A matrix approach to some identities involving Sheffer polynomial sequences

A matrix approach to some identities involving Sheffer polynomial sequences

Applied Mathematics and Computation 253 (2015) 83–101 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homepage...

623KB Sizes 0 Downloads 1 Views

Recommend Documents

No documents
Applied Mathematics and Computation 253 (2015) 83–101

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

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.

D.S. Kim, T. Kim / Applied Mathematics and Computation 253 (2015) 83–101

101

[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)).