General Mathematics Vol. 18, No. 1 (2010), 131–137
Some properties from Laguerre
polynomials 1
Ioan Ţincu
Abstract
In this paper, we demonstrate some properties of Laguerre polynomials. Using the Beta function we prove a projection inequality
(α)
x
and an inequality of the form |Ln (x)| < an e 2 , x ≥ 0. The last
(α)
result consist of the representation of polynomial L n depending on
Hermite polynomials.
2000 Mathematics Subject Classification: 33C05, 33C45
Key words and phrases: Laguerre polynomials, Hermite polynomials,
Feldheim formula, Beta function, Mehler-Hermite quadrature formula.
1
Received 11 October, 2009
Accepted for publication (in revised form) 22 November, 2009
131
132
1
I. Ţincu
Introduction
(α)
Let Ln (x), α > −1, x ≥ 0, n ∈ N Laguerre polynomials on degree n and
order α.
(α)
The polynomial Ln (x) verifies the following equalities:


n
X
n+α
(−x)k


(1)
L(α)
(x)
=
·
,
n
k!
n
−
k
k=0
L(α+β+1)
(x
n
(2)
n
X
+ y) =
(α)
(β)
Lk (x)Ln−k (y),
k=0
L(α)
n (tx)
(3)
=
n
X
k=0
(4)
(5)
(− 21 )
Ln
(x) =
L(α)
n (x)
Γ(n + α + 1)
(α)
tk (1 − t)n−k Lk (x),
(n − k)!Γ(α + k + 1)
√
√
(−1)n
(−1)n
( 21 )
√
H
(
(x)
=
x),
L
H
(
x),
n
2n
2n+1
n!4n
n!22n+1 x
(−1)n Γ(n + α + 1)
= √
π(2n)!Γ(α + 21 )
Z
1
√
1
1
(1 − t2 )α− 2 H2n (t x)dt, α > − ,
2
−1
Hn (x) being Hermite polynomial on degree n.
From (1) for x = 0 we obtain


L(α)
n (0) =
(α)
(α)
We denote ln (x) =
Ln (x)
n+α
n


Laguerre polynomial of α order normalized
(α)
Ln (0)
(α)
through the condition ln (0) = 1
133
Some properties from Laguerre polynomials
(α)
(α)
Using ln (x) and using properties (1)-(5), we proof that ln (x) verifies:


n
k
X
n+α
1
 (−x) ,


(6)
ln(α) (x) = 
k!
n−k
n+α
k=0


n
(7) lnα+β+1 (x+y) =
n
X
k=0
(8)


n

(β)
 B(k + α + 1, n − k + β + 1) lk(α) (x)ln−k
(y),
B(α
+
1,
β
+
1)
k
ln(α) (tx) =
n
X
k=0


n
k

 tk (1 − t)n−k lk(α) (x),
√
(−1)n · n!
H2n ( x),
(2n)!
Z 1
1
(α+β+1)
00
(8 ) lm+n (x) =
l(α) (xt)ln(β) ((1−t)x)tα (1−t)β dt (FeldB(α + 1, β + 1) 0 m
heim)
(80 )
2
(− 12 )
ln
(x) =
Main results
Theorem 1. If β > α > −1 and x ≥ 0 then
Z 1
1
(β)
(9)
ln (x) =
tα (1 − t)β−α−1 ln(α) (tx)dt,
B(α + 1, β − α) 0
B(x, y) being beta function.
Proof. From (8) we have:
tα (1 − t)β−α−1 ln(α) (tx) =
n
X
k=0


n
k

 tα+k (1 − t)n−k+β−α−1 lk(α) (x),
134
(10)
I. Ţincu
Z
n
X
1
α
0
t (1 −
t)β−α−1 ln(α) (tx)dt
=
k=0
·(1 − t)n−k+β−α−1 dt =
n
X


n


n
k

 lk(α) (x)
Z
1
0
tα+k ·

 B(α + k + 1, n − k + β − α)lk(α) (x).
k
In (7) we consider y = 0 and β = β − α − 1 and we obtain
(11)
ln(β) (x) =
n
X
k=0


k=0
n

 B(k + α + 1), n − k + β − α lk(α) (x)
B(α + 1, β − α)
k
From (10) and (11) we write
ln(β) (x)
1
B(α + 1, β − α)
Z
1
0
tα (1 − t)β+1 ln(α) (tx)dt,
for β > α > −1, and the proof is complete.
Theorem 2. Let β > α > −1 and Pn (x; α) =
Pn (x; α) ≥ 0, (∀)x ≥ 0 then
(12)
n
X
k=0
(α)
ak lk (x), x ≥ 0. If
Pn (x; β) ≥ 0, (∀)x ≥ 0 and β > α > −1
Proof. From (9) we write: Z
1
1
(β)
(α)
ak lk (x) =
tα (1 − t)β−α−1 ak lk (tx)dt, we make sum
B(α + 1, β − α) 0
over the k, and we obtain:
1
Pn (x; β) =
B(α + 1, β − α)
Z
1
0
tα (1 − t)β−α−1 Pn (xt; α)dt
From Pn (x; α) ≥ 0 (∀)x ≥ 0 results Pn (x; β) ≥ 0, (∀)x ≥ 0.
135
Some properties from Laguerre polynomials
(β)
Theorem 3. If ln (x) is Laguerre normalized polynomial through
(α)
ln (0) = 1, then for β > − 21 and k0 = 1, 086435 we have:
k0 · 2n · n! x
|ln(β) (x)| < p
· e 2 , (∀)x ≥ 0
(2n)!
(13)
Proof. In (9) let α = − 21 ,
ln(β) (x)
1
=
1
B( 2 , β + 12 )
Z
1
0
1
1
(− 12 )
t− 2 (1 − t)β− 2 ln
(tx)dt.
Using (80 ) we obtain
ln(β) (x)
(−1)n · n!
=
(2n)!B( 21 , β + 12 )
Z
1
0
√
1
1
t− 2 (1 − t)β− 2 H2n ( tx)dt.
We know that
p
√
x
|H2n ( x)| < k0 (2n)!2n e 2 , k0 = 1, 086435.
It results,
x
|ln(β) (x)|
k0 2n n!e 2
< p
, (∀)x ≥ 0.
(2n)!
Theorem 4. If n, m, p ∈ N such that m + n ≤ 2p − 1 then
p
(14)
lm+n (x) =
(0)
lm+n (x)
√
(−1)n+m n!m! X
2k − 1
=
H2m ( x cos
π)·
(2n)!(2m)!p k=1
4p
√
2k − 1
·H2 n( x · sin
π), (∀)x > 0.
4p
Proof. In Feldheim formula we consider α = β = − 12 ,
1
lm+n (x) =
π
Z
1
(− 12 )
lm
0
(− 12 )
(xt)ln
dt
((1 − t)x) p
.
t(1 − t)
136
I. Ţincu
From (80 ) we have
(−1)n+m n!m! 1
(15) lm+n (x) =
·
(2n)!(2m)!
π
Z
1
0
p
√
dt
H2m ( xt)H2n ( (1 − t)x) p
t(1 − t)
We observe that in integral we have a polynomial of m + n degree in t. In
the following we use the Mehler-Hermite cuadrature formula:
Z 1
p
π
2k − 1
dt
πX
=
f (xk )+ 2p−1
f (2p) (ξ), xk = cos
, ξ ∈ [−1, 1].
f (t) √
2 (2p)!
2p
1−t2 p k=1
−1
For f ∈ Π2p−1 we obtain:
Z
1
−1
f (t) √
p
dt
πX
f (xk ).
=
p k=1
1 − t2
Making variable substitution: t = 2u − 1, it results
Z
1
0
p
du
πX
f (2u − 1) p
=
f (xk ).
p
u(1 − u)
k=1
We put: f (2u − 1) = g(u).
we obtain f (y) = g( 1+y
),
For u = 1+y
2
2
Z 1
Z 1
p
du
du
πX
1 + xk
f (2u − 1) p
g(u) p
g
,
=
=
p
2
u(1 − u)
u(1 − u)
0
0
k=1
Z
1
0
p
πX
2 2k − 1
g(t) p
g cos
π , g ∈ Π2p−1 .
=
p k=1
4p
t(1 − t)
dt
Using (15) we have
s
!
p
X
(−1)
n!m! 1
2k − 1
lm+n (x) =
·
H2m
x cos2
π
(2n)!(2m)!
p k=1
4p
s
!
(−1)n+m n!m!
2 2k − 1
·H2n
x sin
π =
4p
(2n)!(2m)!
n+m
Some properties from Laguerre polynomials
137
p
√
√
1X
2k − 1
2k − 1
·
H2m
x cos
π H2n
x sin
π
p k=1
4p
4p
References
[1] Richard Askey, George Gasper, Positive Jacobi polynomial sums, II,
Amer. J. Math. 98, 1976, 709-737.
[2] E. Feldheim, Développment en série de polynomes d’Hermite et de Laguerre a l’aide des transformations de Gauss et de Hankel, I, II, Konin.
Neder. Akad. van Weten. 43, 1940, 224-248.
[3] G. Palamă, Sulli unlimite inferioare della distanza di due zeri consecutivi
di Hn (x) e su di una limitazione di Hn2 (x) − Hn1 (x)Hn+1 (x), Boll. Un.
Mat. Hol., VII, 3, 1952, 311-315.
[4] G. Szegö, Orthogonal polynomials, Amer. Math. Soc. Colloq. Publications, 23, 1978.
Ioan Ţincu
”Lucian Blaga” University
Department of Mathematics
Sibiu, Romania
e-mail: tincuioan@yahoo.com