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arXiv:1009.0097v1 [math.NT] 1 Sep 2010
A NOTE ON q-BERNSTEIN POLYNOMIALS
Taekyun Kim
Abstract. Recently, Simsek-Acikgoz([17]) and Kim-Jang-Yi([9]) have studied the qextension of Bernstein polynomials. In this paper we propose the q-extension of Bernstein polynomials of degree n, which are different q-Bernstein polynomials of SimsekAcikgoz([17]) and Kim-Jang-Yi([9]). From these q-Bernstein polynomials, we derive
some fermionic p-adic integral representations of several q-Bernstein type polynomials.
Finally, we investigate some identities between q-Bernstein polynomials and q-Euler
numbers.
§1. Introduction
Let C[0, 1] denote the set of continuous function on [0, 1]. For f ∈ C[0, 1], Bernstein
introduced the following well known linear operators (see [1, 3]):
(1)
Bn (f |x) =
n
X
k=0
n
X
k
k n k
n−k
x (1 − x)
=
f ( )Bk,n (x).
f( )
n k
n
k=0
Here Bn (f |x) is called Bernstein operator of order n for f . For k, n ∈ Z+ (= N ∪ {0}),
the Bernstein polynomials of degree n is defined by
n k
x (1 − x)n−k , (see [1, 2, 3]).
(2)
Bk,n (x) =
k
A Bernoulli trial involves performing an experiment once and noting whether a particular event A occurs. The outcome of Bernoulli trial is said to be “success” if A
occurs and a “failure” otherwise. Let k be the number of successes in n independent
Bernoulli trials, the probabilities of k are given by the binomial probability law:
n k
p (1 − p)n−k , for k = 0, 1, · · · , n,
pn (k) =
k
Key words and phrases. q-Bernstein polynomials, q-Euler numbers , q-Stirling numbers, fermionic
p-adic integrals.
2000 AMS Subject Classification: 11B68, 11S80, 60C05, 05A30
The present Research has been conducted by the research Grant of Kwangwoon University in 2010
Typeset by AMS-TEX
1
where pn (k) is the probability of k successes in n trials. For example, a communication
system transmit binary information over channel that introduces random bit errors
with probability ξ = 10−3 . The transmitter transmits each information bit three
times, an a decoder takes a majority vote of the received bits to decide on what the
transmitted bit was. The receiver can correct a single error, but it will make the wrong
decision if the channel introduces two or more errors. If we view each transmission
as a Bernoulli trial in which a “success” corresponds to the introduction of an error,
then the probability of two or more errors in three Bernoulli trial is
3
3
2
(0.001)3 ≈ 3(10−6 ), see [18].
(0.001) (0.999) +
p(k ≥ 2) =
3
2
By the definition of Bernstein polynomials(see Eq.(1) and Eq.(2)), we can see that
Bernstein basis is the probability mass function of binomial distribution. In the reference [15] and [16], Phillips proposed a generalization of classical Bernstein polynomials
based on q-integers. In the last decade some new generalizations of well known positive linear operators based on q-integers were introduced and studied by several aux
thors(see [1-21]). Let 0 < q < 1. Define the q-numbers of x by [x]q = 1−q
1−q (see [1-21]).
Recently, Simsek-Acikgoz([17]) and Kim-Jang-Yi([9]) have studied the q-extension of
Bernstein polynomials, which are different Phillips q-Bernstein polynomials. Let p be
a fixed odd prime number. Throughout this paper Zp , Qp , and Cp denote the rings of
p-adic integers, the fields of p-adic rational numbers, and the completion of algebraic
closure of Qp , respectively. The p-adic absolute value in Cp is normalized in such way
that |p|p = p1 . As well known definition, Euler polynomials are defined by
∞
X
2
tn
xt
e
=
E
(x)
, (see [1-14]).
n
et + 1
n!
n=0
(3)
In the special case, x = 0, En (0) = En are called the n-th Euler numbers. By (3), we
see that the recurrence formula of Euler numbers is given by
(4)
E0 = 1, and (E + 1)n + En = 0 if n > 0, (see [12]),
with the usual convention of replacing E n by En . When one talks of q-analogue, q is
variously considered as an indterminate, a complex number q ∈ C, or a p-adic number
q ∈ Cp . If q ∈ C, we normally assume |q| < 1. If q ∈ Cp , we normally always assume
that |1 − q|p < 1. As the q-extension of (4), author defined the q-Euler numbers as
follows:
(5)
E0,q = 1, and (qEq + 1)n + En,q = 0 if n > 0, ( see [21]),
with the usual convention of replacing Eqn by En,q . Let U D(Zp ) be the space of
uniformly differentiable function on Zp . For f ∈ U D(Zp ), the fermionic p-adic qintegral was defined by
N
(6)
pX
−1
1
f (x)(−q)x , (see [12]).
Iq (f ) =
f (x)dµ−q (x) = lim
pN
N→∞
1
+
q
Zp
x=0
2
Z
In the special case, q = 1, I1 (f ) is called the fermionic p-adic integral on Zp (see [12,
21]). By (6) and the definition of I1 (f ), we see that
(7)
I1 (f1 ) + I1 (f ) = 2f (0), where f1 (x) = f (x + 1).
For n ∈ N, let fn (x) = f (x + n). Then we can also see that
(8)
I1 (fn ) + (−1)
n−1
I1 (f ) = 2
n−1
X
(−1)n−l−1 f (l), (see [21]).
l=0
From (5), (7) and (8), we note that
(9)
Z
∞
X
∞
X
tn
[x]q t
e
dµ−1 (x) =
En,q =
n! n=0
Zp
n=0
Thus we have
En,q
n X
n (−1)l
2
l 1 + ql
(1 − q)n
l=0
!
tn
.
n!
n X
n (−1)l
2
, (see [21]).
=
l 1 + ql
(1 − q)n
l=0
In [21], the q-Euler polynomials are defined by
(10)
En,q (x) =
Z
[y +
x]nq dµ−1 (x)
Zp
n lx
X
n
1
l q
(−1)
.
=
l
(1 − q)n
1 + ql
l=0
By (9) and (100, we get
(11)
En,q (x) =
n X
n
l=0
l
q lx El,q = (q x Eq + 1)n ,
with the usual convention of replacing Eqn by En,q . In this paper we firstly consider the
q-Bernstein polynomials of degree n in R, which are different q-Bernstein polynomials
of Simsek-Acikgoz([17]) and Kim-Jang-Yi([9]). From these q-Bernstein polynomials,
we try to study for the fermionic p-adic integral representations of the several qBernstein type polynomials on Zp . Finally, we give some interesting identities between
q-Bernstein polynomials and q-Euler numbers.
§2. q-Bernstein Polynomials
For n, k ∈ Z+ , the generating function for Bk,n (x) is introduced by Acikgoz and
Araci as follows:
(12)
F
(k)
∞
X
te(1−x)t xk
tn
(t, x) =
Bk,n (x) , (see [1, 9, 10, 17]).
=
k!
n!
n=0
3
For k, n ∈ Z+ , 0 < q < 1 and x ∈ [0, 1], consider the q-extension of (12) as follows:
(13)


n−k
k
[1−x] 1 t
∞ [1 − x]n
∞
k
n![x]
[1
−
x]
k
n
1
1
X
X
q
q
[x]q
(t[x]q ) e
q n+k
q
t

Fq(k) (t, x) =
=
t
=
k!
k! n=0
n!
(n − k)!k!
n!
n=k
=
∞
X
n=k
∞
n
X
tn
n
n−k t
k
Bk,n (x, q) .
=
[x]q [1 − x] 1
q
n!
n!
k
n=k
Because Bk,0 (x, q) = Bk,1 (x, q) = · · · = Bk,k−1 (x, q) = 0, we obtain the following
generating function for Bk,n (x, q):
[1−x] 1 t
Fq(k) (t, x)
(t[x]q )k e
=
k!
q
=
∞
X
Bk,n (x, q)
n=0
tn
, where k ∈ Z+ and x ∈ [0, 1].
n!
Thus, for k, n ∈ Z+ , we note that
(14)
n
[x]kq [1 − x]n−k
Bk,n (x, q) =
, if n ≥ k ,
1
q
k
= 0, if k < n.
By (14), we easily get limq→1 Bk,n (x, q) = Bk,n (x). For 0 ≤ k ≤ n, we have
[1 − x] q1 Bk,n−1 (x, q) + [x]q Bk−1,n−1 (x, q)
n−1
n−1
n−k−1
k
[x]q [1 − x] 1
[x]k−1
[1 − x]n−k
+ [x]q
= [1 − x] q1
1
q
q
q
k
k−1
n−1
n
n−1
[x]kq [1 − x]n−k
[x]kq [1 − x]n−k
+
=
,
[x]kq [1 − x]n−k
=
1
1
1
q
q
q
k−1
k
k
and the derivative of the q-Bernstein polynomials of degree n are also polynomials of
degree n − 1.
d
Bk−1,n (x, q)
dx n
n
log q
log q
n−k
n−k−1
k−1
k
x
[x]q [1 − x] 1
=k
[x]q (n − k)[1 − x] 1
q +
qx
q
q
k
k
q−1
1−q
n−1
log q x
n−1
[x]k−1
[1 − x]n−k
q n
[x]kq [1 − x]n−1−k
=
−n
1
1
q
q
q
k−1
q−1
k
log q x
= n (Bk−1,n−1 (x, q) − Bk,n−1 (x, q))
q .
q−1
Therefore, we obtain the following theorem.
4
Theorem 1. For k, n ∈ Z+ and x ∈ [0, 1], we have
[1 − x] 1q Bk,n−1 (x, q) + [x]q Bk−1,n−1 (x, q) = Bk,n (x, q),
and
log q x
d
Bk,n (x, q) = n (Bk−1,n−1 (x, q) − Bk,n−1 (x, q))
q .
dx
q−1
Let f be a continuous function on [0, 1]. Then the q-Bernstein operator of order n
for f is defined by
(15)
Bn,q (f |x) =
n
X
k=0
k
f ( )Bk,n (x, q), where 0 ≤ x ≤ 1 and n ∈ Z+ .
n
By (14) and (15), we see that
Bn,q (1|x) =
n
X
k=0
n n
X
n
n−k
k
= 1.
[x]q [1 − x] 1 = [x]q + [1 − x] 1q
Bk,n (x, q) =
q
k
k=0
Also, we get from (15) that for f (x) = x,
n
n−1
X
X n − 1
k n
n−k
k
[x]q [1 − x] 1 =
Bn,q (x|x) =
= [x]q .
[x]k+1
[1 − x]n−k−1
1
q
q
q
n k
k
k=0
k=0
The q-Bernstein polynomials are symmetric polynomials in the following sense:
n
1
[x]kq = Bk,n (x, q).
[1 − x]n−k
Bn−k,n (1 − x, ) =
1
q
n−k
q
Thus, we obtain the following theorem.
Theorem 2. For n, k ∈ Z+ and (x ∈ [0, 1], we have
1
Bn−k,n (1 − x, ) = Bk,n (x, q).
q
Moreover, Bn,q (1|x) = 1 and Bn,q (x|x) = [x]q .
From (15), we note that
(16)
n
X
n
X
k
k n
[x]kq [1 − x]n−k
Bn,q (f |x) =
f ( )Bk,n (x, q) =
f( )
1
q
n
n k
k=0
k=0
n−k
n
X n − k
X k n
k
(−1)j [x]jq .
[x]q
=
f( )
j
n k
j=0
k=0
5
By the definition of binomial coefficient, we easily get
k+j
n
n n−k
.
=
k
k+j
j
k
Let k + j = m. Then we have
m
n
n n−k
.
=
(17)
k
m
j
k
From (16) and (17), we have
(18)
m n X
X
k
m
n
m
(−1)m−k f ( ).
[x]q
Bn,q (f |x) =
k
m
n
m=0
k=0
Therefore, we obtain the following theorem.
Theorem 3. For f ∈ C[0, 1] and n ∈ Z+ , we have
m n X
X
k
m
n
m
(−1)m−k f ( ).
[x]q
Bn,q (f |x) =
k
m
n
m=0
k=0
It is well known that the second kind stirling numbers are defined by
(19)
∞
k X
tn
1 X k
(et − 1)k
k−l lt
(−1) e =
s(n, k) , for k ∈ N, (see [12, 21]).
=
l
k!
k!
n!
n=0
l=0
Let ∆ be the shift difference operator with ∆f (x) = f (x + 1) − f (x). By iterative
process, we easily get
n X
n
n
(−1)n−k f (k).
(20)
∆ f (0) =
k
k=0
From (19) and (20), we can easily derive the following equation (21).
(21)
1 k n
∆ 0 = s(n, k).
k!
By (18) and (20) we obtain the following theorem.
Theorem 4. For f ∈ C[0, 1] and n ∈ Z+ , we have
n X
0
n
[x]kq ∆k f ( ).
Bn,q (f |x) =
k
n
k=0
In the special case, f (x) = xm (m ∈ Z+ ), we have the following corollary.
6
Corollary 5. For x ∈ [0, 1] and m, n ∈ Z+ , we have
n X
n
[x]kq ∆k 0m ,
n Bn,q (x |x) =
k
m
m
k=0
and
n X
n
[x]kq k!s(m, k).
n Bn,q (x |x) =
k
m
m
k=0
For x, t ∈ C and n ∈ Z+ with n ≥ k, consider
n!
2πi
(22)
Z
C
([x]q t)k ([1−x] q1 t) dt
e
,
k!
tn+1
where C is a circle around the origin and integration is in the positive direction. We
see from the definition of the q-Bernstein polynomials and the basic theory of complex
analysis including Laurent series that
(23)
Z
C
∞ Z
X
Bk,m (x, q)tm dt
Bk,n (x, q)
([x]q t)k ([1−x] q1 t) dt
.
e
=
= 2πi
n+1
k!
tn+1
m!
t
n!
C
m=0
We get from (22) and (23) that
n!
2πi
(24)
Z
C
([x]q t)k ([1−x] q1 t) dt
e
= Bk,n (x, q),
k!
tn+1
and
Z
C
(25)
!
Z
∞
[1 − x]m
1
[x]kq X
([x]q t)k ([1−x] q1 t) dt
q
tm−n−1+k dt
e
=
k!
tn+1
k! m=0
m!
C


[x]kq [1 − x]n−k
1
2πi
n
q
=
= 2πi 
[x]kq [1 − x]n−k
.
1
q
k!(n − k)!
n! k
By (22) and (25), we see that
(26)
n!
2πi
Z
C
([x]q t)k ([1−x] q1 t) dt
e
=
k!
tn+1
n
[x]kq [1 − x]n−k
.
1
q
k
From (24) and (26), we note that
n
.
[x]kq [1 − x]n−k
Bk,n (x, q) =
1
q
k
7
By the definition of q-Bernstein polynomials, we easily get
n−k
k+1
Bk,n (x, q) +
Bk+1,n (x, q)
n
n
(n − 1)!
(n − 1)!
n−k
k
[x]q [1 − x] 1 +
[x]k+1
[1 − x]n−k−1
=
1
q
q
q
k!(n − k − 1)!
k!(n − k − 1)!
= [1 − x] q1 + [x]q Bk,n−1 (x, q) = Bk,n−1 (x, q).
Therefore, we can write q-Bernstein polynomials as a linear combination of polynomials of higher order.
Theorem 6. For k, n ∈ Z+ and x ∈ [0, 1], we have
n+1−k
n+1
Bk,n+1 (x, q) +
k+1
n+1
Bk+1,n+1 (x, q) = Bk,n (x, q).
We easily get from (14) that for n, k ∈ N,
!
n−k+1
[x]q
Bk−1,n (x, q)
k
[1 − x] 1q
!
n−k+1
n
[x]q
=
[x]k−1
[1 − x]n−k+1
1
q
q
k−1
k
[1 − x] q1
n!
=
[x]kq [1 − x]n−k
= Bk,n (x, q).
1
q
k!(n − k)!
Therefore, we obtain the following corollary.
Corollary 7. For k, n ∈ N and x ∈ [0, 1], we have
n−k+1
k
[x]q
[1 − x] 1q
!
Bk−1,n (x, q) = Bk,n (x, q).
By (14) and binomial theorem, we easily see that
n n−k
X
X n − k n
l
n
l
l
k
(−1)l−k [x]lq .
(−1) [x]q =
[x]q
Bk,n (x, q) =
k
k
l
k
l=0
Therefore, we obtain the following theorem.
8
l=k
Theorem 8. For k, n ∈ Z+ and x ∈ [0, 1], we have
Bk,n (x, q) =
n X
n
l
k
l=k
k
(−1)l−k [x]lq .
It is possible to write [x]kq as a linear combination of the q-Bernstein polynomials
by using the degree evaluation formulae and mathematical induction. We easily see
from the property of the q-Bernstein polynomials that
n X
k
k=1
n
Bk,n (x, q) =
n−1
X
n−1
[x]k+1
[1 − x]n−k−1
= [x]q ,
1
q
q
k
k=0
and that
n
X
k=2
k
2
n Bk,n (x, q)
2
=
n−2
X
k=0
n−2
[x]k+2
[1 − x]n−2−k
= [x]2q .
1
q
q
k
Continuing this process, we get
n
X
k=j
k
j
n Bk,n (x, q)
j
= [x]jq , for j ∈ Z+ .
Therefore, we obtain the following theorem.
Theorem 9. For n, j ∈ Z+ and x ∈ [0, 1], we have
n
X
k=j
k
j
n Bk,n (x, q)
j
= [x]jq .
In [7], the q-stirling numbers of the second kind are defined by
k
k
j
q −(2) X
j ( 2) k
sq (n, k) =
[k − j]nq ,
(−1) q
j q
[k]q ! j=0
(27)
where
(28)
k
j q
=
[k]q !
[j]q ![k−j]q !
[x]nq
and [k]q ! =
=
n
X
k=0
Qk
i=1 [i]q .
For n ∈ Z+ , it is known that
k
x
(
)
2
[k]q !sq (n, k), (see [7, 21]).
q
k q
By (27), (28) and Theorem 7, we obtain the following corollary.
9
Corollary 10. For n, j ∈ Z+ and x ∈ [0, 1], we have
n
X
k=j
k
j
n Bk,n (x, q)
j
=
j
X
k=0
k
x
(
)
[k]q !sq (j, k).
q 2
k q
§3. On fermionic p-adic integral representations of q-Bernstein polynomials
In this section we assume that q ∈ Cp with |1 − q|p < 1. From (10) we note that
(29)
Z
Z
n n
n
[x + x1 ]nq dµ−1 (x1 ), (see [21]).
[1 − x + x1 ] 1 dµ−1 (x1 ) = (−1) q
En, q1 (1 − x) =
q
Zp
Zp
From (29) we have
Z
Z
n
n
n
[1 − x] 1 dµ−1 (x) = q (−1)
[x − 1]nq dµ−1 (x)
q
Zp
=
Z
Zp
n
Zp
(1 − [x]q ) dµ−1 (x) = (−1)n q n En, 1q (−1) = En,q (2).
By (5) and (10), we easily get
En,q (2) = 2 + En,q , if n > 0.
Thus, we obtain the following theorem.
Theorem 11. For n ∈ N, we have
Z
Z
Z
n
n
(1 − [x]q ) dµ−1 (x) = 2 +
[x]nq dµ−1 (x).
[1 − x] 1 dµ−1 (x) =
Zp
q
Zp
Zp
By using Theorem 11, we derive our main results in this section. Taking the
fermionic p-adic integral on Zp for one q-Bernstein polynomials in (14), we get
Z
n
[x]kq [1 − x]n−k
Bk,n (x, q)dµ−1 (x) =
dµ−1 (x)
1
q
k
Zp
Zp
n−k
Z
n X n−k
l
[x]qk+l dµ−1 (x)
(−1)
=
l
k
Zp
l=0
n−k
n X n−k
(−1)l Ek+l,q .
=
l
k
l=0
10
Z
(30)
From (14) and Theorem 2, we note that
Z
Z
1
Bk,n (x, q)dµ−1 (x) =
Bn−k,n (1 − x, )dµ−1 (x)
q
Zp
Zp
X
(31)
Z
k
k
n
k+j
[1 − x]n−j
(−1)
dµ−1 (x).
=
1
q
k j=0 j
Zp
For n > k, by (31) and Theorem 11, we get
X
k k
n
(−1)k+j
Bk,n (x, q)dµ−1 (x) =
j
k
Zp
j=0
Z
(32)
2+
Z
[x]n−j
dµ−1 (x)
q
Zp
!
= 2 + En,q , if k = 0
X
k k
n
(−1)k+j En−j,q , if k > 0.
=
j
k
j=0
From m, n, k ∈ Z+ with m + n > 2k, the fermionic p-adic integral for multiplication
of two q-Bernstein polynomials on Zp can be given by the following relation:
Z
Z
n m
n+m−2k
[x]2k
Bk,n (x, q)Bk,m(x, q)dµ−1 (x) =
dµ−1 (x)
q [1 − x] 1
q
k
k
Zp
Zp
X
Z
2k 2k
n m
j+2k
dµ−1 (x)
[1 − x]n+m−j
(−1)
1
(33) = k
q
j
k
Z
p
j=0
!
X
Z
2k
n m
[x]qn+m−j dµ−1 (x) .
(−1)j+2k 2 +
=
k
k
Zp
j=0
From (33), we have
Z
Bk,n (x, q)Bk,m(x, q)dµ−1 (x) = 2 + En+m,q , if k = 0
Zp
X
2k 2k
n m
(−1)j+2k En+m−j,q , if k > 0.
=
k j=0 j
k
For m, k ∈ Z+ , it is difficult to show that
(34)
n+m−2k
Z
X n + m − 2k n m
(−1)j Ej+2k,q .
Bk,n (x, q)Bk,m (x, q)dµ−1 (x) =
j
k
k
Zp
j=0
Continuing this process we obtain the following theorem.
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Theorem 12. (I). For n1 , · · · , ns , k ∈ Z+ (s ∈ N) with n1 + · · · + ns > sk, we have
!
Z
s
Y
Bk,ni (x, q) dµ−1 (x) = 2 + En1 +···+ns ,q , if k = 0,
Zp
and
Z
s
Y
Zp
i=1
!
Bk,ni (x, q) dµ−1 (x) =
i=1
sk s X
Y
sk
ni
k
i=1
j=0
j
(−1)sk−j En1 +···+ns −j,q , if k > 0.
(II). Let k, n1 , · · · ns ∈ Z+ (s ∈ N). Then we have
!
Z
s
Y
Bk,ni (x, q) dµ−1 (x)
Zp
=
i=1
s Y
ni
k
i=1
! Ps
ni −sk Ps
X
i=1
i=1
j=0
ni − sk
(−1)j Ej+sk,q .
j
By Theorem 12, we obtain the following corollary.
Corollary 13. For n1 , · · · , ns , k ∈ Z+ (s ∈ N) with n1 + · · · + ns > sk, we have
Ps
ni −sk Ps
X
i=1
i=1
j=0
and
Ps
ni − sk
(−1)j Ej+sk,q = 2 + En1 +···+ns ,q , if k = 0,
j
ni −sk Ps
X
i=1
i=1
j=0
=
sk X
sk
j=0
j
ni − sk
(−1)j Ej+sk,q
j
(−1)sk−j En1 +···+ns −j,q , if k > 0.
Let m1 , · · · , ms , n1 , · · · , ns , k ∈ Z+ (s ∈ N) with m1 n1 + · · · + ms ns > (m1 + · · · +
ms
ms )k. By the definition of Bk,n
(x, q), we can also easily see that
s
!
Z
s
Y
mi
Bk,ni (x, q) dµ−1 (x)
Zp
=
i=1
i=1
i=1
=
Ps
Ps
s mi k X
i=1 mi
Y
k
ni
k
j=0
Ps
j
Ps
s mi k X
i=1 mi
Y
k
ni
i=1
i=1
k
j=0
mi
j
Z
Ps
P
n m −j
k si=1 mi −j
[1 − x] 1 i=1 i i dµ−1 (x)
(−1)
Zp
mi
q
Ps
(−1)k i=1 mi −j 2 + EPsi=1 mi ni −j,q .
12
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Taekyun Kim
Division of General Education-Mathematics, Kwangwoon University, Seoul
139-701, S. Korea e-mail: [email protected]
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