# A Note on Eulerian Polynomials

Authors: D. S. Kim; T. Kim; W. J. Kim; D. V. Dolgy
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study Genocchi, Euler, and tangent numbers. From those numbers we derive some identities on Eulerian polynomials in connection with Genocchi and tangent numbers.


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## Body

## 1. Introduction

As is well known, the Eulerian polynomials,An(t), are defined by generating function as follows:
(1.1)1-texp⁡(x(t-1))-t=eA(t)x=∑n=0∞‍An(t)xnn!,
with the usual convention about replacing An(t) by An(t) (see [1–18]). From (1.1), we note that
(1.2)(A(t)+(t-1))n-tAn(t)=(1-t)δ0,n,
where δn,k is the Kronecker symbol (see [3]).Thus, by (1.2), we get
(1.3)A0(t)=1,An(t)=1t-1∑l=0n-1‍(nl)Al(t)(t-1)n-l,(n≥1).By (1.1), (1.2), and (1.3), we see that
(1.4)∑i=1m‍inti=∑l=1n‍(-1)n+l(nl)tm+1An-l(t)(t-1)n-l+1ml+(-1)nt(tm-1)(t-1)n+1An(t),
where m≥1 and n≥0 (see [1]).The Genocchi polynomials are defined by(1.5)2tet+1ext=eG(x)t=∑n=0∞‍Gn(x)tnn!,(see [6–18]). In the special case, x=0, Gn(0)=Gn are called the nth Genocchi numbers (see [14, 17, 18]).It is well known that the Euler polynomials are also defined by(1.6)2et+1ext=eE(x)t=∑n=0∞‍En(x)tnn!,(see [1–5, 19–24]). Here x=0, then En(0)=En is called the nth Euler number. From (1.6), we have
(1.7)E0=1,(E+1)n+En=2δ0,n,(see [3–5, 19–23]).As is well known, the Bernoulli numbers are defined by(1.8)B0=1,(B+1)n-Bn=δ0,n,(see [5, 18, 19]), with the usual convention about replacing Bn by Bn.From (1.8), we note that the Bernoulli polynomials are also defined as
(1.9)Bn(x)=∑l=0n‍(nl)Blxn-l=(B+x)n,(see [5, 18, 19]).The tangent numbersT2n-1(n≥1) are defined as the coefficients of the Taylor expansion of tanx:
(1.10)tanx=∑n=1∞‍T2n-1(2n-1)!x2n-1=x1!+x33!2+x55!16+⋯,(see [1–3, 5]).In this paper, we give some identities on the Eulerian polynomials att=-1 associated with Genocchi, Euler, and tangent numbers.


## 2. Witt's Formula for Eulerian Polynomials

In this section, we assume thatℤp, ℚp, and ℂp will, respectively, denote the ring of p-adic integers, the field of p-adic numbers, and the completion of algebraic closure of ℚp. The p-adic norm is normalized so that |p|p=1/p.Letq be an indeterminate with |1-q|p<1. Then the q-number is defined by
(2.1)[x]q=1-qx1-q,[x]-q=1-(-q)x1+q,(see [6–18]).LetC(ℤp) be the space of continuous functions on ℤp. For f∈C(ℤp), the fermionic p-adic q-integral on ℤp is defined by
(2.2)I-q(f)=∫Zp‍f(x)dμ-q(x)=lim⁡N→∞1[pN]-q∑x=0pN-1‍f(x)(-q)x,
(see [7, 10–13]). From (2.2), we can derive the following:
(2.3)q-1I-q-1(f1)+I-q-1(f)=[2]q-1f(0),
where f1(x)=f(x+1).Let us takef(x)=e-x(1+q)t. Then, by (2.3), we get
(2.4)(q+e-(1+q)tq)∫Zp‍e-x(1+q)tdμ-q-1(x)=[2]q-1.Thus, from (2.4), we have
(2.5)∫Zp‍e-x(1+q)tdμ-q-1(x)=1+qe-(1+q)t+q=∑n=0∞‍An(-q)tnn!.By Taylor expansion on the left-hand side of (2.5), we get
(2.6)∑n=0∞‍(-1)n∫Zp‍xndμ-q-1(x)(1+q)ntnn!=∑n=0∞‍An(-q)tnn!.Comparing coefficients on the both sides of (2.6), we have
(2.7)∫Zp‍xndμ-q-1(x)=(-1)n(1+q)nAn(-q).
Therefore, by (2.7), we obtain the following theorem.Theorem 2.1.
Forn∈ℤ+, one has
(2.8)∫Zp‍xndμ-q-1(x)=(-1)n(1+q)nAn(-q),
where An(-q) is an Eulerian polynomials.It seems interesting to study Theorem2.1 at q=1. By (2.3), we get
(2.9)I-1(f1)+I-1(f)=2f(0),
where f1(x)=f(x+1). From (2.9), we can derive the following equation:
(2.10)∫Zp‍f(x+n)dμ-1(x)+(-1)n-1∫Zp‍f(x)dμ-1(x)=2∑l=0n-1‍(-1)n-l+1f(l),
where n∈ℤ+ (see [5–13]).From (2.9), we can derive the following:
(2.11)0=∫ℤp‍sina(x+1)dμ-1(x)+∫ℤp‍sinaxdμ-1(x)=(cosa+1)∫ℤp‍sinaxdμ-1(x)+sina∫ℤp‍cosaxdμ-1(x),2=∫Zp‍cos⁡a(x+1)dμ-1(x)+∫Zp‍cos⁡axdμ-1(x)=(cos⁡a+1)∫Zp‍cos⁡axdμ-1(x)-sin⁡a∫Zp‍sin⁡axdμ-1(x).
By (2.11), we get
(2.12)∫Zp‍sin⁡axdμ-1(x)=-sin⁡acos⁡a+1=-tana2.
From (1.10) and (2.12), we have
(2.13)∑n=1∞‍T2n-1(2n-1)!(a2)2n-1=-∫Zp‍sin⁡axdμ-1(x)=∑n=1∞‍(-1)na2n-1(2n-1)!∫Zp‍x2n-1dμ-1(x).
By comparing coefficients on the both sides of (2.13), we get
(2.14)∫Zp‍x2n-1dμ-1(x)=(-1)nT2n-122n-1,forn∈N,
where T2n-1 is the (2n-1)th tangent number.Therefore, by (2.14), we obtain the following theorem.Theorem 2.2.
Forn∈ℕ, one has
(2.15)∫Zp‍x2n-1dμ-1(x)=(-1)nT2n-122n-1,
where T2n-1 is the (2n-1)th tangent numbers.From Theorem2.1, one has
(2.16)∫Zp‍xndμ-1(x)=(-1)n2nAn(-1).
Therefore, by Theorem 2.2 and (2.16), we obtain the following corollary.Corollary 2.3.
Forn∈ℕ, one has
(2.17)A2n-1(-1)=(-1)n-1T2n-1.From (1.6) and (2.9), we have
(2.18)∫Zp‍extdμ-1(x)=2et+1=∑n=0∞‍Entnn!,
(see [5]). Thus, by (2.16) and (2.18), we get
(2.19)∫Zp‍x2n-1dμ-1(x)=E2n-1=(-1)nT2n-122n-1.
Therefore, by Corollary 2.3 and (2.19), we obtain the following corollary.Corollary 2.4.
Forn∈ℕ, one has
(2.20)E2n-1=(-1)nT2n-122n-1=-A2n-1(-1)22n-1.By (1.5) and (2.9), we get
(2.21)t∫Zp‍extdμ-1(x)=2te2t-1et-2te2t-1=∑n=0∞‍Bn(12)2ntnn!-∑n=0∞‍2nBnn!tn=∑n=0∞‍(Bn(12)-Bn)2ntnn!.By (2.21), we get
(2.22)∫Zp‍xndμ-1(x)=(Bn+1(1/2)-Bn+1)n+12n+1.Thus, from (2.19), Theorem 2.2 and Corollary 2.3, we have
(2.23)(B2n(1/2)-B2n)22n2n=(-1)nT2n-122n-1=-A2n-1(-1)22n-1.Therefore, by (2.23), we obtain the following theorem.Theorem 2.5.
Forn∈ℕ, one has
(2.24)(B2n(1/2)-B2n)22nn=(-1)nT2n-122n-2=-A2n-1(-1)22n-2.From (1.5), we note that
(2.25)t∫Zp‍extdμ-1(x)=2tet+1=∑n=0∞‍Gntnn!
(see [13, 14]). Thus, by (2.25), we get
(2.26)G0=0,(G+1)n+Gn=2δ1,n,
(see [13, 14]), with the usual convention about replacing Gn by Gn.From (1.5) and (2.9), one has
(2.27)t∫Zp‍extdμ-1(x)=2(tet-1-2te2t-1)=2∑n=0∞‍(Bn-2nBn)tnn!.
Thus, by (2.27), we get
(2.28)∫Zp‍xndμ-1(x)=2(Bn+1-2n+1Bn+1n+1).From (2.28), we have
(2.29)G2n2n=∫Zp‍x2n-1dμ-1(x)=B2n-22nB2nn,for∈N.
Therefore, by (2.19), Corollary 2.3 and (2.29), we obtain the following theorem.Theorem 2.6.
Forn∈ℕ, we have
(2.30)G2n=2(B2n-22nB2n).
In particular,(2.31)-122n-1(A2n-1(-1))=((-1)nT2n-1)122n-1=G2n2n.


## 3. Further Remark

In complex plane, we note that(3.1)tanx=1i(eix-e-ixeix+e-ix)=1i(1-2e-ixeix+e-ix)=1i(1-∑n=0∞‍Enn!2ninxn)=1i(-∑n=1∞‍Enn!2ninxn)=∑n=1∞‍(-1)n(2n-1)!E2n-122n-1x2n-1.
By (1.10) and (3.1), we also get
(3.2)T2n-1=(-1)nE2n-122n-1,forn∈N.
From (1.5), we have
(3.3)∑n=1∞‍t2n(2n)!G2n=∑n=1∞‍(it)2n(2n)!(-1)nG2n=2it1+eit-it=it(1-eit)1+eit=it(e-it/2-eit/2)eit/2+e-it/2=t((eit/2-e-it/2)/2i(eit/2+e-it/2)/2)=ttan(t2).
Thus, by (1.10) and (3.3), we get
(3.4)∑n=1∞‍t2n(2n)!G2n=ttan(t2)=t∑n=1∞‍(t/2)2n-1(2n-1)!T2n-1=∑n=1∞‍t2n(2n-1)!22n-1T2n-1.From (3.4), we have
(3.5)nT2n-1=22n-2G2n=22n-1(1-22n)B2n.
By (1.1), we see that
(3.6)21+e-2it=∑n=0∞‍An(-1)intnn!.
Thus, we note that
(3.7)∑n=1∞‍in-1An(-1)tnn!=1i(21+e-2it-1)=1-e-2it(1+e-2it)i=((eit-e-it)/2)((eit+e-it)/2)i=tant=∑n=1∞‍T2n-1t2n-1(2n-1)!.From (3.7), we have
(3.8)A2n(-1)=0,A2n-1(-1)=(-1)n-1T2n-1,(n≥1).
It is easy to show that
(3.9)∑k=1m‍kn(-1)k=(-1)m∑k=0n‍(nk)Ak(-1)2k+1mn-k-{(-1)m-1}2n+1An(-1).
For simple calculation, we can derive the following equation:
(3.10)itanx=eix-e-ixeix+e-ix=1-2e2ix-1+4e4ix-1.
By (3.10), we get
(3.11)xtanx=-ix+2ixe2ix-1-4ixe4ix-1=∑n=1∞‍(-1)nB2n4n(1-4n)(2n)!x2n.
Thus, from (3.11),we have
(3.12)tanx=∑n=1∞‍(-1)nB2n4n(1-4n)(2n)!x2n-1.By (1.10) and (3.12), we get
(3.13)T2n-1=(-1)nB2n4n(1-4n)2n,forn∈N.
From Corollary 2.3 and (3.13), we can derive the following identity:
(3.14)A2n-1(-1)=-B2n22n-1(1-4n)n.



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*Source: 269640-2012-07-17.xml*