# q-Bernstein Polynomials Associated with q-Stirling Numbers and Carlitz's q-Bernoulli Numbers

Authors: T. Kim; J. Choi; Y. H. Kim
Journal: Abstract and Applied Analysis
 (2010)

---

## Abstract

Recently, Kim (2011) introducedq-Bernstein polynomials which are
different q-Bernstein polynomials of  Phillips (1997). In this paper, we give a p-adic
q-integral representation for q-Bernstein type polynomials and investigate some
interesting identities of q-Bernstein type polynomials associated with q-extensions
of the binomial distribution, q-Stirling numbers, and Carlitz's q-Bernoulli numbers.


---

## Body

## 1. Introduction

Letp be a fixed prime number. Throughout this paper, ℤp, ℚp, ℂ, and ℂp denote the ring of p-adic integers, the field of p-adic rational numbers, the complex number field, and the completion of the algebraic closure of ℚp, respectively. Let ℕ be the set of natural numbers and ℤ+=ℕ∪{0}. Let νp be the normalized exponential valuation of ℂp with |p|p=p-νp(p)=1/p.When one talks ofq-extensions, q is variously considered as an indeterminate, a complex number q∈ℂ, or a p-adic number q∈ℂp. If q∈ℂ then one normally assumes |q|<1, and if q∈ℂp then one normally assumes |1-q|p<1.Theq-bosonic natural numbers are defined by [n]q=(1-qn)/(1-q)=1+q+q2+⋯+qn-1 for n∈ℕ, and the q-factorial is defined by [n]q!=[n]q[n-1]q⋯[2]q[1]q (see [1–3]). For the q-extension of binomial coefficients, we use the following notation in the form of (1.1)(nk)q=[n]q![n-k]q![k]q!=[n]q[n-1]q⋯[n-k+1]q[k]q!.LetC[0,1] denote the set of continuous functions on the real interval [0,1]. The Bernstein operator for f∈C[0,1] is defined by (1.2)𝔹n(f∣x)=∑k=0nf(kn)(nk)xk(1-x)n-k=∑k=0nf(kn)Bk,n(x),
where n,k∈ℤ+. The polynomials Bk,n(x)=(nk)xk(1-x)n-k are called Bernstein polynomials of degree n (see [4–8]). For f∈C[0,1], q-Bernstein type operator of order n for f is defined by(1.3)𝔹n,q(f∣x)=∑k=0nf(kn)(nk)[x]qk[1-x]1/qn-k=∑k=0nf(kn)Bk,n(x,q),
where n,k∈ℤ+. Here Bk,n(x,q)=(nk)[x]qk[1-x]1/qn-k are called q-Bernstein type polynomials of degree n (see [9]).We say thatf is uniformly differentiable function at a point a∈ℤp and write f∈UD(ℤp), if the difference quotient Ff(x,y)=(f(x)-f(y))/(x-y) has a limit f'(a) as (x,y)→(a,a). For f∈UD(ℤp), the p-adic q-integral on ℤp is defined by (1.4)Iq(f)=∫ℤpf(x)dμq(x)=limN→∞1[pN]q∑x=0pN-1f(x)qx
(see [10]). Carlitz's q-Bernoulli numbers can be represented by a p-adic q-integral on ℤp as follows:(1.5)∫ℤp[x]qndμq(x)=limN→∞1[pN]q∑x=0pN-1[x]qnqx=βn,q
(see [10, 11]). The kth order factorial of [x]q is defined by (1.6)[x]k,q=[x]q[x-1]q⋯[x-k+1]q=(1-qx)(1-qx-1)⋯(1-qx-k+1)(1-q)k
and is called the q-factorial of x of order k (see [10]).In this paper, we give ap-adic q-integral representation for q-Bernstein type polynomials and derive some interesting identities for the q-Bernstein type polynomials associated with the q-extension of binomial distributions, q-Stirling numbers, and Carlitz's q-Bernoulli numbers.


## 2.q-Bernstein Polynomials

In this section, we assume that0<q<1. Let ℙn,q={∑iai[x]qi∣ai∈ℝ} be the space of q-polynomials of degree less than or equal to n.We claim that theq-Bernstein type polynomials of degree n defined by (1.3) are a basis for ℙn,q.First, we see that theq-Bernstein type polynomials of degree n span the space of q-polynomials. That is, any q-polynomials of degree less than or equal to n can be written as a linear combination of the q-Bernstein type polynomials of degree n.Forn,k∈ℤ+ and x∈[0,1], we have(2.1)Bk,n(x,q)=∑l=kn(nl)(lk)(-1)l-k[x]ql
(see [9]). If there exist constants C0,C1,…,Cn such that C0B0,n(x,q)+C1B1,n(x,q)+⋯+CnBn,n(x,q)=0 holds for all x, then we can derive the following equation from (2.1): (2.2)0=C0B0,n(x,q)+C1B1,n(x,q)+⋯+CnBn,n(x,q)=C0∑i=0n(-1)i(ni)(i0)[x]qi+C1∑i=1n(-1)i-1(ni)(i1)[x]qi+⋯+Cn∑i=nn(-1)i-n(ni)(in)[x]qi=C0+{∑i=01Ci(-1)i-1(n1)(1i)}[x]q+⋯+{∑i=0nCi(-1)i-n(nn)(ni)}[x]qn.
Since the power basis is a linearly independent set, it follows that (2.3)C0=0,∑i=01Ci(-1)i-1(n1)(1i)=0,⋮⋮∑i=0nCi(-1)i-n(nn)(ni)=0,
which implies that C0=C1=⋯=Cn=0 (C0 is clearly zero, substituting this in the second equation gives C1=0, substituting these two into the third equation gives C2=0, and so on). Hence, we have the following theorem.Theorem 2.1.
Theq-Bernstein type polynomials of degree n are a basis for ℙn,q.Let us consider aq-polynomial Pq(x)∈ℙn,q as a linear combination of q-Bernstein type basis functions as follows:(2.4)Pq(x)=C0B0,n(x,q)+C1B1,n(x,q)+⋯+CnBn,n(x,q).
We can write (2.4) as a dot product of two values:(2.5)Pq(x)=(B0,n(x,q),B1,n(x,q),…,Bn,n(x,q))(C0C1⋮Cn).
From (2.5), we can derive the following equation: (2.6)Pq(x)=(1,[x]q,…,[x]qn)(b0000⋯0b10b110⋯0⋮⋮⋮⋱⋮bn0bn1bn2⋯bnn)(C0C1⋮Cn),
where the bij are the coefficients of the power basis that are used to determine the respective q-Bernstein type polynomials.From (1.3) and (2.1), we note that (2.7)B0,2(x,q)=[1-x]1/q2=∑l=02(2l)(-1)l[x]ql=1-2[x]q+[x]q2,B1,2(x,q)=(21)[x]q[1-x]1/q=2[x]q-2[x]q2,B2,2(x,q)=(22)[x]q2=[x]q2.In the quadratic case (n=2), the matrix representation is (2.8)Pq(x)=(1,[x]q,[x]q2)(100-2201-21)(C0C1C2).
In the cubic case (n=3), the matrix representation is (2.9)Pq(x)=(1,[x]q,[x]q2,[x]q3)(1000-33003-630-13-31)(C0C1C2C3).
In many applications of q-Bernstein polynomials, a matrix formulation for the q-Bernstein type polynomials seems to be useful.Remark 2.2 (see [12]).
All results of this section forq=1 are well known in classical case (see Bernstein Polynomials by Joy).


## 3.q-Bernstein Polynomials, q-Stirling Numbers, and q-Bernoulli Numbers

In this section, we assume thatq∈ℂp with |1-q|p<1.Forf∈UD(ℤp), let us consider the p-adic analogue of q-Bernstein type operator of order n on ℤp as follows: (3.1)𝔹n,q(f∣x)=∑k=0nf(kn)(nk)[x]qk[1-x]1/qn-k=∑k=0nf(kn)Bk,n(x,q).
Here Bk,n(x,q) is the q-Bernstein type polynomials of degree n on ℤp defined by (3.2)Bk,n(x,q)=(nk)[x]qk[1-x]1/qn-k,
for n,k∈ℤ+ and x∈ℤp.Let(Eh)(x)=h(x+1) be the shift operator. Then the q-difference operator is defined by(3.3)Δqn∶=(E-I)qn=∏i=1n(E-qi-1I),
where (Ih)(x)=h(x). From (3.3), we derive the following equation:(3.4)Δqnf(0)=∑k=0n(nk)q(-1)kq(k2)f(n-k).
By (3.4), we easily see that (3.5)f(x)=∑n≥0(xn)qΔqnf(0)
(see [10, 11]).Theq-Stirling number of the first kind is defined by(3.6)∏k=1n(1+[k]qz)=∑k=0nS1,q(n,k)zk,
and the q-Stirling number of the second kind is also defined by(3.7)∏k=1n(11+[k]qz)=∑k=0nS2,q(n,k)zk.
By (3.3), (3.4), (3.6), and (3.7), we get (3.8)S2,q(n,k)=q-(k2)[k]q!∑j=0k(-1)jq(j2)(kj)q[k-j]qn=q-(k2)[k]q!Δqk0n,
for n,k∈ℤ+ (see [10, 13]).From the definition ofq-Bernstein type polynomials of degree n on ℤp, we easily see that(3.9)∫ℤpBk,n(x,q)dμq(x)=∑l=0n-k(n-kl)(nk)(-1)l∫ℤp[x]ql+kdμq(x).
By (1.5) and (3.9), we obtain the following proposition.Proposition 3.1.
Forn,k∈ℤ+, one has
(3.10)∫ℤpBk,n(x,q)dμq(x)=∑l=0n-k(n-kl)(nk)(-1)lβl+k,q,
where βl+k,q are the (l+k)th Carlitz's q-Bernoulli numbers.From the definition ofq-Bernstein polynomial, we note that(3.11)∑k=in(ki)(ni)Bk,n(x,q)=∑k=0iq(k2)(xk)q[k]q!S2,q(k,i-k),
where i∈ℕ. From the definition of q-binomial coefficient, we have(3.12)(n+1k)q=(nk-1)q+qk(nk)q=qn-k(nk-1)q+(nk)q.
By (3.12), we see that(3.13)∫ℤp(xn)qdμq(x)=(-1)n[n+1]qq(n+1)-(n+12)
(see [10, 11]). From (1.5), (3.11), and (3.13), we obtain the following theorem.Theorem 3.2.
Forn,k∈ℤ+ and i∈ℕ, one has
(3.14)∑k=in∑l=0n-k(ki)(ni)(n-kl)(nk)(-1)lβl+k,q=∑k=0iq(k2)[k]q!S2,q(k,i-k)(-1)k[k+1]qq(k+1)-(k+12).It is easy to see that, fori∈ℕ,(3.15)∑k=in(ki)(ni)Bk,n(x,q)=[x]qi.
By (3.11) and (3.15), we easily get (3.16)[x]qi=∑k=0iq(k2)(xk)q[k]q!S2,q(k,i-k)
(see [10]). Thus, we have(3.17)∫ℤp[x]qidμq(x)=∑k=0iq(k2)[k]q!S2,q(k,i-k)∫ℤp(xk)qdμq(x)=q∑k=0i[k]q!S2,q(k,i-k)(-1)k[k+1]q.
By (1.5) and (3.17), we obtain the following corollary.Corollary 3.3.
Forn,k∈ℤ+ and i∈ℕ, one has
(3.18)βi,q=q∑k=0i[k]q!S2,q(k,i-k)(-1)k[k+1]q.It is known that(3.19)S2,q(n,k)=1(1-q)k∑j=0k(-1)k-j(k+nk-j)(j+nj)q
(see [10]) and (3.20)(nk)q=∑j=0n(nj)(q-1)j-kS2,q(k,j-k).
By a simple calculation, we have that(3.21)qnx=∑k=0n(q-1)kq(k2)(nk)q[x]k,q=∑m=0n{∑k=mn(q-1)k(nk)qS1,q(k,m)}[x]qm,qnx=∑m=0n(nm)(q-1)m[x]qm.
From (3.21), we note that (3.22)(nm)=∑k=mn(q-1)-m+k(nk)qS1,q(k,m)
(see [10]).Thus, we obtain the following proposition.Proposition 3.4.
Forn,k∈ℤ+, one has
(3.23)Bk,n(x,q)=(nk)[x]qk[1-x]1/qn-k=∑m=kn(q-1)-k+m(nm)qS1,q(m,k)[x]qk[1-x]1/qn-k.From the definition of theq-Stirling numbers of the first kind, we get(3.24)q(n2)(xn)q[n]q!=[x]n,qq(n2)=∑k=0nS1,q(n,k)[x]qk.
By (3.11) and (3.24), we obtain the following theorem.Theorem 3.5.
Forn,k∈ℤ+ and i∈ℕ, one has
(3.25)∑k=in(ki)(ni)Bk,n(x,q)=∑k=0i∑l=0kS1,q(k,l)S2,q(k,i-k)[x]ql.By (3.15) and Theorem 3.5, we obtain the following corollary.Corollary 3.6.
Fori∈ℤ+, one has
(3.26)βi,q=∑k=0i∑l=0kS1,q(k,l)S2,q(k,i-k)βl.q.Theq-Bernoulli polynomials of order k∈ℤ+ are defined by (3.27)βn,q(k)(x)=1(1-q)n∑i=0n(ni)(-1)iqix∫ℤp⋯∫ℤpq∑l=1k(k-l+i)xldμq(x1)⋯dμq(xk).
Thus, we have (3.28)βn,q(k)(x)=1(1-q)n∑i=0n(-1)i(ni)(i+k)⋯(i+1)[i+k]q⋯[i+1]qqix
(see [10]). The inverse q-Bernoulli polynomials of order k are defined by(3.29)βn,q(-k)(x)=1(1-q)n∑i=0n(-1)i(ni)qix∫ℤp⋯∫ℤpq∑l=1k(k-l+i)xldμq(x1)⋯dμq(xk).
In the special case x=0, βn,q(k)(0)=βn,q(k) are called the nth q-Bernoulli numbers of order k, and βn,q(-k)(0)=βn,q(-k) are also called the inverse q-Bernoulli numbers of order k (see [10]).From (3.29), we have(3.30)βk,q(-n)=1(1-q)k∑j=0k(-1)j(kj)[j+n]q⋯[j+1]q(j+n)⋯(j+1)=1(1-q)k∑j=0k(-1)j(k+nn-j)(k+nn)(j+nn)q[n]q!n!=[n]q!(k+nn)n!{1(1-q)k∑j=0k(-1)j(k+nn-j)(j+nn)q}.
By (3.19) and (3.30), we get(3.31)n![n]q!(k+nn)βk,q(-n)=S2,q(n,k).
Therefore, by (3.11) and (3.31), we obtain the following theorem.Theorem 3.7.
Fori,n,k∈ℤ+, one has
(3.32)∑k=in(ki)(ni)Bk,n(x,q)=∑k=0iq(k2)k!(ik)(xk)qβi-k,q(-k).It is easy to show that(3.33)q(n2)(xn)q=1[n]q!∏k=0n-1([x]q-[k]q)=1[n]q!∑k=0n(-1)k[x]qn-kS1,q(n-1,k).
Thus, we have that (3.34)∑k=in(ki)(ni)Bk,n(x,q)=∑k=0i∑j=0k(-1)j[x]qk-jS1,q(k-1,j)k![k]q!(ik)βi-k,q(-k),
where n,k,i∈ℤ+.



---

*Source: 150975-2011-01-24.xml*