General Mathematics Vol. 14, No. 4 (2006), 47–50
The properties of Laguerre polynomials
1
Ioan Ţincu
In memoriam of Associate Professor Ph. D. Luciana Lupaş
Abstract
In this paper we prove a property of the Laguerre polynomials
Lαn using the interpolation polynomial of Hermite.
2000 Mathematics Subject Classification: 33C45
Γ(α + 1) x −α −x n+α (n)
e x (e x ) be the
Γ(n + α + 1)
(α)
polynomials of degree n normalized by Ln (0) = 1.
(α)
In the following we shall use the notation Ln (x) = Ln (x). The following
formulas are known:
Let α > −1, x ≥ 0 and Lαn (x) =
xy”(x) + (1 + α − x)y ′ (x) + ny(x) = 0,
(1)
y(x) = Ln (x) ,
(2)
(n + α + 1)Ln+1 (x) + (x − α − 2n − 1)Ln (x) + nLn−1 (x) = 0 ,
d
x Ln (x) = nLn (x) − nLn−1 (x), (∀) α > −1, (∀) x ≥ 0 .
(3)
dx
Theorem 1.Let ω(x) = λ(x − x1 )(x − x2 ) . . . (x − xn ) by ω(0) = 1 and
xi 6= xj for i 6= j.
If
n
X
1
1
+ (α + 1 − x)
+n=0 ,
(x
−
x
)(x
−
x
)
x
−
x
i
j
k
1≤i<j≤n
k=1
X
2x
is verified, then
ω(x) = Ln (x) .
1
Received ...., 2006
Accepted for publication (in revised form) ....., 2006
47
α > −1
48
Ioan Ţincu
Proof. We consider ∆2n (x) = L2n (x) − Ln+1 (x)Ln−1 (x) and observe that
∆2n (0) = 0. According to Hermite interpolation formula
∆2n (x) = H2n (x1 x1 , x2 x2 , . . . , xn xn , c; ∆2n (x)) =
·
¸2
n
X
Ln (x)
ϕk (x)
=
∆2n (c) + (x − c)
Bk (∆2n ; x)
Ln (c)
xk − c
k=1
where x1 , x2 , . . . , xn are the roots of Ln (x) and
¸2
Ln (x)
,
ϕk (x) =
(x − xk )L′n (xk )
¸
·
1
L′′n (xk )
′
∆2n (xk ) −
∆2n (xk ) .
Bk (∆2n ; x) = ∆2n (xk )+(x−xk ) ∆2n (xk ) − ′
Ln (xk )
xk − c
·
For c = 0, we obtain
∆2n (0) = 0 ,
¸2
n ·
X
Ln (x)
1
2
∆2n (x) = Ln (x)−Ln+1 (x)Ln−1 (x) = x
· ·Bk (∆2n ; x)−
′
(x − xk )Ln (xk ) xk
k=1
¸2
n ·
X
Bk (∆2n ; x)
l
Ln+1 (x) Ln−1 (x)
.
·
=x
·
−1 −
′
Ln (x)
Ln (x)
(x − xk )Ln (xk )
xk
k=1
Further, we investigate Bk (∆2n ; x). Observe that
∆2n (xk ) = −Ln+1 (xk )Ln−1 (xk ) .
From (2) we have
(4)
Ln+1 (xk ) = −
(5)
∆2n (xk ) = −
n
Ln−1 (xk )
n+α+1
n
L2 (xk )
n + α + 1 n−1
∆′2n (xk ) = −L′n+1 (xk )Ln−1 (xk ) − Ln+1 (xk )L′n−1 (xk ) .
Using (2) and (3) one finds
L′n+1 (xk ) =
n+1
Ln+1 (xk ) ,
xk
L′n−1 (xk ) =
xk − α − n
Ln−1 (xk ) .
xk
49
The properties of Laguerre polynomials
Therefore
(6)
n
xk − α + 1 2
·
Ln−1 (xk ) .
n+α+1
xk
From (1), (5) and (6) we obtain
∆′2n (xk ) =
xk − α + 1
∆′2n (xk )
=
∆2n xk
xk
(7)
and
1 + α − xk
L′′n (xk )
.
=−
′
Ln (xk )
xk
(8)
By means (7), (8) we have
¸¾
½
· ′
1
∆2n (xk ) L′′n (xk )
−
−
,
Bk (∆2n ; x) = ∆2n (xk ) 1 + (x − xk )
∆2n (xk ) L′n (xk ) xk
Bk (∆2n ; x) =
(9)
x
∆2n (xk ) .
xk
Therefore
∆2n (x) =
L2n (x)
− Ln+1 (x)Ln−1 (x) = x
n
X
k=1
1−
L2n (x)
x
2 · 2 ∆2n (xk ) ,
′
[(x − xk )Ln (xk )] xk
n
X
Ln+1 (x) Ln−1 (x)
x
1
· 2 ∆2n (xk )
·
=x
2
Ln (x)
Ln (x)
[(x − xk )L′n (xk )] xk
k=1
From (3), we have
(10)
(11)
(12)
x L′ (x)
Ln−1 (x)
=1− · n
Ln (x)
n Ln (x)
¸
· ′
Ln+1 (x)
Ln (x)
x
=1+
−1 .
Ln (x)
n + α + 1 Ln (x)
xk
Ln−1 (xk )
=−
′
Ln (xk )
n
· ′
¾
¸¾ ½
½
Ln (x)
x L′n (x)
x
=
−1 · 1− ·
1− 1+
n + α + 1 Ln (x)
n Ln (x)
·
¸2
n
X
n
Ln−1 (xk )
1
2
=
x
·
,
nα + 1 k=1 (x − xk )2
L′n (xk )
50
Ioan Ţincu
·
¸2
n
xX
1
α + 1 − x L′n (x) x L′n (x)
+1=
·
+
,
n
L(x)
n Ln (x)
n k=1 (x − xk )2
à n
!2
n
n
α+1−xX 1
1
xX
x X 1
+
+1=
,
2
n k=1 x − xk
n
x
−
x
n
(x
−
x
k
k)
k=1
k=1
n
X
1
1
+ (α + 1 − x)
+n=0 .
2x
(x
−
x
)(x
−
x
)
x
−
x
i
j
k
1≤i<j≤n
k=1
X
In conclusion ω(x) = Ln (x).
References
[1] G. Gasper, On the estension of Turan, s inequality to Jacobi polynomials,
Duke Math. J. 38(1971), 415-428.
[2] A. Lupaş, On the inequality of P. Turan for ultraspherical polynomials,
Seminar of numerical and statistical calculus, University of Cluj-Napoca,
Faculty of Mathematics, Research Seminaries, Preprint nr.4, 1985, 8287.
[3] G. Szegö, Orthogonal Polynomials , Amer. Math. Soc. Providence, R.I.
1985.
[4] I. Ţincu, A proof of Turan, s inequality for Laguerre polynomials, The
5th Romanian - German Seminar on Approximation Theory and its Applications, RoGer 2002, Sibiu.
”Lucian Blaga” University of Sibiu
Faculty of Sciences
Department of Mathematics
Str. Dr. I. Raţiu, no. 5–7
550012 Sibiu - Romania
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