# Some Identities on the High-Orderq-Euler Numbers and Polynomials with Weight 0

Authors: Jongsung Choi; Hyun-Mee Kim; Young-Hee Kim
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We construct theNth order nonlinear ordinary differential equation related to the generating function of q-Euler numbers with weight 0. From this, we derive some identities on q-Euler numbers and polynomials of higher order with weight 0.


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## Body

## 1. Introduction

As a well-known definition, the Euler polynomialEn(x) is given by
(1)2et+1ext=∑n=0∞‍En(x)tnn!.
In the special case, x=0, En(0)=En is the nth Euler number.From (1), we note that
(2)E0=1,(E+1)n+En=0,ifn>0,
with the usual convention of replacing En by En (see [1–16]).In the viewpoint of theq-extension of (1) and (2), let us consider the following q-Euler number and polynomial:
(3)2qet+1ext=∑n=0∞‍E~n,q(x)tnn!,(4)E~0,q=21+q,q(E~q+1)n+E~n,q=0,ifn>0,
with the usual convention of replacing E~qn by E~n,q.Equation (3) is called the generating function of q-Euler polynomial with weight 0. In the case x=0, E~n,q(0)=E~n,q is the nth q-Euler number with weight 0 (see [5, 11]).Throughout this paper, letq be a complex number with |q|<1. As q→1, we obtain (1) and (2) from (3) and (4).The generating function of Eulerian polynomialHn(x∣u) is defined by
(5)1-uet-uext=∑n=0∞‍Hn(x∣u)tnn!,
where u∈ℂ with u≠1. In the special case, x=0, Hn(0∣u)=Hn(u) is called the nth Eulerian number (see [1–3]). Sometimes that is called the nth Frobenius-Euler number (see [9–11, 15]).From (1) and (5), we note that Hn(x∣-1)=En(x). From (5), we have
(6)H0(u)=1,Hn(1∣u)-uHn(u)=(1-u)δ0,n,
where δn,k is Kronecker symbol (see [9–11]).ForN∈ℕ, the q-Euler polynomial of order N is defined by the generating function as follows:
(7)GqN(t,x)=(2qet+1)×⋯×(2qet+1)︸N-timesext=∑n=0∞‍E~n,q(N)(x)tnn!.
In the special case, x=0, E~n,q(N)(0)=E~n,q(N) is called the nth q-Euler number of order N with weight 0 (see [5, 11]).In [9], Kim derived some identities between the sums of products of Frobenius-Euler polynomials and Frobenius-Euler polynomials of higher order. The main idea is to construct nonlinear ordinary differential equations with respect to t which are closely related to the generating function of Frobenius-Euler polynomial. In [3], Choi considered nonlinear ordinary differential equations with respect to u not t.In this paper, we construct nonlinear ordinary differential equations with respect toq. The purpose of this paper is to give some new identities on the high order q-Euler numbers and polynomials with weight 0 by using the differential equations of q.


## 2. Construction of Nonlinear Differential Equations

We define(8)G=G(q)=1qet+1,GN(t,x)=G×⋯×G︸N-timesextforN∈ℕ.
From (7) and (8), we note that
(9)GqN(t,x)=2NGN(t,x)=2NGNext.
By differentiating (8) with respect to q, we get
(10)G(1)=dGdq=-qet+1-1q(qet+1)2=-Gq+G2q,qG(1)+G=G2.
By differentiating (10) with respect to q, we get
(11)q2G(2)+4qG(1)+2G=2!G3,
where G(N)=dNG/dqN.By the derivative of (11) with respect to q, we have
(12)q3G(3)+9q2G(2)+18qG(1)+3!G=3!G4.
Continuing this process, we get
(13)(N-1)!GN=∑k=0N-1‍ak(N)qkG(k).Let us consider the derivative of (13) with respect to q to find the coefficient ak(N) in (13).By (10), we get
(14)qddq((N-1)!GN)=N!GN-1qG(1)=N!GN-1(-G+G2)=N!GN+1-N(N-1)!GN.
From (13) and (14), we get
(15)N!GN+1=N(N-1)!GN+∑k=0N-1‍kak(N)qkG(k)+∑k=1N‍ak-1(N)qkG(k),
where N!GN+1=∑k=0N‍ak(N+1)qkG(k).By comparing coefficients on both sides of (15), we obtain the following recurrence relations:
(16)a0(N+1)=Na0(N),aN(N+1)=aN-1(N),(17)ak(N+1)=Nak(N)+kak(N)+ak-1(N),
for 1≤k≤N-1 and ak(N)=0.From the first part of (16), we have
(18)a0(N+1)=Na0(N)=N(N-1)a0(N-1)=⋯=N!a0(2).
By (10) and (13), we have
(19)qG(1)+G=G2=∑k=01‍ak(2)qkG(k)=a0(2)G+a1(2)qG(1).
From (18) and (19), we get
(20)a0(2)=1,a1(2)=1,a0(N)=(N-1)!.
From the second part of (16), we have
(21)aN(N+1)=aN-1(N)=⋯=a1(2)=1.To findak(N) in (13) from (17), we set
(22)g(t,s)=∑N≥1‍∑0≤k≤N-1‍ak(N)tNN!sk,
where |t|<1 (see [9]).From (17) and (22), we have
(23)∑N≥1‍∑0≤k≤N-1‍ak+1(N+1)tNN!sk=∑N≥1‍∑0≤k≤N-1‍Nak-1(N)tNN!sk+∑N≥1‍∑0≤k≤N-1‍(k+1)ak+1(N)tNN!sk+g(t,s).
From the left hand side of (23), we have
(24)∑N≥1‍∑0≤k≤N-1‍ak+1(N+1)tNN!sk=1s∑N≥2‍∑1≤k≤N-1‍ak(N)tN-1(N-1)!sk=1s∑N≥2‍(∑0≤k≤N-1‍ak(N)tN-1(N-1)!sk-a0(N)tN-1(N-1)!)=1s(∑N≥1‍∑0≤k≤N-1‍ak(N)tN-1(N-1)!sk-a0(1)-∑N≥2‍tN-1)=1s(gt+1t-1),
where gt=∂g/∂t. From the first term of the right hand side of (23), we have
(25)∑N≥1‍∑0≤k≤N-1‍Nak+1(N)tNN!sk=ts∑N≥1‍∑1≤k≤N-1‍ak(N)tN-1(N-1)!sk=ts(∑N≥1‍∑0≤k≤N-1‍ak(N)tN-1(N-1)!sk-∑N≥1‍a0(N)(N-1)!tN-1)=ts(gt+1t-1).
From the second term of the right hand side of (23), we have
(26)∑N≥1‍∑0≤k≤N-1‍(k+1)ak+1(N)tNN!sk=∑N≥1‍∑1≤k≤N‍kak(N)tNN!sk-1=∑N≥1‍∑0≤k≤N-1‍kak(N)tNN!sk-1=gs,
where gs=∂g/∂s.From (22)–(26), we obtain the following initial value problem quasilinear first-order partial differential equation:
(27)(t-1)gt+sgs=-sg-1,|t|<1,g(0,s)=0,s∈ℝ.We consider Cauchy problem for the following first-order quasilinear partial differential equation:(28)P(x,y,z)zx+Q(x,y,z)zy=R(x,y,z),z(x0(t),y0(t))=z0(t),t∈I,
where I is some interval.We know that (28) has a unique solution under some conditions as follows.Theorem A (see [17, page 65]).
Suppose thatP,Q, and R are of class C1 in a domain Ω of ℝ3 containing the point (x0,y0,z0) and suppose that
(29)P(x0,y0,z0)dy0(t0)dt-Q(x0,y0,z0)dx0(t0)dt≠0.
Then in a neighborhood U of (x0,y0) there exists a unique solution of (28) at every point of initial curve contained in U.Since (27) satisfies (29) and regularity conditions, there exists a unique solution of (27).It is customary to write (27) in the form
(30)dtt-1=dss=dg-sg-1,(31)t=0,s=p,g=0.
Since dt/(t-1)=ds/s is separable, we get
(32)u1(t,s,g)=1-ts.u1 is a solution of partial differential equation of (27).From (30), we get the linear equation
(33)dgds=-g-1s.
By the integrating factor method, we have
(34)u2(t,s,g)=esg+Ei(s).
The exponential integral Ei(s) is defined by
(35)Ei(s)=∫-∞s‍errdr=γ+ln|s|+∑n=1∞‍snn·n!,(s∈ℝ,s≠0),
where γ is Euler constant.u
2 is another solution of partial differential equation of (27), and u1 and u2 are linearly independent.From the parameterized initial conditions (31), (33), and (34), we get
(36)u2=Ei(1u1),esg+Ei(x)=Ei(s1-t).
Thus, from (35) and (36), we obtain the following unique solution of (27):
(37)g(t,s)=e-s(-ln|1-t|+∑n=1∞‍snn·n!((11-t)n-1)).
Moreover, if we choose another initial condition
(38)g(t,0)=∑N≥1∞‍a0(N)tNN!=∑N≥1∞‍tNN
from (20) and (22), then (37) satisfies it.We note that(39)(11-t)n-1=(∑l1≥0‍tl1)×⋯×(∑ln≥0‍tln)︸n-times-1=∑N≥1‍(∑l1+⋯+ln=N‍tN)=∑N≥1‍(n+N-1N)tN.
By (37) and (39), we get
(40)g(t,s)=(∑k≥0‍(-1)kk!sk)(∑N≥1‍tNN)+(∑k≥0‍(-1)kk!sk)×(∑n≥1‍snn·n!∑N≥1‍(n+N-1N)tN)=∑N≥1‍∑0≤k≤N-1‍(-1)kN·k!tNsk+∑N≥1‍∑k≥N‍(-1)kN·k!tNsk+∑N≥1‍∑1≤k≤N-1‍(∑l=1k‍(-1)k-l(k-l)!l·l!(l+N-1N))tNsk+∑N≥1‍∑k≥N‍(∑l=1k‍(-1)k-l(k-l)!l·l!(l+N-1N))tNsk.
It is known that
(41)(-1)l(l+N-1N)=lN(-Nl),∑l=0k‍(kl)(Nl)=(k+Nk).
In the case of k≥N in (40), from (41), we get
(42)(-1)kN·k!+∑l=1k‍(-1)k-l(k-l)!l·l!(l+N-1N)=(-1)k(1N·k!+1N·k!∑l=1k‍(kl)(-Nl))=(-1)k1N·k!(1+(k-Nk)-1)=0.
By (40) and (41), we get
(43)g(t,s)=∑N≥1‍∑1≤k≤N-1‍(-1)k(N-1)!k!(k-Nk)tNN!sk+∑N≥1‍(N-1)!tNN!=∑N≥1‍∑0≤k≤N-1‍(N-k-1)!(N-1k)2tNN!sk,
where (k-Nk)=(-1)k(N-1k). Thus, by (22) and (43), we get
(44)ak(N)=(N-k-1)!(N-1k)2.
Therefore, by (13) and (44), we obtain the following theorem.Theorem 1.
Forq∈ℂ with |q|<1 and N∈ℕ, one can consider the following nonlinear (N-1)th order ordinary differential equation with respect to q:
(45)GN(q)=1(N-1)!∑k=0N-1‍(N-k-1)!(N-1k)2qkG(k)=∑k=0N-1‍1k!(N-1k)qkG(k),
where G(k)=dkG(q)/dqk and GN(q)=G(q)×⋯×G(q)︸N-times. Then G(q)=1/(qet+1) is a solution of (45).Let us defineG(k)(t,x)=G(k)(q)ext. Then we obtain the following corollary.Corollary 2.
ForN∈ℕ, one considers
(46)GN(t,x)=1(N-1)!∑k=0N-1‍(N-k-1)!(N-1k)2qkG(k)(t,x)=∑k=0N-1‍1k!(N-1k)qkG(k)(t,x).
Then G(t,x)=ext/(qet+1) is a solution of (46).


## 3. Identities on the High-Orderq-Euler Numbers and Polynomials with Weight 0

From (3), (7), and (8), we get
(47)GN(q)=12N(2qet+1)×⋯×(2qet+1)︸N-times=12N∑n=0∞‍E~n,q(N)tnn!,G(q)=122qet+1=12∑n=0∞‍E~n,qtnn!.
From (47), we note that
(48)G(k)=dkG(q)dqk=12∑n=0∞‍dkE~n,qdqktnn!.
Therefore, by (47), (48), and (45), we obtain the following theorem.Theorem 3.
ForN∈ℕ and n∈ℕ∪{0}, one has
(49)E~n,q(N)=2N-1∑k=0N-1‍1k!(N-1k)qkdkE~n,qdqk.From (48), we get
(50)G(k)(t,x)=G(k)(q)ext=(12∑n=0∞‍dkE~n,qdqktnn!)(∑n=0∞‍xntnn!)=∑n=0∞‍(∑l=0n‍12(nl)xn-ldkE~l,qdqk)tnn!.
Therefore, by (7), (47), and (50), we obtain the following corollary.Corollary 4.
ForN∈ℕ and n∈ℕ∪{0}, one has
(51)E~n,q(N)(x)=2N-1∑k=0N-1‍1k!(N-1k)qk∑l=0n‍(nl)xn-ldkE~l,qdqk.From (3) and (7), we get
(52)∑n=0∞‍E~n,q(N)tnn!=(2qet+1)×⋯×(2qet+1)︸N-times=(∑l1=0∞‍E~l1,qtl1l1!)×⋯×(∑lN=0∞‍E~lN,qtlNlN!)=∑n=0∞‍(∑l1+⋯+lN=n‍n!E~l1,q⋯E~lN,ql1!⋯lN!)tnn!=∑n=0∞‍(∑l1+⋯+lN=n‍(nl1,…,lN)E~l1,q⋯E~lN,q)tnn!.
Therefore, by (49) and (52), we obtain the following corollary.Corollary 5.
ForN∈ℕ and n∈ℕ∪{0}, one has
(53)∑l1+⋯+lN=n‍(nl1,…,lN)E~l1,q⋯E~lN,q=2N-1∑k=0N-1‍1k!(N-1k)qkdkE~n,qdqk.



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