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A matrix approach to some identities involving Sheffer polynomial sequences
Applied Mathematics and Computation 253 (2015) 83–101
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
A matrix approach to some identities involving Sheffer polynomial sequences Dae San Kim a, Taekyun Kim b,⇑ a b
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
a r t i c l e
i n f o
a b s t r a c t A simple but elegant method was adopted in Youn and Yang (2011) in order to derive a differential equation and recursive formulas for Sheffer polynomials. Namely, they used the so called the generalized Pascal functional matrix of an analytic function and the Wronskian matrix of several analytic functions. In this paper, we will use their method to find some identities satisfied by Sheffer polynomials. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Sheffer sequences Umbral calculus Generalized Pascal functional matrix Wronskian matrix
1. Introduction n o P tk Let F ¼ hðtÞ ¼ 1 k¼0 ak k! ak 2 C be the C-algebra of formal power series. For hðtÞ 2 F , the generalized Pascal functional matrix of hðtÞ, denoted by Pn ½hðtÞ, is an ðn þ 1Þ ðn þ 1Þ lower triangular matrix given by
8 > < i hðijÞ ðtÞ; if i P j Pn ½hðtÞij ¼ j > : 0; if i < j:
ð1:1Þ ðiÞ
i
Here we remind the reader of the fact that h ðtÞ stands for the ith order derivative of hðtÞ and h ðtÞ for the ith power of ð0Þ 0 hðtÞ. In particular, h ðtÞ ¼ hðtÞ, h ðtÞ ¼ 1. Also, for h1 ðtÞ; h2 ðtÞ; . . . ; hm ðtÞ 2 F , the nth order Wronskian matrix of h1 ðtÞ; h2 ðtÞ; . . . ; hm ðtÞ, denoted by W n ½h1 ðtÞ; h2 ðtÞ; . . . ; hm ðtÞ, is the ðn þ 1Þ m matrix given by
2
h1 ðtÞ 6 h0 ðtÞ 6 1 W n ½h1 ðtÞ; h2 ðtÞ; . . . ; hm ðtÞ ¼ 6 6 .. 4. ðn Þ
h2 ðtÞ
. . . hm ðtÞ
0 h2 ðtÞ
0 hm ðtÞ
.. .
...
.. . ðn Þ
ðnÞ
3 7 7 7: 7 5
h1 ðtÞ h2 ðtÞ . . . hm ðtÞ The following proposition, stated as Property 1 in [17], is of fundamental use throughout this paper. Proposition (a) For hðtÞ; lðtÞ 2 F ; a; b 2 C, we have ⇑ Corresponding author. E-mail addresses: [email protected] (D.S. Kim), [email protected] (T. Kim). http://dx.doi.org/10.1016/j.amc.2014.12.048 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
ð1:2Þ
84
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Pn ½ahðtÞ þ blðtÞ ¼ aPn ½hðtÞ þ bP n ½lðtÞ; W n ½ahðtÞ þ blðtÞ ¼ aW n ½hðtÞ þ bW n ½lðtÞ: (b) For hðtÞ; lðtÞ 2 F , we have
Pn ½lðtÞPn ½hðtÞ ¼ P n ½hðtÞPn ½lðtÞ ¼ P n ½hðtÞlðtÞ: (c) For hðtÞ; lðtÞ 2 F , we have
Pn ½lðtÞW n ½hðtÞ ¼ P n ½hðtÞW n ½lðtÞ ¼ W n ½lðtÞhðtÞ: 0
(d) For hðtÞ; lðtÞ 2 F , with lð0Þ ¼ 0; l ð0Þ – 0, we have
h i 2 n W n ½hðlðtÞÞt¼0 ¼ W n 1; lðtÞ; l ðtÞ; . . . ; l ðtÞ
t¼0
X1 n W n ½hðtÞt¼0 ;
where Xn ¼ diag ½0!; 1!; . . . ; n! is the diagonal matrix with the diagonal entries 0!; 1!; . . . ; n!. Let gðtÞ be an invertible series, and let f ðtÞ be a delta series. Then there is a unique sequence of polynomials sn ðxÞ such that D E gðtÞf ðtÞk sn ðxÞ ¼ n!dn;k , ðn; k 0Þ. The sequence sn ðxÞ is called Sheffer polynomial sequence for ðgðtÞ; f ðtÞÞ which is denoted by sn ðxÞ ðgðtÞ; f ðtÞÞ. It is known that 1 X 1 tk sn ðxÞ ðgðtÞ; f ðtÞÞ () exf ðtÞ ¼ s k ð xÞ ; k! g f ðtÞ k¼0
ð1:3Þ
where f ðtÞ is the compositional inverse of f ðtÞ determined by f f ðtÞ ¼ f ðf ðtÞÞ ¼ t (see [14]). Thus, by (1.3), we get
k d 1 xf ðtÞ sk ðxÞ ¼ e dt g f ðtÞ
;
ðk 0Þ:
t¼0
Sheffer polynomial sequences arise in numerous problems of applied mathematics, theoretical physics, approximation theory, and several other mathematical branches (see [5]). In this past few decades, there has been a renewed interest in Sheffer polynomials (see [1–18]). The following lemma was shown by Youn and Yang. Lemma [17]. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have
h i 2 n W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT X1 n ¼ W n 1; f ðtÞ; f ðtÞ ; . . . ; f ðtÞ
t¼0
X1 n Pn
1 gðtÞ
t¼0
Pn ext t¼0 ;
T
where W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞ is the transpose of W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞ. A simple but elegant method was adopted in [17] in order to derive a differential equation and recursive formulas for Sheffer polynomial sequences. Namely, they used the so called the generalized Pascal functional matrix of an analytic function and the Wronskian matrix of several analytic functions. In this paper, we will use their method to find some identities of special polynomials satisfied by Sheffer polynomial sequences. 2. A matrix approach to some identities involving Sheffer polynomial sequences From the method of Youn and Yang in [17], we derive some identities of special polynomials satisfied by Sheffer polynomial sequences in this section. Theorem 2.1. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the recursive formula:
1 ; and g ð0Þ n n X X n 1 a0 snþ1 ðxÞ ¼ bk x bkþ1 sðnkÞ ðxÞ ankþ1 sk ðxÞ; k! k1 k¼0 k¼1 s 0 ð xÞ ¼
0 ðkÞ where ak ¼ f f ðtÞ g f ðtÞ
t¼0
and bk ¼
d k dt
gðtÞ
t¼0
ðn 0Þ;
.
Proof. Now, we consider
" Wn
d exf ðtÞ f f ðtÞ g f ðtÞ dt g f ðtÞ 0
!# : t¼0
ð2:1Þ
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
85
On one hand, by (c) of the Proposition, we get
" !# 0
0
d exf ðtÞ Pn f f ðtÞ g f ðtÞ t¼0 W n ¼ Pn f f ðtÞ g f ðtÞ t¼0 ½s1 ðxÞ; s2 ðxÞ; . . . ; snþ1 ðxÞT : dt g f ðtÞ t¼0
ð2:2Þ
It is easy to show that
d exf ðtÞ dt g f ðtÞ
!
! g 0 f ðtÞ 1 exf ðtÞ x 0 : g f ðtÞ f f ðtÞ g f ðtÞ
¼
ð2:3Þ
So, on the other hand, by (a)–(d) of Proposition, we obtain
# ! xt h i g 0 f ðtÞ e 0 x exf ðtÞ ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 W ð xgðtÞ g ðtÞ Þ n n t¼0 gðtÞ g f ðtÞ t¼0 t¼0 h i
1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 Pn ext t¼0 W n ½xgðtÞ g 0 ðtÞt¼0 n Pn t¼0 gðtÞ t¼0 ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT X1 xW n ½gðtÞt¼0 W n ½g 0 ðtÞt¼0 : n
" Wn
ð2:4Þ
Equating (2.2) and (2.4), we get
2
a0 0 a0
6a 6 1 6 6 6 a2 6 6 6. 6. 6. 6 4
a1
1
... 0
0
2
an
3 3 2 s 0 ð xÞ 2 s1 ðxÞ ... 0 7 7 76 s2 ðxÞ 7 6 s 1 ð xÞ 76 7 6 7 6 6 ... 0 7 6 . 76 7 6 s ð xÞ .. 76 7¼6 2 7 6 6 . . .. 7 .. 76 7 6 .. 5 4. . . 74 . 7 5 ð x Þ s nþ1 s n ð xÞ . . . a0
0
a0
.. . n
.. . n
1
2
an1
an2
0 s01 ðxÞ 1! s02 ðxÞ 1!
0 0 s002 ðxÞ 2!
.. .
.. .
s0n ðxÞ 1!
s00n ðxÞ 2!
32 3 b x b1 76 0 . . . 0 76 b1 x b2 7 7 76 7 b xb 7 7 . . . 0 76 2 3 7: 76 7 76 .. 7 6 . . .. 74 . 5 . . 5 ðnÞ sn ðxÞ x b b n nþ1 ... n! ... 0
Comparing the nth rows, we obtain n X n
ank skþ1 ðxÞ ¼
k
k¼0
n X 1
k! k¼0
sðnkÞ ðxÞ bk x bkþ1 :
ð2:5Þ
Equivalently, n X 1
a0 snþ1 ðxÞ ¼
k¼0
k!
n X bk x bkþ1 sðnkÞ ðxÞ k¼1
0
Remark. Note that a0 ¼ f ð0Þg ð0Þ. Also, if gðtÞ ¼
n k1
P1
ankþ1 sk ðxÞ:
tk k¼0 ak k! ,
ð2:6Þ
then bk ¼ g ðkÞ ð0Þ ¼ ak .
Theorem 2.2. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the following identity: n X ð1Þk
k!
k¼0
where ak ¼
sðnkÞ ðxÞxk ¼ an ; ðkÞ
1 g ðf ðtÞÞ
.
t¼0
Proof. Let’s consider W n
½a0 ; a1 ; . . . ; an T
1 g ðf ðtÞÞ
. On one hand, it is equal to
t¼0
ð2:7Þ
On the other hand, by (b)–(d) of the Proposition, it is equal to
" Wn
# " # xt h i 1 exf ðtÞ xf ðtÞ e 2 n 1 xt ¼ W ¼ W f ðtÞ; f ðtÞ f ðtÞ e 1; ; . . . ; X W e n n n n t¼0 gðtÞ gf ðtÞ g f ðtÞ t¼0 h i
1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 Pn ext t¼0 W n ext t¼0 n Pn t¼0 gðtÞ t¼0 h iT ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT X1 1; x; ðxÞ2 ; . . . ; ðxÞn : n
ð2:8Þ
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D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Equating the nth rows of (2.7) and (2.8), we get
an ¼
n X 1 k¼0
k!
sðnkÞ ðxÞðxÞk ¼
n X ð1Þk
k!
k¼0
sðnkÞ ðxÞxk :
ð2:9Þ
0 Theorem 2.3. Let sn ðxÞ ðgðtÞ; f ðtÞÞ, rn ðxÞ f ðtÞ; f ðtÞ . Then we have the following identity:
a0 snþ1 ðxÞ þ
n X k¼1
n
anþ1k sk ðxÞ ¼ ðx þ b0 Þrn ðxÞ þ
k1
where
ðkÞ ak ¼ g f ðtÞ
t¼0
bk ¼
;
!ðkÞ g 0 f ðtÞ g f ðtÞ
n1 X n bnk r k ðxÞ; k k¼0
ðn 1Þ;
:
t¼0
xf ðtÞ Proof. Here we consider W n g f ðtÞ dtd g e f ðtÞ . On one hand, by (c) of the Proposition, it is equal to ð Þ t¼0
"
d exf ðtÞ dt g f ðtÞ
Pn
!#
W n g f ðtÞ t¼0 :
ð2:10Þ
t¼0
On the other hand, by (a), (c) of the Proposition, it is equal to
" Wn
! xf ðtÞ # g 0 f ðtÞ e x 0 g f ðtÞ f f ðtÞ
"
" # # g 0 f ðtÞ exf ðtÞ ¼ Pn x W n 0 g f ðtÞ t¼0 f f ðtÞ t¼0 " # ! g 0 f ðtÞ ¼ xI þ Pn ½r 0 ðxÞ; r 1 ðxÞ; . . . ; r n ðxÞT : g f ðtÞ
t¼0
ð2:11Þ
t¼0
ðkÞ Now, let ak ¼ g f ðtÞ
2
s 1 ð xÞ
6 s ð xÞ 6 2 6 6 6 s 3 ð xÞ 6 6 6. 6. 6. 6 4 snþ1 ðxÞ
t¼0
, bk ¼
g 0 ðf ðtÞÞ g ðf ðtÞÞ
ðkÞ
. Equating (2.10) and (2.11), we have
t¼0
0
0
... 0
s 1 ð xÞ 2 s 2 ð xÞ 1 .. . n s n ð xÞ 1
0 2 s 1 ð xÞ 2 .. . n sn1 ðxÞ 2
... 0
3
2
3
2
x þ b0
7 a0 6b 7 6 1 76 7 6 a 76 1 7 6 76 7 6 b 2 ... 0 76 a2 7 6 76 7 ¼ 6 76 . 7 6 . . . .. 76 . 7 6 . 74 . 5 6 . . . 7 6 5 an 4 bn . . . ð n Þs1 ðxÞ
0
0
... 0
x þ b0 2 b1 1 .. . n bn1 1
0
... 0
x þ b0
... 0
.. . n bn2 2
..
Equating the nth rows, we have n X n k¼0
k
ak snþ1k ðxÞ ¼
n X n k¼0
k
bnk rk ðxÞ þ xr n ðxÞ:
Equivalently,
a0 snþ1 ðxÞ þ
n X k¼1
n k1
anþ1k sk ðxÞ ¼ ðx þ b0 Þrn ðxÞ þ
n1 X n bnk r k ðxÞ: k k¼0
0
Remark. a0 ¼ g ð0Þ; b0 ¼ ggðð00ÞÞ. Theorem 2.4. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the following recursive formula:
1 ; and g ð0Þ n n X X n 1 a0 snþ1 ðxÞ ¼ ðbk x þ rk ÞsðnkÞ ðxÞ anþ1k sk ðxÞ; k! k1 k¼0 k¼1 s 0 ð xÞ ¼
ðkÞ where ak ¼ g f ðtÞ
ðkÞ , bk ¼ fgðtÞ 0 ðtÞ t¼0
, t¼0
rk
0 ðtÞ ðkÞ ¼ g f 0 ðtÞ
t¼0
.
ðn 0Þ;
.
.. .
. . . x þ b0
3 3 2 7 r 0 ðxÞ 7 76 r ðxÞ 7 76 1 7 76 7 76 r 2 ðxÞ 7 76 7: 76 . 7 76 . 7 74 . 5 7 5 r n ðxÞ
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
87
xf ðtÞ Proof. As in the proof of Theorem 3, we consider W n g f ðtÞ dtd g e f ðtÞ . On one hand, by (c) of the Proposition, it is equal ð Þ t¼0 to
" Pn
d exf ðtÞ dt g f ðtÞ
!#
W n g f ðtÞ t¼0 :
ð2:12Þ
t¼0
On the other hand, by (a)–(d) of the Proposition, it is equal to
"
! xf ðtÞ # xt h i g f ðtÞ g 0 f ðtÞ e e gðtÞ g 0 ðtÞ x 0 0 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 W x n 0 0 n t¼0 gðtÞ f ðtÞ f ðtÞ t¼0 f f ðtÞ f f ðtÞ g f ðtÞ t¼0 h i 1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 n Pn t¼0 gðtÞ t¼0
gðtÞ g 0 ðtÞ P n ext t¼0 W n x 0 0 f ðtÞ f ðtÞ t¼0 ( 0 ) gðtÞ g ðtÞ : ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT X1 xW þ W n n 0 0 n f ðtÞ t¼0 f ðtÞ t¼0
Wn
ð2:13Þ
Equating (2.12) and (2.13), we get n X n k¼0
ak snþ1k ðxÞ ¼
k
n X 1 k¼0
k!
sðnkÞ ðxÞðbk x þ rk Þ:
Equivalently, we have
a0 snþ1 ðxÞ ¼
n X 1 k¼0
k!
ðbk x þ rk ÞsðnkÞ ðxÞ
n X k¼1
n
k1
anþ1k sk ðxÞ:
Theorem 2.5. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the identity:
sn ðxÞ ¼ where ak ¼ f
nþ1 1 X ak ðkÞ s ðxÞ; n þ 1 k¼1 k! nþ1
ðkÞ
ðn 0Þ;
ð0Þ.
xf ðtÞ Proof. Here we consider W n t g e f ðtÞ . On one hand, by (c) of the Proposition, it is equal to ð Þ t¼0
" Pn ½t t¼0 W n
1 exf ðtÞ g f ðtÞ
# ð2:14Þ
: t¼0
On the other hand, by (b)–(d) of the Proposition, it is equal to
" # exf ðtÞ
ext W n f f ðtÞ ¼ W n 1; f ðtÞ; . . . ; f ðtÞn t¼0 X1 W f ðtÞ n n gðtÞ t¼0 g f ðtÞ t¼0 h i
1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . f ðtÞn X1 P Pn ext t¼0 n n t¼0 gðtÞ t¼0 ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT Equating the nth rows of (2.14) and (2.15), we have
nsn1 ðxÞ ¼
n X 1 k¼0
k!
ak sðnkÞ ðxÞ:
Replacing n by n þ 1, we get the desired result. h
X1 n W n ½f ðtÞt¼0 :
W n ½f ðtÞt¼0 ð2:15Þ
88
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Theorem 2.6. Let sn ðxÞ ðgðtÞ; f ðtÞÞ, rn ðxÞ ð1; f ðtÞÞ. Then we have the following identity:
r n ð xÞ ¼
n X n
k
k¼0
ank sk ðxÞ ¼
ðkÞ where ak ¼ g f ðtÞ
t¼0
n X k ðnÞ xk f ðtÞ ð0Þ ; k! k¼1
ðn P 1Þ;
.
Proof. Here we consider
h i W n exf ðtÞ
t¼0
" # exf ðtÞ ¼ W n g f ðtÞ : g f ðtÞ t¼0
On one hand, by (c) of the Proposition, it is equal to
Pn g f ðtÞ t¼0 W n
"
exf ðtÞ g f ðtÞ
#
¼ P n g f ðtÞ t¼0 ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT :
ð2:16Þ
t¼0
On the other hand, as exf ðtÞ ¼
P1
tk k¼0 r k ðxÞ k! ,
it is equal to
½r 0 ðxÞ; r 1 ðxÞ; . . . ; r n ðxÞT : Also, it is equal to
h i W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn
xt X1 n Wn e
t¼0
t¼0
ð2:17Þ
:
Equating (2.16) and (2.17), we get
2
a0 0 6a a 6 1 0 6 6 6 a2 6 6 6. 6. 6. 6 4
an
... 0
0
2 1
3 2 1 3 2 7 s ð x Þ 6 0 ... 0 7 76 s ðxÞ 7 6 0 76 1 7 6 7 6 6 ... 0 7 6 76 s2 ðxÞ 7 6 0 76 7¼6 7 6 6 . . .. 7 76 .. 7 6. 5 6 .. . . 74 . 7 4 5 s n ð xÞ . . . a0 0
0
a1
a0
.. . n
.. . n
1
2
an1
an2
0 f ð1Þ ð0Þ 1! f ð2Þ ð0Þ 1!
.. . f ðnÞ ð0Þ 1!
3 0 ... 0 2 3 ð1Þ ð1Þ 1 2 n ð f Þ ð0 Þ ð f Þ ð0 Þ 7 7 . . . 76 x 7 2! n! 76 7 ð2Þ ð2Þ 6 7 ðf 2 Þ ð0Þ ðf n Þ ð0Þ 7 7 6 x2 7 . . . 2! n! 76 7 76 .. 7 .. . . .. 74 . 5 7 . . . 5 xn ð n Þ ð n Þ ðf 2 Þ ð0Þ ðf n Þ ð0Þ ... 2! n!
Equating the nth rows, we have
r n ð xÞ ¼
n X n k¼0
k
ank sk ðxÞ ¼
n X 1 k ðnÞ f ðtÞ ð0Þxk : k! k¼1
Theorem 2.7. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the following identity:
n n X X n1 1 n ank sk ðxÞ ¼ ðxbk þ rk ÞsðnkÞ ðxÞ; k! k 1 k¼1 k¼1
0 ðkÞ where ak ¼ f f ðtÞ
t¼0
, bk ¼
ðkÞ
f ðtÞ f 0 ðtÞ
,
rk ¼
t¼0
ðkÞ
g 0 ðtÞf ðtÞ gðtÞf 0 ðtÞ
. t¼0
xf ðtÞ 0 Proof. Let us consider W n tf f ðtÞ dtd g e f ðtÞ . On one hand, by (b) and (c) of the Proposition, it is equal to ð Þ t¼0
0
Pn ½t t¼0 P n f f ðtÞ t¼0
" Wn
d exf ðtÞ dt g f ðtÞ
!# ð2:18Þ
: t¼0
On the other hand, by (a)–(d) of the Proposition,
" Wn
# ! h i g 0 f ðtÞ f f ðtÞ exf ðtÞ g 0 ðtÞ f ðtÞ ext x 0 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 x 0 n Wn t¼0 gðtÞ f ðtÞ gðtÞ t¼0 g f ðtÞ f f ðtÞ g f ðtÞ t¼0 h i ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 n t¼0
1 f ðtÞ g 0 ðtÞf ðtÞ Pn Pn ext t¼0 W n x 0 0 gðtÞ t¼0 f ðtÞ gðtÞf ðtÞ t¼0 T
¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞ
T X1 n ½b0 x þ r0 ; b1 x þ r1 ; . . . ; bn x þ rn :
ð2:19Þ
89
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Equating the nth rows of (2.18) and (2.19), we obtain
n n X X n1 1 n ank sk ðxÞ ¼ ðxbk þ rk ÞsðnkÞ ðxÞ: k! k 1 k¼1 k¼0
ð2:20Þ
Noting that b0 ¼ r0 ¼ 0, we have the desired result.
h
Theorem 2.8. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the following identity:
n n X X n1 1 n ank sk ðxÞ ¼ ðxbk þ rk ÞsðnkÞ ðxÞ; k! k 1 k¼1 k¼1
ðkÞ where ak ¼ g f ðtÞ
t¼0
, bk ¼
ðkÞ
gðtÞf ðtÞ f 0 ðtÞ
,
rk ¼
t¼0
g 0 ðtÞf ðtÞ f 0 ðtÞ
ðkÞ
.
t¼0
xf ðtÞ Proof. Here we consider W n tg f ðtÞ dtd g e f ðtÞ . On one hand, by (b) and (c) of the Proposition, it is equal to ð Þ t¼0
Pn ½t t¼0
Pn g f ðtÞ t¼0
" Wn
d exf ðtÞ dt g f ðtÞ
!# ð2:21Þ
: t¼0
On the other hand, by (a)–(d) of the Proposition, it is equal to
" Wn
# h i f f ðtÞ exf ðtÞ f ðtÞ ext 0 xg f ðtÞ g 0 f ðtÞ 0 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 W ð xgðtÞ g ðtÞ Þ n 0 n t¼0 f ðtÞ gðtÞ t¼0 f f ðtÞ g f ðtÞ t¼0 h i 1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 Pn n t¼0 gðtÞ t¼0 ( 0 ) xt
gðtÞf ðtÞ g ðtÞf ðtÞ P n e t¼0 xW n þ Wn 0 0 f ðtÞ t¼0 f ðtÞ t¼0 ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT
T X1 n ½xb0 þ r0 ; xb1 þ r1 ; . . . ; xbn þ rn :
ð2:22Þ Equating the nth rows of (2.21) and (2.22), we have
n n X X n1 1 n ank sk ðxÞ ¼ ðxbk þ rk ÞsðnkÞ ðxÞ: k! k1 k¼1 k¼0 Noting that b0 ¼ r0 ¼ 0, we get the desired result.
h
3. Examples 3.1. Examples of Theorem 2.1 For any sequence sn ðxÞ ðgðtÞ; tÞ, we have ak ¼ bk ¼ g ðkÞ ð0Þ. Applying Theorem 2.1 to Bernoulli polynomials et 1 Appell 1 Bn ðxÞ t ; t , we have ak ¼ bk ¼ kþ1 and hence get
Bnþ1 ðxÞ ¼
n n X X n 1 1 1 1 BðnkÞ ðxÞ x Bk ðxÞ; k! k þ 1 k þ 2 n kþ2 k 1 k¼0 k¼1
ð3:1Þ
d k where BðnkÞ ðxÞ ¼ dx Bn ðxÞ. Thus, by (3.1), we get
n n X X n n 1 1 1 Bnk ðxÞ x Bk ðxÞ k þ 1 k þ 2 n kþ2 k k 1 k¼0 k¼1 n n X X n n 1 nþ1 1 nþ1 Bk ðxÞ ¼ Bnk ðxÞ: x x ¼ nkþ2 kþ2 k nkþ1 k kþ1 k¼0 k¼0
Bnþ1 ðxÞ ¼
ð3:2Þ
90
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
We now apply Theorem 2.1 to Euler polynomials En ðxÞ we have
et þ1 2
; t . Here a0 ¼ b0 ¼ 1 and ak ¼ bk ¼ 12, for all k 1, and hence
n n X n 1 1 1 ðkÞ 1X En ðxÞ þ ðx 1Þ Ek ðxÞ; Enþ1 ðxÞ ¼ x E n ð xÞ 2 2 k! 2 k¼1 k 1 k¼1 where EðnkÞ ðxÞ ¼
d k dx
ð3:3Þ
En ðxÞ.
Thus, by (3.3), we get
! ! n n X n n 1 1 1X Enk ðxÞ Ek ðxÞ En ðxÞ þ ðx 1Þ Enþ1 ðxÞ ¼ x 2 2 2 k¼1 k 1 k k¼1 ! n1 X n 1 1 1 Ek ðxÞ ¼ xEn ðxÞ þ ðx 1ÞEn ðxÞ þ ðx 1Þ 2 2 2 k k¼0 ! n n 1X E k ð xÞ 2 k¼0 k 1 ! ! n n X n n 1 1 1X k Ek ðxÞ E k ð xÞ ¼ xEn ðxÞ þ ðx 1Þ 2 2 2 k¼0 n k þ 1 k k k¼0 ! n n 1 1X nþ1 x Ek ðxÞ ¼ xEn ðxÞ þ 2 2 k¼0 k nkþ1 ! n n 1 1X nþ1 Enk ðxÞ: ¼ xEn ðxÞ þ x 2 2 k¼0 k kþ1
ð3:4Þ
ð3:5Þ
ð3:6Þ
t Here we apply Theorem 2.1 to Bernoulli polynomials of the second kind bn ðxÞ gðtÞ ¼ et 1 ; f ðtÞ ¼ et 1 , which are also ðkÞ called the Cauchy polynomials of the first kind and denoted by C n ðxÞ. Note that bk ¼ g ð0Þ ¼ Bk where Bk is the kth Bernoulli number. As is known, the Daehee numbers of the first kind Dn are given by the generating function 1 log ð1 þ tÞ X tk ¼ Dk : t k! k¼0
Observing that 1 X log ð1 þ t Þ tm 0 ðDm þ mDm1 Þ ; f f ðtÞ g f ðtÞ ¼ ð1 þ t Þ ¼ D0 þ t m! m¼1
ðkÞ ak ¼ f 0 f ðtÞ g f ðtÞ
t¼0
¼
1;
if k ¼ 0
Dk þ kDk1 ; if k > 0:
So we obtain
bnþ1 ðxÞ ¼
n X 1 k¼0
where
ðkÞ bn ðxÞ
¼
d k dx
k!
ðkÞ
ðBk x Bkþ1 Þbn ðxÞ
n X k¼1
n k1
ðDnkþ1 þ ðn k þ 1ÞDnk Þbk ðxÞ;
bn ðxÞ.
Let us consider the Bell polynomials Beln ðxÞ which are given by the generating function
exðe 1Þ ¼ t
1 X tk Belk ðxÞ : k! k¼0
ð3:7Þ
Thus, by (3.7), we get
Beln ðxÞ ð1; log ð1 þ t ÞÞ:
ð3:8Þ
From Theorem 2.1, we have b0 ¼ 1; bk ¼ 0, for k > 0 and ak ¼ ð1Þk . By Theorem 2.1, we get
Belnþ1 ðxÞ ¼ xBeln ðxÞ þ
n X ð1Þnk k¼1
n k1
Belk ðxÞ:
ð3:9Þ
91
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
3.2. Examples of Theorem 2.2 Applying to Bernoulli polynomials
t e 1 ;t ; t
Bn ðxÞ we have
n X n Bnk ðxÞxk ¼ Bn ; ð1Þk k k¼0
ðn 0Þ:
Applying to Euler polynomials En ðxÞ
n X n Enk ðxÞxk ¼ En ; ð1Þk k k¼0
et þ1 2
; t , we have
ðn 0Þ:
ð3:10Þ
The Laguerre polynomials LðnaÞ ðxÞ is defined by
LðnaÞ ðxÞ gðtÞ ¼ ð1 tÞa1 ; f ðtÞ ¼
As is known,
ak ¼
d ðaÞ L ðxÞ dx n
ðaþ1Þ
!ðkÞ 1 g f ðtÞ
¼ nLn1 ðxÞ. So
t : t1 d k dx
ðkÞ ¼ ð1 tÞa1
t¼0
ðaþkÞ
LðnaÞ ðxÞ ¼ ð1Þk ðnÞk Lnk ðxÞ, where ðaÞn ¼ aða 1Þ . . . ða n þ 1Þ. Also,
t¼0
¼ ða þ 1Þða þ 2Þ . . . ða þ kÞð1 t Þak1
t¼0
¼ ða þ kÞk :
Thus, by Theorem 2.2, we get n X n ðaþkÞ L ðxÞxk ¼ ða þ nÞn ; k nk k¼0
ðn 0Þ:
ð3:11Þ
3.3. Examples of Theorem 2.3 For any Appell polynomial sn ðxÞ ðgðtÞ; t Þ, Theorem 2.3 is reduced to
a0 snþ1 ðxÞ þ
n X k¼1
n k1
anþ1k sk ðxÞ ¼ ðx þ b0 Þxn þ
n1 X n bnk xk ; k k¼0
ðn 1Þ;
ð3:12Þ
where
ak ¼ g ðkÞ ð0Þ; bk ¼
0 ðkÞ g ðtÞ gðtÞ
:
t¼0
For Bn ðxÞ
et 1 t
1 ; t ; ak ¼ kþ1 . Noting that
1 0 g 0 ðtÞ et 1 1 X Bnþ1 tn 0 ¼ ðlog gðtÞÞ ¼ log et 1 log t ¼ t þ ¼ þ : gðtÞ 2 n¼1 n þ 1 n! e 1 t
Thus, we have
( bk ¼
12 ; Bkþ1 kþ1
if k ¼ 0 ; if k > 0:
By Theorem 2.3, we get
Bnþ1 ðxÞ þ
n
n1 n n n Bkþ1 nk 1 n X Bnkþ1 k 1 n X k1 x x x ¼ x x ; Bk ðxÞ ¼ x 2 2 nþ2k k nkþ1 k kþ1 k¼1 k¼0 k¼1
n X
ðn 1Þ: ð3:13Þ
For En ðxÞ
et þ1 2
; t ; a0 ¼ 1 and ak ¼ 12 for k > 0. Note that
k X k 1 1 X g 0 ðtÞ et 1 2et 1 t 1 t ¼ t ¼ ¼ E ð 1 Þ ¼ E d : k k 0;k gðtÞ 2 et þ 1 k¼0 2 e þ1 k! k¼0 2 k!
92
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Thus,
1 bk ¼ Ek d0;k ¼ 2
(
12 1 E 2 k
if k ¼ 0 if k > 0:
By Theorem 2.3, we get
Enþ1 ðxÞ þ
n n1 n n 1X 1 n 1X E k ð xÞ ¼ x x þ Enk xk ; 2 k¼1 k 1 2 2 k¼0 k
ðn 1Þ:
ð3:14Þ
ðbÞ
The actuarial polynomials an ðxÞ are Sheffer polynomial sequences for (gðtÞ ¼ ð1 tÞb ; f ðtÞ ¼ log ð1 tÞ). So
ð1Þ ð1 þ t Þ1 ; log ð1 þ t Þ , and the inverse wn ðxÞ of an ðxÞ under umbral composition is given by wn ðxÞ ðet ; et 1Þ. t Also, the Cauchy polynomials of the first kind C n ðxÞ are given by C n ðxÞ et 1 ; et 1 . Thus, we can apply Theorem 2.3 with sn ðxÞ ¼ C n ðxÞ, rn ðxÞ ¼ wn ðxÞ. Here ð1Þ an ðxÞ
1 log ð1 þ t Þ X tk g f ðtÞ ¼ ¼ Dk : t k! k¼0
So, ak ¼ Dk , for k 0. Also, we note that
! ! 1 1 X X g 0 f ðtÞ 1 t 1 tk 1 1 tk ¼ 1þt 1þt Ck 1þt1 t Ck ¼ ¼ t log ð1 þ t Þ t t 2 k! k! g f ðtÞ k¼0 k¼2 ! 1 1 k k X t 1 1 1 X C kþ1 t ¼ ¼ t Ck : t 2 2 k¼1 k þ 1 k! k! k¼2
ð3:15Þ
Thus, by (3.15), we get
! k g 0 f ðtÞ d C kþ1 bk ¼ ; ¼ dt k þ1 g f ðtÞ t¼0
1 b0 ¼ ; 2
for k > 0:
By Theorem 2.3, we get
n X n 1 Dnþ1k C k ðxÞ ¼ Dk C nþ1k ðxÞ ¼ x þ wn ðxÞ 2 k1 k k¼1 k¼0 n1 n X X n n C kþ1 C nkþ1 1 wn ðxÞ wk ðxÞ ¼ x þ wnk ðxÞ; 2 n k þ 1 k k kþ1 k¼0 k¼1
C nþ1 ðxÞ þ
n X
n
ð3:16Þ
where n 1. t Let wn ðxÞ ðet ; et 1Þ. The Daehee polynomials of the first kind Dn ðxÞ are given by Dn ðxÞ ðe 1 ; et 1Þ. Note that t ðkÞ a ¼ gðf ðtÞÞ ¼ C . Now, we observe that k
k
t¼0 ! t log ð1 þ tÞ 1þt g 0 f ðtÞ dt dg f ðtÞ 1 log ð1 þ tÞ 1 t ¼ ð1 þ t Þ ¼ 1 t þ ¼ 2 dt t t log ð1 þ tÞ g f ðtÞ df ðtÞ g f ðtÞ ðlog ð1 þ t ÞÞ ! 1 1 X 1 1 tk 1 X C kþ1 t k ¼ þ 1 t þ 1 þ t þ Ck : ¼ t 2 2 k¼1 k þ 1 k! k! k¼2
So, b0 ¼ 12, and
bk ¼
! k g 0 f ðtÞ d C kþ1 ; ¼ dt kþ1 g f ðtÞ t¼0
for k > 0:
Hence, by Theorem 2.3, we get
n X n 1 C nþ1k Dk ðxÞ ¼ C k Dnkþ1 ðxÞ ¼ x wn ðxÞ 2 k1 k k¼1 k¼0 n1 n X X n n C kþ1 C nkþ1 1 wn ðxÞ þ wk ðxÞ ¼ x wnk ðxÞ; þ 2 n k þ 1 k k kþ1 k¼0 k¼1
Dnþ1 ðxÞ þ
n X
n
ð3:17Þ
where n 1. Accidently, by comparing (3.16) and (3.17), we obtain the following duality result between the Cauchy and Daehee polynomials: n n X X n n Dk C nþ1k ðxÞ þ C k Dnþ1k ðxÞ ¼ 2xwn ðxÞ ¼ 2xðx 1Þn ¼ 2ðxÞnþ1 : k k k¼0 k¼0
ð3:18Þ
93
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Note here that 1 1 X X tk 1 tk wk ðxÞ ¼ ð1 þ t Þx ¼ ð1 þ t Þx1 ¼ ðx 1Þk : k! t þ 1 k! k¼0 k¼0
3.4. Examples of Theorem 2.4 We apply Theorem 2.4 to
Bn ðxÞ
t e 1 ;t : t
P t 0 k 1 t 1 Here, g f ðtÞ ¼ gðtÞ=f ðtÞ ¼ e 1 ¼ 1 k¼0 kþ1 k! . So, ak ¼ bk ¼ kþ1. Also, t
! ! 1 1 k 1 1 k g 0 ðtÞ 1 et 1 1 X 1 tk X t 1 X 1 tk X t t ¼ e ¼ ¼ 0 t t t k¼0 k þ 1 k! k¼0 k! t k¼1 k þ 1 k! k¼1 k! f ðtÞ ! k 1 1 1 kþ1 kþ1 X 1 t X 1 X 1 t 1 t ¼ : ¼ t k¼0 ðk þ 2Þðk þ 1Þ k! k þ 1 k þ 2 k! k! k¼0 k¼0
So,
rk ¼
d k g0 ðtÞ dt f 0 ðtÞ
t¼0
1 ¼ kþ2 . Hence, by Theorem 2.4, we get
n n n X X X n n n 1 1 1 1 Bnk ðxÞ Bnþ1 ðxÞ ¼ x B k ð xÞ Bk ðxÞ ¼ x k þ 1 k þ 2 n þ 2 k n kþ1 k k 1 k k¼0 k¼1 k¼0 n n n X X X n n n 1 1 Bk ðxÞ B k ð xÞ Bk ðxÞ ¼ x k nkþ2 k1 nkþ2 k nkþ1 k¼0 k¼0 k¼0 n n n X X n n n nþ2 B k ð xÞ Bk ðxÞ 1 X þ ¼x Bk ðxÞ: nkþ2 k k1 k n k þ 1 n þ 2 k¼0 k k¼0 k¼0
ð3:19Þ
Now, we apply Theorem 2.4 to
En ðxÞ
t e þ1 ;t : 2
0 ðtÞ t 0 Note that g f ðtÞ ¼ gðtÞ=f ðtÞ ¼ e 2þ1. So, a0 ¼ b0 ¼ 1, and ak ¼ bk ¼ 12, for k > 0. Also, g ¼ 12 et . So f 0 ðtÞ Hence, by Theorem 2.4, we get
n n X n n 1 1 1X 1 En ðxÞ þ ðx 1Þ Enk ðxÞ E k ð xÞ ¼ x En ðxÞ x 2 2 2 k¼1 k 1 2 k k¼1 n1 n n1 n n n n 1 X 1X 1 1 X E k ð xÞ þ Ek ðxÞ ¼ x En ðxÞ þ x Ek ðxÞ þ x 2 k¼0 k 2 k¼1 2 2 k¼0 k k k1 n n n nþ1 n nþ1 1X 1 1 1 X 1X Ek ðxÞ ¼ x En ðxÞ xEn þ x Ek ðxÞ Ek ðxÞ 2 k¼1 2 2 2 k¼0 k 2 k¼1 k k n n n nþ1 1 1 X 1X Ek ðxÞ Ek ðxÞ: ¼ ðx 1ÞEn ðxÞ þ x 2 2 k¼0 k 2 k¼1 k
rk ¼ 12, for all k 0.
Enþ1 ðxÞ ¼
We apply Theorem 2.4 to the Laguerre polynomial
LðnaÞ ðxÞ gðtÞ ¼ ð1 tÞa1 ; f ðtÞ ¼
t : t1
ðkÞ Thus, we have g f ðtÞ ¼ ð1Þk ða þ 1Þk ð1 t Þakþ1 . Hence,
ðk Þ ak ¼ g f ðtÞ ð0Þ ¼ ð1Þk ða þ 1Þk : We observe that
ðkÞ k gðtÞ d ¼ ð1 t Þaþ1 ¼ ða þ k 2Þk ð1 tÞakþ1 : 0 dt f ðtÞ
So, bk ¼
ðkÞ
gðtÞ f 0 ðtÞ
¼ ða þ k 2Þk . It is easy to show that t¼0
0 ðkÞ k g ðtÞ d ¼ ða þ 1Þð1 tÞa ¼ ða þ 1Þaða þ 1Þ . . . ða þ k 1Þð1 tÞak : 0 dt f ðtÞ
ð3:20Þ
94
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
So,
rk
k g0 ðtÞ ¼ dtd f 0 ðtÞ
t¼0
n X 1
ðaÞ
Lnþ1 ðxÞ ¼
k!
k¼0
¼ ða þ 1Þða þ k 1Þk . From Theorem 2.4, we note that
n X ðaþkÞ ða þ k 2Þk x þ ða þ 1Þða þ k 1Þk ð1Þk ðnÞk Lnk ðxÞ k¼1
n k1
ðaÞ ð1Þnþ1k ða þ 1Þnþ1k Lk ðxÞ: ð3:21Þ
By (3.21), we get
n n X X ðaþkÞ n n ðaÞ ð1Þk1 ða þ k 2Þk x ða þ 1Þða þ k 1Þk Lnk ðxÞ þ ð1Þk1 ða þ 1Þk Lnþ1k ðxÞ: k k k¼0 k¼1
ðaÞ
Lnþ1 ðxÞ ¼
ð3:22Þ
We apply Theorem 2.4 to the Cauchy polynomials of the first kind
C n ð xÞ
t t : ; e 1 et 1
P tk Note that g f ðtÞ ¼ log ðt1þtÞ ¼ 1 k¼0 Dk k! . Thus, we have
ðkÞ ak ¼ g f ðtÞ
t¼0
¼
k d g f ðtÞ t¼0 ¼ Dk ; dt
ðk 0Þ:
We observe that 1 X gðtÞ tet tk ¼ Bk ð1Þ : ¼ t 0 k! f ðtÞ e 1 k¼0
So, bk ¼
ðkÞ
gðtÞ f 0 ðtÞ
¼ Bk ð1Þ, and we have
t¼0
g 0 ðtÞ 1 0 ¼ f ðtÞ t
(
t et 1
2
tet t e 1
)
0 0 1 1 k 1 k k 1X BXB C Ct ¼ @ @ ABl Bkl Bk ð1ÞA t k¼0 l¼0 k! 1
0 0 1 1 0 0 1 1 k kþ1 1 k 1 kþ1 k1 k X X X X t 1 B B C C B B C Ct ¼ ¼ @ @ ABl Bkl Bk ð1ÞA @ @ ABl Bkþ1l Bkþ1 ð1ÞA : k þ 1 k! k! k¼1 l¼0 k¼0 l¼0 1 1 Thus, we get
rk
! 0 ðkÞ kþ1 X kþ1 g ðtÞ 1 Bl Bkþ1l Bkþ1 ð1Þ : ¼ ¼ 0 k þ 1 l¼0 1 f ðtÞ t¼0
From Theorem 2.4, we can derive the following recursive formula:
1 Pkþ1 k þ 1 Bl Bkþ1l Bkþ1 ð1ÞC n B n l¼0 X X n 1 C ðkÞ BBk ð1Þ Dnþ1k C k ðxÞ; C nþ1 ðxÞ ¼ xþ C C n ð xÞ B A @ k! ðk þ 1Þ! k1 k¼0 k¼1 0
where C ðnkÞ ðxÞ ¼
d k dx
ð3:23Þ
C n ðxÞ.
3.5. Examples of Theorem 2.5 Now, we apply Theorem 2.5 to
ðxÞn 1; et 1 : Note that
ak ¼ f ðkÞ ð0Þ ¼
0
if k ¼ 0
1 if k > 0:
So,
ðxÞn ¼ ðkÞ
nþ1 1 X 1 ðk Þ ð xÞ ; n þ 1 k¼1 k! nþ1
where ðxÞnþ1 ¼
d k dx
ðxÞnþ1 .
ð3:24Þ
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
95
It is known that
ðxÞnþ1 ¼
nþ1 X S1 ðn þ 1; lÞxl :
ð3:25Þ
l¼0
Thus, by (3.25), we get ðk Þ
ðxÞnþ1 ¼
nþ1 X ðlÞk S1 ðn þ 1; lÞxlk :
ð3:26Þ
l¼k
From (3.24) and (3.26), we have
ðxÞn ¼
nþ1 X nþ1 nþ1 nþ1k X 1 þ k 1 1 X 1 X S1 ðn þ 1; lÞxlk ¼ S1 ðn þ 1; l þ kÞxl n þ 1 k¼1 l¼k k n þ 1 k¼1 l¼0 k n X nk n nl X lþkþ1 lþkþ1 1 X 1 X S1 ðn þ 1; l þ k þ 1Þxl ¼ S1 ðn þ 1; l þ k þ 1Þxl ¼ n þ 1 k¼0 l¼0 n þ 1 l l l¼0 k¼0 n n X kþ1 1 X l S1 ðn þ 1; k þ 1Þx : ¼ n þ 1 k¼l 1 l¼0
ð3:27Þ
By (3.25) and (3.27), we get
S1 ðn; lÞ ¼
n kþ1 1 X S1 ðn þ 1; k þ 1Þ; n þ 1 k¼l 1
for 0 6 l 6 n:
We apply Theorem 2.5 to Bell polynomials
Beln ðxÞ ð1; log ð1 þ t ÞÞ: Note that f ðtÞ ¼ log ð1 þ tÞ; f
ak ¼ f ðkÞ ð0Þ ¼
ðkÞ
ðtÞ ¼ ð1Þk1 ðk 1Þ!ð1 þ t Þk , for k > 0. So,
0;
if k ¼ 0
ð1Þk1 ðk 1Þ!; if k > 0:
From Theorem 2.5, we note that
Beln ðxÞ ¼
nþ1 1 X ð1Þk1 ðkÞ Belnþ1 ðxÞ; n þ 1 k¼1 k
ðkÞ
where Belnþ1 ðxÞ ¼
Belnþ1 ðxÞ ¼
d k dx
ð3:28Þ
Belnþ1 ðxÞ. Since,
nþ1 X
S2 ðn þ 1; lÞxl :
l¼0
Thus, we have ðk Þ
Belnþ1 ðxÞ ¼
nþ1 nþ1k X X ðlÞk S2 ðn þ 1; lÞxlk ¼ ðl þ kÞk S2 ðn þ 1; l þ kÞxl : l¼k
ð3:29Þ
l¼0
By (3.28) and (3.29), we get
Beln ðxÞ ¼ ¼ ¼
nþ1 nþ1k n X nk X ðl þ k þ 1Þkþ1 1 X ðl þ kÞk 1 X ð1Þk1 ð1Þk S2 ðn þ 1; l þ k þ 1Þxl S2 ðn þ 1; l þ kÞxl ¼ n þ 1 k¼1 l¼0 n þ 1 k¼0 l¼0 k kþ1 n X nl ðl þ k þ 1Þkþ1 1 X ð1Þk S2 ðn þ 1; l þ k þ 1Þxl n þ 1 l¼0 k¼0 kþ1 n X l¼0
n ðk þ 1Þklþ1 1 X ð1Þkl S2 ðn þ 1; k þ 1Þxl : n þ 1 k¼l klþ1
ð3:30Þ
As we noted in the above,
Beln ðxÞ ¼
n X S2 ðn; lÞxl : l¼0
ð3:31Þ
96
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Therefore, by (3.30) and (3.31), we obtain the following identity:
S2 ðn; lÞ ¼
n kþ1 1 X S2 ðn þ 1; k þ 1Þ; ð1Þkl ðk lÞ! n þ 1 k¼l 1
ð3:32Þ
where 0 6 l 6 n. We apply Theorem 2.5 to the Laguerre polynomials LðnaÞ ðxÞ of order a which are given by
LðnaÞ ðxÞ ð1 t Þa1 ;
t : t1
ð3:33Þ
From f ð0Þ ¼ 0, we have a0 ¼ 0. By f
ðkÞ
ðtÞ ¼ ð1Þk k!ðt 1Þðkþ1Þ , ðk > 0Þ, we get
ak ¼ f ðkÞ ð0Þ ¼ ð1Þk k!ð1Þkþ1 ¼ k!; ðk > 0Þ: From Theorem 2.5, we have
LðnaÞ ðxÞ ¼
nþ1 nþ1 nþ1 ðkÞ 1 X 1 X 1 X ðaÞ ðaþkÞ ðaþkÞ Lnþ1 ðxÞ ¼ ð1Þk ðn þ 1Þk Lnþ1k ðxÞ ¼ ð1Þk1 ðn þ 1Þk Lnþ1k ðxÞ: n þ 1 k¼1 n þ 1 k¼1 n þ 1 k¼1
ð3:34Þ
We apply Theorem 2.5 to the Abel polynomials An ðx; aÞ; ða – 0Þ which are given by
An ðx; aÞ 1; teat :
ðkÞ k1 ðkÞ k1 From f ðtÞ ¼ teat , we have a0 ¼ f ð0Þ ¼ 0, f ðtÞ ¼ ka þ ak t eat , for k > 0. So, ak ¼ f ð0Þ ¼ ka , for k > 0. Now, we 0 at observe that f ðtÞAn ðx; aÞ ¼ te An ðx; aÞ ¼ nAn1 ðx; aÞ. Thus, we have An ðx þ a; aÞ ¼ nAn1 ðx; aÞ. By replacing x by x a, we get
A0n ðx; aÞ ¼ nAn1 ðx a; aÞ: Continuing this process, we have
AðnkÞ ðx; aÞ ¼ ðnÞk Ank ðx ka; aÞ;
for all k 0:
From Theorem 2.5, we note that
An ðx; aÞ ¼
k1 nþ1 nþ1 1 X ka 1 X k1 n þ 1 Anþ1k ðx ka; aÞ: ðn þ 1Þk Anþ1k ðx ka; aÞ ¼ ka n þ 1 k¼1 k! n þ 1 k¼1 k
ð3:35Þ
For sn ðxÞ ðgðtÞ; f ðtÞÞ, r n ðxÞ ð1; f ðtÞÞ, let
rn ðxÞ ¼
n X C n;k sk ðxÞ:
ð3:36Þ
k¼0
Then the first identity in Theorem 2.6 follows also from the formula for C n;k in [14, p.132]. Indeed,
1 k n 1 k n 1 g f ðtÞ t x ¼ g f ðtÞ t x ¼ g f ðtÞ ðnÞk xnk ¼ k! k! k! n ðnkÞ n g f ðtÞ ¼ ank : ¼ k k t¼0
C n;k ¼
n nk g f ðtÞ x k ð3:37Þ
So,
r n ð xÞ ¼
n X n k¼0
k
ank sk ðxÞ:
Also, the second identity in Theorem 2.6 follows from the conjugation representation in [14, p.108].
rn ðxÞ ¼
n n n X X 1 D k n E k X 1 k ðnÞ 1 k ðnÞ f ðtÞ x x ¼ f ðtÞ f ðtÞ ð0Þxk ¼ ð0Þxk ; k! k! k! k¼0 k¼0 k¼1
3.6. Examples of Theorem 2.7 Apply Theorem 2.7 to
Bn ðxÞ
t e 1 ;t : t
ðn 1Þ:
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
97
0 0 Note that f f ðtÞ ¼ f ðtÞ ¼ 1. Thus, we get
ðkÞ ak ¼ f 0 f ðtÞ
t¼0
and
f ðtÞ f 0 ðtÞ
1; if k ¼ 0
¼
¼ f ðtÞ ¼ t. So, bk ¼
0;
ðkÞ
f ðtÞ f 0 ðtÞ
if k > 0
¼ t¼0
1; 0;
if k ¼ 1 . Now, we observe that if k – 1:
1 1 X X g ðtÞf ðtÞ g ðtÞt ðte þ et 1Þ=t tet tk tk ¼ ¼ 1 ¼ 1 B ð 1 Þ ¼ B ð 1 Þ : ¼ k k 0 gðtÞ ðet 1Þ=t et 1 k! k! gðtÞf ðtÞ k¼0 k¼1 0
t
0
Thus, we get
rk ¼
ðkÞ 0 if k ¼ 0 g 0 ðtÞf ðtÞ ¼ 0 Bk ð1Þ if k > 0: gðtÞf ðtÞ t¼0
From Theorem 2.7, we have
nBn ðxÞ ¼ xBðn1Þ ðxÞ
n X B k ð 1Þ
k!
k¼1
where BðnkÞ ðxÞ ¼
d k dx
BðnkÞ ðxÞ ¼ nxBn1 ðxÞ
n X n Bk ð1ÞBnk ðxÞ; k k¼1
ð3:38Þ
Bk ðxÞ. So by (3.38), we get
n n 1 1X Bn1 ðxÞ Bk Bnk ðxÞ; Bn ðxÞ ¼ x 2 n k¼2 k
ðn 1Þ:
ð3:39Þ
Apply Theorem 2.7 to
t e þ1 ;t : 2
En ðxÞ
Thus, we have
ak ¼
1; if k ¼ 0 0;
if k > 0;
and bk ¼
1 if k ¼ 1; 0
if k – 1:
We observe that
g 0 ðtÞf ðtÞ g 0 ðtÞt tet =2 2et t ¼ ¼ ¼ ¼ 0 gðtÞ 2 ðet þ 1Þ=2 et þ 1 gðtÞf ðtÞ k 1 X 1 t kEk1 ð1Þ : ¼ 2 k! k¼1
! X kþ1 1 1 X tk t 1 t ¼ Ek ð1Þ E k ð 1Þ 2 2 k! k! k¼0 k¼0
Thus, we have
(
rk ¼
0;
if k ¼ 0
12 kEk1 ð1Þ;
if k > 0:
From Theorem 2.7, we have
nEn ðxÞ ¼ xEðn1Þ ðxÞ þ where EðnkÞ ðxÞ ¼
d k dx
n X 1 1 kEk1 ð1Þ EðnkÞ ðxÞ; k! 2 k¼1
ð3:40Þ
En ðxÞ. Note that
1 1 1 X X X tk 2et 2 tk tk Ek ð1Þ ¼ t ¼2 t ¼2 Ek ¼ 1 Ek : e þ1 k! e þ 1 k! k! k¼0 k¼0 k¼1
ð3:41Þ
Hence, by (3.41), we get
E k ð 1Þ ¼
1;
if k ¼ 0
Ek ; if k > 0:
ð3:42Þ
By (3.40) and (3.42), we see that
n n n 1 1X Ek1 1 1X En1 ðxÞ þ Ek1 Enk ðxÞ: nEn ðxÞ ¼ nxEn1 ðxÞ nEn1 ðxÞ þ ðnÞk Enk ðxÞ ¼ n x k 2 2 k¼2 ðk 1Þ! 2 2 k¼2 k
ð3:43Þ
98
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Thus, from (3.43), we have
E n ð xÞ ¼
n n 1 1 X En1 ðxÞ þ Ek1 Enk ðxÞ: x k 2 2n k¼2 k
ð3:44Þ
Apply Theorem 2.7 to the Mittag-Leffler polynomials
et 1 : Mn ðxÞ 1; f ðtÞ ¼ t e þ1 0 t 0 Note that f ðtÞ ¼ log 1þt . So, f f ðtÞ ¼ , f ðtÞ ¼ ðet2e 1t þ1Þ2
ðkÞ ak ¼ f 0 f ðtÞ
t¼0
8 1 if k ¼ 0 > <2 ¼ 1 if k ¼ 2 > : 0 if k – 0; 2:
2ð1þt 1tÞ 1t 2 . Thus, we have 2 ¼ 2 þ1Þ ð1þt 1t
Note that
t 2 f ðtÞ ðet þ 1Þ e 1 1 ¼ et et : ¼ 0 t t þ1 e 2 2e f ðtÞ Thus, we have bk ¼
ðkÞ
f ðtÞ f 0 ðtÞ
0; 1;
¼ t¼0
if k even and if k odd
rk ¼
ðkÞ
g 0 ðtÞf ðtÞ gðtÞf 0 ðtÞ
¼ 0. t¼0
For n 3, by Theorem 2.7, we get
n1 1 M n2 ðxÞ ¼ n M n ð xÞ 2 n3 where M ðnkÞ ðxÞ ¼
1 M n ðxÞ 2
d k dx
X 16k6n k1 mod 2
1 xM ðnkÞ ðxÞ; k!
ð3:45Þ
M n ðxÞ. Thus, by (3.45), we get
n1 x Mn2 ðxÞ ¼ n 2
X 16k6n k1 mod 2
1 ðk Þ M ðxÞ; k! n
ðn 3Þ:
ð3:46Þ
Now, we apply Theorem 2.7 to the Laguerre polynomials
t : LðnaÞ ðxÞ gðtÞ ¼ ð1 t Þa1 ; f ðtÞ ¼ 1t Note that
f ðtÞ ¼ t ; 1t
0
f ðtÞ ¼
1 ð1 t Þ2
g 0 ðtÞ ¼ ða þ 1Þð1 t Þa2 :
;
ð3:47Þ
Thus, we have
1 0 2 f f ðtÞ ¼ ¼ ð1 tÞ ; t 2 1 þ 1t
f ðtÞ ¼ t t2 0 f ðtÞ
ð3:48Þ
and
g 0 ðtÞf ðtÞ g 0 ðtÞ f ðtÞ ða þ 1Þð1 t Þa2 tð1 t Þ ¼ ða þ 1Þt: ¼ ¼ 0 gðtÞ f 0 ðtÞ gðtÞf ðtÞ ð1 t Þa1
ð3:49Þ
From (3.47)–(3.49), we note that
ðkÞ ak ¼ f 0 f ðtÞ
t¼0
ðkÞ f ðtÞ bk ¼ 0 f ðtÞ t¼0
8 1; > > > < 2; ¼ > 2; > > : 0;
8 0; > > > < 1; ¼ > 2; > > : 0;
if k ¼ 0 if k ¼ 1 if k ¼ 2 if k > 2; if k ¼ 0 if k ¼ 1 if k ¼ 2 if k > 2:
ð3:50Þ
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
99
Finally, we see that
rk ¼
ðkÞ ða þ 1Þ; if k ¼ 1 g 0 ðtÞf ðtÞ ¼ 0 0; if k – 1: gðtÞf ðtÞ t¼0
ð3:51Þ
For n 3, by (3.50), (3.51) and Theorem 2.7, we get
n o ðaÞ ðaÞ ðaþ1Þ ðaþ2Þ n LðnaÞ ðxÞ 2ðn 1ÞLn1 ðxÞ þ ðn 1Þðn 2ÞLn2 ðxÞ ¼ nðx a 1ÞLn1 ðxÞ nðn 1ÞxLn2 ðxÞ:
ð3:52Þ
That is, ðaÞ
ðaÞ
ðaþ1Þ
ðaþ2Þ
LðnaÞ ðxÞ 2ðn 1ÞLn1 ðxÞ þ ðn 1Þðn 2ÞLn2 ðxÞ ¼ ðx a 1ÞLn1 ðxÞ þ ðn 1ÞxLn2 ðxÞ:
ð3:53Þ
3.7. Examples of Theorem 2.8 Here we apply Theorem 2.8 to the Bernoulli polynomials
Bn ðxÞ
t e 1 ;t : t
It is easy to show 1 et 1 X 1 tk g f ðtÞ ¼ ¼ ; t k þ 1 k! k¼0
1 k X gðtÞf ðtÞ t ¼ et 1 ¼ 0 k! f ðtÞ k¼1
ð3:54Þ
and
k 1 g 0 ðtÞf ðtÞ et 1 X 1 t ¼ 1 : ¼ et þ 0 t k þ 1 k! f ðtÞ k¼0
ð3:55Þ
By (3.54) and (3.55), we get
ak ¼
1 ; kþ1
rk ¼
1 k 1¼ ; kþ1 kþ1
bk ¼
0; if k ¼ 0 1; if k > 0:
ð3:56Þ
From Theorem 2.8 and (3.56), we note that
n n n X X X n1 n 1 1 k k x Bnk ðxÞ: n Bk ðxÞ ¼ x ðnÞk Bnk ðxÞ ¼ k! kþ1 kþ1 k1 nkþ1 k k¼1 k¼1 k¼1
ð3:57Þ
We apply Theorem 2.8 to the Euler polynomials
En ðxÞ
t e þ1 ;t : 2
It is not difficult to show that
( ) ( ) 1 1 1 X X X et þ 1 1 1 k 1 1 k 1 tk g f ðtÞ ¼ ¼ t ¼ t ¼1þ ; 1þ 2þ 2 2 k! 2 k! 2 k! k¼0 k¼1 k¼1
ð3:58Þ
1 1 1 X X X gðtÞf ðtÞ et þ 1 1 kþ1 1 k tk t¼tþ t ¼tþ tk ¼ t þ ¼ 0 2 2k! 2ðk 1Þ! 2 k! f ðtÞ k¼1 k¼2 k¼2
ð3:59Þ
1 X g 0 ðtÞf ðtÞ 1 t k tk e t ¼ : ¼ 0 2 2 k! f ðtÞ k¼1
ð3:60Þ
and
Thus, by (3.58)–(3.60), we get
(
ak ¼
1; if k ¼ 0 1 ; 2
if k > 0;
8 > < 0; if k ¼ 0 bk ¼ 1; if k ¼ 1 > :k ; if k 2; 2
(
rk ¼
0;
if k ¼ 0
k ; 2
if k > 0;
ð3:61Þ
100
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
From Theorem 2.8 and (3.61), we have
( ) n1 n n X n1 n n 1X ðx 1Þ X 1 E k ð xÞ þ E n ð xÞ ¼ Enk ðxÞ þ n x En1 ðxÞ ðxbk þ rk ÞEnk ðxÞ ¼ k 2 k¼1 k 1 2 k¼2 2 k k k¼1 n n1 nðx 1Þ X 1 Enk ðxÞ þ n x En1 ðxÞ: ¼ 2 2 k1 k¼2
n
ð3:62Þ
Thus, by (3.62), we get
n1 n n1 n1 1X x 1X 1 Ek ðxÞ þ En ðxÞ ¼ Enk ðxÞ þ x En1 ðxÞ: 2 k¼1 k 1 2 k¼2 k 1 2
ð3:63Þ
Finally, we apply Theorem 2.8 to the Mittag-Leffler polynomials
et 1 : Mn ðxÞ 1; f ðtÞ ¼ t e þ1 0
Note that f ðtÞ ¼
2et 2. ðet þ1Þ
ðk Þ ak ¼ g f ðtÞ
So,
¼
t¼0
ðkÞ gðtÞf ðtÞ bk ¼ 0 f ðtÞ
1; if k ¼ 0 0; if k > 0;
¼
t¼0
1 t e et 2
ðkÞ
¼ t¼0
0; if k even 1; if k odd
and
rk
0 ðkÞ g ðtÞf ðtÞ ¼ ¼ 0: 0 gðtÞf ðtÞ t¼0
From Theorem 2.8, we have
X
nMn ðxÞ ¼ x
16k6n k1 mod 2
where M ðnkÞ ðxÞ ¼
d k dx
1 ðk Þ M ðxÞ; k! n
ðn 1Þ;
ð3:64Þ
M n ðxÞ.
Remark. It is easy to see that
log ð1 þ tÞ t t log ð1 þ t Þ ð1 þ t Þx þ ð1 þ t Þx ¼ 2ð1 þ tÞx : t log ð1 þ t Þ log ð1 þ t Þ t
ð3:65Þ
Thus, by (3.65), we get
nþ1 nþ1 X X nþ1 nþ1 Dk C nþ1k ðxÞ þ C k Dnþ1k ðxÞ ¼ 2ðxÞnþ1 ; k k k¼0 k¼0 which is different from the formula in (3.18). References [1] S. Araci, Novel identities involving Genocchi numbers and polynomials arising from applications of umbral calculus, Appl. Math. Comput. 233 (2014) 599–607 (MR 3215012). [2] A. Bayad, J. Chikhi, Non linear recurrences for Apostol–Bernoulli–Euler numbers of higher order, Adv. Stud. Contemp. Math. (Kyungshang) 22 (1) (2012) 1–6 (MR 2931600). [3] R.B. Corcino, C.B. Corcino, R. Aldema, Asymptotic normality of the ðr; bÞ-Stirling numbers, Ars Combin. 81 (2006) 81–96 (MR 2267804 (2008a:05010)). [4] R. Dere, Y. Simsek, Applications of umbral algebra to some special polynomials, Adv. Stud. Contemp. Math. (Kyungshang) 22 (3) (2012) 433–438 (MR 2976601). [5] A. Di B., D. Loeb, A selected survey of umbral calculus, Electron. J. Combin. 2 (1995) (Dynamic Survey 3, p. 28 (electronic). MR 1659320 (99j:05017)). [6] D. Ding, J. Yang, Some identities related to the Apostol–Euler and Apostol–Bernoulli polynomials, Adv. Stud. Contemp. Math. (Kyungshang) 20 (1) (2010) 7–21 (MR 2597988 (2011k:05030)). [7] S. Gaboury, R. Tremblay, B.-J. Fugère, Some explicit formulas for certain new classes of Bernoulli, Euler and Genocchi polynomials, Proc. Jangjeon Math. Soc. 17 (1) (2014) 115–123 (MR 3184467). [8] M.X. He, P.E. Ricci, Differential equation of Appell polynomials via the factorization method, J. Comput. Appl. Math. 139 (2) (2002) 231–237. MR1876637 (2003b:33017). [9] D.S. Kim, T. Kim, Higher-order Cauchy of the second kind and poly-Cauchy of the second kind mixed type polynomials, Ars Combin. 115 (2014) 435– 451.
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[10] D.S. Kim, T. Kim, J.J. Seo, Higher-order Daehee polynomials of the first kind with umbral calculus, Adv. Stud. Contemp. Math. (Kyungshang) 24 (1) (2014) 5–18 (MR 3157404). [11] T. Kim, Identities involving Laguerre polynomials derived from umbral calculus, Russ. J. Math. Phys. 21 (1) (2014) 36–45 (MR 3182545). [12] D.H. Lehmer, A new approach to Bernoulli polynomials, Am. Math. Mon. 95 (10) (1988) 905–911. MR979133 (90c:11014). [13] Q.-M. Luo, F. Qi, Relationships between generalized Bernoulli numbers and polynomials and generalized Euler numbers and polynomials, Adv. Stud. Contemp. Math. (Kyungshang) 7 (1) (2003) 11–18 (MR 1981601). [14] S. Roman, The Umbral Calculus, Pure and Applied Mathematics, vol. 111, Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York, 1984 (MR 741185 (87c:05015)). [15] K.H. Rosen, J.G. Michaels, J.L. Gross, J.W. Grossman, D.R. Shier, Handbook of Discrete and Combinatorial Mathematics, CRC Press, Boca Raton, FL, 2000 (MR 1725200 (2000g:05001)). [16] Y. Yang, C. Micek, Generalized Pascal functional matrix and its applications, Linear Algebra Appl. 423 (2–3) (2007) 230–245 (MR 2312403 (2007m:15008)). [17] H. Youn, Y. Yang, Differential equation and recursive formulas of Sheffer polynomial sequences, ISRN Discrete Math. 2011 (2011) 16 (no. Article ID476462). [18] Z. Zhang, H. Yang, Some closed formulas for generalized Bernoulli–Euler numbers and polynomials, Proc. Jangjeon Math. Soc. 11 (2) (2008) 191–198 (MR 2482602 (2010a:11036)).
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
A matrix approach to some identities involving Sheffer polynomial sequences Dae San Kim a, Taekyun Kim b,⇑ a b
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
a r t i c l e
i n f o
a b s t r a c t A simple but elegant method was adopted in Youn and Yang (2011) in order to derive a differential equation and recursive formulas for Sheffer polynomials. Namely, they used the so called the generalized Pascal functional matrix of an analytic function and the Wronskian matrix of several analytic functions. In this paper, we will use their method to find some identities satisfied by Sheffer polynomials. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Sheffer sequences Umbral calculus Generalized Pascal functional matrix Wronskian matrix
1. Introduction n o P tk Let F ¼ hðtÞ ¼ 1 k¼0 ak k! ak 2 C be the C-algebra of formal power series. For hðtÞ 2 F , the generalized Pascal functional matrix of hðtÞ, denoted by Pn ½hðtÞ, is an ðn þ 1Þ ðn þ 1Þ lower triangular matrix given by
8 > < i hðijÞ ðtÞ; if i P j Pn ½hðtÞij ¼ j > : 0; if i < j:
ð1:1Þ ðiÞ
i
Here we remind the reader of the fact that h ðtÞ stands for the ith order derivative of hðtÞ and h ðtÞ for the ith power of ð0Þ 0 hðtÞ. In particular, h ðtÞ ¼ hðtÞ, h ðtÞ ¼ 1. Also, for h1 ðtÞ; h2 ðtÞ; . . . ; hm ðtÞ 2 F , the nth order Wronskian matrix of h1 ðtÞ; h2 ðtÞ; . . . ; hm ðtÞ, denoted by W n ½h1 ðtÞ; h2 ðtÞ; . . . ; hm ðtÞ, is the ðn þ 1Þ m matrix given by
2
h1 ðtÞ 6 h0 ðtÞ 6 1 W n ½h1 ðtÞ; h2 ðtÞ; . . . ; hm ðtÞ ¼ 6 6 .. 4. ðn Þ
h2 ðtÞ
. . . hm ðtÞ
0 h2 ðtÞ
0 hm ðtÞ
.. .
...
.. . ðn Þ
ðnÞ
3 7 7 7: 7 5
h1 ðtÞ h2 ðtÞ . . . hm ðtÞ The following proposition, stated as Property 1 in [17], is of fundamental use throughout this paper. Proposition (a) For hðtÞ; lðtÞ 2 F ; a; b 2 C, we have ⇑ Corresponding author. E-mail addresses: [email protected] (D.S. Kim), [email protected] (T. Kim). http://dx.doi.org/10.1016/j.amc.2014.12.048 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
ð1:2Þ
84
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Pn ½ahðtÞ þ blðtÞ ¼ aPn ½hðtÞ þ bP n ½lðtÞ; W n ½ahðtÞ þ blðtÞ ¼ aW n ½hðtÞ þ bW n ½lðtÞ: (b) For hðtÞ; lðtÞ 2 F , we have
Pn ½lðtÞPn ½hðtÞ ¼ P n ½hðtÞPn ½lðtÞ ¼ P n ½hðtÞlðtÞ: (c) For hðtÞ; lðtÞ 2 F , we have
Pn ½lðtÞW n ½hðtÞ ¼ P n ½hðtÞW n ½lðtÞ ¼ W n ½lðtÞhðtÞ: 0
(d) For hðtÞ; lðtÞ 2 F , with lð0Þ ¼ 0; l ð0Þ – 0, we have
h i 2 n W n ½hðlðtÞÞt¼0 ¼ W n 1; lðtÞ; l ðtÞ; . . . ; l ðtÞ
t¼0
X1 n W n ½hðtÞt¼0 ;
where Xn ¼ diag ½0!; 1!; . . . ; n! is the diagonal matrix with the diagonal entries 0!; 1!; . . . ; n!. Let gðtÞ be an invertible series, and let f ðtÞ be a delta series. Then there is a unique sequence of polynomials sn ðxÞ such that D E gðtÞf ðtÞk sn ðxÞ ¼ n!dn;k , ðn; k 0Þ. The sequence sn ðxÞ is called Sheffer polynomial sequence for ðgðtÞ; f ðtÞÞ which is denoted by sn ðxÞ ðgðtÞ; f ðtÞÞ. It is known that 1 X 1 tk sn ðxÞ ðgðtÞ; f ðtÞÞ () exf ðtÞ ¼ s k ð xÞ ; k! g f ðtÞ k¼0
ð1:3Þ
where f ðtÞ is the compositional inverse of f ðtÞ determined by f f ðtÞ ¼ f ðf ðtÞÞ ¼ t (see [14]). Thus, by (1.3), we get
k d 1 xf ðtÞ sk ðxÞ ¼ e dt g f ðtÞ
;
ðk 0Þ:
t¼0
Sheffer polynomial sequences arise in numerous problems of applied mathematics, theoretical physics, approximation theory, and several other mathematical branches (see [5]). In this past few decades, there has been a renewed interest in Sheffer polynomials (see [1–18]). The following lemma was shown by Youn and Yang. Lemma [17]. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have
h i 2 n W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT X1 n ¼ W n 1; f ðtÞ; f ðtÞ ; . . . ; f ðtÞ
t¼0
X1 n Pn
1 gðtÞ
t¼0
Pn ext t¼0 ;
T
where W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞ is the transpose of W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞ. A simple but elegant method was adopted in [17] in order to derive a differential equation and recursive formulas for Sheffer polynomial sequences. Namely, they used the so called the generalized Pascal functional matrix of an analytic function and the Wronskian matrix of several analytic functions. In this paper, we will use their method to find some identities of special polynomials satisfied by Sheffer polynomial sequences. 2. A matrix approach to some identities involving Sheffer polynomial sequences From the method of Youn and Yang in [17], we derive some identities of special polynomials satisfied by Sheffer polynomial sequences in this section. Theorem 2.1. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the recursive formula:
1 ; and g ð0Þ n n X X n 1 a0 snþ1 ðxÞ ¼ bk x bkþ1 sðnkÞ ðxÞ ankþ1 sk ðxÞ; k! k1 k¼0 k¼1 s 0 ð xÞ ¼
0 ðkÞ where ak ¼ f f ðtÞ g f ðtÞ
t¼0
and bk ¼
d k dt
gðtÞ
t¼0
ðn 0Þ;
.
Proof. Now, we consider
" Wn
d exf ðtÞ f f ðtÞ g f ðtÞ dt g f ðtÞ 0
!# : t¼0
ð2:1Þ
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
85
On one hand, by (c) of the Proposition, we get
" !# 0
0
d exf ðtÞ Pn f f ðtÞ g f ðtÞ t¼0 W n ¼ Pn f f ðtÞ g f ðtÞ t¼0 ½s1 ðxÞ; s2 ðxÞ; . . . ; snþ1 ðxÞT : dt g f ðtÞ t¼0
ð2:2Þ
It is easy to show that
d exf ðtÞ dt g f ðtÞ
!
! g 0 f ðtÞ 1 exf ðtÞ x 0 : g f ðtÞ f f ðtÞ g f ðtÞ
¼
ð2:3Þ
So, on the other hand, by (a)–(d) of Proposition, we obtain
# ! xt h i g 0 f ðtÞ e 0 x exf ðtÞ ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 W ð xgðtÞ g ðtÞ Þ n n t¼0 gðtÞ g f ðtÞ t¼0 t¼0 h i
1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 Pn ext t¼0 W n ½xgðtÞ g 0 ðtÞt¼0 n Pn t¼0 gðtÞ t¼0 ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT X1 xW n ½gðtÞt¼0 W n ½g 0 ðtÞt¼0 : n
" Wn
ð2:4Þ
Equating (2.2) and (2.4), we get
2
a0 0 a0
6a 6 1 6 6 6 a2 6 6 6. 6. 6. 6 4
a1
1
... 0
0
2
an
3 3 2 s 0 ð xÞ 2 s1 ðxÞ ... 0 7 7 76 s2 ðxÞ 7 6 s 1 ð xÞ 76 7 6 7 6 6 ... 0 7 6 . 76 7 6 s ð xÞ .. 76 7¼6 2 7 6 6 . . .. 7 .. 76 7 6 .. 5 4. . . 74 . 7 5 ð x Þ s nþ1 s n ð xÞ . . . a0
0
a0
.. . n
.. . n
1
2
an1
an2
0 s01 ðxÞ 1! s02 ðxÞ 1!
0 0 s002 ðxÞ 2!
.. .
.. .
s0n ðxÞ 1!
s00n ðxÞ 2!
32 3 b x b1 76 0 . . . 0 76 b1 x b2 7 7 76 7 b xb 7 7 . . . 0 76 2 3 7: 76 7 76 .. 7 6 . . .. 74 . 5 . . 5 ðnÞ sn ðxÞ x b b n nþ1 ... n! ... 0
Comparing the nth rows, we obtain n X n
ank skþ1 ðxÞ ¼
k
k¼0
n X 1
k! k¼0
sðnkÞ ðxÞ bk x bkþ1 :
ð2:5Þ
Equivalently, n X 1
a0 snþ1 ðxÞ ¼
k¼0
k!
n X bk x bkþ1 sðnkÞ ðxÞ k¼1
0
Remark. Note that a0 ¼ f ð0Þg ð0Þ. Also, if gðtÞ ¼
n k1
P1
ankþ1 sk ðxÞ:
tk k¼0 ak k! ,
ð2:6Þ
then bk ¼ g ðkÞ ð0Þ ¼ ak .
Theorem 2.2. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the following identity: n X ð1Þk
k!
k¼0
where ak ¼
sðnkÞ ðxÞxk ¼ an ; ðkÞ
1 g ðf ðtÞÞ
.
t¼0
Proof. Let’s consider W n
½a0 ; a1 ; . . . ; an T
1 g ðf ðtÞÞ
. On one hand, it is equal to
t¼0
ð2:7Þ
On the other hand, by (b)–(d) of the Proposition, it is equal to
" Wn
# " # xt h i 1 exf ðtÞ xf ðtÞ e 2 n 1 xt ¼ W ¼ W f ðtÞ; f ðtÞ f ðtÞ e 1; ; . . . ; X W e n n n n t¼0 gðtÞ gf ðtÞ g f ðtÞ t¼0 h i
1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 Pn ext t¼0 W n ext t¼0 n Pn t¼0 gðtÞ t¼0 h iT ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT X1 1; x; ðxÞ2 ; . . . ; ðxÞn : n
ð2:8Þ
86
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Equating the nth rows of (2.7) and (2.8), we get
an ¼
n X 1 k¼0
k!
sðnkÞ ðxÞðxÞk ¼
n X ð1Þk
k!
k¼0
sðnkÞ ðxÞxk :
ð2:9Þ
0 Theorem 2.3. Let sn ðxÞ ðgðtÞ; f ðtÞÞ, rn ðxÞ f ðtÞ; f ðtÞ . Then we have the following identity:
a0 snþ1 ðxÞ þ
n X k¼1
n
anþ1k sk ðxÞ ¼ ðx þ b0 Þrn ðxÞ þ
k1
where
ðkÞ ak ¼ g f ðtÞ
t¼0
bk ¼
;
!ðkÞ g 0 f ðtÞ g f ðtÞ
n1 X n bnk r k ðxÞ; k k¼0
ðn 1Þ;
:
t¼0
xf ðtÞ Proof. Here we consider W n g f ðtÞ dtd g e f ðtÞ . On one hand, by (c) of the Proposition, it is equal to ð Þ t¼0
"
d exf ðtÞ dt g f ðtÞ
Pn
!#
W n g f ðtÞ t¼0 :
ð2:10Þ
t¼0
On the other hand, by (a), (c) of the Proposition, it is equal to
" Wn
! xf ðtÞ # g 0 f ðtÞ e x 0 g f ðtÞ f f ðtÞ
"
" # # g 0 f ðtÞ exf ðtÞ ¼ Pn x W n 0 g f ðtÞ t¼0 f f ðtÞ t¼0 " # ! g 0 f ðtÞ ¼ xI þ Pn ½r 0 ðxÞ; r 1 ðxÞ; . . . ; r n ðxÞT : g f ðtÞ
t¼0
ð2:11Þ
t¼0
ðkÞ Now, let ak ¼ g f ðtÞ
2
s 1 ð xÞ
6 s ð xÞ 6 2 6 6 6 s 3 ð xÞ 6 6 6. 6. 6. 6 4 snþ1 ðxÞ
t¼0
, bk ¼
g 0 ðf ðtÞÞ g ðf ðtÞÞ
ðkÞ
. Equating (2.10) and (2.11), we have
t¼0
0
0
... 0
s 1 ð xÞ 2 s 2 ð xÞ 1 .. . n s n ð xÞ 1
0 2 s 1 ð xÞ 2 .. . n sn1 ðxÞ 2
... 0
3
2
3
2
x þ b0
7 a0 6b 7 6 1 76 7 6 a 76 1 7 6 76 7 6 b 2 ... 0 76 a2 7 6 76 7 ¼ 6 76 . 7 6 . . . .. 76 . 7 6 . 74 . 5 6 . . . 7 6 5 an 4 bn . . . ð n Þs1 ðxÞ
0
0
... 0
x þ b0 2 b1 1 .. . n bn1 1
0
... 0
x þ b0
... 0
.. . n bn2 2
..
Equating the nth rows, we have n X n k¼0
k
ak snþ1k ðxÞ ¼
n X n k¼0
k
bnk rk ðxÞ þ xr n ðxÞ:
Equivalently,
a0 snþ1 ðxÞ þ
n X k¼1
n k1
anþ1k sk ðxÞ ¼ ðx þ b0 Þrn ðxÞ þ
n1 X n bnk r k ðxÞ: k k¼0
0
Remark. a0 ¼ g ð0Þ; b0 ¼ ggðð00ÞÞ. Theorem 2.4. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the following recursive formula:
1 ; and g ð0Þ n n X X n 1 a0 snþ1 ðxÞ ¼ ðbk x þ rk ÞsðnkÞ ðxÞ anþ1k sk ðxÞ; k! k1 k¼0 k¼1 s 0 ð xÞ ¼
ðkÞ where ak ¼ g f ðtÞ
ðkÞ , bk ¼ fgðtÞ 0 ðtÞ t¼0
, t¼0
rk
0 ðtÞ ðkÞ ¼ g f 0 ðtÞ
t¼0
.
ðn 0Þ;
.
.. .
. . . x þ b0
3 3 2 7 r 0 ðxÞ 7 76 r ðxÞ 7 76 1 7 76 7 76 r 2 ðxÞ 7 76 7: 76 . 7 76 . 7 74 . 5 7 5 r n ðxÞ
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
87
xf ðtÞ Proof. As in the proof of Theorem 3, we consider W n g f ðtÞ dtd g e f ðtÞ . On one hand, by (c) of the Proposition, it is equal ð Þ t¼0 to
" Pn
d exf ðtÞ dt g f ðtÞ
!#
W n g f ðtÞ t¼0 :
ð2:12Þ
t¼0
On the other hand, by (a)–(d) of the Proposition, it is equal to
"
! xf ðtÞ # xt h i g f ðtÞ g 0 f ðtÞ e e gðtÞ g 0 ðtÞ x 0 0 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 W x n 0 0 n t¼0 gðtÞ f ðtÞ f ðtÞ t¼0 f f ðtÞ f f ðtÞ g f ðtÞ t¼0 h i 1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 n Pn t¼0 gðtÞ t¼0
gðtÞ g 0 ðtÞ P n ext t¼0 W n x 0 0 f ðtÞ f ðtÞ t¼0 ( 0 ) gðtÞ g ðtÞ : ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT X1 xW þ W n n 0 0 n f ðtÞ t¼0 f ðtÞ t¼0
Wn
ð2:13Þ
Equating (2.12) and (2.13), we get n X n k¼0
ak snþ1k ðxÞ ¼
k
n X 1 k¼0
k!
sðnkÞ ðxÞðbk x þ rk Þ:
Equivalently, we have
a0 snþ1 ðxÞ ¼
n X 1 k¼0
k!
ðbk x þ rk ÞsðnkÞ ðxÞ
n X k¼1
n
k1
anþ1k sk ðxÞ:
Theorem 2.5. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the identity:
sn ðxÞ ¼ where ak ¼ f
nþ1 1 X ak ðkÞ s ðxÞ; n þ 1 k¼1 k! nþ1
ðkÞ
ðn 0Þ;
ð0Þ.
xf ðtÞ Proof. Here we consider W n t g e f ðtÞ . On one hand, by (c) of the Proposition, it is equal to ð Þ t¼0
" Pn ½t t¼0 W n
1 exf ðtÞ g f ðtÞ
# ð2:14Þ
: t¼0
On the other hand, by (b)–(d) of the Proposition, it is equal to
" # exf ðtÞ
ext W n f f ðtÞ ¼ W n 1; f ðtÞ; . . . ; f ðtÞn t¼0 X1 W f ðtÞ n n gðtÞ t¼0 g f ðtÞ t¼0 h i
1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . f ðtÞn X1 P Pn ext t¼0 n n t¼0 gðtÞ t¼0 ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT Equating the nth rows of (2.14) and (2.15), we have
nsn1 ðxÞ ¼
n X 1 k¼0
k!
ak sðnkÞ ðxÞ:
Replacing n by n þ 1, we get the desired result. h
X1 n W n ½f ðtÞt¼0 :
W n ½f ðtÞt¼0 ð2:15Þ
88
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Theorem 2.6. Let sn ðxÞ ðgðtÞ; f ðtÞÞ, rn ðxÞ ð1; f ðtÞÞ. Then we have the following identity:
r n ð xÞ ¼
n X n
k
k¼0
ank sk ðxÞ ¼
ðkÞ where ak ¼ g f ðtÞ
t¼0
n X k ðnÞ xk f ðtÞ ð0Þ ; k! k¼1
ðn P 1Þ;
.
Proof. Here we consider
h i W n exf ðtÞ
t¼0
" # exf ðtÞ ¼ W n g f ðtÞ : g f ðtÞ t¼0
On one hand, by (c) of the Proposition, it is equal to
Pn g f ðtÞ t¼0 W n
"
exf ðtÞ g f ðtÞ
#
¼ P n g f ðtÞ t¼0 ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT :
ð2:16Þ
t¼0
On the other hand, as exf ðtÞ ¼
P1
tk k¼0 r k ðxÞ k! ,
it is equal to
½r 0 ðxÞ; r 1 ðxÞ; . . . ; r n ðxÞT : Also, it is equal to
h i W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn
xt X1 n Wn e
t¼0
t¼0
ð2:17Þ
:
Equating (2.16) and (2.17), we get
2
a0 0 6a a 6 1 0 6 6 6 a2 6 6 6. 6. 6. 6 4
an
... 0
0
2 1
3 2 1 3 2 7 s ð x Þ 6 0 ... 0 7 76 s ðxÞ 7 6 0 76 1 7 6 7 6 6 ... 0 7 6 76 s2 ðxÞ 7 6 0 76 7¼6 7 6 6 . . .. 7 76 .. 7 6. 5 6 .. . . 74 . 7 4 5 s n ð xÞ . . . a0 0
0
a1
a0
.. . n
.. . n
1
2
an1
an2
0 f ð1Þ ð0Þ 1! f ð2Þ ð0Þ 1!
.. . f ðnÞ ð0Þ 1!
3 0 ... 0 2 3 ð1Þ ð1Þ 1 2 n ð f Þ ð0 Þ ð f Þ ð0 Þ 7 7 . . . 76 x 7 2! n! 76 7 ð2Þ ð2Þ 6 7 ðf 2 Þ ð0Þ ðf n Þ ð0Þ 7 7 6 x2 7 . . . 2! n! 76 7 76 .. 7 .. . . .. 74 . 5 7 . . . 5 xn ð n Þ ð n Þ ðf 2 Þ ð0Þ ðf n Þ ð0Þ ... 2! n!
Equating the nth rows, we have
r n ð xÞ ¼
n X n k¼0
k
ank sk ðxÞ ¼
n X 1 k ðnÞ f ðtÞ ð0Þxk : k! k¼1
Theorem 2.7. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the following identity:
n n X X n1 1 n ank sk ðxÞ ¼ ðxbk þ rk ÞsðnkÞ ðxÞ; k! k 1 k¼1 k¼1
0 ðkÞ where ak ¼ f f ðtÞ
t¼0
, bk ¼
ðkÞ
f ðtÞ f 0 ðtÞ
,
rk ¼
t¼0
ðkÞ
g 0 ðtÞf ðtÞ gðtÞf 0 ðtÞ
. t¼0
xf ðtÞ 0 Proof. Let us consider W n tf f ðtÞ dtd g e f ðtÞ . On one hand, by (b) and (c) of the Proposition, it is equal to ð Þ t¼0
0
Pn ½t t¼0 P n f f ðtÞ t¼0
" Wn
d exf ðtÞ dt g f ðtÞ
!# ð2:18Þ
: t¼0
On the other hand, by (a)–(d) of the Proposition,
" Wn
# ! h i g 0 f ðtÞ f f ðtÞ exf ðtÞ g 0 ðtÞ f ðtÞ ext x 0 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 x 0 n Wn t¼0 gðtÞ f ðtÞ gðtÞ t¼0 g f ðtÞ f f ðtÞ g f ðtÞ t¼0 h i ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 n t¼0
1 f ðtÞ g 0 ðtÞf ðtÞ Pn Pn ext t¼0 W n x 0 0 gðtÞ t¼0 f ðtÞ gðtÞf ðtÞ t¼0 T
¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞ
T X1 n ½b0 x þ r0 ; b1 x þ r1 ; . . . ; bn x þ rn :
ð2:19Þ
89
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Equating the nth rows of (2.18) and (2.19), we obtain
n n X X n1 1 n ank sk ðxÞ ¼ ðxbk þ rk ÞsðnkÞ ðxÞ: k! k 1 k¼1 k¼0
ð2:20Þ
Noting that b0 ¼ r0 ¼ 0, we have the desired result.
h
Theorem 2.8. Let sn ðxÞ ðgðtÞ; f ðtÞÞ. Then we have the following identity:
n n X X n1 1 n ank sk ðxÞ ¼ ðxbk þ rk ÞsðnkÞ ðxÞ; k! k 1 k¼1 k¼1
ðkÞ where ak ¼ g f ðtÞ
t¼0
, bk ¼
ðkÞ
gðtÞf ðtÞ f 0 ðtÞ
,
rk ¼
t¼0
g 0 ðtÞf ðtÞ f 0 ðtÞ
ðkÞ
.
t¼0
xf ðtÞ Proof. Here we consider W n tg f ðtÞ dtd g e f ðtÞ . On one hand, by (b) and (c) of the Proposition, it is equal to ð Þ t¼0
Pn ½t t¼0
Pn g f ðtÞ t¼0
" Wn
d exf ðtÞ dt g f ðtÞ
!# ð2:21Þ
: t¼0
On the other hand, by (a)–(d) of the Proposition, it is equal to
" Wn
# h i f f ðtÞ exf ðtÞ f ðtÞ ext 0 xg f ðtÞ g 0 f ðtÞ 0 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 W ð xgðtÞ g ðtÞ Þ n 0 n t¼0 f ðtÞ gðtÞ t¼0 f f ðtÞ g f ðtÞ t¼0 h i 1 ¼ W n 1; f ðtÞ; f ðtÞ2 ; . . . ; f ðtÞn X1 Pn n t¼0 gðtÞ t¼0 ( 0 ) xt
gðtÞf ðtÞ g ðtÞf ðtÞ P n e t¼0 xW n þ Wn 0 0 f ðtÞ t¼0 f ðtÞ t¼0 ¼ W n ½s0 ðxÞ; s1 ðxÞ; . . . ; sn ðxÞT
T X1 n ½xb0 þ r0 ; xb1 þ r1 ; . . . ; xbn þ rn :
ð2:22Þ Equating the nth rows of (2.21) and (2.22), we have
n n X X n1 1 n ank sk ðxÞ ¼ ðxbk þ rk ÞsðnkÞ ðxÞ: k! k1 k¼1 k¼0 Noting that b0 ¼ r0 ¼ 0, we get the desired result.
h
3. Examples 3.1. Examples of Theorem 2.1 For any sequence sn ðxÞ ðgðtÞ; tÞ, we have ak ¼ bk ¼ g ðkÞ ð0Þ. Applying Theorem 2.1 to Bernoulli polynomials et 1 Appell 1 Bn ðxÞ t ; t , we have ak ¼ bk ¼ kþ1 and hence get
Bnþ1 ðxÞ ¼
n n X X n 1 1 1 1 BðnkÞ ðxÞ x Bk ðxÞ; k! k þ 1 k þ 2 n kþ2 k 1 k¼0 k¼1
ð3:1Þ
d k where BðnkÞ ðxÞ ¼ dx Bn ðxÞ. Thus, by (3.1), we get
n n X X n n 1 1 1 Bnk ðxÞ x Bk ðxÞ k þ 1 k þ 2 n kþ2 k k 1 k¼0 k¼1 n n X X n n 1 nþ1 1 nþ1 Bk ðxÞ ¼ Bnk ðxÞ: x x ¼ nkþ2 kþ2 k nkþ1 k kþ1 k¼0 k¼0
Bnþ1 ðxÞ ¼
ð3:2Þ
90
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
We now apply Theorem 2.1 to Euler polynomials En ðxÞ we have
et þ1 2
; t . Here a0 ¼ b0 ¼ 1 and ak ¼ bk ¼ 12, for all k 1, and hence
n n X n 1 1 1 ðkÞ 1X En ðxÞ þ ðx 1Þ Ek ðxÞ; Enþ1 ðxÞ ¼ x E n ð xÞ 2 2 k! 2 k¼1 k 1 k¼1 where EðnkÞ ðxÞ ¼
d k dx
ð3:3Þ
En ðxÞ.
Thus, by (3.3), we get
! ! n n X n n 1 1 1X Enk ðxÞ Ek ðxÞ En ðxÞ þ ðx 1Þ Enþ1 ðxÞ ¼ x 2 2 2 k¼1 k 1 k k¼1 ! n1 X n 1 1 1 Ek ðxÞ ¼ xEn ðxÞ þ ðx 1ÞEn ðxÞ þ ðx 1Þ 2 2 2 k k¼0 ! n n 1X E k ð xÞ 2 k¼0 k 1 ! ! n n X n n 1 1 1X k Ek ðxÞ E k ð xÞ ¼ xEn ðxÞ þ ðx 1Þ 2 2 2 k¼0 n k þ 1 k k k¼0 ! n n 1 1X nþ1 x Ek ðxÞ ¼ xEn ðxÞ þ 2 2 k¼0 k nkþ1 ! n n 1 1X nþ1 Enk ðxÞ: ¼ xEn ðxÞ þ x 2 2 k¼0 k kþ1
ð3:4Þ
ð3:5Þ
ð3:6Þ
t Here we apply Theorem 2.1 to Bernoulli polynomials of the second kind bn ðxÞ gðtÞ ¼ et 1 ; f ðtÞ ¼ et 1 , which are also ðkÞ called the Cauchy polynomials of the first kind and denoted by C n ðxÞ. Note that bk ¼ g ð0Þ ¼ Bk where Bk is the kth Bernoulli number. As is known, the Daehee numbers of the first kind Dn are given by the generating function 1 log ð1 þ tÞ X tk ¼ Dk : t k! k¼0
Observing that 1 X log ð1 þ t Þ tm 0 ðDm þ mDm1 Þ ; f f ðtÞ g f ðtÞ ¼ ð1 þ t Þ ¼ D0 þ t m! m¼1
ðkÞ ak ¼ f 0 f ðtÞ g f ðtÞ
t¼0
¼
1;
if k ¼ 0
Dk þ kDk1 ; if k > 0:
So we obtain
bnþ1 ðxÞ ¼
n X 1 k¼0
where
ðkÞ bn ðxÞ
¼
d k dx
k!
ðkÞ
ðBk x Bkþ1 Þbn ðxÞ
n X k¼1
n k1
ðDnkþ1 þ ðn k þ 1ÞDnk Þbk ðxÞ;
bn ðxÞ.
Let us consider the Bell polynomials Beln ðxÞ which are given by the generating function
exðe 1Þ ¼ t
1 X tk Belk ðxÞ : k! k¼0
ð3:7Þ
Thus, by (3.7), we get
Beln ðxÞ ð1; log ð1 þ t ÞÞ:
ð3:8Þ
From Theorem 2.1, we have b0 ¼ 1; bk ¼ 0, for k > 0 and ak ¼ ð1Þk . By Theorem 2.1, we get
Belnþ1 ðxÞ ¼ xBeln ðxÞ þ
n X ð1Þnk k¼1
n k1
Belk ðxÞ:
ð3:9Þ
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D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
3.2. Examples of Theorem 2.2 Applying to Bernoulli polynomials
t e 1 ;t ; t
Bn ðxÞ we have
n X n Bnk ðxÞxk ¼ Bn ; ð1Þk k k¼0
ðn 0Þ:
Applying to Euler polynomials En ðxÞ
n X n Enk ðxÞxk ¼ En ; ð1Þk k k¼0
et þ1 2
; t , we have
ðn 0Þ:
ð3:10Þ
The Laguerre polynomials LðnaÞ ðxÞ is defined by
LðnaÞ ðxÞ gðtÞ ¼ ð1 tÞa1 ; f ðtÞ ¼
As is known,
ak ¼
d ðaÞ L ðxÞ dx n
ðaþ1Þ
!ðkÞ 1 g f ðtÞ
¼ nLn1 ðxÞ. So
t : t1 d k dx
ðkÞ ¼ ð1 tÞa1
t¼0
ðaþkÞ
LðnaÞ ðxÞ ¼ ð1Þk ðnÞk Lnk ðxÞ, where ðaÞn ¼ aða 1Þ . . . ða n þ 1Þ. Also,
t¼0
¼ ða þ 1Þða þ 2Þ . . . ða þ kÞð1 t Þak1
t¼0
¼ ða þ kÞk :
Thus, by Theorem 2.2, we get n X n ðaþkÞ L ðxÞxk ¼ ða þ nÞn ; k nk k¼0
ðn 0Þ:
ð3:11Þ
3.3. Examples of Theorem 2.3 For any Appell polynomial sn ðxÞ ðgðtÞ; t Þ, Theorem 2.3 is reduced to
a0 snþ1 ðxÞ þ
n X k¼1
n k1
anþ1k sk ðxÞ ¼ ðx þ b0 Þxn þ
n1 X n bnk xk ; k k¼0
ðn 1Þ;
ð3:12Þ
where
ak ¼ g ðkÞ ð0Þ; bk ¼
0 ðkÞ g ðtÞ gðtÞ
:
t¼0
For Bn ðxÞ
et 1 t
1 ; t ; ak ¼ kþ1 . Noting that
1 0 g 0 ðtÞ et 1 1 X Bnþ1 tn 0 ¼ ðlog gðtÞÞ ¼ log et 1 log t ¼ t þ ¼ þ : gðtÞ 2 n¼1 n þ 1 n! e 1 t
Thus, we have
( bk ¼
12 ; Bkþ1 kþ1
if k ¼ 0 ; if k > 0:
By Theorem 2.3, we get
Bnþ1 ðxÞ þ
n
n1 n n n Bkþ1 nk 1 n X Bnkþ1 k 1 n X k1 x x x ¼ x x ; Bk ðxÞ ¼ x 2 2 nþ2k k nkþ1 k kþ1 k¼1 k¼0 k¼1
n X
ðn 1Þ: ð3:13Þ
For En ðxÞ
et þ1 2
; t ; a0 ¼ 1 and ak ¼ 12 for k > 0. Note that
k X k 1 1 X g 0 ðtÞ et 1 2et 1 t 1 t ¼ t ¼ ¼ E ð 1 Þ ¼ E d : k k 0;k gðtÞ 2 et þ 1 k¼0 2 e þ1 k! k¼0 2 k!
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D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Thus,
1 bk ¼ Ek d0;k ¼ 2
(
12 1 E 2 k
if k ¼ 0 if k > 0:
By Theorem 2.3, we get
Enþ1 ðxÞ þ
n n1 n n 1X 1 n 1X E k ð xÞ ¼ x x þ Enk xk ; 2 k¼1 k 1 2 2 k¼0 k
ðn 1Þ:
ð3:14Þ
ðbÞ
The actuarial polynomials an ðxÞ are Sheffer polynomial sequences for (gðtÞ ¼ ð1 tÞb ; f ðtÞ ¼ log ð1 tÞ). So
ð1Þ ð1 þ t Þ1 ; log ð1 þ t Þ , and the inverse wn ðxÞ of an ðxÞ under umbral composition is given by wn ðxÞ ðet ; et 1Þ. t Also, the Cauchy polynomials of the first kind C n ðxÞ are given by C n ðxÞ et 1 ; et 1 . Thus, we can apply Theorem 2.3 with sn ðxÞ ¼ C n ðxÞ, rn ðxÞ ¼ wn ðxÞ. Here ð1Þ an ðxÞ
1 log ð1 þ t Þ X tk g f ðtÞ ¼ ¼ Dk : t k! k¼0
So, ak ¼ Dk , for k 0. Also, we note that
! ! 1 1 X X g 0 f ðtÞ 1 t 1 tk 1 1 tk ¼ 1þt 1þt Ck 1þt1 t Ck ¼ ¼ t log ð1 þ t Þ t t 2 k! k! g f ðtÞ k¼0 k¼2 ! 1 1 k k X t 1 1 1 X C kþ1 t ¼ ¼ t Ck : t 2 2 k¼1 k þ 1 k! k! k¼2
ð3:15Þ
Thus, by (3.15), we get
! k g 0 f ðtÞ d C kþ1 bk ¼ ; ¼ dt k þ1 g f ðtÞ t¼0
1 b0 ¼ ; 2
for k > 0:
By Theorem 2.3, we get
n X n 1 Dnþ1k C k ðxÞ ¼ Dk C nþ1k ðxÞ ¼ x þ wn ðxÞ 2 k1 k k¼1 k¼0 n1 n X X n n C kþ1 C nkþ1 1 wn ðxÞ wk ðxÞ ¼ x þ wnk ðxÞ; 2 n k þ 1 k k kþ1 k¼0 k¼1
C nþ1 ðxÞ þ
n X
n
ð3:16Þ
where n 1. t Let wn ðxÞ ðet ; et 1Þ. The Daehee polynomials of the first kind Dn ðxÞ are given by Dn ðxÞ ðe 1 ; et 1Þ. Note that t ðkÞ a ¼ gðf ðtÞÞ ¼ C . Now, we observe that k
k
t¼0 ! t log ð1 þ tÞ 1þt g 0 f ðtÞ dt dg f ðtÞ 1 log ð1 þ tÞ 1 t ¼ ð1 þ t Þ ¼ 1 t þ ¼ 2 dt t t log ð1 þ tÞ g f ðtÞ df ðtÞ g f ðtÞ ðlog ð1 þ t ÞÞ ! 1 1 X 1 1 tk 1 X C kþ1 t k ¼ þ 1 t þ 1 þ t þ Ck : ¼ t 2 2 k¼1 k þ 1 k! k! k¼2
So, b0 ¼ 12, and
bk ¼
! k g 0 f ðtÞ d C kþ1 ; ¼ dt kþ1 g f ðtÞ t¼0
for k > 0:
Hence, by Theorem 2.3, we get
n X n 1 C nþ1k Dk ðxÞ ¼ C k Dnkþ1 ðxÞ ¼ x wn ðxÞ 2 k1 k k¼1 k¼0 n1 n X X n n C kþ1 C nkþ1 1 wn ðxÞ þ wk ðxÞ ¼ x wnk ðxÞ; þ 2 n k þ 1 k k kþ1 k¼0 k¼1
Dnþ1 ðxÞ þ
n X
n
ð3:17Þ
where n 1. Accidently, by comparing (3.16) and (3.17), we obtain the following duality result between the Cauchy and Daehee polynomials: n n X X n n Dk C nþ1k ðxÞ þ C k Dnþ1k ðxÞ ¼ 2xwn ðxÞ ¼ 2xðx 1Þn ¼ 2ðxÞnþ1 : k k k¼0 k¼0
ð3:18Þ
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D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Note here that 1 1 X X tk 1 tk wk ðxÞ ¼ ð1 þ t Þx ¼ ð1 þ t Þx1 ¼ ðx 1Þk : k! t þ 1 k! k¼0 k¼0
3.4. Examples of Theorem 2.4 We apply Theorem 2.4 to
Bn ðxÞ
t e 1 ;t : t
P t 0 k 1 t 1 Here, g f ðtÞ ¼ gðtÞ=f ðtÞ ¼ e 1 ¼ 1 k¼0 kþ1 k! . So, ak ¼ bk ¼ kþ1. Also, t
! ! 1 1 k 1 1 k g 0 ðtÞ 1 et 1 1 X 1 tk X t 1 X 1 tk X t t ¼ e ¼ ¼ 0 t t t k¼0 k þ 1 k! k¼0 k! t k¼1 k þ 1 k! k¼1 k! f ðtÞ ! k 1 1 1 kþ1 kþ1 X 1 t X 1 X 1 t 1 t ¼ : ¼ t k¼0 ðk þ 2Þðk þ 1Þ k! k þ 1 k þ 2 k! k! k¼0 k¼0
So,
rk ¼
d k g0 ðtÞ dt f 0 ðtÞ
t¼0
1 ¼ kþ2 . Hence, by Theorem 2.4, we get
n n n X X X n n n 1 1 1 1 Bnk ðxÞ Bnþ1 ðxÞ ¼ x B k ð xÞ Bk ðxÞ ¼ x k þ 1 k þ 2 n þ 2 k n kþ1 k k 1 k k¼0 k¼1 k¼0 n n n X X X n n n 1 1 Bk ðxÞ B k ð xÞ Bk ðxÞ ¼ x k nkþ2 k1 nkþ2 k nkþ1 k¼0 k¼0 k¼0 n n n X X n n n nþ2 B k ð xÞ Bk ðxÞ 1 X þ ¼x Bk ðxÞ: nkþ2 k k1 k n k þ 1 n þ 2 k¼0 k k¼0 k¼0
ð3:19Þ
Now, we apply Theorem 2.4 to
En ðxÞ
t e þ1 ;t : 2
0 ðtÞ t 0 Note that g f ðtÞ ¼ gðtÞ=f ðtÞ ¼ e 2þ1. So, a0 ¼ b0 ¼ 1, and ak ¼ bk ¼ 12, for k > 0. Also, g ¼ 12 et . So f 0 ðtÞ Hence, by Theorem 2.4, we get
n n X n n 1 1 1X 1 En ðxÞ þ ðx 1Þ Enk ðxÞ E k ð xÞ ¼ x En ðxÞ x 2 2 2 k¼1 k 1 2 k k¼1 n1 n n1 n n n n 1 X 1X 1 1 X E k ð xÞ þ Ek ðxÞ ¼ x En ðxÞ þ x Ek ðxÞ þ x 2 k¼0 k 2 k¼1 2 2 k¼0 k k k1 n n n nþ1 n nþ1 1X 1 1 1 X 1X Ek ðxÞ ¼ x En ðxÞ xEn þ x Ek ðxÞ Ek ðxÞ 2 k¼1 2 2 2 k¼0 k 2 k¼1 k k n n n nþ1 1 1 X 1X Ek ðxÞ Ek ðxÞ: ¼ ðx 1ÞEn ðxÞ þ x 2 2 k¼0 k 2 k¼1 k
rk ¼ 12, for all k 0.
Enþ1 ðxÞ ¼
We apply Theorem 2.4 to the Laguerre polynomial
LðnaÞ ðxÞ gðtÞ ¼ ð1 tÞa1 ; f ðtÞ ¼
t : t1
ðkÞ Thus, we have g f ðtÞ ¼ ð1Þk ða þ 1Þk ð1 t Þakþ1 . Hence,
ðk Þ ak ¼ g f ðtÞ ð0Þ ¼ ð1Þk ða þ 1Þk : We observe that
ðkÞ k gðtÞ d ¼ ð1 t Þaþ1 ¼ ða þ k 2Þk ð1 tÞakþ1 : 0 dt f ðtÞ
So, bk ¼
ðkÞ
gðtÞ f 0 ðtÞ
¼ ða þ k 2Þk . It is easy to show that t¼0
0 ðkÞ k g ðtÞ d ¼ ða þ 1Þð1 tÞa ¼ ða þ 1Þaða þ 1Þ . . . ða þ k 1Þð1 tÞak : 0 dt f ðtÞ
ð3:20Þ
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D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
So,
rk
k g0 ðtÞ ¼ dtd f 0 ðtÞ
t¼0
n X 1
ðaÞ
Lnþ1 ðxÞ ¼
k!
k¼0
¼ ða þ 1Þða þ k 1Þk . From Theorem 2.4, we note that
n X ðaþkÞ ða þ k 2Þk x þ ða þ 1Þða þ k 1Þk ð1Þk ðnÞk Lnk ðxÞ k¼1
n k1
ðaÞ ð1Þnþ1k ða þ 1Þnþ1k Lk ðxÞ: ð3:21Þ
By (3.21), we get
n n X X ðaþkÞ n n ðaÞ ð1Þk1 ða þ k 2Þk x ða þ 1Þða þ k 1Þk Lnk ðxÞ þ ð1Þk1 ða þ 1Þk Lnþ1k ðxÞ: k k k¼0 k¼1
ðaÞ
Lnþ1 ðxÞ ¼
ð3:22Þ
We apply Theorem 2.4 to the Cauchy polynomials of the first kind
C n ð xÞ
t t : ; e 1 et 1
P tk Note that g f ðtÞ ¼ log ðt1þtÞ ¼ 1 k¼0 Dk k! . Thus, we have
ðkÞ ak ¼ g f ðtÞ
t¼0
¼
k d g f ðtÞ t¼0 ¼ Dk ; dt
ðk 0Þ:
We observe that 1 X gðtÞ tet tk ¼ Bk ð1Þ : ¼ t 0 k! f ðtÞ e 1 k¼0
So, bk ¼
ðkÞ
gðtÞ f 0 ðtÞ
¼ Bk ð1Þ, and we have
t¼0
g 0 ðtÞ 1 0 ¼ f ðtÞ t
(
t et 1
2
tet t e 1
)
0 0 1 1 k 1 k k 1X BXB C Ct ¼ @ @ ABl Bkl Bk ð1ÞA t k¼0 l¼0 k! 1
0 0 1 1 0 0 1 1 k kþ1 1 k 1 kþ1 k1 k X X X X t 1 B B C C B B C Ct ¼ ¼ @ @ ABl Bkl Bk ð1ÞA @ @ ABl Bkþ1l Bkþ1 ð1ÞA : k þ 1 k! k! k¼1 l¼0 k¼0 l¼0 1 1 Thus, we get
rk
! 0 ðkÞ kþ1 X kþ1 g ðtÞ 1 Bl Bkþ1l Bkþ1 ð1Þ : ¼ ¼ 0 k þ 1 l¼0 1 f ðtÞ t¼0
From Theorem 2.4, we can derive the following recursive formula:
1 Pkþ1 k þ 1 Bl Bkþ1l Bkþ1 ð1ÞC n B n l¼0 X X n 1 C ðkÞ BBk ð1Þ Dnþ1k C k ðxÞ; C nþ1 ðxÞ ¼ xþ C C n ð xÞ B A @ k! ðk þ 1Þ! k1 k¼0 k¼1 0
where C ðnkÞ ðxÞ ¼
d k dx
ð3:23Þ
C n ðxÞ.
3.5. Examples of Theorem 2.5 Now, we apply Theorem 2.5 to
ðxÞn 1; et 1 : Note that
ak ¼ f ðkÞ ð0Þ ¼
0
if k ¼ 0
1 if k > 0:
So,
ðxÞn ¼ ðkÞ
nþ1 1 X 1 ðk Þ ð xÞ ; n þ 1 k¼1 k! nþ1
where ðxÞnþ1 ¼
d k dx
ðxÞnþ1 .
ð3:24Þ
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
95
It is known that
ðxÞnþ1 ¼
nþ1 X S1 ðn þ 1; lÞxl :
ð3:25Þ
l¼0
Thus, by (3.25), we get ðk Þ
ðxÞnþ1 ¼
nþ1 X ðlÞk S1 ðn þ 1; lÞxlk :
ð3:26Þ
l¼k
From (3.24) and (3.26), we have
ðxÞn ¼
nþ1 X nþ1 nþ1 nþ1k X 1 þ k 1 1 X 1 X S1 ðn þ 1; lÞxlk ¼ S1 ðn þ 1; l þ kÞxl n þ 1 k¼1 l¼k k n þ 1 k¼1 l¼0 k n X nk n nl X lþkþ1 lþkþ1 1 X 1 X S1 ðn þ 1; l þ k þ 1Þxl ¼ S1 ðn þ 1; l þ k þ 1Þxl ¼ n þ 1 k¼0 l¼0 n þ 1 l l l¼0 k¼0 n n X kþ1 1 X l S1 ðn þ 1; k þ 1Þx : ¼ n þ 1 k¼l 1 l¼0
ð3:27Þ
By (3.25) and (3.27), we get
S1 ðn; lÞ ¼
n kþ1 1 X S1 ðn þ 1; k þ 1Þ; n þ 1 k¼l 1
for 0 6 l 6 n:
We apply Theorem 2.5 to Bell polynomials
Beln ðxÞ ð1; log ð1 þ t ÞÞ: Note that f ðtÞ ¼ log ð1 þ tÞ; f
ak ¼ f ðkÞ ð0Þ ¼
ðkÞ
ðtÞ ¼ ð1Þk1 ðk 1Þ!ð1 þ t Þk , for k > 0. So,
0;
if k ¼ 0
ð1Þk1 ðk 1Þ!; if k > 0:
From Theorem 2.5, we note that
Beln ðxÞ ¼
nþ1 1 X ð1Þk1 ðkÞ Belnþ1 ðxÞ; n þ 1 k¼1 k
ðkÞ
where Belnþ1 ðxÞ ¼
Belnþ1 ðxÞ ¼
d k dx
ð3:28Þ
Belnþ1 ðxÞ. Since,
nþ1 X
S2 ðn þ 1; lÞxl :
l¼0
Thus, we have ðk Þ
Belnþ1 ðxÞ ¼
nþ1 nþ1k X X ðlÞk S2 ðn þ 1; lÞxlk ¼ ðl þ kÞk S2 ðn þ 1; l þ kÞxl : l¼k
ð3:29Þ
l¼0
By (3.28) and (3.29), we get
Beln ðxÞ ¼ ¼ ¼
nþ1 nþ1k n X nk X ðl þ k þ 1Þkþ1 1 X ðl þ kÞk 1 X ð1Þk1 ð1Þk S2 ðn þ 1; l þ k þ 1Þxl S2 ðn þ 1; l þ kÞxl ¼ n þ 1 k¼1 l¼0 n þ 1 k¼0 l¼0 k kþ1 n X nl ðl þ k þ 1Þkþ1 1 X ð1Þk S2 ðn þ 1; l þ k þ 1Þxl n þ 1 l¼0 k¼0 kþ1 n X l¼0
n ðk þ 1Þklþ1 1 X ð1Þkl S2 ðn þ 1; k þ 1Þxl : n þ 1 k¼l klþ1
ð3:30Þ
As we noted in the above,
Beln ðxÞ ¼
n X S2 ðn; lÞxl : l¼0
ð3:31Þ
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D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Therefore, by (3.30) and (3.31), we obtain the following identity:
S2 ðn; lÞ ¼
n kþ1 1 X S2 ðn þ 1; k þ 1Þ; ð1Þkl ðk lÞ! n þ 1 k¼l 1
ð3:32Þ
where 0 6 l 6 n. We apply Theorem 2.5 to the Laguerre polynomials LðnaÞ ðxÞ of order a which are given by
LðnaÞ ðxÞ ð1 t Þa1 ;
t : t1
ð3:33Þ
From f ð0Þ ¼ 0, we have a0 ¼ 0. By f
ðkÞ
ðtÞ ¼ ð1Þk k!ðt 1Þðkþ1Þ , ðk > 0Þ, we get
ak ¼ f ðkÞ ð0Þ ¼ ð1Þk k!ð1Þkþ1 ¼ k!; ðk > 0Þ: From Theorem 2.5, we have
LðnaÞ ðxÞ ¼
nþ1 nþ1 nþ1 ðkÞ 1 X 1 X 1 X ðaÞ ðaþkÞ ðaþkÞ Lnþ1 ðxÞ ¼ ð1Þk ðn þ 1Þk Lnþ1k ðxÞ ¼ ð1Þk1 ðn þ 1Þk Lnþ1k ðxÞ: n þ 1 k¼1 n þ 1 k¼1 n þ 1 k¼1
ð3:34Þ
We apply Theorem 2.5 to the Abel polynomials An ðx; aÞ; ða – 0Þ which are given by
An ðx; aÞ 1; teat :
ðkÞ k1 ðkÞ k1 From f ðtÞ ¼ teat , we have a0 ¼ f ð0Þ ¼ 0, f ðtÞ ¼ ka þ ak t eat , for k > 0. So, ak ¼ f ð0Þ ¼ ka , for k > 0. Now, we 0 at observe that f ðtÞAn ðx; aÞ ¼ te An ðx; aÞ ¼ nAn1 ðx; aÞ. Thus, we have An ðx þ a; aÞ ¼ nAn1 ðx; aÞ. By replacing x by x a, we get
A0n ðx; aÞ ¼ nAn1 ðx a; aÞ: Continuing this process, we have
AðnkÞ ðx; aÞ ¼ ðnÞk Ank ðx ka; aÞ;
for all k 0:
From Theorem 2.5, we note that
An ðx; aÞ ¼
k1 nþ1 nþ1 1 X ka 1 X k1 n þ 1 Anþ1k ðx ka; aÞ: ðn þ 1Þk Anþ1k ðx ka; aÞ ¼ ka n þ 1 k¼1 k! n þ 1 k¼1 k
ð3:35Þ
For sn ðxÞ ðgðtÞ; f ðtÞÞ, r n ðxÞ ð1; f ðtÞÞ, let
rn ðxÞ ¼
n X C n;k sk ðxÞ:
ð3:36Þ
k¼0
Then the first identity in Theorem 2.6 follows also from the formula for C n;k in [14, p.132]. Indeed,
1 k n 1 k n 1 g f ðtÞ t x ¼ g f ðtÞ t x ¼ g f ðtÞ ðnÞk xnk ¼ k! k! k! n ðnkÞ n g f ðtÞ ¼ ank : ¼ k k t¼0
C n;k ¼
n nk g f ðtÞ x k ð3:37Þ
So,
r n ð xÞ ¼
n X n k¼0
k
ank sk ðxÞ:
Also, the second identity in Theorem 2.6 follows from the conjugation representation in [14, p.108].
rn ðxÞ ¼
n n n X X 1 D k n E k X 1 k ðnÞ 1 k ðnÞ f ðtÞ x x ¼ f ðtÞ f ðtÞ ð0Þxk ¼ ð0Þxk ; k! k! k! k¼0 k¼0 k¼1
3.6. Examples of Theorem 2.7 Apply Theorem 2.7 to
Bn ðxÞ
t e 1 ;t : t
ðn 1Þ:
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
97
0 0 Note that f f ðtÞ ¼ f ðtÞ ¼ 1. Thus, we get
ðkÞ ak ¼ f 0 f ðtÞ
t¼0
and
f ðtÞ f 0 ðtÞ
1; if k ¼ 0
¼
¼ f ðtÞ ¼ t. So, bk ¼
0;
ðkÞ
f ðtÞ f 0 ðtÞ
if k > 0
¼ t¼0
1; 0;
if k ¼ 1 . Now, we observe that if k – 1:
1 1 X X g ðtÞf ðtÞ g ðtÞt ðte þ et 1Þ=t tet tk tk ¼ ¼ 1 ¼ 1 B ð 1 Þ ¼ B ð 1 Þ : ¼ k k 0 gðtÞ ðet 1Þ=t et 1 k! k! gðtÞf ðtÞ k¼0 k¼1 0
t
0
Thus, we get
rk ¼
ðkÞ 0 if k ¼ 0 g 0 ðtÞf ðtÞ ¼ 0 Bk ð1Þ if k > 0: gðtÞf ðtÞ t¼0
From Theorem 2.7, we have
nBn ðxÞ ¼ xBðn1Þ ðxÞ
n X B k ð 1Þ
k!
k¼1
where BðnkÞ ðxÞ ¼
d k dx
BðnkÞ ðxÞ ¼ nxBn1 ðxÞ
n X n Bk ð1ÞBnk ðxÞ; k k¼1
ð3:38Þ
Bk ðxÞ. So by (3.38), we get
n n 1 1X Bn1 ðxÞ Bk Bnk ðxÞ; Bn ðxÞ ¼ x 2 n k¼2 k
ðn 1Þ:
ð3:39Þ
Apply Theorem 2.7 to
t e þ1 ;t : 2
En ðxÞ
Thus, we have
ak ¼
1; if k ¼ 0 0;
if k > 0;
and bk ¼
1 if k ¼ 1; 0
if k – 1:
We observe that
g 0 ðtÞf ðtÞ g 0 ðtÞt tet =2 2et t ¼ ¼ ¼ ¼ 0 gðtÞ 2 ðet þ 1Þ=2 et þ 1 gðtÞf ðtÞ k 1 X 1 t kEk1 ð1Þ : ¼ 2 k! k¼1
! X kþ1 1 1 X tk t 1 t ¼ Ek ð1Þ E k ð 1Þ 2 2 k! k! k¼0 k¼0
Thus, we have
(
rk ¼
0;
if k ¼ 0
12 kEk1 ð1Þ;
if k > 0:
From Theorem 2.7, we have
nEn ðxÞ ¼ xEðn1Þ ðxÞ þ where EðnkÞ ðxÞ ¼
d k dx
n X 1 1 kEk1 ð1Þ EðnkÞ ðxÞ; k! 2 k¼1
ð3:40Þ
En ðxÞ. Note that
1 1 1 X X X tk 2et 2 tk tk Ek ð1Þ ¼ t ¼2 t ¼2 Ek ¼ 1 Ek : e þ1 k! e þ 1 k! k! k¼0 k¼0 k¼1
ð3:41Þ
Hence, by (3.41), we get
E k ð 1Þ ¼
1;
if k ¼ 0
Ek ; if k > 0:
ð3:42Þ
By (3.40) and (3.42), we see that
n n n 1 1X Ek1 1 1X En1 ðxÞ þ Ek1 Enk ðxÞ: nEn ðxÞ ¼ nxEn1 ðxÞ nEn1 ðxÞ þ ðnÞk Enk ðxÞ ¼ n x k 2 2 k¼2 ðk 1Þ! 2 2 k¼2 k
ð3:43Þ
98
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
Thus, from (3.43), we have
E n ð xÞ ¼
n n 1 1 X En1 ðxÞ þ Ek1 Enk ðxÞ: x k 2 2n k¼2 k
ð3:44Þ
Apply Theorem 2.7 to the Mittag-Leffler polynomials
et 1 : Mn ðxÞ 1; f ðtÞ ¼ t e þ1 0 t 0 Note that f ðtÞ ¼ log 1þt . So, f f ðtÞ ¼ , f ðtÞ ¼ ðet2e 1t þ1Þ2
ðkÞ ak ¼ f 0 f ðtÞ
t¼0
8 1 if k ¼ 0 > <2 ¼ 1 if k ¼ 2 > : 0 if k – 0; 2:
2ð1þt 1tÞ 1t 2 . Thus, we have 2 ¼ 2 þ1Þ ð1þt 1t
Note that
t 2 f ðtÞ ðet þ 1Þ e 1 1 ¼ et et : ¼ 0 t t þ1 e 2 2e f ðtÞ Thus, we have bk ¼
ðkÞ
f ðtÞ f 0 ðtÞ
0; 1;
¼ t¼0
if k even and if k odd
rk ¼
ðkÞ
g 0 ðtÞf ðtÞ gðtÞf 0 ðtÞ
¼ 0. t¼0
For n 3, by Theorem 2.7, we get
n1 1 M n2 ðxÞ ¼ n M n ð xÞ 2 n3 where M ðnkÞ ðxÞ ¼
1 M n ðxÞ 2
d k dx
X 16k6n k1 mod 2
1 xM ðnkÞ ðxÞ; k!
ð3:45Þ
M n ðxÞ. Thus, by (3.45), we get
n1 x Mn2 ðxÞ ¼ n 2
X 16k6n k1 mod 2
1 ðk Þ M ðxÞ; k! n
ðn 3Þ:
ð3:46Þ
Now, we apply Theorem 2.7 to the Laguerre polynomials
t : LðnaÞ ðxÞ gðtÞ ¼ ð1 t Þa1 ; f ðtÞ ¼ 1t Note that
f ðtÞ ¼ t ; 1t
0
f ðtÞ ¼
1 ð1 t Þ2
g 0 ðtÞ ¼ ða þ 1Þð1 t Þa2 :
;
ð3:47Þ
Thus, we have
1 0 2 f f ðtÞ ¼ ¼ ð1 tÞ ; t 2 1 þ 1t
f ðtÞ ¼ t t2 0 f ðtÞ
ð3:48Þ
and
g 0 ðtÞf ðtÞ g 0 ðtÞ f ðtÞ ða þ 1Þð1 t Þa2 tð1 t Þ ¼ ða þ 1Þt: ¼ ¼ 0 gðtÞ f 0 ðtÞ gðtÞf ðtÞ ð1 t Þa1
ð3:49Þ
From (3.47)–(3.49), we note that
ðkÞ ak ¼ f 0 f ðtÞ
t¼0
ðkÞ f ðtÞ bk ¼ 0 f ðtÞ t¼0
8 1; > > > < 2; ¼ > 2; > > : 0;
8 0; > > > < 1; ¼ > 2; > > : 0;
if k ¼ 0 if k ¼ 1 if k ¼ 2 if k > 2; if k ¼ 0 if k ¼ 1 if k ¼ 2 if k > 2:
ð3:50Þ
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
99
Finally, we see that
rk ¼
ðkÞ ða þ 1Þ; if k ¼ 1 g 0 ðtÞf ðtÞ ¼ 0 0; if k – 1: gðtÞf ðtÞ t¼0
ð3:51Þ
For n 3, by (3.50), (3.51) and Theorem 2.7, we get
n o ðaÞ ðaÞ ðaþ1Þ ðaþ2Þ n LðnaÞ ðxÞ 2ðn 1ÞLn1 ðxÞ þ ðn 1Þðn 2ÞLn2 ðxÞ ¼ nðx a 1ÞLn1 ðxÞ nðn 1ÞxLn2 ðxÞ:
ð3:52Þ
That is, ðaÞ
ðaÞ
ðaþ1Þ
ðaþ2Þ
LðnaÞ ðxÞ 2ðn 1ÞLn1 ðxÞ þ ðn 1Þðn 2ÞLn2 ðxÞ ¼ ðx a 1ÞLn1 ðxÞ þ ðn 1ÞxLn2 ðxÞ:
ð3:53Þ
3.7. Examples of Theorem 2.8 Here we apply Theorem 2.8 to the Bernoulli polynomials
Bn ðxÞ
t e 1 ;t : t
It is easy to show 1 et 1 X 1 tk g f ðtÞ ¼ ¼ ; t k þ 1 k! k¼0
1 k X gðtÞf ðtÞ t ¼ et 1 ¼ 0 k! f ðtÞ k¼1
ð3:54Þ
and
k 1 g 0 ðtÞf ðtÞ et 1 X 1 t ¼ 1 : ¼ et þ 0 t k þ 1 k! f ðtÞ k¼0
ð3:55Þ
By (3.54) and (3.55), we get
ak ¼
1 ; kþ1
rk ¼
1 k 1¼ ; kþ1 kþ1
bk ¼
0; if k ¼ 0 1; if k > 0:
ð3:56Þ
From Theorem 2.8 and (3.56), we note that
n n n X X X n1 n 1 1 k k x Bnk ðxÞ: n Bk ðxÞ ¼ x ðnÞk Bnk ðxÞ ¼ k! kþ1 kþ1 k1 nkþ1 k k¼1 k¼1 k¼1
ð3:57Þ
We apply Theorem 2.8 to the Euler polynomials
En ðxÞ
t e þ1 ;t : 2
It is not difficult to show that
( ) ( ) 1 1 1 X X X et þ 1 1 1 k 1 1 k 1 tk g f ðtÞ ¼ ¼ t ¼ t ¼1þ ; 1þ 2þ 2 2 k! 2 k! 2 k! k¼0 k¼1 k¼1
ð3:58Þ
1 1 1 X X X gðtÞf ðtÞ et þ 1 1 kþ1 1 k tk t¼tþ t ¼tþ tk ¼ t þ ¼ 0 2 2k! 2ðk 1Þ! 2 k! f ðtÞ k¼1 k¼2 k¼2
ð3:59Þ
1 X g 0 ðtÞf ðtÞ 1 t k tk e t ¼ : ¼ 0 2 2 k! f ðtÞ k¼1
ð3:60Þ
and
Thus, by (3.58)–(3.60), we get
(
ak ¼
1; if k ¼ 0 1 ; 2
if k > 0;
8 > < 0; if k ¼ 0 bk ¼ 1; if k ¼ 1 > :k ; if k 2; 2
(
rk ¼
0;
if k ¼ 0
k ; 2
if k > 0;
ð3:61Þ
100
D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101
From Theorem 2.8 and (3.61), we have
( ) n1 n n X n1 n n 1X ðx 1Þ X 1 E k ð xÞ þ E n ð xÞ ¼ Enk ðxÞ þ n x En1 ðxÞ ðxbk þ rk ÞEnk ðxÞ ¼ k 2 k¼1 k 1 2 k¼2 2 k k k¼1 n n1 nðx 1Þ X 1 Enk ðxÞ þ n x En1 ðxÞ: ¼ 2 2 k1 k¼2
n
ð3:62Þ
Thus, by (3.62), we get
n1 n n1 n1 1X x 1X 1 Ek ðxÞ þ En ðxÞ ¼ Enk ðxÞ þ x En1 ðxÞ: 2 k¼1 k 1 2 k¼2 k 1 2
ð3:63Þ
Finally, we apply Theorem 2.8 to the Mittag-Leffler polynomials
et 1 : Mn ðxÞ 1; f ðtÞ ¼ t e þ1 0
Note that f ðtÞ ¼
2et 2. ðet þ1Þ
ðk Þ ak ¼ g f ðtÞ
So,
¼
t¼0
ðkÞ gðtÞf ðtÞ bk ¼ 0 f ðtÞ
1; if k ¼ 0 0; if k > 0;
¼
t¼0
1 t e et 2
ðkÞ
¼ t¼0
0; if k even 1; if k odd
and
rk
0 ðkÞ g ðtÞf ðtÞ ¼ ¼ 0: 0 gðtÞf ðtÞ t¼0
From Theorem 2.8, we have
X
nMn ðxÞ ¼ x
16k6n k1 mod 2
where M ðnkÞ ðxÞ ¼
d k dx
1 ðk Þ M ðxÞ; k! n
ðn 1Þ;
ð3:64Þ
M n ðxÞ.
Remark. It is easy to see that
log ð1 þ tÞ t t log ð1 þ t Þ ð1 þ t Þx þ ð1 þ t Þx ¼ 2ð1 þ tÞx : t log ð1 þ t Þ log ð1 þ t Þ t
ð3:65Þ
Thus, by (3.65), we get
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