Analysis of Bell Based Euler Polynomials and Thier Application

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2. Preliminaries

In the present paper, we take the symbols Z, N, N0, R, and C to be set of integer, set of natural number, set of non negative integers, set of real number and set of complex number respectively. The bivariate Bell polynomials are describe by the following generating function, which is defined as follows: n t t B (x; y) = ext ey(e −1). (2.1) n n! n≥0 X When we take x = 0, Bn(0; y)= Bn(y) are called classical Bell polynomials (or exponential polynomials) which is describe by following generating function (see [2, 3, 4, 13, 17]) which is defined as follows: n t t B (y) = ey(e −1). (2.2) n n! n≥0 X If we take y = 1 in (2.2) i.e. Bn(0;1) = Bn(1) = Bn are called Bell number which is defined as follows (see [2, 3, 4, 13, 17]): n t t B = e(e −1). (2.3) n n! n≥0 X The generating function of an Euler polynomials of order α (see[8, 9, 10, 11, 18, 19]) is as folows: tn 2 α E (α)(x) = ext (|t| < 2π). (2.4) n n! et +1 n≥0 X (α) (α) If we take x = 0 in (2.4) i.e. En (0) = En are called Euler number (see[8, 9, 10, 11, 18, 19]) which is shown as follows: tn 2 α E (α) = . (2.5) n n! et +1 n≥0 X The generating function of second kind string polynomials S2(n, k; x) and string number S2(n, k) are defined as (see[2, 3]): tn et − 1 k S (n, k; x) = etx. (2.6) 2 n! k! n≥0 X When x = 0 in (2.6) i.e. S2(n, k;0) = S2(n, k) are called string number and defined by following exponential generating function (see[2, 3]): tn et − 1 k S (n, k) = . (2.7) 2 n! k! n≥0 X ANALYSIS OF BELL BASED EULER POLYNOMIAL AND THEIR APPLICATION 3

3. Bell based Euler polynomials and Number In this segment, we introduced Bell based Euler polynomials of order α and investigate numerous correlation formulae like implicit summation formulae, derivative formulae. For any n ∈ N and α ∈ C, we deﬁne Bell based Euler polynomial of order α as:

n α (α) t 2 xt+y(et−1) BE (x; y) = e (|t| < 2π). (3.1) n n! et +1 n≥0 X If x=0 and y=1 in (3.1) then we get a Bell Based Euler number of order α, which is defined as follows: n α (α) t 2 (et−1) BE = e (|t| < 2π). (3.2) n n! et +1 n≥0 X 3.1. Special Cases: In this section, we introduce some special types of Bell based Euler polynomials of order α, which is obtain by putting particular value in (3.1) and defined as follows: (1) If we choose x=0 in (3.1), we get Bell based Euler polynomials of order α, which are an extension of Euler polynomials of order α defined in (2.4) as follows: n α (α) t 2 y(et−1) BE (y) = e . n n! et +1 n≥0 X (2) In case y=0 in (3.1) the Bell based Euler polynomials of order α reduced (α) to the famililar Euler polynomials En (x) of order α defined in (2.4)

n α (α) t 2 xt BE (x) = e . n n! et +1 n≥0 X (α) (3) In case y=0 and α=1 in (3.1) the Bell Based Euler polynomials BEn (x; y) reduced to usual Euler polynomials En(x) deﬁned as: n t 2 xt BE (x) = e . n n! et +1 n≥0 X Theorem 3.1. The following relation hold true for α ∈ C and n ∈ N;

n n E (α)(x; y)= E (α)(x)B (y). (3.3) B n k k n−k k X=0 Proof. By using relation (3.1), we have 4 N. U. KHAN AND S. HUSAIN

n α (α) t 2 xt+y(et−1) BE (x; y) = e n n! et +1 n≥0 X α 2 t = ext ey(e −1) et +1 n o tk tn = E (α)(x) B (y) . k k! n n! (k≥ ) (n≥0 ) X0 X Now, using series rearrangement method, we get

n n n (α) t n (α) t BE (x; y) = E (x)B − (y) . n n! k k n k n! n≥0 n≥0 (k ) X X X=0 By equating same power of t both side, we get desired result (3.3).

Theorem 3.2. Bell based Euler polynomials of order α satisfy following relation for any α ∈ C and n ∈ N:

n n E (α)(x; y)= E (α)B (x; y). (3.4) B n k k n−k k X=0 Proof. By using generating function (3.1), we have

n α (α) t 2 xt+y(et−1) BE (x; y) = e n n! et +1 n≥0 X α 2 t = ext+y(e −1) et +1 n o tk tn = E (α) B (x; y) k k! n n! (k≥ ) (n≥0 ) X0 X After applying series rearrangement technique and compare same power of t, we get desired result (3.4).

Theorem 3.3. If α ∈ C and n ∈ N, following relation hold true;

n n E (α)(x; y)= E (α)(y) xn−k. (3.5) B n k k k X=0 ANALYSIS OF BELL BASED EULER POLYNOMIAL AND THEIR APPLICATION 5

Proof. Using relation (3.1), we have

n α (α) t 2 xt+y(et−1) BE (x; y) = e n n! et +1 n≥0 X α 2 t = ey(e −1) ext et +1 k n (α) t (xt) = BE (y) k k! n! (k≥ ) (n≥0 ) X0 X n n+k (α) x t = BE (y) . k n! k! (n≥0 k≥ ) X X0 By using series rearrangement, we get

n n n (α) t n (α) n−k t BE (x; y) = BE (y) x . n n! k k n! n≥0 n≥0 (k ) X X X=0 By equating same power of t both side, we get desired result (3.5).

4. Implicit summation formulae In this section, we discus some useful implicit summation formulae for Bell Based Euler polynomials of order α, which is deﬁned in following theorem as follows:

Theorem 4.1. If α1,α2 ∈ C and n ∈ N, following relation hold true;

n n E (α1+α2)(x + x ; y + y )= E (α1)(x ; y ) E (α2)(x ; y ). (4.1) B n 1 2 1 2 k B k 1 1 B n−k 2 2 k X=0 Proof. Using the following identity

α1+α2 2 t e(x1+x2)t+(y1+y2)(e −1) et +1 2 α1 2 α2 = ex1t + y (et − 1) ex2t + y (et − 1) . et +1 1 et +1 2 By using generating function (3.1), we have

n (α1+α2) t BE (x + x ; y + y ) n 1 2 1 y n! n≥0 X 6 N. U. KHAN AND S. HUSAIN

α1+α2 2 t = e(x1+x2)t+(y1+y2)(e −1) et +1 2 α1 2 α2 = ex1t + y (et − 1) ex2t + y (et − 1) et +1 1 et +1 2 k n (α1) t (α2) t = BE (x ; y ) BE (x ; y ) k 1 1 k! n 2 2 n! (k≥ ) (n≥0 ) X0 X n+k (α1) (α2) t = BE (x ; y ) BE (x ; y ) k 1 1 n 2 2 n!k! (n≥0 k≥ ) X X0 Using series rearrangement technique, we obtain

n n n (α1+α2) t n (α1) (α2) t BE (x + x ; y + y ) = BE (x ; y ) BE (x ; y ) n 1 2 1 y n! k k 1 1 n−k 2 2 n! n≥0 n≥0 (k ) X X X=0 Now, equating same power of t both side, we get desired result (4.1).

Remark 4.1. In case, if we choose α1 = α, α2 = 0, x1 = x, x2 = 1, y1 = y and y2 =0 in (4.1), we have

n n E (α)(x + 1; y)= E (α)(x; y), (4.2) B n k B k k X=0 which is an extension of Euler polynomials deﬁned by

n n E (x +1)= E (x; y). (4.3) n k k k X=0 Theorem 4.2. If α ∈ C and n ∈ N, Bell bassed Euler polynomials of order α satisfy following summation formula:

n n +1 E (α) (x + 1; y) − E (α) (x; y)= E (α)(x; y) (4.4) B n+1 B n+1 k B k k X=0 Proof. Using relation deﬁned in (3.1), we get

n n (α) t (α) t BE (x + 1; y) − BE (x; y) n n! n n! n≥0 n≥0 X X ANALYSIS OF BELL BASED EULER POLYNOMIAL AND THEIR APPLICATION 7

α α 2 t 2 t = e(x+1)t+y(e −1) − ext+y(e −1) et +1 et +1 α 2 t = ext+y(e −1) (et − 1) et +1 k n+1 (α) t t = BE (x; y) . k k! (n + 1)! (k≥ ) (n≥0 ) X0 X After, series rearrangement to compare both side the same power of t, we obtain a desired result (4.4) Theorem 4.3. If α = 1 and n ∈ N then Bell based Euler polynomials satisfy following relation: E (x + 1; y)+ E (x; y) B (x; y)= B n B n . (4.5) n 2 Proof. Using generating function (3.1) for α = 1 and deﬁnition of bivariate Bell polynomials, we get n t t B (x; y) =ext+y(e −1) n n! n≥0 X ∞ et +1 = BE (x; y) 2 k (k=0 ) t X e +1 2 t = ext+y(e −1) 2 et +1 1 2 t 2 t = e(x+1)t+y(e −1) + ext+y(e −1) 2 et +1 et +1 1 tn tn = BE (x + 1; y) + BE (x; y) . 2 n n! n n! (n≥0 n≥0 ) X X By equating same power of t both side, we get desired result (4.5). Remark 4.2. The formula (4.5) is generalized form of Euler polynomials deﬁned as E (x +1)+ E (x) xn = n n . (4.6) 2 Theorem 4.4. If n ≥ 0, then n n E (α)(x; y)= (x) S (j, k) E (α)(y). (4.7) B n j k 2 B n j=0 k≥ X X0 8 N. U. KHAN AND S. HUSAIN

Proof. By using (3.1), we have

n α (α) t 2 xt+y(et−1) BE (x; y) = e n n! et +1 n≥0 X α 2 t = ey(e −1) ext et +1 α 2 t = ey(e −1) (1 + et − 1)x et +1 n t k (α) t (e − 1) = BE (x; y) (x) n n! k k! (n≥0 ) (k≥ ) X X0 n j (α) t t = BE (x; y) (x) S (j, k) n n! k 2 j! (n≥0 ) (k≥ j≥0 ) X X0 X By using series rearrangement technique and equating same power of t both side, we get desired result (4.7).

5. Deriavtive formula Theorem 5.1. The diﬀerential operator formula for the bell based Euler polynomials of order α w.r.t. x deﬁned as follows: ∂ E (α)(x; y)= n E (α) (x; y), (5.1) ∂x B n B n−1 hold for all n ∈ N.

Proof. we know that

∂ t t ext+y(e −1) = t ext+y(e −1) (5.2) ∂x By using definition (3.1) in (5.2), we obtain desired result (5.1). Theorem 5.2. The difference operator formula for Bell based Euler polynomials of order α w.r.t. y defined as: ∂ E (α)(x; y)=(−2) E (α)(x; y) − E (α−1)(x; y) , (5.3) ∂y B n B n B n hold for all n ∈ N.

Proof. By using well known derivative properties

∂ t t ext+y(e −1) =(et − 1) ext+y(e −1) (5.4) ∂y ANALYSIS OF BELL BASED EULER POLYNOMIAL AND THEIR APPLICATION 9

Now, using deﬁnition (3.1), we get

n α ∂ (α) t ∂ 2 xt+y(et−1) BE (x; y) = e ∂y n n! ∂y et +1 (n≥0 ) X α 2 t = ext+y(e −1) (et − 1) et +1 α 2 t =(−2+ et + 1) ext+y(e −1) et +1 t α e +1 2 t =(−2) 1 − ext+y(e −1) 2 et +1 Now, using (3.1) l.h.s of above equation and equating same power of t both side, we get desired result (5.3).

6. Application of Bell based Euler polynomials from umbral calculus In this section, we discus the concept of umbral calculus (see[5, 6, 7, 11, 12, 14, 16]) and by using the concept of umbral calculus we drive some useful relation of Bell Based Euler polynomilas. Let G be set of all formal power series in the variable t over complex ﬁeld C with c G = g|g(t)= k tk s.t. c ∈ C . (6.1) k! k ( k≥ ) X0 Let P be set of polynomials in the single vriable t and P∗ be set of vector space of all linear functional on P. In the umbral calculus, we denote hT | q(x)i be linear functional T on the polynomials q(x). Now, we deﬁned the vector space operations on P∗ as follows:

The formal power series c g(t)= k tk ∈ G (6.2) k! k≥ X0 deﬁned a linear functional on P as n hg(t)| x i = cn (6.3) for all n ∈ N ∪{0}. 10 N. U. KHAN AND S. HUSAIN

If we choose g(t)= tk in (6.2) and (6.3), we get k n t | x = n!δn,k, (6.4) for all n, k ∈ N ∪{0} and 0 if n =6 k δ = . n,k 1, if n = k Since any linear functional T in P∗ has in the form of (6.2) i.e., k k t gT (t)= T| x k! k≥0 X and n n hgT (t)| x i = hT | x i .

So, the linear functional T = gT (t). We know that, the map T → gT (t) is a vector space isomorphism from set of vector space of all linear functional on P onto set of formal series G. Therefore, set of formal series G have vector space of all linear functionals on P, also G have algebra of formal power series and so for g(t) ∈ G will be treatrd as both a formal power series and a linear functional. From (6.3), we get eyt| xn = yn (6.5) and so, we have eyt| q(x) = q(y) (6.6) for all q(x) ∈ P. We know that the order of power series g(t) (i.e. o(g(t)) will be the smallest positive integer k such that the coeﬃcient of tk does not vanish. We know that g(t) is invertible series if the order of formal power series g(t) is zero, also the formal power series g(t) is a delta series if the order of g(t) is one (i.e. o(g(t) = 1). (see[5, 6, 7, 11, 12, 14, 16]).

For g1(t), ..., gm(t) ∈ G, then

Therfore, ∀ g(t) ∈ G tk g(t)= g(t)| xk , (6.8) k! k≥0 X and ∀ polynomials q(x) tk| q(x) q(x)= tk. (6.9) k! k≥0 X By using (6.9), we get tl| q(x) k qk(x)= Dkq(x)= xl−k (l − s + 1) (6.10) l! k≥0 s=1 X Y From (6.10), we have qk(0) = tk| q(x) and qk(0) = 1| qk(x) (6.11) By (6.11), we note that tkq(x)= qk(x). (6.12)

Let g(t) and h(t) be an element of formal powr series G such that g(t) be a delta and h(t) be an invertible series. Then ∃ an unique sequence Sn(x) of the polynomials with following properties k h(t)g(t) | Sn(x) = n!δn,k (6.13) For all n, k ∈ N ∪{0}, which is orthogonality condition for the Sheﬀer sequence (see[5, 6, 7, 11, 12, 14, 16]).

The sequence Sn(x) is said to be Sheffer sequence for (h(t), g(t)), which is de- noted by Sn(x) ∼ (h(t), g(t)). Let Sn(x) is the Sheffer sequence for (h(t), g(t)). Then for f(t) ∈ G and for q(x), we have the the following relation; hf(t)| S (x)i f(t)= k h(t)g(t)k (6.14) k! k≥ X0 and h(t)g(t)k| q(x) q(x)= S (x) (6.15) k! k k≥0 X and the sequence Sn(x) be a Sheffer sequence for (h(t), g(t)), iff 1 S (y) ey¯g(t) = k tk, ∀ y ∈ C. (6.16) h(g¯(t)) k! k≥ X0 Here, g¯(t) is composition inverse of g(t) i.e. g¯(g(t)) = g(g¯(t)) = t. 12 N. U. KHAN AND S. HUSAIN

Suppose that Sn(x) be an appell sequence for h(t). From (6.16), we get

1 n S (x)= x ⇔ tS (x)= nS − (x). (6.17) n h(t) n n 1

Recently, many author have studied Euler polynomilas, Bernoulli polynomials and Bell polynomials under theory of umbral calculus (see[5, 6, 7, 11, 12, 14, 15, 16]) . Recall from (3.1), we have

n (α) t 2 xt+y(et−1) BE (x; y) = e (|t| < 2π). (6.18) n n! et +1 n≥0 X

As t goes to zero in (6.18) gives BEn(x; y) equal to one (i.e. BEn(x; y) = 1) which 2 xt+y(et−1) means that o et+1 e = 0, which implies that (3.1) is an invertible series and treated as a Sheffer sequence. We defined an important and useful properties of Bell based Euler polynomials by using the definition of umbral calculus. Which is defined as follows: From (6.16) and (6.18), we have

t e +1 −y(et−1) BE (x; y) ∼ e , t (6.19) n 2 and

t BEn(x; y)= n BEn−1(x; y) (6.20)

Then from (6.19) and (6.20) we say that BEn(x; y) is an appell sequence for et+1 −y(et−1) 2 e .

Theorem 6.1. If q(x) ∈ P, there exist a constant b0, b1, ..., bn such that

q(x)= bk BEn(x; y), (6.21) k X=0 where t 1 e +1 t b = e−y(e −1)tk | q(x) (6.22) k k! 2 Proof. From (6.13), (6.16) and (6.19), we noted that

t e +1 −y(et−1) k e t | BE (x; y) = n!δ n, k ∈ N ∪{0}, 2 n n,k ANALYSIS OF BELL BASED EULER POLYNOMIAL AND THEIR APPLICATION 13

By using (6.21), we obtain

t t n e +1 −y(et−1) k e +1 −y(et−1) k e t | q(x) = e t | b BE (x; y) 2 2 l n * l + X=0 n t e +1 −y(et−1) k = b e t | BE (x; y) l 2 n l=0 Xn

= bll!δl,k = k!bk l X=0 Which is desired result (6.22).

Theorem 6.2. If n ∈ N, we have

x+z ezt − 1 E (v; y) dv = E (x; y) (6.23) B n t B n Zx Proof. By using (6.20), we have

x+z 1 BE (v; y) dv = {BE (x + z; y) − BE (x; y)} n n +1 n+1 n+1 Zx 1 n +1 k = BE − (x; y) z n +1 k n+l k k≥ X1 k 1 z k = t − 1 BE (x; y) n +1 k! n k≥ ! X0 ezt − 1 = E (x; y) t B n we get desired result.

Corollary 6.1. If n ∈ N ∪{0}, we have

z ezt − 1 E (v; y) dv = | E (y) (6.24) B n t B n Z0 Proof. From (6.20), we get

t BE (x; y)= BE (x; y) (6.25) n n +1 n+1 14 N. U. KHAN AND S. HUSAIN and by using (6.25), we have ezt − 1 ezt − 1 t | BE (y) = | BE (y) t n t n +1 n+1 zt 1 = e − 1 | BE (y) n +1 n+1 1 = {BE (z; y) − BE (0; y)} n +1 n+1 n+1 z = BEn(v; y) dv Z0

For any µ ∈ N ∪{0} from (3.1) the Bell based Euler polynomials of order µ are given as: n µ (µ) t 2 xt+y(et−1) BE (x; y) = e (|t| < 2π). (6.26) n n! et +1 n≥0 X (µ) 2 µ xt+y(et−1) When t goes to zero then BEn (x; y) = 1, which means that o et+1 e = 0. Hence the generating function (6.26) of order µ is an invertible and it will be treated as Sheﬀer sequence. Suppose that t µ (e + 1) t gµ(t; y)= e−y(e −1) 2µ µ (µ) We know that g (t; y) is an invertible series. From (6.26) we says that BEn (x; y) is an appell sequence for gµ(t; y). Hence from (6.17), we have

(µ) 1 n BE (x; y)= x , n gµ(t; y) and µ µ t BEn (x; y)= n BEn−1(x; y) Thus, we have t µ µ (e + 1) −y(et−1) BE (x; y) ∼ e , t n 2µ Now, using above result we discuse some interesting theorem. Theorem 6.3. If n ≥ 0, then

µ−1 (µ) n BEn (y)= BEiµ (y) Eij (6.27) i1, ..., iµ i1+...+i =n j=1 Xµ Y ANALYSIS OF BELL BASED EULER POLYNOMIAL AND THEIR APPLICATION 15

Proof. By using (6.3) and (6.26), we have

µ n 2 zt+y(et−1) n (µ) n (µ) l e | x = BE (z; y)= BE (y) z (6.28) (et + 1)µ n l n−l l X=0 and µ 2 t 2 2 t ezt+y(e −1)| xn = × ... × ey(e −1)| xn (et + 1)µ et +1 et +1 n (µ) = BEiµ (y) ×Ei1 × ... ×Eiµ−1 . i1, ..., iµ i1+...+i =n Xµ (6.29) From (6.28) and (6.29), we get result (6.27).

Theorem 6.4. If q(x) ∈ Pn, then n µ (µ) q(x)= bk BEn (x; y) ∈ Pn k X=0 where 1 t bµ = (et + 1)µe−y(e −1)tk| q(x) (6.30) k 2µk! D E Proof. Let as assume that n µ (µ) q(x)= bk BEn (x; y) ∈ Pn (6.31) k X=0 with the help of (6.31), we can write

t µ t µ n e +1 −y(et−1) k e +1 −y(et−1) k µ µ e t | q(x) = e t | b BE (x; y) 2 2 l n * l + X=0 n t µ µ e +1 −y(et−1) k = b e t | BE (x; y) l 2 n l=0 Xn µ µ = bl l!δl,k = k!bk , l=0 X (6.32) and we know that t µ e +1 t 1 t e−y(e −1)tk | q(x) = (et + 1)µe−y(e −1)tk| q(x) . (6.33) 2 2µ D E Hence from (6.32) and (6.33), we obtain the result (6.30). 16 N. U. KHAN AND S. HUSAIN

7. Conclusions In this paper, we introduced the Bell based Euler polynomials of order α and study their various correlation, implicit sumation and derivative formula. Also, we investigate various application of Bell based Euler polynomials of order α by using defintion of umbral calculus. Now, it is very useful because polynomials is solution of various differential equation and also play an importnat role in multifarious area like mathematics, physics and engineering sciences. Conflicts of Interest: The authors declare no conflict of interest. Ethical approval: This article does not contain any studies with human par- ticipants or animals performed by any of the authors. Author Contributions: All authors contributed equally to this manuscript. Acknowledgments: The authors express their deep gratitude to the anony- mous referees for their critical comments and suggestions to improve this paper to its current form. Availability of data and material: Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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Nabiullah Khan: Department of Applied Mathematics, Faculty of Engineering and Technology, Aligarh Muslim University, Aligarh 202002, India Email address: [email protected] Saddam Husain: Department of Applied Mathematics, Faculty of Engineering and Technology, Aligarh Muslim University, Aligarh 202002, India Email address: [email protected]