Umbral calculus and special polynomials

arXiv:1301.7520v1 [math.NT] 31 Jan 2013

UMBRAL CALCULUS AND SPECIAL POLYNOMIALS TAEKYUN KIM1 AND DAE SAN KIM2

Abstract. In this paper, we consider several special polynomials related to associated sequences of polynomials. Finally, we give some new and interesting identities of those polynomials arising from transfer formula for the associated sequences.

1. Introduction In this paper, we assume that λ ∈ C with λ 6= 1. For α ∈ R, the Frobenius-Euler polynomials are defined by the generating function to be α ∞ X tn 1−λ xt (α) e = H (x|λ) , (see [1,5,13,15,20,22,23]). (1.1) n et − λ n! n=0 (α)

(α)

In the special case, x = 0, Hn (0|λ) = Hn (λ) are called the n-th Frobenius-Euler numbers of order α. As is well known, the Bernoulli polynomials of order α are given by α ∞ X tn t xt e = Bn(α) (x) , (see [2,3,4,6,14,15,19,21]). (1.2) t e −1 n! n=0 (α)

(α)

In the special case, x = 0, Bn (0) = Bn are called the n-th Bernoulli numbers of order α. For n ≥ 0, the Stirling numbers of the second kind are defined by generating function to be ∞ X tl t n (1.3) (e − 1) = n! S2 (l, n) , (see [8-12,17,18]), l! l=n

and the Stirling numbers of the first kind are given by (x)n = x(x − 1) · · · (x − n + 1) =

n X

S1 (n, l)xl , (see [7,8,10,17,18]).

(1.4)

l=0

Let F be the set of all formal power series in the variable t over C with ( ) ∞ X ak k F = f (t) = t ak ∈ C . k!

(1.5)

k=0

Let P be the algebra of polynomials in the variable x over C and P∗ be the vector space of all linear functionals on P. As a notation, the action of the linear functional 1991 Mathematics Subject Classification. 05A10, 05A19. Key words and phrases. Bernoulli polynomial, Euler polynomial, Abel polynomial. 1

TAEKYUN KIM1 AND DAE SAN KIM2

2

k

P∞

ak tk! ∈ F . Then

hf (t)|xn i = an , (n ≥ 0), (see [10,12,16,17,18]).

(1.6)

L on a polynomial p(x) is denoted by hL | p(x)i. Let f (t) = we define the linear functional f (t) on P by

k=0

From (1.6), we note that

k n t |x = n!δn,k , (n, k ≥ 0),

(1.7)

where δn,k is the Kronecker symbol (see [8, 10, 11, 17, 18]). P∞ hL|xk i Let fL (t) = k=0 k! tk . Then, by (1.7), we get hfL (t)|xn i = hL|xn i. The map L 7→ fL (t) is a vector space isomorphism from P∗ onto F . Henceforth, F thought of as both a formal power series and a linear functional. We shall call F the umbral algebra. The umbral calculus is the study of umbral algebra (see [10, 16, 17, 18]). The order o(f (t)) of the non-zero power series f (t) is the smallest integer k for which the coefficient of tk does not vanish (see [10, 11, 12, 17, 18]). If o(f (t)) = 1, then f (t) is called a delta series, and if o(f (t)) = 0, then f (t) is called an invertible series. Let o(f (t)) = 1 and o(g(t)) = 0. Then

there exists a unique sequence Sn (x) of polynomials such that g(t)f (t)k |Sn (x) = n!δn,k where n, k ≥ 0. The sequence Sn (x) is called Sheffer sequence for (g(t), f (t)), which is denoted by Sn (x) ∼ (g(t), f (t)). If Sn (x) ∼ (1, f (t)), then Sn (x) is called the associated sequence for f (t) (see [10, 16, 17, 18]). From (1.7), we note that heyt | p(x)i = p(y). Let f (t) ∈ F and p(x) ∈ P. Then we have ∞

∞ k X X f (t)|xk k t |p(x) k f (t) = t , p(x) = x , (see [17,18]). (1.8) k! k! k=0

k=0

From (1.9), we can derive the following equation: D E

p(k) (0) = tk |p(x) and 1 p(k) (x) = p(k) (0), (see [10,16,17,18]).

for k ≥ 0, by (1.9), we easily see that tk p(x) = p(k) (x) = Let Sn (x) ∼ (g(t), f (t)). Then we see that

(1.9)

dk p(x) dxk .

∞

X Sk (x) 1 ¯ y f(t) tk , for all y ∈ C, e = k! g(f¯(t)) k=0

(1.10)

where f¯(t) is the compositional inverse of f (t) (see [17, 18]). Let pn (x) ∼ (1, f (t)), qn (x) ∼ (1, g(t)). Then, the transfer formula for the associated sequence is given by n f (t) qn (x) = x x−1 pn (x), (see [11,12,16,17,18]). (1.11) g(t) For n ≥ 0, b 6= 0, the Abel sequences are given by An (x; b) = x(x − bn)n−1 ∼ 1, tebt .

(1.12)

In this paper, we consider several special polynomials related to associated sequences of polynomials. Finally, we give some new and interesting identities of those polynomials arising from transfer formula for the associated sequences.

UMBRAL CALCULUS AND SPECIAL POLYNOMIALS

3

2. Umbral calculus and special polynomials From (1.1), we note that Hn(α) (x|λ)

∼

et − λ 1−λ

α

,t .

(2.1)

1−λ et − λ

α

xn .

(2.2)

Thus, we get Hn(α) (x|λ) = Let us assume that

pn (x) ∼ 1, t(et − λ) , qn (x) ∼

t a e −λ 1, t , (a 6= 0). 1−λ

From xn ∼ (1, t), (1.11) and (2.3), we note that n n x t 1−λ −1 n pn (x) = x x x = xn−1 t(et − λ) (1 − λ)n et − λ 1 (n) = xHn−1 (x|λ). (1 − λ)n and qn (x) = x

1−λ et − λ

na

(an)

x−1 xn = xHn−1 (x|λ).

(2.3)

(2.4)

(2.5)

From (1.11), (2.3), (2.4) and (2.5), we can derive 1 (n) xHn−1 (x|λ) (1 − λ)n  t a n −λ t e1−λ (an)   x−1 xHn−1 =x (x|λ) t(et − λ)

(a−1)n (an) x et − λ Hn−1 (x|λ) (1 − λ)an (a−1)n X (a − 1)n x (an) (−λ)(a−1)n−l elt Hn−1 (x|λ) = (1 − λ)an l =

(2.6)

l=0

(a−1)n

X x = an (1 − λ) l=0

(a − 1)n (an) (−λ)(a−1)n−l Hn−1 (x + l|λ), l

where a, n ∈ N. Therefore, by (2.6), we obtain the following theorem. Theorem 2.1. For a, n ∈ N, we have (n)

Hn−1 (x|λ) =

(a−1)n X (a − 1)n 1 (an) (−λ)(a−1)n−l Hn−1 (x + l|λ). l (1 − λ)(a−1)n l=0

Let us consider the following associated sequences: a 1−λ 1 (n) t t , (a 6= 0). (2.7) xHn−1 (x|λ) ∼ 1, t(e − λ) , pn (x) ∼ 1, t (1 − λ)n e −λ


4

For xn ∼ (1, t), by (1.11) and (2.7), we get n



an t t e −λ a  x−1 xn = x xn−1 pn (x) = x  1 − λ 1−λ t et −λ an X 1 an =x (−λ)an−l (x + l)n−1 . (1 − λ)an l

(2.8)

l=0

For n ≥ 1, by (1.11) and (2.7), we get 

t

n

t (e − λ)  −1 x (n) a pn (x) = x  x (x|λ) H n n−1 (1 − λ) 1−λ t et −λ (a+1)n (a+1)n (n) 1 =x et − λ Hn−1 (x|λ). 1−λ

(2.9)

By (2.8) and (2.9), we get an X an (−λ)an−l (x + l)n−1 l l=0

(a+1)n (n) 1 et − λ Hn−1 (x|λ) n (1 − λ) (a+1)n X (a + 1)n 1 (n) (−λ)(a+1)n−l Hn−1 (x + l|λ). = (1 − λ)n l

=

(2.10)

l=0

Therefore, by (2.10), we obtain the following theorem. Theorem 2.2. For n ≥ 1 and a ∈ Z+ = N ∪ {0}, we have an X an l=0

l

−l

(−λ)

(x + l)n−1 =

(a+1)n X (a + 1)n 1 (n) (−λ)n−l Hn−1 (x + l|λ). (1 − λ)n l l=0

Let us consider the following associated sequences: t

(x)n ∼ 1, e − 1 ,

(an) xHn−1 (x|λ)

a t e −λ , (a 6= 0). ∼ 1, t 1−λ

(2.11)

By (1.11) and (2.11), we get 

t

n

e − 1  −1 (an) a xHn−1 (x|λ) = x  x (x)n t −λ t e1−λ n an t 1−λ e −1 (x − 1)n−1 . =x t et − λ

(2.12)


Replacing x by x + 1, we have n an n−1 t X 1−λ e −1 (an) S1 (n − 1, l)xl Hn−1 (x + 1|λ) = t t e −λ l=0 t n n−1 X e −1 (an) = S1 (n − 1, l)Hl (x|λ) t l=0

n−1 l XX

5

(2.13)

n! (an) = S1 (n − 1, l) S2 (k + n, n)(l)k Hl−k (x|λ) (k + n)! l=0 k=0 n−1 l l XX (an) k = S1 (n − 1, l)S2 (k + n, n) k+n Hl−k (x|λ). n

l=0 k=0

Therefore, by (2.13), we obtain the following theorem. Theorem 2.3. For n ≥ 1, a ∈ Z+ , we have (an) Hn−1 (x

+ 1|λ) =

n−1 l XX

S1 (n − 1, l)S2 (k + n, n)

l=0 k=0

Let

l (an) k Hl−k (x|λ) k+n n

t a (an) −λ , (a 6= 0), xHn−1 (x|λ) ∼ 1, t e1−λ t (x)n ∼ (1, e − 1) .

Then, by (1.11) and (2.14), we get  t a n −λ t e1−λ (an)  x−1 xHn−1  (x|λ) (x)n = x et − 1 n t an t e −λ (an) =x t Hn−1 (x|λ) e −1 1−λ n t xn−1 =x t e −1

(2.14)

(2.15)

(n)

= xBn−1 (x). and (x)n =

n X

S1 (n, l)xl = x

l=0

Therefore, by (2.15) and (2.16), we get

n−1 X

S1 (n, l + 1)xl , (n ≥ 1).

(2.16)

l=0

Theorem 2.4. For n ≥ 1, 0 ≤ l ≤ n − 1, we have n−1 (n) S1 (n, l + 1) = Bn−1−l . l From (2.15), we note that t n e −1 (an) (x − 1)n−1 = (1 − λ)−an (et − λ)an Hn−1 (x|λ), (n ≥ 1). t

(2.17)


6

LHS of (2.17) =

et − 1 t

n−1 X

n n−1 X

S1 (n − 1, l), (x − 1)l

l=0

l X

n!l! S2 (k + n, n)(x − 1)l−k (k + n)!(l − k)! l=0 k=0 l n−1 l X X k S1 (n − 1, l) = S2 (k + n, n)(x − 1)l−k , k+n =

S1 (n − 1, l)

k=0

l=0

n

(2.18)

and an X an (an) RHS of (2.17) = (1 − λ) (−λ)an−l elt Hn−1 (x|λ) l l=0 an X an (an) = (1 − λ)−an (−λ)an−l Hn−1 (x + l|λ). l −an

(2.19)

l=0

Therefore, by (2.17), (2.18) and (2.19), we obtain the following theorem. Theorem 2.5. For n ≥ 1, a ∈ Z+ , we have

−an

(1 − λ)

an X an l=0

=

n−1 l XX

l=0 k=0

l

l k S1 (n k+n n

(an)

(−λ)an−l Hn−1 (x + l|λ)

− 1, l)S2 (k + n, n)(x − 1)l−k .

Let

pn (x) ∼

a 1−λ t , (x)n ∼ 1, et − 1 , (a 6= 0). 1, t e −λ

(2.20)

By (2.8), we have

an X an an x (−λ)an−l (x + l)n−1 pn (x) = l l=0 an X an n−1 X n − 1 an 1 an−k x (−λ) k n−1−l xl . = 1−λ k l

1 1−λ

k=0

l=0

(2.21)


7

From (1.11) and (2.20), we have a n  t e1−λ t −λ  x−1 pn (x) (x)n = x  t e −1

n an an X an n−1 X ann − 1 t 1−λ 1 k n−1−l (−λ)an−k xl et − 1 et − λ 1−λ k l k=0 l=0 an n X an n−1 X an n − 1 1 t (an) = x t k n−1−l (−λ)an−k Hl (x|λ) 1−λ e −1 k l k=0 l=0 an X an n−1 l X X an n − 1 l 1 (an) (n) x k n−1−l (−λ)an−k Hl−m (λ)Bm (x) = k 1−λ l m k=0 l=0 m=0 an X m l X an n−1 XX 1 an n − 1 l m n−1−l (an) (n) = k (−λ)an−k Hl−m (λ)Bm−p xp x k l m p 1−λ k=0 l=0 m=0 p=0   an n−1 l an n−1   X X X X 1 an n − 1 l m n−1−l (an) (n) = x k (−λ)an−k Hl−m (λ)Bm−p xp .   k 1−λ l m p

=x

k=0 l=p m=p

p=0

(2.22)

Therefore, by (2.16) and (2.22), we obtain the following theorem. Theorem 2.6. For n ≥ 1, a ∈ Z+ and 0 ≤ p ≤ n − 1, we have S1 (n, p+1) =

1 1−λ

an X an n−1 l XX an n − 1 l m n−1−l (an) (n) k (−λ)an−k Hl−m (λ)Bm−p . k l m p m=p k=0 l=p

Theorem 2.7. For n ≥ 0, we have  ∞ n X k X X n  e−xt (et − λ)n = (1 − λ)n−l l j=0 k=0

l=0

Proof. Note that e

−xt

t

n

(e − λ) = e

−xt

t

n

(e − 1 + 1 − λ) =

k j S2 (j j+l l

n X n l=0

l



+ l, l)(−1)k−j xk−j 

tk+l . k!

(1 − λ)n−l (et − 1)l e−xt , (2.23)

and (et − 1)l e−xt

 tk+l l!k! S2 (j + l, l)(−1)k−j xk−j  = (j + l)!(k − j)! k! j=0 k=0   k ∞ k k+l X X j k−j k−j  t  = S (j + l, l)(−1) x . 2 j+l k! l j=0 ∞ X

 k X 

(2.24)

k=0

From (2.23) and (2.24), we can derive Theorem 2.7.


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By (1.12) and Theorem 2.7, we get an 1 (an) e−nbt (et − λ)an Hn−1 (x|λ) 1−λ an k an n−1−l k X X X l j (n − 1)k+l (an) (1 − λ)an−l S2 (j + l, l)(−1)k−j (nb)k−j Hn−1−l−k (x|λ). j+l

An (x; b) = x(x − bn)n−1 = x =

x (1 − λ)an

l=0

l

k=0 j=0

(2.25)

Therefore, by (2.25), we obtain the following theorem. Theorem 2.8. For n ≥ 1, a ∈ Z+ and b 6= 0, we have n−1

(x−bn)

=

an k (n−1)k+l k! l j j+l (1 − λ)l l

an n−1−l k X X X l=0

k=0 j=0

(an)

S2 (j+l, l)(−1)k−j (nb)k−j Hn−1−l−k (x|λ).

Let us consider the Changhee polynomials of the second kind as follows: ∞ X

Ck (x|λ)

k=0

tk 1 = (1 + t)x . k! 1 + λ(1 + t)

(2.26)

From (1.10) and (2.26), we note that Ck (x|λ) ∼ 1 + λet , et − 1 .

(2.27)

(1 + λet )Cn (x|λ) = (x)n ∼ 1, et − 1 ,

(2.28)

Hence λ ∈ C with λ 6= −1. Thus, by (2.27), we get

and (x)n = x

t t e −1

n

−1 n

t t e −1

n

(n)

xn−1 = xBn−1 (x).

(2.29)

n X 1 (n) (n) xB (x) = (−λ)l elt xBn−1 (x) n−1 t λe + 1

(2.30)

x

x =x

Thus, by (2.28) and (2.29), we get Cn (x|λ) =

l=0

=

n X

l

(−λ) (x +

(n) l)Bn−1 (x

+ l).

l=0

Let tn (x|λ) ∼

t 1, 1 + λ(1 + t)

.

(2.31)


9

Then, by (1.11) and (2.31), we get !n t x−1 xn = x(1 + λ(1 + t))n xn−1 tn (x|λ) = x t 1+λ(1+t)

n−1 n n X X n a X a b n−1 n l l n−1 λ tx =x λ (1 + t) x =x a b l a=0 b=0 l=0 n n−1 n X n X X na X a a n−b a n = λ (n − 1)b x = λ (n − 1)n−b xb a b a n − b a=0 b=0 a=0 b=1 n n XX n a (n − 1)! b = λa x . a n − b (b − 1)! a=0 b=1 (2.32)

Let us also consider the following associated sequence: t , (µ ∈ N). Sn (x|µ) ∼ 1, (1 + t)µ Then, by (1.11) and (2.33), we easily get n X µn (n − 1)! k Sn (x|µ) = x . n − k (k − 1)!

(2.33)

(2.34)

k=1

From (1.11), (2.32) and (2.33), we can derive !n n t (1 + t)µ 1+λ(1+t) −1 x tn (x|λ) = x x−1 tn (x|λ) Sn (x|µ) = x t 1 + λ(1 + t) (1+t)µ !n ∞ X Cl (µ|λ) l x−1 tn (x|λ) t =x l! l=0 ( !) Y ∞ n X X tl −1 l =x Cli (µ|λ) x tn (x|λ) l1 , . . . , ln l! i=1 l=0 l1 +···+ln =l ) ( !) ( n n Y ∞ n X X a (n − 1)! b−1 tl X X a n l λ =x x Cli (µ|λ) l! a=0 a n − b (b − 1)! l1 , . . . , ln i=1 b=1 l=0 l1 +···+ln =l ! Y n X n X b−1 n X X a (n − 1)! (b − 1)l b−1−l l a n =x Cli (µ|λ) λ x a n − b (b − 1)! l1 , . . . , ln l! a=0 b=1 l=0 l1 +···+ln =l i=1 ! Y n X n X b−1 n X X a l (n − 1)! b − 1 b−l a n x = Cli (µ|λ) λ l a n − b (b − 1)! l1 , . . . , ln a=0 b=1 l=0 l1 +···+ln =l i=1 ) ( n n ! Y n n X X XX a b−k b − 1 (n − 1)! a n xk . = Cli (µ|λ) λ (b − 1)! a n − b l , . . . , l k − 1 1 n a=0 i=1 k=1

b=k l1 +···+ln =b−k

(2.35)

Therefore, by (2.34) and (2.35), we obtain the following theorem.


10

Theorem 2.9. For n ≥ 1, 1 ≤ k ≤ n, b 6= 0 and µ, a ∈ Z+ , we have µn n X n X X n a b−k 1 b−1 a n−k = λ (k − 1)! a=0 (b − 1)! a n − b l1 , . . . , ln k−1 b=k l1 +···+ln =b−k

n Y

!

Cli (µ|λ) .

i=1

Remark. From (1.1), we note that l t ∞ X 1−λ 1−λ 1 e −1 l = = = (−1) t −1 et − λ et − 1 + 1 − λ 1−λ 1 + e1−λ l=0 ! ∞ k l X X tk 1 l!S2 (k, l) , = (−1)l 1−λ k! k=0

and

(2.36)

l=0

∞ X 1−λ tn = H (λ) , n et − λ n=0 n!

(2.37)

where Hn (λ) are the Frobenius-Euler numbers. By (2.36) and (2.37), we get l k X 1 l!S2 (k, l). (2.38) Hk (λ) = λ−1 l=0

Let us consider the following associated sequences: 1−λ t , xn ∼ (1, t). pn (x) ∼ 1, t e −λ

(2.39)

Then, by (1.11) and (2.39), we get n n X n 1 n 1 t n n−1 (e − λ) x = x (−λ)n−k (x + k)n−1 , pn (x) = x 1−λ 1−λ k k=0 (2.40) and n  1−λ n t t −λ e  x−1 pn (x) = x 1 − λ x−1 pn (x). (2.41) xn = x  t et − λ

Thus, by (2.40) and (2.41), we get ( ) ) ( n n−1 n−1 X X X n l tl n−1 n−k (x + k) x = Hl1 (λ) · · · Hln (λ) × (−λ) l1 , . . . , ln l! (1 − λ)n k l=0 l1 +···+ln =l k=0 ! Y n n−1 n X X X 1 n n−1 l = (λ) H (−λ)n−k (x + k)n−1−l . li (1 − λ)n k l1 , . . . , ln l i=1 k=0 l=0 l1 +···+ln =l

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Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea. E-mail address: [email protected]