Identities involving Laguerre polynomials derived from umbral calculus

0.15 MB·English

by  Taekyun Kim

IDENTITIES INVOLVING LAGUERRE POLYNOMIALS
DERIVED FROM UMBRAL CALCULUS
3
1 TAEKYUNKIM
0
2

n
a
Abstract. Inthispaper,weinvestigatesomeidentitiesofLaguerrepolynomi-
J
als involvingBernoulliandEulerpolynomialswhicharederivedfromumbral
5
calculus.
1

] 1. Introduction
T
N For λ(6= 1) ∈ C, the Frobenius-Euler polynomials are given by the generating
function to be
.
h
mat e1t−−λλ ext =eH(x|λ)t = ∞ Hn(x|λ)tnn! (see [6,7,13]) (1.1)
(cid:18) (cid:19) n=0
X
[ withtheusualconventionaboutreplacingHn(x|λ)byH (x|λ). Inthespecialcase,
n
1 x=0, Hn(0|λ)=Hn(λ) arecalledthe n-th Frobenius-Euler numbers. By (1.1), we
v easily see that
3 n
n
9 H (x|λ)= H (λ)xn−k, (see [6]). (1.2)
n k
k
7
k=0(cid:18) (cid:19)
3 X
As is wellknown, the Bernoulli polynomials are defined by the generating function
.
1 to be
0 t ∞ tn
3 ext =eB(x)t = B (x) , (1.3)
1 et−1 n n!
(cid:18) (cid:19) n=0
: X
v with the usual convention about replacing Bn(x) by B (x). In the special case,
n
Xi x=0, Bn(0)=Bn are called the n-th Bernoulli numbers. From (1.3), we have
r n n
a B (x)= B xn−k. (1.4)
n k
k
k=0(cid:18) (cid:19)
X
The Euler polynomials are also defined by the generating function to be
2 ∞ tn
ext = E (x) , (see [2,3,8,10]). (1.5)
et+1 n n!
(cid:18) (cid:19) n=0
X
The Euler numbers, E , are defined by E (0)=E . By (1.5), we get
n n n
n
n
E (x)= E xn−k, (see [8,15,16,17]). (1.6)
n k
k
k=0(cid:18) (cid:19)
X
1991 Mathematics Subject Classification. 05A10,05A19.
Key words and phrases. Bernoullipolynomial,Eulerpolynomial,Abelpolynomial.
1
2 TAEKYUNKIM
As is well known, Laguerre polynomials are defined by the generating function to
be
exp − xt ∞
1−t = L (x)tn, (see[1,9]). (1.7)
1(cid:16)−t (cid:17) n
n=0
X
From (1.7), we note that
∞ ∞ (−1)rxr
L (x)tn = (1−t)−r−1tr
n
r!
n=0 r=0
X X (1.8)
∞ n (−1)r n
= r xr tn.
r!
n=0 r=0 (cid:0) (cid:1) !
X X
Thus, by (1.8), we get
n (−1)r n
L (x)= xr. (1.9)
n
r! r
r=0 (cid:18) (cid:19)
X
Thus, by (1.9), we note that L (x) is a polynomial of degree n with rational co-
n
efficient and leading coefficient (−1)n. From (1.9), we note that u = L (x) is a
n! n
solution of the following differential equation of order 2:
′′ ′
xu (x)+(1−x)u (x)+nu(x)=0. (1.10)
The Rodrigues’ formula for L (x) is given by
n
1 dn
L (x)= ex e−xxn , (see [1,9]). (1.11)
n n! dxn
(cid:18) (cid:19)
By (1.11), we easily see that
∞
e−xL (x)L (x)dx=δ , (m,n∈N∪{0}), (1.12)
m n m,n
Z0
where δ is the Kroneckersymbol.
m,n
LetCbethecomplexnumberfieldandletF bethesetofallformalpowerseries
in the variable t over C with
∞ a
F = f(t)= ktk a ∈C .
k
( k! (cid:12) )
Xk=0 (cid:12)
(cid:12)
Let P = C[x]. Then P∗ will be defined by the(cid:12)vector space of all linear function-
(cid:12)
als on P. As is known, hL | p(x)i denotes the action of a linear functional on a
polynomial p(x), and we remind that the vector space structure on P∗ are defined
by hL+M|p(x)i = hL|p(x)i+hM|p(x)i and hcL|p(x)i = chL|p(x)i, where c is a
complex constant. The formal power series
∞ tk
f(t)= a ∈F
k
k!
k=0
X
defines a linear functional on P by setting
hf(t)|xni=a for all n≥0 (see [4,5,7,13]). (1.13)
n
Thus, by (1.13), we get
tk|xn =n!δ , (n,k≥0). (1.14)
n,k
(cid:10) (cid:11)
IDENTITIES INVOLVING LAGUERRE POLYNOMIALS DERIVED FROM UMBRAL CALCULUS3
Let f (t) = ∞ hL|xkitk. Then, by (1.14), we see that hf (t)|xni = hL|xni and
L k=0 k! L
so as a linear functional L = f (t). Thus we note that the map L 7→ f (t) is a
L L
vector spacePisomorphism from P∗ onto F. Henceforth, F will denote both the
algebra of formal power series in t and the vector space of all linear functionals on
P,andsoanelementf(t)ofF willbe thoughtofasbothaformalpowerseriesand
a linear functional. We shall call F the umbral algebra. The umbral calculus is the
study of umbral algebra.
The order o(f(t)) of the non-zero power series f(t) is the smallest integer k for
which the coefficient of tk does not vanish (see [13, 14]). If o(f(t)) = 1, then f(t)
is called a delta series and if o(f(t)) = 0, then f(t) is called an invertible series.
From (1.13) and (1.14), we note that heyt|xni = yn and so heyt|p(x)i = p(y). Let
f(t)∈F and p(x)∈P. Then we have the following equations (see [11, 12, 13]):
∞ f(t)|xk ∞ tk|p(x)
f(t)= tk, p(x)= xk. (1.15)
k! k!
k=0(cid:10) (cid:11) k=0(cid:10) (cid:11)
X X
It is easy to show that
hf(t)g(t)|p(x)i=hf(t)|g(t)p(x)i=hg(t)|f(t)p(x)i,
where f(t),g(t) ∈ F and p(x) ∈ P. From (1.15), we can derive the following
equations:
p(k)(0)= tk|p(x) and t0|p(k)(x) =p(k)(0). (1.16)
D E
(cid:10) (cid:11)
Thus, by (1.16), we get
dkp(x)
tkp(x)=p(k)(x)= , eytp(x)=p(x+y). (1.17)
dxk
Let o(f(t)) = 1 and o(g(t)) = 0. Then there exists a unique sequence S (x) such
n
that
g(t)f(t)k|S (x) =n!δ , (n,k ≥0). (1.18)
n n,k
This sequence S (x) is(cid:10)calledthe Sheffe(cid:11)r sequencefor (g(t),f(t)), whichis denoted
n
by S (x) ∼ (g(t),f(t)). Let S (x) ∼ (g(t),f(t)). Then we have the following
n n
equations (see [6, 7, 13]) :
∞ hh(t)|S (x)i ∞ g(t)f(t)k|p(x)
h(t)= k g(t)f(t)k, p(x)= S (x), (1.19)
k
k! k!
k=0 k=0(cid:10) (cid:11)
X X
f(t)S (x)=nS (x), hf(t)|p(αx)i=hf(αt)|p(x)i, (1.20)
n n−1
and
1 ∞ S (y)
eyf¯(t) = k tk, for all y ∈C, (1.21)
g(f¯(t)) k!
k=0
X
where f¯(t) is the compositional inverse of f(t).
In this paper, we investigate some identities of Laguerre polynomials involving
Bernoulli and Euler polynomials which are derived from umbral calculus.
4 TAEKYUNKIM
2. Some identities of Laguerre polynomials
Let
P ={p(x)∈C[x]|degp(x)≤n}. (2.1)
n
Then we note that P is an inner product space with weighted inner product
n
∞
hp(x),q(x)i= e−xp(x)q(x)dx, (2.2)
Z0
wherep(x),q(x)∈P . By(1.12),(2.1)and(2.2),weeasilyweethatL (x),L (x),...,L (x)
n 0 1 n
are orthogonal basis for P .
n
For p(x)∈P , let us assume that polynomial p(x) defined on [0,∞) is given by
n
∞
p(x)= C L (x). (2.3)
k k
k=0
X
Then, by (1.12) and (2.3), we get
∞
C =hp(x),L (x)i= e−xp(x)L (x)dx
k k k
Z0 (2.4)
1 ∞ dk
= e−xxk p(x)dx.
k! dxk
Z0 (cid:18) (cid:19)
Let us take p(x)=xn (n≥0). Then, by (2.4), we get
1 ∞ dk
C = e−xxk xndx
k k! dxk
Z0 (cid:18) (cid:19) (2.5)
(n) ∞ n
=(−1)k k e−xxndx=(−1)k n!.
k! k
Z0 (cid:18) (cid:19)
From (2.3) and (2.5), we have
n
n
xn =n! (−1)k L (x). (2.6)
k
k
k=0 (cid:18) (cid:19)
X
Note that
(x) =x(x−1)···(n−k+1)∼ 1,et−1 . (2.7)
k
By (1.19) and (2.6), we easily get
(cid:0) (cid:1)
n (et−1)k xn n
xn = (x) = S (n,k)(x) , (2.8)
k! k 2 k
kX=0 (cid:10) (cid:12)(cid:12) (cid:11) kX=0
where S (n,k) is the Stirling number of the second kind.
2
From (2.6) and (2.8), we have
(−1)k
L (x)= S (n,k)(x) ,
k n! n 2 k
k
where n,k∈Z+ with n≥k. (cid:0) (cid:1)
Theorem 2.1. For n,k∈Z with n≥k, we have
+
(−1)k
L (x)= S (n,k)(x) .
k n! n 2 k
k
(cid:0) (cid:1)
IDENTITIES INVOLVING LAGUERRE POLYNOMIALS DERIVED FROM UMBRAL CALCULUS5
For p (x) and q (x) = n q xk sequences of polynomials, we define the
n n k=0 n,k
umbral composition of q (x) with p (x) to be the sequences
n n
P
n
(q ◦p)(x)= q p (x), (see [3,6,7,13]). (2.9)
n n,k k
k=0
X
From (1.7) and (1.21), we note that
t
L (x)∼ 1−t, . (2.10)
n
1−t
(cid:18) (cid:19)
Let s (x)∼(g(t),f(t)), r (x)∼(h(t),l(t)). Then we have
n n
(r ◦s)(x)=(g(t)h(f(t)),l(f(t))), (see [13]). (2.11)
n
From (2.10) and (2.11), we can derive
(L ◦L)(x)∼(1,t). (2.12)
n
By (1.21) and (2.12), we get
(L ◦L)(x)=xn. (2.13)
n
Therefore, by (2.6) and (2.13), we obtain the following theorem.
Theorem 2.2. For n≥0, we have
n
1 n
(L ◦L)(x)= (−1)k L (x).
n k
n! k
k=0 (cid:18) (cid:19)
X
For p(x)=E (x)∈P , let us assume that
n n
n
E (x)= C L (x). (2.14)
n k k
k=0
X
By (2.4) and (2.14), we get
1 ∞ dk
C = e−xxk E (x)dx
k k! dxk n
Z0 (cid:18) (cid:19)
n ∞
=(−1)k e−xxkE (x)dx
k n−k
(cid:18) (cid:19)Z0
n n−k n−k ∞ (2.15)
=(−1)k e−xxk+ldx
k l
(cid:18) (cid:19) l=0 (cid:18) (cid:19)Z0
X
n−k k+l E
=n! (−1)k n−k−l .
l (n−k−l)!
l=0 (cid:18) (cid:19)
X
From (1.5), we have
2 et+1 −1 et−1 −1 ∞ et−1 j
= = 1+ = (−1)j . (2.16)
et+1 2 2 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) j=0 (cid:18) (cid:19)
X
6 TAEKYUNKIM
Note that (et−1)j is a delta series and E (x) ∼ et+1,t . By (1.5), (1.17) and
n 2
(1.21), we get (cid:16) (cid:17)
∞ tn 2 ∞ et−1 j
E (x) = ext = (−1)j ext
n n! et+1 2
n=0 (cid:18) (cid:19) j=0 (cid:18) (cid:19)
X X
∞ n et−1 j tn
= (−1)j xn (2.17)
 2  n!
n=0 j=0 (cid:18) (cid:19)
X X
 
∞ n 1 j tn
= − (et−1)jxn .
 2  n!
n=0 j=0(cid:18) (cid:19)
X X
 
By comparing the coefficients on the both sides of (2.17), we get
n j
1
E (x)= − (et−1)jxn
n
2
j=0(cid:18) (cid:19)
X
n 1 j ∞ tk
= − j! S (k,j) xn
2 2 k!
j=0(cid:18) (cid:19) k=j
X X
n 1 j ∞ n (2.18)
= − j! S (k,j) xn−k
2 2 k
j=0(cid:18) (cid:19) k=j (cid:18) (cid:19)
X X
n n k 1 j
= − j!S (k,j) xn−k.
k  2 2 
Xk=0(cid:18) (cid:19)Xj=0(cid:18) (cid:19) 
From (1.6) and (2.18), we have
 
k j
1
E = − j!S (k,j). (2.19)
k 2
2
j=0(cid:18) (cid:19)
X
By (2.15) and (2.19), we get
n−k k+l 1 n−k−l 1 j
C =n! (−1)k − j!S (n−k−l,j)
k l (n−k−l)! 2 2
l=0 (cid:18) (cid:19) j=0 (cid:18) (cid:19)
X X (2.20)
n−kn−k−l
k+l j!S (n−k−l,j)
=n! (−1)k+l 2 .
l 2j(n−k−l)!
l=0 j=0 (cid:18) (cid:19)
X X
Therefore, by (2.14) and (2.20), we obtain the following theorem.
Theorem 2.3. For n≥0, we have
n n−kn−k−l
k+l j!S (n−k−l,j)
E (x)=n! (−1)k+l 2 L (x).
n l 2j(n−k−l)! k
k=0 l=0 j=0 (cid:18) (cid:19)
XX X
By (1.3), we see that
t et−1 −1 et−t−1 −1 ∞ et−t−1 j
= = 1+ = (−1)j .
et−1 t t t
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) j=0 (cid:18) (cid:19)
X
(2.21)
IDENTITIES INVOLVING LAGUERRE POLYNOMIALS DERIVED FROM UMBRAL CALCULUS7
Note that et−t−1 j = et−1 −1 j is a delta series in F and B (x) ∼ et−1,t .
t t n t
By (1.3), (1(cid:16).21) an(cid:17)d (2.2(cid:16)1), we get(cid:17) (cid:16) (cid:17)
∞ tn et−t−1 −1 ∞ et−t−1 j
B (x) = 1+ ext = (−1)j ext
n
n! t t
n=0 (cid:18) (cid:19) j=0 (cid:18) (cid:19)
X X
(2.22)
∞ n et−t−1 j tn
= (−1)j xn .
 t n!
n=0 j=0 (cid:18) (cid:19)
X X
 
By comparing the coefficients on the both sides of (2.22), we get
n et−t−1 j
B (x)= (−1)j xn. (2.23)
n
t
j=0 (cid:18) (cid:19)
X
From (1.15), we have
et−t−1 j n−j tk et−tt−1 jxn
t xn = (cid:28) (cid:12)(cid:12)(cid:16) k! (cid:17) (cid:29)xk
(cid:18) (cid:19) Xk=0 (cid:12)(cid:12)
n−j et−tt−1 j tkxn
= Xk=0 (cid:28)(cid:16) k(cid:17)! (cid:12)(cid:12)(cid:12)(cid:12) (cid:29)xk (2.24)
n−j n et−t−1 j
= xn−k xk
k t
Xk=0(cid:18) (cid:19)*(cid:18) (cid:19) (cid:12)(cid:12) +
n−j n j j (cid:12)(cid:12) et−1 l
= (−1)j−l (cid:12) t0 xn−k xk.
k l t
Xk=0(cid:18) (cid:19)Xl=0(cid:18) (cid:19) * (cid:12)(cid:12)(cid:18) (cid:19) +
(cid:12)
(cid:12)
It is easy to show that (cid:12)
et−1 l 1 ∞ tm
xn−k = l! S (m,l) xn−k
t tl 2 m!
(cid:18) (cid:19) m=l
X
n−k l!
= S (m+l,l) tmxn−k (2.25)
2 (m+l)!
m=0
X
n−k
l!
= S (m+l,l) (n−k) xn−k−m.
2 (m+l)! m
m−0
X
Thus, by (2.24) and (2.25), we get
et−t−1 j n−j j n j S (n−k+l,l)
xn = (−1)j−l 2 xk. (2.26)
t k l n−k+l
(cid:18) (cid:19) k=0l=0(cid:18) (cid:19)(cid:18) (cid:19) l
XX
(cid:0) (cid:1)
8 TAEKYUNKIM
Thus, by (2.23) and (2.26), we get
n n−j j
n j S (n−k+l,l)
B (x)= (−1)l 2 xk
n k l n−k+l
j=0k=0l=0(cid:18) (cid:19)(cid:18) (cid:19) l
XXX
n n−k j (cid:0) (cid:1)
n j S (n−k+l,l)
= (−1)l 2 xk (2.27)
k  l n−k+l 
Xk=0(cid:18) (cid:19)Xj=0Xl=0(cid:18) (cid:19) l 
n k j (cid:0) (cid:1)
n  j S (k+l,l) 
= (−1)l 2 (−1)l xn−k.
k  l k+l 
Xk=0(cid:18) (cid:19)Xj=0Xl=0 (cid:18) (cid:19) l 
(cid:0) (cid:1)
By (1.4) and (2.27), we get 
k j
j S (k+l,l)
B = (−1)l 2 . (2.28)
k l k+l
j=0l=0 (cid:18) (cid:19) l
XX
From (1.4), we note that B (x)∈P . So, let us a(cid:0)ssum(cid:1)e that
n n
n
B (x)= C L (x). (2.29)
n k k
k=0
X
By (2.4), we get
1 ∞ dk
C = e−xxk B (x)dx
k k! dxk n
Z0 (cid:18) (cid:19)
n ∞
=(−1)k e−xxkB (x)dx
k n−k
(cid:18) (cid:19)Z0
n n−k n−k ∞
=(−1)k B e−xxk+ldx
k l n−k−l (2.30)
(cid:18) (cid:19) l=0 (cid:18) (cid:19) Z0
X
n n−k n−k
=(−1)k B (k+l)!
k l n−k−l
(cid:18) (cid:19) l=0 (cid:18) (cid:19)
X
n−k(−1)kB k+l
=n! n−k−l .
(n−k−l)! l
l=0 (cid:18) (cid:19)
X
From (2.28) and (2.30), we have
n−kn−k−l j (−1)m+k j S (n−k−l+m,m)
C =n! m 2 . (2.31)
k (n−k−l)! n−k−l+m
l=0 j=0 m=0 (cid:0) (cid:1) m
X X X
(cid:0) (cid:1)
Therefore, by (2.29) and (2.31), we obtain the following theorem.
Theorem 2.4. For n≥0, we have
n n−kn−k−l j (−1)m+k j S (n−k−l+m,m)
B (x)=n! m 2 L (x).
n  (n−k−l)! n−k−l+m  k
Xk=0Xl=0 Xj=0 mX=0 (cid:0) (cid:1) m 
(cid:0) (cid:1)
 
IDENTITIES INVOLVING LAGUERRE POLYNOMIALS DERIVED FROM UMBRAL CALCULUS9
From (1.1), we note that
1−λ et−1 −1 ∞ 1 j
= 1+ = (−1)j (et−1)j. (2.32)
et−λ 1−λ 1−λ
(cid:18) (cid:19) j=0 (cid:18) (cid:19)
X
Notethat(et−1)j is adelta seriesinF andH (x|λ)∼ et−λ,t . By (1.1),(1.21)
n 1−λ
and (2.32), we get (cid:16) (cid:17)
n j
1
H (x|λ)= (−1)j (et−1)jxn, (2.33)
n
1−λ
j=0 (cid:18) (cid:19)
X
and
n tk
(et−1)jxn =j! S (k,j) xn
2 k!
k=j
X (2.34)
n
n
=j! S (k,j) xn−k.
2 k
k=j (cid:18) (cid:19)
X
From (2.33) and (2.34), we can derive the following equation (2.35):
n n
n j!
H (x|λ)= (−1)j S (k,j)xn−k
n k (1−λ)j 2
j=0k=j(cid:18) (cid:19)
XX
(2.35)
n k
n j!
= S (k,j) xn−k,
k  (λ−1)j 2 
kX=0(cid:18) (cid:19)Xj=0 
 
and
n
n
H (x|λ)= H (λ)xn−k. (2.36)
n k
k
k=0(cid:18) (cid:19)
X
thus, by (2.35) and (2.36), we get
k
j!
H (λ)= S (k,j). (2.37)
k (λ−1)j 2
j=0
X
By (2.36), we see that H (x|λ)∈P . Let us assume that
n n
n
H (x|λ)= C L (x). (2.38)
n k k
k=0
X
10 TAEKYUNKIM
From (2.4), we have
1 ∞ dk
C = e−xxk H (x|λ)dx
k k! dxk n
Z0 (cid:18) (cid:19)
n ∞
=(−1)k e−xxkH (x|λ)dx
k n−k
(cid:18) (cid:19)Z0
n n−k n−k ∞
=(−1)k H (λ) e−xxk+ldx
k l n−k−l (2.39)
(cid:18) (cid:19) l=0 (cid:18) (cid:19) Z0
X
n−k
n n−k
=(−1)k H (λ)(k+l)!
k l n−k−l
(cid:18) (cid:19) l=0 (cid:18) (cid:19)
X
n−k
k+l H (λ)
=n! (−1)k n−k−l .
l (n−k−l)!
l=0 (cid:18) (cid:19)
X
By (2.37) and (2.39), we get
n−kn−k−l (−1)k k+l j!
C =n! l S (n−k−l,j). (2.40)
k (n−k−l)!(λ−1)j 2
l=0 j=0 (cid:0) (cid:1)
X X
Therefore, by (2.38) and (2.40), we obtain the following theorem.
Theorem 2.5. For n≥0, we have
n n−kn−k−l
k+l j!S (n−k−l,j)
H (x|λ)=n! (−1)k 2 L (x).
n  l (n−k−l)!(λ−1)j k
Xk=0Xl=0 Xj=0 (cid:18) (cid:19) 
For sn(x)∼(g(t),f(t)) and rn(x)∼(h(t),l(t)), let us assume that
n
s (x)= C r (x). (2.41)
n n,k k
k=0
X
Then, we note that
1 h(f¯(t))
C = l(f¯(t))k xn (2.42)
n,k k! g(f¯(t))
(cid:28) (cid:12) (cid:29)
(cid:12)
where n,k≥0 (see [6, 7, 13]). From (1.5), (1.7) and(cid:12)(1.21), we not that
(cid:12)
t 1+et
L (x)∼ 1−t, , E (x)∼ ,t . (2.43)
n n
t−1 2
(cid:18) (cid:19) (cid:18) (cid:19)
Now, we assume that
n
E (x)= C L (x). (2.44)
n n,k k
k=0
X
By (2.42) and (2.43), we get
1 2 tk
C =− xn
n,k k! et+1 (t−1)k−1
(cid:28)(cid:18) (cid:19) (cid:12) (cid:29)
n 2 1 k−(cid:12)(cid:12)1 (2.45)
=− (cid:12) xn−k .
k et+1 t−1
(cid:18) (cid:19)(cid:28) (cid:12)(cid:18) (cid:19) +
(cid:12)
(cid:12)
(cid:12)