Bell-Based Bernoulli Polynomials with Applications

Home , Bell polynomials, Bernoulli polynomials, Stirling polynomials, Umbral calculus

axioms

Article Bell-Based Bernoulli Polynomials with Applications

Ugur Duran 1 , Serkan Araci 2,* and Mehmet Acikgoz 3

1 Department of Basic Sciences of Engineering, Faculty of Engineering and Natural Sciences, Iskenderun Technical University, TR-31200 Hatay, Turkey; [email protected] 2 Department of Economics, Faculty of Economics, Administrative and Social Sciences, Hasan Kalyoncu University, TR-27410 Gaziantep, Turkey 3 Department of Mathematics, Faculty of Arts and Science, University of Gaziantep, TR-27310 Gaziantep, Turkey; [email protected] * Correspondence: [email protected]

Abstract: In this paper, we consider Bell-based Stirling polynomials of the second kind and derive some useful relations and properties including some summation formulas related to the Bell polynomials and Stirling numbers of the second kind. Then, we introduce Bell-based Bernoulli polynomials of order α and investigate multifarious correlations and formulas including some summation formulas and derivative properties. Also, we acquire diverse implicit summation formulas and symmetric identities for Bell-based Bernoulli polynomials of order α. Moreover, we attain several interesting formulas of Bell-based Bernoulli polynomials of order α arising from umbral calculus.

Keywords: Bernoulli polynomials; bell polynomials; mixed-type polynomials; stirling numbers of the second kind; umbral calculus; summation formulas; derivative properties

1. Introduction Special polynomials and numbers possess much importance in multifarious areas of sciences such as physics, mathematics, applied sciences, engineering and other related Citation: Duran, U.; Araci, S.; research ﬁelds covering differential equations, number theory, functional analysis, quan- Acikgoz, M. Bell-Based Bernoulli tum mechanics, mathematical analysis, mathematical physics and so on, cf.[1–25] and see Polynomials with Applications. also each of the references cited therein. For example; Bernoulli polynomials and num- Axioms 2021, 10, 29. https://doi.org/ 10.3390/axioms10010029 bers are closely related to the Riemann zeta function, which possesses a connection with the distribution of prime numbers, cf. [22,24]. Some of the most signiﬁcant polynomials

Academic Editor: Clemente Cesarano in the theory of special polynomials are the Bell, Euler, Bernoulli, Hermite, and Genoc- Received: 5 February 2021 chi polynomials. Recently, the aforesaid polynomials and their diverse generalizations Accepted: 24 February 2021 have been densely considered and investigated by many physicists and mathematicians, Published: 2 March 2021 cf.[1–22,26] and see also the references cited therein. In recent years, properties of special polynomials arising from umbral calculus have Publisher’s Note: MDPI stays neu- been studied and examined by several mathematicians. For instance, Dere et al. [7] tral with regard to jurisdictional clai- considered Hermite base Bernoulli type polynomials and, by applying the umbral algebra ms in published maps and institutio- to these polynomials, gave new identities for the Bernoulli polynomials of higher order, nal afﬁliations. the Hermite polynomials and the Stirling numbers of the second kind. Kim et al. [11] acquired several new formulas for the Bernulli polynomials based upon the theory of the umbral calculus. Kim et al. [12] derived some identities of Bernoulli, Euler and Abel polynomials arising from umbral calculus. Kim et al. [14] studied partially degenerate Copyright: © 2021 by the authors. Li- censee MDPI, Basel, Switzerland. Bell numbers and polynomials by using umbral calculus and derived some new identities. This article is an open access article Kim et al. [16] investigated some properties and new identities for the degenerate ordered distributed under the terms and con- Bell polynomials associated with special polynomials derived from umbral calculus. ditions of the Creative Commons At- In this paper, we consider Bell-based Stirling polynomials of the second kind and tribution (CC BY) license (https:// derive some useful relations and properties including some summation formulas related creativecommons.org/licenses/by/ to the Bell polynomials and Stirling numbers of the second kind. Then, we introduce 4.0/). Bell-based Bernoulli polynomials of order α and investigate multifarious correlations

Axioms 2021, 10, 29. https://doi.org/10.3390/axioms10010029 https://www.mdpi.com/journal/axioms Axioms 2021, 10, 29 2 of 23

and formulas including some summation formulas and derivative properties. Also, we acquire diverse implicit summation formulas and symmetric identities for Bell-based Bernoulli polynomials of order α. Moreover, we analyze some special cases of the results. Furthermore, we attain several interesting formulas of Bell-based Bernoulli polynomials of order α arising from umbral calculus to have alternative ways of deriving our results.

2. Preliminaries

Throughout this paper, the familiar symbols C, R, Z, N and N0 refer to the set of all complex numbers, the set of all real numbers, the set of all integers, the set of all natural numbers and the set of all non-negative integers, respectively. The Stirling polynomials S2(n, k : x) and numbers S2(n, k) of the second kind are given by the following exponential generating functions (cf.[3,8,13,15,26]):

∞ n t k ∞ n t k t e − 1 tx t e − 1 ∑ S2(n, k : x) = e and ∑ S2(n, k) = . (1) n=0 n! k! n=0 n! k!

In combinatorics, Stirling numbers of the second kind S2(n, k) counts the number of ways in which n distinguishable objects can be partitioned into k indistinguishable subsets when each subset has to contain at least one object. The Stirling numbers of the second kind can also be derived by the following recurrence relation for ζ ∈ N0 (cf.[3,8,13,15,26]):

n n x = ∑ S2(n, k)(x)k, (2) k=0

where (x)n = x(x − 1)(x − 2) ··· (x − (n − 1)) for n ∈ N with (x)0 = 1 (see [4,18,19]). n k For each integer k ∈ N0, Sk(n) = ∑l=0 l is named the sum of integer powers. The exponential generating function of Sk(n) is as follows (cf.[20]):

∞ tk e(n+1)t − 1 S (n) = . (3) ∑ k t − k=0 k! e 1

The bivariate Bell polynomials are deﬁned as follows:

∞ n t y(et−1) xt ∑ Beln(x; y) = e e . (4) n=0 n!

When x = 0, Beln(0; y) := Beln(y) is called the classical Bell polynomials (also called exponential polynomials) given by means of the following generating function (cf.[3,4,9,26]):

∞ n t y(et−1) ∑ Beln(y) = e . (5) n=0 n!

The Bell numbers Beln are attained by taking y = 1 in (5), that is Beln(0; 1) = Beln(1) := Beln and are given by the following exponential generating function (cf.[3,4,9,26]):

∞ n t (et−1) ∑ Beln = e . (6) n=0 n!

The Bell polynomials considered by Bell [26] appear as a standard mathematical tool and arise in combinatorial analysis. Since the ﬁrst consideration of the Bell polynomials, these polynomials have been intensely investigated and studied by several mathematicians, cf. [2,3,8,12–16,22,26] and see also the references cited therein. Axioms 2021, 10, 29 3 of 23

The usual Bell polynomials and Stirling numbers of the second kind satisfy the following relation (cf.[9]) n m Beln(y) = ∑ S2(n, m)y . (7) m=0 (α) The Bernoulli polynomials Bn (x) of order α are defined as follows (cf.[1,7,11,12,18,21]):

∞ n α (α) t t B (x) = ext (|t| < 2π). (8) ∑ n t − n=0 n! e 1

(α) (α) Setting x = 0 in (8), we get Bn (0) := Bn known as the Bernoulli numbers of order α. We (α) (α) also note that when α = 1 in (8), the polynomials Bn (x) and numbers Bn reduce to the classical Bernoulli polynomials Bn(x) and numbers Bn.

3. Bell-Based Stirling Polynomials of the Second Kind In this section, we introduce the Bell-based Stirling polynomials of the second kind and analyze their elementary properties and relations. Here is the deﬁnition of the Bell-based Stirling polynomials of the second kind as follows.

Deﬁnition 1. The Bell-based Stirling polynomials of the second kind are introduced by the following generating function:

∞ n t k t e − 1 xt+y(et−1) ∑ BelS2(n, k : x, y) = e . (9) n=0 n! k!

Diverse special circumstances of BelS2(n, k : x, y) are discussed below:

Remark 1. Replacing x = 0 in (9), we acquire Bell-Stirling polynomials BelS2(n, k : y) of the second kind, which are also a new generalization of the usual Stirling numbers of the second kind in (1), as follows: ∞ n t k t e − 1 y(et−1) ∑ BelS2(n, k : y) = e . (10) n=0 n! k!

Remark 2. Substituting y = 0 in (9), we get the Stirling polynomials of the second kind given by (1), cf. [11,21,22].

Remark 3. Upon setting x = y = 0 in (9), the Bell-based Stirling polynomials of the second kind reduce to the classical Stirling numbers of the second kind S2(n, k) by (1), cf. [9,22,24].

We now ready to derive some properties of the BelS2(n, k : x, y).

Theorem 1. The following correlation

n n BelS2(n, k : x, y) = ∑ S2(u, k)Beln−u(x; y) (11) u=0 u

holds for non-negative integer n. Axioms 2021, 10, 29 4 of 23

Proof. This theorem is proved by (1), (4), and (9), as follows:

∞ n t k t e − 1 xt+y(et−1) ∑ BelS2(n, k : x, y) = e n=0 n! k! ∞ tn ∞ tn = ∑ S2(n, k) ∑ Beln(x; y) n=k n! n=0 n! ∞ n n tn = ∑ ∑ S2(u, k)Beln−u(x; y) , n=0 u=0 u n!

which provides the desired result (11).

Remark 4. Theorem1 gives the following formula including the Stirling numbers of the second kind, the Bell-Stirling polynomials of the second kind, and Bell polynomials:

n n BelS2(n, k : y) = ∑ S2(u, k)Beln−u(y). u=0 u

Theorem 2. The following relations

n n n−l BelS2(n, k : x, y) = ∑ BelS2(l, k : y)x (12) l=0 l

and n n BelS2(n, k : x, y) = ∑ S2(l, k : x)Beln−l(y) (13) l=0 l hold for non-negative integers n and k with n ≥ k.

Proof. The proofs are similar to Theorem1.

Theorem 3. The following summation formulae for Bell-based Stirling polynomials of the second kind n n n−u BelS2(n, k : x1 + x2, y) = ∑ BelS2(u, k : x1, y)x2 (14) u=0 u and n n BelS2(n, k : x, y1 + y2) = ∑ BelS2(u, k : x, y1)Beln−u(y2) (15) u=0 u hold for non-negative integers n and k with n ≥ k.

Proof. Using the following equalities

t t t t t e(x1+x2)t+y(e −1) = ex1t+y(e −1)ex2t and ext+(y1+y2)(e −1) = ext+y1(e −1)ey2(e −1),

the proofs are similar to Theorem1. So, we omit them.

Theorem 4. The following relation

k !k ! n n S (n, k + k : x, y) = 1 2 S (u, k : x, y)S (n − u, k ) (16) Bel 2 1 2 ( + ) ∑ Bel 2 1 2 2 k1 k2 ! u=0 u

is valid for non-negative integer n. Axioms 2021, 10, 29 5 of 23

Proof. In view of (1) and (9), we have

∞ n t k1+k2 t e − 1 t S (n, k + k : x, y) = ext+y(e −1) ∑ Bel 2 1 2 ( + ) n=0 n! k1 k2 ! t k1 t k2 k !k ! e − 1 t e − 1 = 1 2 ext+y(e −1) (k1 + k2)! k1 k2! k !k ! ∞ n n tn = 1 2 S (u, k : x, y)S (n − u, k ) , ( + ) ∑ ∑ Bel 2 1 2 2 k1 k2 ! n=0 u=0 u n!

which gives the asserted result (16).

Theorem 5. The following relation

n n S2(n, k) = ∑ BelS2(u, k : x, y)Beln−u(−x; −y) (17) u=0 u

holds for non-negative integer n.

Proof. Utilizing the following equality

t k t k e − 1 e − 1 t t = ext+y(e −1)e−xt−y(e −1), k! k! it is similar to Theorem1. So, we omit the proof.

Remark 5. Theorem5 gives the following formula including the Stirling numbers of the second kind, the Bell-Stirling polynomials of the second kind, and Bell polynomials:

n n S2(n, k) = ∑ BelS2(u, k : −y)Beln−u(y). u=0 u

4. Bell-Based Bernoulli Polynomials and Numbers of Order α In this section, we introduce Bell-based Bernoulli polynomials of order α and investigate multifarious correlations and formulas including summation formulas, derivation rules, and correlations with the Bell-based Stirling numbers of the second kind. We now introduce Bell-based Bernoulli polynomials of order α as follows.

Deﬁnition 2. The Bell-based Bernoulli polynomials of order α are deﬁned by the following exponential generating function:

∞ n α (α) t t t B (x; y) = ext+y(e −1). (18) ∑ Bel n t − n=0 n! e 1

Some special cases of the Bell-based Bernoulli polynomials of order α are analyzed below.

(α) Remark 6. In the special case x = 0 in (18), we acquire Bell-Bernoulli polynomials Bel Bn (y) of order α, which are also new extensions of the Bernoulli numbers of order α in (8), as follows:

∞ n α (α) t t t B (y) = ey(e −1). (19) ∑ Bel n t − n=0 n! e 1

(α) Remark 7. Upon letting y = 0 in (18), the Bell-based Bernoulli polynomials Bel Bn (x; y) of order (α) α reduce to the familiar Bernoulli polynomials Bn (x) of order α in (8). Axioms 2021, 10, 29 6 of 23

(α) Remark 8. When y = 0 and α = 1, the polynomials Bel Bn (x; y) reduce to the usual Bernoulli polynomials Bn(x).

We also note that (1) Bel Bn (x; y) := Bel Bn(x; y), which we call the Bell-based Bernoulli polynomials. We now perform to derive some properties of the Bell-based Bernoulli polynomials of order α and we ﬁrst provide the following theorem.

Theorem 6. Each of the following summation formulae

n n (α) = (α) Bel Bn (x; y) ∑ Bk Beln−k(x; y) (20) k=0 k n n (α) = (α) Bel Bn (x; y) ∑ Bk (x)Beln−k(y) (21) k=0 k n n (α) = (α) n−k Bel Bn (x; y) ∑ Bel Bk (y)x (22) k=0 k

hold for n ∈ N0.

Proof. They are similar to Theorem1. So, we omit them.

We provide an implicit summation formula for the Bell-based Bernoulli polynomials by the following theorem.

Theorem 7. The following relationship

n n (α1+α2) + + = (α1) (α2) Bel Bn (x1 x2; y1 y2) ∑ Bel Bk (x1; y1)Bel Bn−k(x2; y2) (23) k=0 k

is valid for n ∈ N0.

Proof. Using the following equality

α +α α α t 1 2 t t 1 t t 2 t (x1+x2)t+(y1+y2)(e −1) x1t+y1(e −1) x2t+y2(e −1) + e = α e α e , (et − 1)α1 α2 (et − 1) 1 (et − 1) 2

the proof is similar to Theorem1. So, we omit it.

One of the special cases of Theorem7 is given, for every n ∈ N0, by

n n (α) + = (α) Bel Bn (x 1; y) ∑ Bel Bk (x; y), (24) k=0 k

which is a generalization of the well-known formula for usual Bernoulli polynomials given by n n Bn(x + 1) = ∑ Bk(x) (cf.[11]). k=0 k (α) We now provide derivative operator properties for the polynomials Bel Bn (x; y) as follows. Axioms 2021, 10, 29 7 of 23

Theorem 8. The difference operator formulas for the Bell-based Bernoulli polynomials

∂ (α) (α) B (x; y) = n B (x; y) (25) ∂x Bel n Bel n−1 and ∂ (α) (α− ) B (x; y) = n B 1 (x; y). (26) ∂y Bel n Bel n−1 hold for n ∈ N.

Proof. Based on the following derivative properties

∂ t t ∂ t t ext+y(e −1) = text+y(e −1) and ext+y(e −1) = et − 1ext+y(e −1), ∂x ∂y

the proof is completed.

A recurrence relation for the Bell-based Bernoulli polynomials is given by the following theorem.

Theorem 9. The following summation formula

n B + (x + 1; y) − B + (x; y) 1 n + 1 Bel (x; y) = Bel n 1 Bel n 1 = B (x; y) (27) n + + ∑ Bel k n 1 n 1 k=0 k

holds for n ∈ N0.

Proof. By means of Deﬁnition2, based on the following equality

t ∞ n xt+y(et−1) e − 1 t e = ∑ Bel Bn(x; y) , t n=0 n!

the proof is done.

Remark 9. The result (27) is an extension of the well-known formula for Bernoulli polynomials given by (cf. [22,23]) B (x + 1) − B (x) xn = n+1 n+1 n + 1

An explicit formula for the Bell-based Bernoulli polynomials is given by the following theorem.

Theorem 10. The following explicit formula

∞ k−1 n+1 k − 1 − − (l + x) B (x; y) = yk (−1)k l 1 Bel n ∑ ∑ + k=0 l=0 l n 1

holds for n ∈ N0. Axioms 2021, 10, 29 8 of 23

Proof. By means of Deﬁnition2, based on the following equality

∞ n xt ∞ t te t k−1 B (x; y) = ey(e −1) = text yket − 1 ∑ Bel n t − ∑ n=0 n! e 1 k=0 ∞ k−1 k − 1 − − = t ∑ ∑ yk (−1)k l 1e(l+x)t k=0 l=0 l ∞ ∞ k−1 n−1 k − 1 − − t = ∑ ∑ ∑ yk (−1)k l 1(l + x)n , n=0 k=0 l=0 l n!

which gives the asserted result.

We give the following theorem.

Theorem 11. The following formula including the Bell-based Bernoulli polynomials of higher-order and Stirling numbers of the second kind

n+k n!k! n + k (k) Bel (x; y) = B (x; y)S (n + k − l, m) (28) n ( + ) ∑ Bel l 2 n k ! l=0 l

is valid for n ∈ N0 and k ∈ N.

Proof. By means of Deﬁnition2, based on the following equality

t k ∞ n xt+y(et−1) −k e − 1 (k) t e = k!t ∑ Bel Bn (x; y) , k! n=0 n!

the proof is completed.

Here, we present the following theorem including the Bell-based Bernoulli polynomials and the Stirling polynomials of the second kind.

Theorem 12. The following correlation

n ∞ n (α) = (α) Bel Bn (x; y) ∑ ∑ (x)kS2(l, k) Bel Bn−l(y) (29) l=0 k=0 l

holds for non-negative integers n.

Proof. By means of Deﬁnition2 and, using (1) and (19), we obtain

∞ n α ( ) t t t x B α (x y) = ey(e −1)et − + ∑ Bel n ; t α 1 1 n=0 n! (e − 1) α ∞ t k t t e − 1 = ey(e −1) (x) t α ∑ k (e − 1) k=0 k! ∞ n ∞ n tn = (α) ∑ ∑ ∑ (x)kS2(l, k) Bel Bn−l(y) , n=0 l=0 k=0 l n!

which gives the asserted result (29).

A correlation including the Bell-based Bernoulli polynomials of order α and the Bell- based Stirling polynomials of the second kind is stated below. Axioms 2021, 10, 29 9 of 23

Theorem 13. The following summation formula

n+k n!k! n + k (k) Bel (x + x ; y + y ) = B (x ; y ) S (n + k − l, k : x , y ) n 1 2 1 2 ( + ) ∑ Bel l 2 2 Bel 2 1 1 n k ! l=0 l (30) holds for non-negative integers k and n with n ≥ k.

Proof. By (4) and (9), we have

∞ n ∞ n k t t t t (k) (x1+x2)t+(y1+y2)(e −1) ∑ BelS2(n, k : x1, y1) ∑ Bel Bn (x2; y2) = e , n=0 n! n=0 n! k!

which implies the claimed result (30).

Recently, implicit summation formulas and symmetric identities for special polynomials have been studied by some mathematicians, cf.[8,20] and see the references cited therein. Now, we investigate some implicit summation formula and symmetric identities for Bell-based Bernoulli polynomials of order α. We note that the following series manipulation formulas hold (cf.[20,24]):

∞ (x + y)N ∞ xn ym ∑ f (N) = ∑ f (n + m) (31) N=0 N! n,m=0 n! m!

and ∞ ∞ k ∑ A(l, k) = ∑ ∑ A(l, k − l). (32) k,l=0 k=0 l=0 We give the following theorem.

Theorem 14. The following implicit summation formula

k,l k l (α) = − n+m (α) Bel Bk+l(x; y) ∑ (x z) Bel Bk+l−n−m(z; y) (33) n,m=0 n m

holds.

Proof. Upon setting t by t + u in (18), we derive

α ∞ k l t + u t+u (α) t u ey(e −1) = e−z(t+u) B (z; y) . t+u − ∑ Bel k+l e 1 k,l=0 k! l!

Again, replacing z by x in the last equation, and using (31), we get

∞ k l α (α) t u t + u t+u e−x(t+u) B (x; y) = ey(e −1) ∑ Bel k+l t+u − k,l=0 k! l! e 1

By the last two equations, we obtain

∞ tk ul ∞ tk ul (α) = (x−z)(t+u) (α) ∑ Bel Bk+l(x; y) e ∑ Bel Bk+l(z; y) , k,l=0 k! l! k,l=0 k! l!

which yields

∞ tk ul ∞ tn um ∞ tk ul (α) = − n+m (α) ∑ Bel Bk+l(x; y) ∑ (x z) ∑ Bel Bk+l(z; y) . k,l=0 k! l! n,m=0 n! m! k,l=0 k! l! Axioms 2021, 10, 29 10 of 23

Utilizing (32), we acquire

∞ k l ∞ k,l n+m (α) (α) t u (x − z) Bel B (z; y) B (x; y) = k+l−n−m tkul, ∑ Bel k+l ∑ ∑ ( − ) ( − ) k,l=0 k! l! k,l=0 n,m=0 n!m! k l ! l m !

which implies the asserted result (33).

Corollary 1. Letting k = 0 in (33), the following implicit summation formula holds:

l l (α) = − m (α) Bel Bl (x; y) ∑ (x z) Bel Bl−m(z; y). m=0 m

Corollary 2. Upon setting k = 0 and replacing x by x + z in (33), we attain

l l (α) + = m (α) Bel Bl (x z; y) ∑ x Bel Bl−m(z; y). m=0 m

Now, we give the following theorem.

Theorem 15. The following symmetric identity

n n n n (α) (α) n−k k = (α) (α) k n−k ∑ Bel Bn−k(bx; y) Bel Bk (ax; y)a b ∑ Bel Bk (bx; y) Bel Bn−k(ax; y)a b (34) k=0 k k=0 k holds for a, b ∈ R and n ≥ 0.

Proof. Let α 2 ! t at bt Υ = e2abxt+y(e −1)+y(e −1). (eat − 1)ebt − 1 Then, the expression for Υ is symmetric in a and b, and we derive the following two expansions of Υ:

∞ n ∞ n (α) (at) (α) (bt) Υ = ∑ Bel Bn (bx; y) ∑ Bel Bn (ax; y) n=0 n! n=0 n! ∞ n n tn = (α) (α) n−k k ∑ ∑ Bel Bn−k(bx; y) Bel Bk (ax; y)a b n=0 k=0 k n!

and, similarly,

∞ n n tn = (α) (α) k n−k Υ ∑ ∑ Bel Bk (bx; y) Bel Bn−k(ax; y)a b , n=0 k=0 k n!

which gives the desired result (34).

(α) Here is another symmetric identity for Bel Bn (x; y) as follows.

Theorem 16. Let a, b ∈ R and n ≥ 0. Then the following identity holds:

n b−1 a−1 n b (α) + + (α) k n−k ∑ ∑ ∑ Bel Bk i j bx1; y Bel Bn−k(ax2; y)a b (35) k=0 i=0 j=0 k a n b−1 a−1 n a = (α) + + (α) k n−k ∑ ∑ ∑ Bel Bk i j ax2; y Bel Bn−k(bx1; y)b a . k=0 i=0 j=0 k b Axioms 2021, 10, 29 11 of 23

Proof. Let

α α (at) (bt) 2 at bt = abt − ab(x1+x2)t+y(e −1)+x(e −1) Ψ + α+1 e 1 e (eat − 1)α 1ebt − 1 α abt ! α abt ! at e − 1 at bt e − 1 bt = eabx1t+y(e −1) eabx2t+y(e −1). (36) eat − 1 eat − 1 ebt − 1 ebt − 1

By (18), the formula (36) can be expanded as follows

α b−1 α a−1 at at bt bt Ψ = eabx1t+y(e −1) eati eabx2t+y(e −1) ebtj at − ∑ bt ∑ e 1 i=0 e − 1 j=0 b−1 a−1 α α at i+ b j+bx at+y eat− bt abx t+y ebt− = e( a 1) ( 1) e 2 ( 1) ∑ ∑ at − bt i=0 j=0 e 1 e − 1 ∞ n b−1 a−1 n b tn = (α) + + (α) k n−k ∑ ∑ ∑ ∑ Bel Bk i j bx1; y Bel Bn−k(ax2; y)a b n=0 k=0 i=0 j=0 k a n!

and similarly,

∞ n b−1 a−1 n a tn = (α) + + (α) n−k k Ψ ∑ ∑ ∑ ∑ Bel Bk i j ax2; x Bel Bn−k(bx1; x)a b , n=0 k=0 i=0 j=0 k b n!

which means the claimed result (35).

Lastly, we provide the following symmetric identity.

Theorem 17. The following symmetric identity

n l nl − (α) (α+1) n+k+1−l l−k ∑ ∑ Sn−l(b 1) Bel Bk (bx1; y) Bel Bl−k (ax2; y)a b l=0 k=0 l k n l nl = − (α) (α+1) n+k+1−l l−k ∑ ∑ Sn−l(a 1) Bel Bk (ax2; y) Bel Bl−k (bx1; y)b a (37) l=0 k=0 l k

holds for a, b ∈ Z and n ≥ 0.

Proof. Let

α+1 α+1 (at) (bt) at bt = abt − ab(x1+x2)t+y(e −1)+y(e −1) Ω + α+1 e 1 e . (eat − 1)α 1ebt − 1

By (3) and (18), we observe that

abt ! α α+1 e − 1 at at bt bt Ω = at eabx1t+y(e −1) eabx2t+y(e −1) eat − 1 eat − 1 ebt − 1 ∞ n ∞ n ∞ n (at) (α) (at) (α+1) (bt) = at ∑ Sn(b − 1) ∑ Bel Bn (bx1; y) ∑ Bel Bn (ax2; y) n=0 n! n=0 n! n=0 n! ∞ n l nl tn−1 = − (α) (α+1) n+k+1−l l−k ∑ ∑ ∑ Sn−l(b 1) Bel Bk (bx1; y) Bel Bl−k (ax2; y)a b n=0 l=0 k=0 l k n! Axioms 2021, 10, 29 12 of 23

and also

∞ n l nl tn−1 = − (α) (α+1) l−k n+k+1−l Ω ∑ ∑ ∑ Sn−l(a 1) Bel Bk (ax2; y) Bel Bl−k (bx1; y)a b , n=0 l=0 k=0 l k n!

which imply the claimed result (37).

5. Applications Arising from Umbral Calculus We now review brieﬂy the concept of umbral calculus. For the properties of umbral calculus, we refer the reader to see the references [1,4–6,9–11,13,15,18]. Let F be the set of all formal power series in the variable t over C with ( ) ∞ tk F = f | f (t) = ∑ ak , (ak ∈ C) . k=0 k!

Let P be the algebra of polynomials in the single variable x over the ﬁeld complex ∗ numbers and let P be the vector space of all linear functionals on P. In the umbral calculus, hL|p(x)i means the action of a linear functional L on the polynomial p(x). This operator ∗ has a linear property on P given by

hL + M|p(x)i = hL|p(x)i + hM|p(x)i

and hcL|p(x)i = chL|p(x)i for any constant c in C. The formal power series ∞ tk f (t) = ∑ ak (38) k=0 k! deﬁnes a linear functional on P by setting

n h f (t)|x i = an (n ≥ 0). (39)

Taking f (t) = tk in (38) and (39) gives

k n ht |x i = n!δn,k, (n, k ≥ 0) (40)

where ( 1, if n = k δn,k = . 0, if n 6= k ∗ Actually, any linear functional L in P has the form (38). That is, since

∞ k k t fL(t) = ∑ hL|x i , k=0 k!

we have n n h fL(t)|x i = hL|x i,

and so as linear functionals L = fL(t). Moreover, the map L → fL(t) is a vector space ∗ isomorphism from P onto F. Henceforth, F will denote both the algebra of formal power series in t and the vector space of all linear functionals on P, and so an element f (t) of F will be thought of as both a formal power series and a linear functional. From (39), we have

eyt|xn = yn (41) Axioms 2021, 10, 29 13 of 23

and so yt e |p(x) = p(y)(p(x) ∈ P). The order o( f (t)) of a power series f (t) is the smallest integer k for which the coefﬁcient of tk does not vanish. If o( f (t)) = 0, then f (t) is called an invertible series. A series f (t) for which o( f (t)) = 1 will be called a delta series (cf. [1,4–6,9–11,13,15,18]). If f1(t), ..., fm(t) are in F, then

n n i1 im h f1(t)... fm(t)|x i = ∑ h f1(t)|x i...h fm(t)|x i, i1,..., im i1+i2+...+im=n

where n n! = . i1, ··· , ir i1! ··· ir! We use the notation tk for the k-th derivative operator on P as follows: ( n! xn−k, k ≤ n tkxn = (n−k)! . 0, k > n

If f (t) and g(t) are in F, then

h f (t)g(t)|p(x)i = h f (t)|g(t)p(x)i = hg(t)| f (t)p(x)i (42)

for all polynomials p(x). Notice that for all f (t) in F, and for all polynomials p(x),

∞ tk ∞ xk f (t) = ∑ h f (t)|xki and p(x) = ∑ htk|p(x)i . (43) k=0 k! k=0 k!

Using (43), we obtain

∞ htl|p(x)i k p(k)(x) := Dk p(x) = ∑ xl−k ∏(l − s + 1) l=k l! s=1 providing p(k)(0) = htk|p(x)i and h1|p(k)(x)i = p(k)(0). (44) Thus, from (44), we note that tk p(x) = p(k)(x). (45) Let f (t) ∈ F be a delta series and let g(t) ∈ F be an invertible series. Then there exists a unique sequence sn(x) of polynomials satisfying the following property:

k hg(t) f (t) |sn(x)i = n!δn,k (n, k ≥ 0), (46)

which is called an orthogonality condition for any Sheffer sequence, cf. [1,4–6,9–11,13,15,18,22]. The sequence sn(x) is called the Sheffer sequence for the pair of (g(t), f (t)), or this sn(x) is Sheffer for (g(t), f (t)), which is denoted by sn(x) ∼ (g(t), f (t)). Let sn(x) be Sheffer for (g(t), f (t)). Then for any h(t) in F, and for any polynomial p(x), we have

∞ ∞ k hh(t)|sk(x)i k hg(t) f (t) |p(x)i h(t) = ∑ g(t) f (t) , p(x) = ∑ sk(x) (47) k=0 k! k=0 k!

and the sequence sn(x) is Sheffer for (g(t), f (t)) if and only if

∞ n 1 x f (t) t e = ∑ sn(x) (48) g( f (t)) n=0 n! Axioms 2021, 10, 29 14 of 23

for all x in C, where f ( f (t)) = f f (t) = t. An important property for the Sheffer sequence sn(x) having (g(t), t) is the Appell sequence. It is also called Appell for g(t) with the following consequence:

1 n s (x) = x ⇔ ts (x) = ns − (x). (49) n g(t) n n 1

Further important property for Sheffer sequence sn(x) is as follows

1 ∞ tn s (x) is Appell for g(t) ⇔ ext = s (x) (x ∈ ). n ( ) ∑ n C g t n=0 n!

For further information about the properties of umbral theory, see [19] and cited references therein. Recently, several authors have studied Bernoulli polynomials, Euler polynomials with various generalizations under the theory of umbral calculus [1,4–6,9–11,13,15,22]. Recall from (18) that

∞ n t t t B (x; y) = ext+y(e −1). (50) ∑ Bel n t − n=0 n! e 1

t ( ) = ( t xt+y(e −1)) = As t approaches to 0 in (50) gives Bel B0 x; y 1 that stands for o et−1 e 0. It means that the generating function of Bell-based Bernoulli polynomials is invertible and thus can be used as an application of Sheffer sequence. Now we list some properties of Bell-based Bernoulli polynomials arising from umbral calculus as follows. From (48) and (49), we have

t e − 1 t B (x; y) ∼ e−y(e −1), t (51) Bel n t

and t Bel Bn(x; y) = n Bel Bn−1(x; y). (52)

et−1 −y(et−1) It follows from (52) that Bel Bn(x; y) is Appell for t e . By (40) and (50), we have

t t t B (x; y) = ey(e −1)xn = ey(e −1)B (x) Bel n et − 1 n n n = ∑ Belk(y)Bn−k(x), k=0 k

which is the special case of the result in (21). By (45) and (50), we also see that

t t t B (x; y) = ey(e −1)xn = Bel (x; y) Bel n et − 1 et − 1 n ∞ n Bk k n = ∑ t Beln(x; y) = ∑ BkBeln−k(x; y). k=0 k! k=0 k

We give the following theorem.

n Theorem 18. For all p(x) ∈ P, there exist constants c0, c1, ... , cn such that p(x) = ∑k=0 ck Bel Bk(x; y), where t 1 e − 1 t c = e−y(e −1)tk | p(x) . (53) k k! t Axioms 2021, 10, 29 15 of 23

Proof. By (46), (48) and (51), we observe that

t e − 1 t e−y(e −1)tk | B (x; y) = n!δ (n, k ≥ 0), t Bel n n,k

which yield the following relation

t n t e − 1 −y(et−1) k e − 1 −y(et−1) k e t | p(x) = ∑ cl e t | Bel Bl(x; y) t l=0 t n = ∑ cll!δl,k = k!ck, l=0

which gives the result in (53).

We give the following theorem.

Theorem 19. For n > 0, we have

∞ k−1 k y k − 1 k−1−l n−1 Bel Bn(x; y) = n ∑ ∑ (−1) (x + l) . (54) k=1 l=0 k! l

Proof. From (40) and (50), we get

∞ t k t t t e − 1 B (x; y) = ey(e −1)xn = yk xn Bel n t − t − ∑ e 1 e 1 k=0 k! ∞ k−1 k y k − 1 − − = t ∑ ∑ (−1)k 1 leltxn k=1 l=0 k! l ∞ k−1 k y k − 1 − − − = n ∑ ∑ (−1)k 1 l(x + l)n 1, k=1 l=0 k! l

which is the claimed result (54).

Here are some integral formulas by the following theorems.

Theorem 20. Let p(x) ∈ P. We have

t ∞ k+1 Z l e − 1 −y(et−1) k k + 1 k−l+1 e p(x) = ∑ ∑ (−y) (−1) p(u)du. t k=0 l=0 l 0

Proof. By (41) and (42), we obtain the following calculations

t t e − 1 t 1 e − 1 t e−y(e −1) | xn = e−y(e −1) | txn+1 t n + 1 t * ∞ + 1 k+1 = (−y)ket − 1 |xn+1 + ∑ n 1 k=0 ∞ k+1 1 k + 1 − + D E = (−y)k (−1)k l 1 elt|xn+1 + ∑ ∑ n 1 k=0 l=0 l ∞ k+1 1 k + 1 − + = (−y)k (−1)k l 1ln+1 + ∑ ∑ n 1 k=0 l=0 l ∞ k+1 l k + 1 − + Z = ∑ ∑ (−y)k (−1)k l 1 xndx. (55) k=0 l=0 l 0 Axioms 2021, 10, 29 16 of 23

Thus, from (55), we arrive at

t ∞ k+1 Z l e − 1 −y(et−1) k k k + 1 k−l+1 e t | p(x) = ∑ ∑ (−y) (−1) p(u)du (p(x) ∈ P). t k=0 l=0 l 0

So, the proof is completed.

Example 1. If we take p(x) = Bel Bn(x; y) in Theorem 20, on the one hand, we derive

∞ k+1 Z l t k k + 1 k−l+1 e − 1 −y(et−1) ∑ ∑ (−y) (−1) Bel Bn(x; y)dx = e Bel Bn(x; y) k=0 l=0 l 0 t t e − 1 t t B + (x; y) = 1 | e−y(e −1) Bel n 1 t n + 1

1 0 n+1 = ht |x i = n!δ + . n + 1 n 1,0 On the other hand,

∞ k+1 Z l k k + 1 k−l+1 (n + 1) ∑ ∑ (−y) (−1) Bel Bn(x; y)dx k=0 l=0 l 0 ∞ k+1 n Z l k k + 1 k−l+1 n u = (n + 1) ∑ ∑ (−y) (−1) ∑ Bel Bn−u(y) x dx k=0 l=0 l u=0 u 0 ∞ k+1 n n + 1 k + 1 − + = B (y) (−y)k(−1)k l 1lu+1, ∑ ∑ ∑ Bel n−u + k=0 l=0 u=0 u 1 l

which yields the following interesting property for n ≥ 0 :

∞ k+1 n n + 1 k + 1 − + B (y) (−y)k(−1)k l 1lu+1 = 0. ∑ ∑ ∑ Bel n−u + k=0 l=0 u=0 u 1 l

Theorem 21. We have

t Z 1 e − 1 −y(et−1) n e x = Beln(u; −y)du. t 0

Proof. From (51) and (52), we write

t t B (x; y) = ey(e −1)xn (n ≥ 0). Bel n et − 1 Axioms 2021, 10, 29 17 of 23

By (41) and (42), we obtain the following calculations

t t e − 1 −y(et−1) n 1 e − 1 −y(et−1) n+1 e x = e | tx t n + 1 t 1 D t E = et − 1e−y(e −1)|xn+1 n + 1 * + 1 ∞ ( Bel (1; −y) − Bel (−y)) = k k tk|xn+1 + ∑ n 1 k=0 k! 1 ∞ ( Bel (1; −y) − Bel (−y)) = k k (n + 1)!δ + ∑ n+1,k n 1 k=0 k! Bel (1; −y) − Bel (−y) = n+1 n+1 n + 1 Z 1 = Beln(u; −y)du. 0 So, the proof is completed.

Theorem 22. Let n be non-negative integer. Then, we have

t ∞ k+1 Z l+1 e − 1 −y(et−1) k k + 1 k−l+1 e Bel Bn(x; y) = ∑ ∑ (−y) (−1) Bel Bn(u + 1; y)du. t k=0 l=0 l l

Proof. By using (22) and (25) for α = 1, we obtain

Z x+z 1 Bel Bn(u; y)du = ( Bel Bn+1(x + z; y) − Bel Bn+1(x; y)). (56) x n + 1 Hence, by utilizing (52), we obtain

t t e − 1 −y(et−1) 1 e − 1 −y(et−1) e Bn(x; y) = e t B + (x; y) t Bel n + 1 t Bel n 1 ∞ k 1 k − D E = (−y)k (−1)k l e(l+1)t − elt B (x; y) + ∑ ∑ Bel n+1 n 1 k=0 l=0 l ∞ k+1 1 k + 1 − + = (−y)k (−1)k l 1 ( B (l + 1; y) − B (l; y)) + ∑ ∑ Bel n+1 Bel n+1 n 1 k=0 l=0 l ∞ k+1 Z l+1 k k + 1 k−l+1 = ∑ ∑ (−y) (−1) Bel Bn(u + 1; y)du. (57) k=0 l=0 l l

Comparing (56) with (57), we complete the proof.

The following theorem is useful for deriving any polynomial as a linear combination of the Bell-based Bernoulli polynomials.

Theorem 23. For q(x) ∈ Pn, let

n q(x) = ∑ bk Bel Bk(x; y), k=0

where ( − ) ( − ) ∞ m q k 1 (l + 1) − q k 1 (l) m m m−l bk = ∑ ∑(−y) (−1) . m=0 l=0 l k! Axioms 2021, 10, 29 18 of 23

Proof. It follows from Theorem 18 that for q(x) ∈ Pn, we have

n q(x) = ∑ bk Bel Bk(x; y), k=0

with t e − 1 t e−y(e −1)tk|q(x) = k!b . (58) t k Thus, from (58), we have

1 D t E b = et − 1e−y(e −1)tk−1|q(x) k k! ∞ m D E 1 m m m−l (l+1)t lt k−1 = ∑ ∑(−y) (−1) e − e t q(x) k! m=0 l=0 l ∞ m D E 1 m m m−l (l+1)t lt (k−1) = ∑ ∑(−y) (−1) e − e q (x) k! m=0 l=0 l ∞ m 1 m − = ∑ ∑(−y)m (−1)m l q(k−1)(l + 1) − q(k−1)(l) . k! m=0 l=0 l

Thus the proof is completed.

When we choose q(x) = En(x), we have the following corollary, which is given by its proof.

Corollary 3. Let n ≥ 1. We have

∞ m m m m−l En(x) = ∑ ∑(−y) (−1) Bel Bn(x; y) m=0 l=0 l n−1 ∞ m (n)k−1 m m−l m n−k+1 +2 ∑ ∑ ∑(−y) (−1) l − En−k+1(l) Bel Bk(x; y). k=0 k! m=0 l=0 l

Proof. Recall that the Euler polynomials En(x) are deﬁned by (cf.[12,23])

∞ tn 2 E (x) = ext ∑ n t + n=0 n! e 1

which yields et + 1 E (x) ∼ , t (n ≥ 0) n 2 and tEn(x) = nEn−1(x). Set q(x) = En(x) ∈ Pn. Then it becomes n En(x) = ∑ bk Bel Bk(x; y). (59) k=0 Axioms 2021, 10, 29 19 of 23

Let us now compute the coefﬁcients bk as follows

1 D t E b = et − 1e−y(e −1)tk−1|E (x) k k! n (n)k−1 D t −y(et−1) E = e − 1 e | E − + (x) k! n k 1 (n) ∞ m D E k−1 m m m−l (l+1)t lt = ∑ ∑(−y) (−1) e − e En−k+1(x) k! m=0 l=0 l ∞ m (n)k−1 m m m−l = ∑ ∑(−y) (−1) (En−k+1(l + 1) − En−k+1(l)). k! m=0 l=0 l

1 Using E1(x) = x − 2 and

n En(x + 1) − En(x) = 2(x − En(x)),

we have

n−1 En(x) = bn Bel Bn(x; y) + ∑ bk Bel Bk(x; y) k=0 ∞ m m m m−l = ∑ ∑(−y) (−1) Bel Bn(x; y) m=0 l=0 l n−1 ∞ m (n)k−1 m m−l m n−k+1 +2 ∑ ∑ ∑(−y) (−1) l − En−k+1(l) Bel Bk(x; y). k=0 k! m=0 l=0 l

Recall from (18) that Bell-based Bernoulli polynomials of order r ∈ N0 are given by the following generating function:

∞ n r ( ) t t t B r (x y) = ext+y(e −1) ∑ Bel n ; t r . (60) n=0 n! (e − 1)

(r) tr xt+y(et−1) If t tends to 0 on the above, we have B (x; y) = 1 that stands for o( r e ) = Bel 0 (et−1) 0. It means that the generating function of Bell-based Bernoulli polynomials of order r is invertible and thus can be used as an application of Sheffer sequence. Let t r e − 1 t gr(t, x) = e−y(e −1). tr r (r) Since g (t, x) is an invertible series. It follows from (60) that Bel Bn (x; y) is Appell for r et−1 −y(et−1) t e . So, by (49), we have

(r) 1 B (x; y) = xn, Bel n gr(t, x)

and (r) (r) t Bel Bn (x; y) = n Bel Bn (x; y). Thus, we have t r ! (r) e − 1 t B (x; y) ∼ e−y(e −1), t . Bel n tr

We give the following theorem.

Theorem 24. Let n be non-negative integer. Then, we have Axioms 2021, 10, 29 20 of 23

r−1 ( ) n B r (x) = B (x) B . Bel n ∑ Bel ir ∏ ij (61) i1, ··· , ir i1+···+ir=n j=1

Proof. By (39) and (60), we get

r n t t ( ) n ( ) ezt+y(e −1)|xn = B r (z y) = B r (y)zl t r Bel n ; ∑ Bel n−l . (62) (e − 1) l=0 l

Here we ﬁnd that

r * y(et−1) + t t t te ey(e −1) | xn = × · · · × | xn (63) (et − 1)r et − 1 et − 1 n = B (y) × B × · · · × B ∑ Bel ir i1 ir−1 . i1, ··· , ir i1+···+ir=n

By using (62), we have

r t y(et−1) n (r) e |x = Bn (y). (64) (et − 1)r Bel

Therefore, by (63) and (64), we arrive at the desired result (61).

(r) By setting q(x) = Bel Bn (x; y) ∈ Pn in Theorem 23, we provide the following Corollary.

Corollary 4. Let n ∈ N0 and r ∈ N0. Then

n ∞ m mn (r) = − m − m−l (r−1) Bel Bn (x; y) ∑ ∑ ∑ ( y) ( 1) Bel Bn−k (l; y) Bel Bk(x; y). k=0 m=0 l=0 l k

Proof. By Theorem 23, we write

n (r) Bel Bn (x; y) = ∑ bk Bel Bk(x; y), (65) k=0

where the coefﬁcient bk is given by

( − ) ( − ) ∞ m q k 1 (l + 1) − q k 1 (l) m m m−l bk = ∑ ∑(−y) (−1) (66) m=0 l=0 l k! ∞ m m (n) = (− )m (− )m−l k−1 (r) ( + ) − (r) ( ) ∑ ∑ y 1 Bel Bn−k+1 l 1; y Bel Bn−k+1 l; y . m=0 l=0 l k!

From (60), we have

∞ n r ( ) ( ) t t t B r (l + y) − B r (l y) = elt+y(e −1)et − ∑ Bel n 1; Bel n ; t r 1 n=0 n! (e − 1) r−1 t lt+y(et−1) = t − e (et − 1)r 1 ∞ n+1 (r−1) t = ∑ Bel Bn (l; y) . n=0 n! Axioms 2021, 10, 29 21 of 23

By comparing the coefﬁcients tn in the above equation, we get

(r) (r) (r−1) Bel Bn (l + 1; y) − Bel Bn (l; y) = n Bel Bn−1 (l; y). (67)

From (65), (66) and (67), the proof is completed.

The following theorem is useful for acquiring any polynomial as a linear combination of the Bell-based Bernoulli polynomials of order r.

Theorem 25. For n ∈ N0, we have

n ( ) = r (r) ∈ q x ∑ bk Bel Bk (x; y) Pn, k=0

where ∞ m+r r 1 m m + r m+r−l (k−r) bk = ∑ ∑ (−y) (−1) q (l). k! m=0 l=0 l

Proof. Let us assume that

n ( ) = r (r) ∈ q x ∑ bk Bel Bk (x; y) Pn. (68) k=0

r We use a similar method in order to ﬁnd the coefﬁcient bk as same as Theorem 23. So we omit the details and give the following equality:

* t r + n * t r + e − 1 t e − 1 t −y(e −1) k| ( ) = r −y(e −1) k (r)( ) r e t q x ∑ bl r e t Bel Bn x; y t l=0 t n r r = ∑ bl l!δl,k = k!bk l=0

which gives for k ≥ r,

* t r + 1 e − 1 t br = e−y(e −1)tk|q(x) k k! tr

* ∞ + 1 m+r = ∑ (−y)met − 1 tk−r|q(x) k! m=0 ∞ m+r 1 m + r + − D E = ∑ ∑ (−y)m (−1)m r l etl|q(k−r)(x) k! m=0 l=0 l ∞ m+r 1 m + r + − = ∑ ∑ (−y)m (−1)m r lq(k−r)(l). k! m=0 l=0 l

r Henceforth, by (68) and coefﬁcient bk, the proof is done. Finally, we state the following Corollary:

Corollary 5. The following equality

n ∞ m+r m + r (n) = − m − m+r−l k−r (r) Bel Bn(x; y) ∑ ∑ ∑ ( y) ( 1) Bel Bn−k+r(l; y) Bel Bk (x; y) k=0 m=0 l=0 l k!

holds for n, r ∈ N0, Axioms 2021, 10, 29 22 of 23

Proof. Let us consider q(x) = Bel Bn(x; y) ∈ Pn. Then, by Theorem 25, we have

n = r (r) Bel Bn(x; y) ∑ bk Bel Bk (x; y). (69) k=0

From Theorem 25 and (69), after some basic computations, we arrive at the claimed result.

6. Conclusions In the present paper, we have considered Bell-based Stirling polynomials of the second kind and derived some useful relations and properties including some summation formulas related to the Bell polynomials and Stirling numbers of the second kind. Then, we have introduced Bell-based Bernoulli polynomials of order α and have investigated multifarious correlations and formulas including some summation formulas and derivative properties. Also, we have acquired diverse implicit summation formulas and symmetric identities for Bell-based Bernoulli polynomials of order α. Moreover, we have analyzed some special cases of the results. Furthermore, we have attained several interesting formulas of Bell- based Bernoulli polynomials of order α arising from umbral calculus to have alternative ways of deriving our results.The results obtained in this paper are generalizations of the many earlier results, some of which are involved in the related references in [1–23]. For future directions, we will consider that the polynomials introduced in this paper can be examined within the context of the monomiality principle.

Author Contributions: All authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Acknowledgments: We would like to thank the reviewers for their careful reading our manuscript, which have improved the paper substantially. Conﬂicts of Interest: The authors declare no conﬂict of interest.

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