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Differential equations for p, q-Touchard polynomials

  • Taekyun Kim , Orli Herscovici , Toufik Mansour and Seog-Hoon Rim EMAIL logo
Published/Copyright: November 11, 2016

Abstract

In this paper, we present differential equation for the generating function of the p, q-Touchard polynomials. An application to ordered partitions of a set is investigated.

MSC 2010: 05A18; 05A48; llB48; llB47

1 Introduction

It is well known that the Touchard polynomials (exponential polynomials or Bell polynomials) Tn(x), play an important role in statistics and probability(see [1-14]). They are given by

Tn(x)=k=0nS(n;k)xk.

where S(n, k) is the Stirling number of second kind (see [6], [14]). Moreover, the Touchard polynomials are given by the generating function (see [6], [13])

ex(et1)=n=0Tn(x)tnn!.

By using the combinatorial definitions of the Stirling and the Bell numbers and the fact that the generating (see [13]) function of the Bell numbers Bn is given by

eet1=n=0Bntnn!,

we have that Bn = Tn(1).

The q-exponential function is given by (see [6], [14])

expq(x)=(1+(1q)x)11q,(1)

which corresponds to the degenerated exponential function of Carlitz [2] with λ = 1−q. Borges [1] showed that this q-exponential function has the following expansion in Taylor series

expq(x)=1+x+k1q(2q1)(3q2)(kq(k1))xk+1(k+1)!.(2)

Following [6] and [3], we define the (p, q)-deformed Touchard polynomials, Tn(x|p, q), for integer n ≥ 0 by means of the generating function

Tp,q(x;t)=expp(x(expq(t)1))=n=0Tn(x|p,q)tnn!,(3)

where deformed exponential functions expp(t) and expq(t) are given by (1). The first five (p, q)-deformed Touchard polynomials are

T0(x|p,q)=1,T1(x|p,q)=x,T2(x|p,q)=xq+x2p,T3(x|p,q)=(2q2q)x+3pqx2+(2p2p)x3,T4(x|p,q)=(6q37q2+2q)x+(11pq24pq)x2+(12p2q6pq)x3+(6p37p2+2p)x4.

We denote the coefficient of xj in Tn(x|p, q) by Sp, q(n, i) (see [3]). For instance, Sp, q(1,1)=1, Sp,q(2,1) = q and Sp, q(2,2) = p. Clearly, Sp, q(n, 1) = q(2q−1)(3q−2)…((n−1)q−(n−2)) and Sp, q(n, n) = p(2p− 1) (3p−2)…((n−1)p−(n−2)).

Note that Kim at al. [7] studied the linear differential equations arising from the Poisson-Charlier, actuarial, and Meixner polynomials. Also, in [11] the authors considered differential equations arising from the generating function of general modified degenerate Euler numbers. New identities were obtained for Laguerre polynomials [10] and for Bernoulli polynomials of the second kind [5] using linear and non-linear differential equations. Using differential equations, new identities for Changee and Euler polynomials were obtained in [8], for Frobenius-Euler polynomials in [4], and for degenerate Euler numbers in [9].

Here, we study the differential equations arising from the generating function of (p, q)-deformed Touchard polynomials. In particular, we show the following result.

Theorem 1.1

LetN ≥ 1. Then the differential equation

F(N)=i=1NSp,q(N,i)xi(1+(1q)t)i1qNF1i(1p),

has a solutionF = Tp, q(x; t), whereF(N)=dNdtNTp,q(x;t).

2 Proofs and Applications

Let F = Tp, q(x;t) and define F(N)=dNdtNTp,q(x;t), for all N > 1. Observe that

F=Tp,q(x;t)=(1(1p)x+(1p)x(1+(1q)t)11q)11p.

Note that ddt(F1p)=(1p)x(1+(1q)t)11q1. Hence,

F(1)=x(1+(1q)t)11q1Fp,F(2)=qx(1+(1q)t)11q2Fp+px2(1+(1q)t)21q2F12(1p).

Let us assume that

F(N)=i=1NaN,i(p,q)xi(1+(1q)t)i1qNF1i(1p).

Clearly, a1,1(p, q) = 1, a2,1(p, q) = q and a2,2(p, q) = p.

By differentiating F(N) with respect to t, we obtain

F(N+1)=ddtF(N)=i=1N(i/(1q)N)(1q)aN,i(p,q)xi(1+(1q)t)i1qN1F1i(1p)+i=1N(1/(1p)i)(1p)xaN,i(p,q)xi(1+(1q)t)i+11qN1F1(i+1)(1p),

which implies

F(N+1)=i=1(iN(1q))aN,i(p,q)xj(1N+(1q)t)i1qN1F1i(1p)+i=2N+1(1i(1p))aN,i1(p,q)xj(1+(1q)t)i1qN1F1i(1p).

Note that we assume that F(N+1)=i=1N+1aN+1,i(p,q)xj(1+(1q)t)i1qN1F1j(1p), so by comparing the coefficients of xi(1+(1q)t)i1qN1F1i(1p), we obtain

aN+1,i(p,q)=(iN(1q))aN,i(p,q)+(1(i1)(1p))aN,i1(p,q)(4)

with the initial conditions a1,1(p, q) = 1 and aN, i(p, q) = 0 whenever N < i or i < 0.

Lemma 2.1

For all 1 ≤ iN, aN, i(p, q) is the coefficient ofxjinTN(x|p, q).

Proof

Define Aj(t)=NiaN,i(p,q)tNN!. Then, by multiplying (4) by tNN!, and summing over Ni, we obtain

(1+(1q)t)ddtAi(t)iAi(t)=Ai1(t)(i1)(1p)Ai1(t)

with A0(t) = 1.

Define A(x;t)=i0Aj(t)xj. By multiplying the last recurrence by xi and summing over i ≥ 0, we have

(1+(1q)t)ddtA(x;t)x(1(1p)x)ddxA(x;t)=xA(x;t)

with A(x; 0) = 1. Solving the partial differentiate equation for A(x; t), gives that there exists a function f(y) such that

A(x;t)=f1(1p)xx(1+(1q)t)11q(1+(1q)t)1(1q)(1p)x11p.

By using the initial condition A(x; 0) = 1, we obtain that f(y)=(y+1p)11p. Hence,

A(x;t)=1(1p)xx(1+(1q)t)11q+1p11p(1+(1q)t)1(1q)(1p)x11p=((1(1p)x)(1+(1q)t)11q+(1p)x)11p(1+(1q)t)1(1q)(1p)=(1(1p)x+(1p)x(1+(1q)t)11q)11p=(1+(1p)x(expq(t)1))11p=expp(x(expq(t)1))=Tp,q(x;t),

which completes the proof. □

Thus, by Lemma 2.1, we obtain that

F(N)=i=1NSp,q(N,i)xiN(1+(1q)t)i1qNF1i(1p),

where Sp, q(N, i) is the coefficient of xi in TN(x|p, q). This completes the proof of Theorem 1.1. □

Note that for p = 1, Theorem 1.1 gives

F(N)|p=1=i=1NS1,qN(N,i)xi(1+(1q)t)i1qNF,

where, by the main result of [12] and (4), we have

S1,q(N,i)=j=1i=1N1(j+(q1))=1,ji(j).

These numbers have the following combinatorial interpretation. An ordered partition of a set [n] = {1, 2,…, n} is a set of nonempty disjoint subsequences B1/B2/…/Bk, called blocks, whose union is [n] (For partitions of a set [n], see [13]). Note that we can assume that min B1 < min B2 < … < min Bk. For example, the ordered partition of the set [3] are given by 123, 132, 213, 231, 312, 321, 12/3, 21/3, 13/2, 31/2, 1/23, 1/32, 1/2/3. We denote the set of all ordered partitions of the set [n] with k blocks by OPn,k. Define OPn=k=0nOPn,k.

Let π be any ordered partition in OPn,k. Let π(i) be the ordered partition that is obtained from π by removing all the elements i + 1, i + 2, …, n from π. An element i is called a closer if there exists a block B in π(i) such that i is the rightmost element of B and B contains at least two elements. Otherwise, it is called an opener. For example, if π = 152/437/68 then 8, 7, 2 is closers and 6, 5, 4, 3, 1 openers.

Let bj(N, m) to be the number of ordered partitions of [N] with exactly m closers and i blocks. We define bN,i(q)=m=0Nbj(N,m)qm.

Lemma 2.2

LetN ≥ 1. Thenb1,1(q) = 1, bN, i(q) = 0 wheneveri > Nori < 0, and

bN+1,i(q)=bN,i1(q)+(i+Nq)bN,i(q),1iN+1.
Proof

Clearly the initial conditions are held. Let π be any ordered partition of [N + 1] with exactly 2 ≤ iN blocks and let us write an equation for bN + 1, i(q). If the element N + 1 is form a block, then it should be the rightmost block of π and in this case we have a contribution of bN, i−1(q) . Otherwise, the element N + 1 belongs to a block, say Bj with 1 ≤ ji. In the case N + 1 is rightmost element of Bj we have a contribution bN, i(q), and in the case N + 1 is not rightmost element of Bj we obtain a contribution of |Bj|qbN, i(q). Hence,

bN+1,i(q)=bN,i1(q)+j=1i(1+q|Bj|)bN,i(q)=bN,i1(q)+(i+Nq)bN,i(q),

as required. □

By (4) and Lemma 2.2, we obtain the following result.

Proposition 2.3

For all 1 ≤ iN, S1,q(N, i) = bN, i(q−1).

Recall that T1,q(x;t)=n0Tn(x|1,q)tnn!. Clearly,

F(N)=dNdtNT1,q(x;t)=n0Tn+N(x|1,q)tnn!.

Hence, by Theorem 1.1, we have

F(N)=i=1N0S1,q(N,i)xj(i1qN)(1q)t!m0Tm(x|1,q)tmm!=n0i=0N=0nni1qN(1q)S1,q(N,i)xjTn(x|1,q)tnn!,

where (c)={c|1}. By comparing the coefficients of tnn!, we obtain the following result.

Corollary 2.4

For allN ≥ 1,

Tn+N(x|1,q)=i=0N=0nni1qN(1q)S1,q(N,i)xjTn(x|1,q).

Acknowledgement

This research was supported by Kyungpook National University Bokhyeon Research Fund, 2015. Taekyun Kim was appointed as a chair professor at Tianjin polytechnic University by Tianjin city in China from Aug 2015 to Aug 2019. The second author was supported by the Ministry of Science and Technology, Israel.

References

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Received: 2016-7-20
Accepted: 2016-10-3
Published Online: 2016-11-11
Published in Print: 2016-1-1

© 2016 Kim et al., published by De Gruyter Open

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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