β + γ + 2n)n+1 (α −1,β−1,γ −1) + φ(x)h (α + n + 1, β + n + 1, γ + n + 1; x) P n+1 (x) (α + β + γ + 2n + 1)n+3 (n + 2)(n + 1)(α + β + γ + 2n + 1)n 2 (α ,β,γ ) − φ (x)h (α + n + 1, β + n + 1, γ + n + 1; x) P n (x). 2(α + β + γ + 2n + 1)n+3
(b) P (α ,β,γ ) (x) = n +1
(c)
Proof. (a) follows from the Rodrigues formula and the identity
(x − a)α +1+n xβ+n (1 − x)γ +n
(n)
(n) (n) = (x − a)α +n xβ+n+1 (1 − x)γ +n − a (x − a)α +n xβ+n (1 − x)γ +n
(n)
(n) (n) = − (x − a)α +n xβ+n (1 − x)γ +n+1 + (x − a)α +n xβ+n (1 − x)γ +n .
and
(x − a)α +n xβ+1+n (1 − x)γ +n
(b) The first equation follows from
(x − a)α +n+1 xβ+n+1 (1 − x)γ +n+1
(n+1)
(n+1) = (x − a) (x − a)α +n xβ+n+1 (1 − x)γ +n+1 (n+1) = (x − a) (x − a)α +n xβ+n+1 (1 − x)γ +n+1 (n) + (n + 1) (x − a)α +n xβ+n+1 (1 − x)γ +n+1 ,
and the other equations can be proved by the same way. (c) From the Leibniz rule, we have
(x − a)α +n+1 xβ+n+1 (1 − x)γ +n+1
(n+3)
(n+2) (x − a)α +n+1 xβ+n+1 (1 − x)γ +n+1 (n+2) = h(x)(x − a)α +n xβ+n (1 − x)γ +n (n+2) = h(x) (x − a)α +n xβ+n (1 − x)γ +n (n+1) + (n + 2)h (x) (x − a)α +n xβ+n (1 − x)γ +n (n) (n + 2)(n + 1) + , h (x) (x − a)α +n xβ+n (1 − x)γ +n =
2
where h(x) = h(α + n + 1, β + n + 1, γ + n + 1; x). If α , β, γ > −1 and n 0, then the Rodrigues formula generates polynomials of degree 2n, that is the Jacobi–Angelesco polynomials, so that the recurrence relation follows. 2 (α ,β,γ )
Theorem 4.2. Let { P n (a)
(α ,β,γ )
Pn
(x) = −
(x)}n∞=0 be the Jacobi–Angelesco polynomials. Then for n 1,
(α +1,β+1,γ +1) (α + β + γ + 2n + 2)n−1 (α +1,β+1,γ +1) φ(x) P n−1 (x) + h(α + 1, β + 1, γ + 1; x) P n−1 (x) . (α + β + γ + 2n + 1)n
D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
1349
(α ,β+1,γ +1) (α + β + γ + 2n + 1)n−1 (α ,β+1,γ +1) φ(x) P n−1 (x) + h(α + n, β + 1, γ + 1; x) P n−1 (x) ; (α + β + γ + 2n + 1)n (α +1,β,γ +1) (α + β + γ + 2n + 1)n−1 (α ,β,γ ) (α +1,β,γ +1) Pn (x) = − (x) + h(α + 1, β + n, γ + 1; x) P n−1 (x) ; φ(x) P n−1 (α + β + γ + 2n + 1)n (α +1,β+1,γ ) (α + β + γ + 2n + 1)n−1 (α ,β,γ ) (α +1,β+1,γ ) Pn (x) = − (x) + h(α + 1, β + 1, γ + n; x) P n−1 (x) . φ(x) P n−1 (α + β + γ + 2n + 1)n (α ,β,γ )
(b)
Pn
(x) = −
Proof. (a) It follows from (α ,β,γ )
(−1)n (α + β + γ + 2n + 1)n (x − a)α xβ (1 − x)γ P n (x) α +n β+n γ +n (n−1) = (x − a) x (1 − x) (α +1,β+1,γ +1) n −1 = (−1) (α + β + γ + 2n + 2)n−1 (x − a)α +1 xβ+1 (1 − x)γ +1 P n−1 (x) . (b) From (a), we have (α −1,β,γ )
Pn
(x) = −
(α ,β+1,γ +1) (α + β + γ + 2n + 1)n−1 (α ,β+1,γ +1) φ(x) P n−1 (x) + h(α , β + 1, γ + 1; x) P n−1 (x) . (α + β + γ + 2n)n (4.1)
Substituting (4.1) into the first equation of Theorem 4.1(b), we obtain the first equation of (b).
2 (α ,β,γ )
Now we are ready to find a third order differential equation for Jacobi–Angelesco polynomials { P n (α ,β,γ )
Theorem 4.3. For α , β, γ > −1 and n 0, the Jacobi–Angelesco polynomial P n
(x)}n∞=0 .
(x) satisfies a differential equation
φ 2 (x) P n (x) + 2φ(x)h(α + 1, β + 1, γ + 1; x) P n (x) + h(α , β, γ ; x)h(α + 1, β + 1, γ + 1; x) + (n − 1)φ(x) (2α + 2β + 2γ + 3n + 6)x − (α + β + n + 2) − a(β + γ + n + 2) P n (x) + n (n + 3)(α + β + γ ) + 2(n + 1)(n + 2) φ(x) + h(α , β, γ ; x) (2α + 2β + 2γ + 3n + 3)x − (α + β + n + 1) − a(β + γ + n + 1) P n (x) = 0, (α ,β,γ )
where P n (x) = P n
(4.2)
(x).
Proof. Applying Theorem 4.2(a), we have for n 0, (α −1,β−1,γ −1)
(x) = −
(α −2,β−2,γ −2)
(x) = −
P n +1
(α + β + γ + 2n + 1)n φ(x) P n (x) + h(α , β, γ ; x) P n (x) (α + β + γ + 2n)n+1
(4.3)
and
P n +2
(α + β + γ + 2n)n+1 (α + β + γ + 2n − 1)n+2 (α −1,β−1,γ −1) (α −1,β−1,γ −1) × φ(x) P n+1 (x) + h(α − 1, β − 1, γ − 1; x) P n+1 (x) .
(4.4)
Substituting Eqs. (4.3) and (4.4) into Theorem 4.1(c), we have for n 0, (α −2,β−2,γ −2)
P n +3
(α + β + γ + 2n + 1)n (α + β + γ + 2n + 1)n+3 × φ 2 (x)h(α + n + 1, β + n + 1, γ + n + 1; x) P n + A P n + B P n ,
(x) = −
(4.5)
where
A = 2φ(x)h(α , β, γ ; x)h(α + n + 1, β + n + 1, γ + n + 1; x)
+ (n + 2)φ 2 (x)h (α + n + 1, β + n + 1, γ + n + 1; x) and
B = h(α + n + 1, β + n + 1, γ + n + 1; x) φ(x)h (α , β, γ ; x) + h(α − 1, β − 1, γ − 1; x)h(α , β, γ ; x)
+ (n + 2)φ(x)h (α + n + 1, β + n + 1, γ + n + 1; x)h(α , β, γ ; x) − (n + 1)(n + 2)(α + β + γ + 3n + 3)φ 2 (x).
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D.W. Lee / Applied Numerical Mathematics 60 (2010) 1342–1351
On the other hand, by applying Theorem 4.2(b) iteratively, we have for n 0, (α −2,β−2,γ −2)
P n +3
(α −2,β−1,γ −1) (α + β + γ + 2n + 1)n+2 φ(x) P n+2 (x) (α + β + γ + 2n + 1)n+3 (α −2,β−1,γ −1) + h(α + n + 1, β − 1, γ − 1; x) P n+2 (x) (α −1,β−1,γ ) (α + β + γ + 2n + 1)n+1 = φ(x) φ(x) P n+1 (x) (α + β + γ + 2n + 1)n+3 (α −1,β−1,γ ) + h(α − 1, β + n + 1, γ ; x) P n+1 (x) (α −1,β−1,γ ) + h(α + n + 1, β − 1, γ − 1; x) φ(x) P n+1 (x) (α −1,β−1,γ ) + h(α − 1, β + n + 1, γ ; x) P n+1 (x) (α + β + γ + 2n + 1)n 3 φ (x) P n (x) =− (α + β + γ + 2n + 1)n+3
(x) = −
+ φ 2 (x)h(3α + n + 3, 3β + n + 3, 3γ + n + 3; x) P n (x) + C P n (x) + D P n (x) ,
(4.6)
where
C = φ(x) φ(x)h (3α , 3β + n + 2, 3γ + 2n + 3; x) + φ (x)h(3α + n + 1, 3β + n + 1, 3γ + n + 1; x)
+ h(2α + n, 2β + n, 2γ − 1; x)h(α , β, γ + n + 1; x)
+ h(α + n + 1, β − 1, γ − 1; x)h(α − 1, β + n + 1, γ ; x) and
D = h(α + n + 1, β − 1, γ − 1; x)h(α − 1, β + n + 1, γ ; x)h(α , β, γ + n + 1; x)
+ φ(x)h(2α + n + 1, 2β + n + 1, 2γ ; x)h (α , β, γ + n + 1; x) + φ(x)h (α − 1, β + n + 1, γ ; x)h(α , β, γ + n + 1; x) − 2(α + β + γ + n + 1)φ 2 (x). Equating (4.5) to (4.6), we have a differential equation
φ 3 (x) P n (x) + φ 2 (x)h(2α + 2, 2β + 2, 2γ + 2; x) P n (x) + (C − A ) P n (x) + ( D − B ) P n (x) = 0, which is (4.2) by tedious calculations.
2
It is worth to remark that the Jacobi–Laguerre polynomials and the Laguerre–Hermite polynomials can be regarded as limiting cases of Jacobi–Angelesco polynomials. More precisely, they satisfy the relations (α ,β)
Ln
(α ,β,γ )
(x) = lim γ 2n P n γ →∞
(x/γ ; a/γ )
and (β)
H n (x) = lim
α →∞
√
αn P n(α ,β,α ) (x/ α ; −1).
Hence, the results in section two and section three can be obtained by taking a limit process on the results in section four. For example, if we put x by x/γ , a by a/γ , the corresponding derivatives, and then taking γ → ∞ on the differential (α ,β)
equation (4.2), we obtain the differential equation (3.1) for L n
(x).
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