About Bernstein polynomials

Annals of the University of Craiova, Math. Comp. Sci. Ser. Volume 35, 2008, Pages 117–121 ISSN: 1223-6934 About Bernstein polynomials Mircea D. Farcaş Abstract. In this article we want to determinate a recursive formula for Bernstein polynomials associated to the functions ep (x) = xp , p ∈ N, and an expresion for the central moments of the Bernstein polinomyals. 2000 Mathematics Subject Classification. 41A10; 41A63. Key words and phrases. Bernstein polynomial, Stirling numbers of first and second kind, central moments. 1. Introduction In this section we recall some notions and results which we will use in this paper. In the following, we note by N the set of positive integer and N0 = N ∪ {0}. Let Bm : C[0, 1] → C[0, 1], m ∈ N be the Bernstein operators defined for any function f ∈ C[0, 1] by µ ¶ m X k , (1) (Bm f )(x) = pm,k (x)f m k=0 where pm,k (x) are the fundamental polynomials, defined by µ ¶ m k pm,k (x) = x (1 − x)m−k , k ∈ {0, 1, . . . , m} k and x ∈ [0, 1]. For the bidimensional case, we have X (Bm f )(x, y) = pm,k,j (x, y)f k+j≥0,k+j≤m µ k j , m m ¶ (2) , (3) (x, y) ∈ ∆2 = {(x, y) ∈ R × R/x, y ≥ 0, x + y ≤ 1}, where pm,k,j (x, y) = m! xk y j (1 − x − y)m−k−j . k!j!(m − k − j)! (4) The operators Bm , m ≥ 1, are named the Bernstein bivariate polynomials, see [2]. For x ∈ R, k ∈ N0 , let x[k] = x(x − 1) . . . (x − k + 1), x[0] = 1. It is well known that (see [4]) k X xk = S(k, ν)x[ν] , x ∈ R, k ∈ N∗ , (5) ν=1 and x[k] = k X s(k, ν)xν , ν=1 117 x ∈ R, k ∈ N∗ , (6) 118 MIRCEA D. FARCAŞ where S(k, ν), ν ∈ {1, 2, . . . , k} are the Stirling numbers of second kind, and s(k, ν), ν ∈ {1, 2, . . . , k} are the Stirling numbers of first kind. These numbers verify the relations S(p, k) = kS(p − 1, k) + S(p − 1, k − 1), S(1, 1) = 1, (7) S(2, 1) = S(2, 2) = 1, S(p, 1) = S(p, p) = 1, for p ∈ N, p ≥ 3, k ∈ {2, 3, . . . , p − 1}, and s(p, k) = s(p − 1, k − 1) − (p − 1)s(p − 1, k), s(1, 1) = 1, (8) s(2, 1) = −1, s(2, 2) = 1, for p ≥ 3, k ∈ {2, 3, . . . , p − 1}. We note S(p, k) = 0 and s(p, k) = 0, from definition, if p, k ∈ N, p < k, or if k = 0. It is not difficult to prove that Proposition 1.1. If m, p ∈ N, p ≥ 3, then p(p − 1) p(p − 1) , s(p, p − 1) = − , 2 2 (p − 2)(p − 1)p(3p − 5) S(p, p − 2) = , 24 (p − 2)(p − 1)p(3p − 1) . s(p, p − 2) = 24 In the paper [3] we proved that (9) S(p, p − 1) = Proposition 1.2. If m, p ∈ N, then (Bm ep )(x) = p 1 X [k] m S(p, k)xk , mp (10) k=1 where ep (x) = xp , x ∈ [0, 1] For the bidimensional case, we have that (see [1]) Proposition 1.3. If m, p, q ∈ N, then (Bm epq )(x, y) = 1 mp+q where epq (x, y) = xp y q , x, y ∈ ∆2 . q p X X m[i+j] S(p, i)S(q, j)xi y j , (11) x(1 − x) (Bm ep )′ (x) m (12) i=1 j=1 2. Main results Theorem 2.1. We have the formula (Bm ep+1 )(x) = x(Bm ep )(x) + for m, p ∈ N, x ∈ [0, 1]. Proof. From the relation (10), we can write: µX p p+1 p X 1 X [k] 1 Bm ep+1 )(x) = p+1 km[k] S(p, k)xk + m S(p+1, k)xk = p+1 S(p, k)· m m k=1 k=1 k=1 µX ¶ p p p X X 1 km[k] S(p, k)xk +mx m[k] S(p, k)xk −x km[k] · ·m[k] (m−k)xk+1 = k+1 m k=1 k=1 k=1 ABOUT BERNSTEIN POLYNOMIALS k · S(p, k)x ¶ = 119 p p mx X [k] x(1 − x) X [k] k−1 km S(p, k)x + m S(p, k)xk = mp+1 mp+1 k=1 k=1 x(1 − x) = (Bm ep )′ (x) + x(Bm ep )(x). m For the bidimensional case, we have ¤ Theorem 2.2. If m, p, q ∈ N and (x, y) ∈ ∆2 , then x(1 − x) ∂ (Bm epq )(x, y)+ m ∂x xy ∂ (Bm epq )(x, y), +x(Bm epq )(x, y) − m ∂y y(1 − y) ∂ (Bm epq+1 )(x, y) = (Bm epq )(x, y)+ m ∂y xy ∂ (Bm epq ) (x, y). +y(Bm epq )(x, y) − m ∂x (Bm ep+1q )(x, y) = (13) (14) Proof. We have (Bm ep+1q )(x, y) = · q p+1 X X m [i+j] 1 mp+q+1 q p+1 X X m[i+j] S(p + 1, i)S(q, j)xi y j = i=1 j=1 i j iS(p, i)S(q, j)x y + i=1 j=1 1 mp+q+1 q p+1 X X 1 · mp+q+1 m[i+j] S(p, i − 1)S(q, j)xi y j = i=1 j=1 q q p p+1 X X 1 1 X X [i+j] x m iS(p, i)S(q, j)xi−1 y j + p+q+1 m[i−1+j] (m − (i − = · p+1 m m m i=1 j=1 i=1 j=1 − 1) − j)S(p, i − 1)S(q, j)xi y j = − i − j)S(p, i)S(q, j)xi+1 y j = ·xi y j − q p X X x ∂ 1 (Bm epq )(x, y) + p+q+1 m[i+j] (m − m ∂x m i=1 j=1 q p x ∂ x X X [i+j] m S(p, i)S(q, j) · (Bm epq )(x, y) + p+q m ∂x m i=1 j=1 µ X q q p X p X X [i+j] i−1 j m[i+j] S(p, i)S(q, j)xi · m S(p, i)S(q, j)ix y +y x p+q+1 x m i=1 j=1 i=1 j=1 ¶ 2 x x xy ∂ ∂ · jy j−1 = l!ef t(Bm epq (x, y) + xBm epq (x, y) − (Bm epq )(x, y) − · m ∂x m ∂x m ∂ x(1 − x) ∂ xy ∂ · (Bm epq )(x, y) = (Bm epq )(x, y)+x(Bm epq )(x, y)− (Bm epq )(x, y). ∂y m ∂x m ∂y We can prove the equality (2.3) analogously. ¤ In the sequel, we will find an expression for the central moments of the Bernstein polynomials. Lemma 2.1. We have (Bm ep )(x) = ap,0 (x) + ap,1 (x)m + . . . + ap,p−1 (x)mp−1 , mp−1 where ap,k (x) = p X j=k+1 for k ∈ {0, 1, . . . , p − 1}. S(p, j)s(j, k + 1)xj , (15) (16) 120 MIRCEA D. FARCAŞ Proof. We have 1 mp−1 p X 1 Ã j p X X m[j] S(p, j)xj = p+1 mp−1 j=1 m j=1   p−1 p X X  S(p, j)s(j, k + 1)xj  mk . (Bm ep )(x) = = 1 k=0 k=1 s(j, k)mk ! S(p, j)xj = j=k+1 ¤ Theorem 2.3. The central moments of the Bernstein polynomials admit the representation bp,0 (x) + bp,1 (x)m + . . . + bp,p−1 (x)mp−1 (Bm (∗ − x)p )(x) = (17) mp−1 where µ ¶ k X j p bp,k (x) = (−1) xj ap−j,k−j (x), (18) j j=0 k ∈ {0, 1, . . . , p − 2} and µ ¶ p−1 X j p xj ap−j,p−j−1 (x) + (−1)p xp . bp,p−1 (x) = (−1) j j=0 µ ¶ µ ¶ p−1 X p j j p xj · x (Bm ep−j )(x) = (−1) Proof. We have (Bm (∗ − x) )(x) = (−1) j j j=0 j=0 µ ¶ µ ¶ p−j−1 p−j−1 p−1 X ap−j,ν (x)mν X X 1 p p p j j · x = p−1 ap−j,ν (x)mν + x m + (−1)p (−1)j p−j−1 p j m m ν=0 ν=0 j=0 p p X j + (−1)p xp , from where the conclusion results. ¤ Lemma 2.2. We have bp,p−1 = 0, p ≥ 1, bp,p−2 = 0, p ≥ 3 bp,p−3 = 0, p ≥ 5. Proof. From the relation (18), we have that µ ¶ µ ¶ p−1 p−2 X X j p p j p j p (−1) (−1) bp,p−1 (x) = x ap−j,p−j−1 (x) + (−1) x , bp,p−2 (x) = xj · j j j=0 j=0 µ ¶ p−3 X p j x ap−j,p−j−3 (x) and after (16) and (9) we · ap−j,p−j−2 (x), bp,p−3 (x) = (−1)j j j=0 can write ap−j,p−j−1 (x) = S(p − j, p − j)s(p − j, p − j)xp−j = xp−j , (p − j)(p − j − 1) p−j−1 (p − j)(p − j − 1) p−j ap−j,p−j−2 (x) = x − x = 2 2 p−j X (p − j)(p − j − 1) p−j (x − xp−j−1 ), ap−j,p−j−3 (x) = S(p − j, ν)s(ν, p − =− 2 ν=p−j−2 −j−2)xν = S(p−j, p−j−2)s(p−j−2, p−j−2)xp−j−2 +S(p−j, p−j−1)s(p−j−1, p−j− (p − j − 2)(p − j − 1)(p − j) − 2)xp−j−1 + S(p − j, p − j)s(p − j, p − j − 2)xp−j = · 24 (p − j − 1)(p − j − 2) (p − j − 1)(p − j) xp−j−1 + · (3p − 3j − 5)xp−j−2 − 2 2 (p − j − 2)(p − j − 1)(p − j)(3p − 3j − 1) p−j (p − j − 2)(p − j − 1)(p − j) x = · + 24 24 ABOUT BERNSTEIN POLYNOMIALS 121 ·((3p−3j −5)xp−j−2 −6(p−j −1)xp−j−1 +(3p−3j −1)xp−j ), so that, using the relation µ ¶ µ ¶ p n X X p p n x = 0, = 0, m, n ∈ N, 0 ≤ m < n, we have bp,p−1 (x) = (−1)j (−1)k k m j k j=0 k=1 µ ¶ p−2 X xp − xp−1 −(p − j)(p − j − 1) p−j j p (x − xp−j−1 ) = − · xj bp,p−2 (x) = (−1) 2 2 j j=0 µ ¶ µ ¶ p−2 p−3 X X p j j p (p − j − 2)(p − j − 1) · · (−1) (p−j)(p−j−1) = 0 and bp,p−3 (x) = (−1) 24 j j j=0 j=0 · (p − j)((3p − 3j − 1)xp − 6(p − j − 1)xp−1 + (3p − 3j − 5)xp−2 ) = 0. ¤ Theorem 2.4. We have lim (Bm (∗ − x)p )(x) = 0, p ≥ 1 m→∞ lim m(Bm (∗ − x)p )(x) = 0, p ≥ 3 m→∞ and lim m2 (Bm (∗ − x)p )(x) = 0, p ≥ 4 m→∞ Proof. The proof follows immediately from the Lemma 2.2. ¤ References [1] M. D. Farcaş, About coefficients of Bernstein multivariate polynomial, (to appear in Creative Math.) [2] G. G. Lorentz, Bernstein polynomials, University of Toronto Press, Toronto, 1953 [3] O. T. Pop, M. D. Farcaş, About Bernstein polynomial and the Stirling numbers of second type, Creative Math., 14 (2005), 53-56 [4] I. Tomescu, Probleme de combinatorică şi teoria grafurilor, E.D.P. Bucureşti, 1981 (Romanian) (Mircea D. Farcaş) National College ”Mihai Eminescu” 5 Mihai Eminescu Street Satu Mare 440014, Romania E-mail address: [email protected]