Return to Article Details Approximation operators constructed by means of Sheffer sequences

REVUE D'ANALYSE NUMÉRIQUE ET DE THÉORIE DE L'APPROXIMATION

Rev. Anal. Numér. Théor. Approx., vol. 30 (2001) no. 2, pp. 135-150
ictp.acad.ro/jnaat

APPROXIMATION OPERATORS
CONSTRUCTED BY MEANS OF SHEFFER SEQUENCES

MARIA CRǍCIUN

Abstract

In this paper we introduce a class of positive linear operators by using the "umbral calculus", and we study some approximation properties of it. Let Q Q QQQ be a delta operator, and S S SSS an invertible shift invariant operator. For f C [ 0 , 1 ] f C [ 0 , 1 ] f in C[0,1]f \in C[0,1]fC[0,1] we define ( L n Q , S f ) ( x ) = 1 s n ( 1 ) k = 0 n ( n k ) p k ( x ) s n k ( 1 x ) f ( k n ) , L n Q , S f ( x ) = 1 s n ( 1 ) k = 0 n ( n k ) p k ( x ) s n k ( 1 x ) f k n , (L_(n)^(Q,S)f)(x)=(1)/(s_(n)(1))sum_(k=0)^(n)((n)/(k))p_(k)(x)s_(n-k)(1-x)f((k)/(n)),\left(L_{n}^{Q, S} f\right)(x)=\frac{1}{s_{n}(1)} \sum_{k=0}^{n}\binom{n}{k} p_{k}(x) s_{n-k}(1-x) f\left(\frac{k}{n}\right),(LnQ,Sf)(x)=1sn(1)k=0n(nk)pk(x)snk(1x)f(kn), where ( p n ) n 0 p n n 0 (p_(n))_(n >= 0)\left(p_{n}\right)_{n \geq 0}(pn)n0 is a binomial sequence which is the basic sequence for Q Q QQQ, and ( s n ) n 0 s n n 0 (s_(n))_(n >= 0)\left(s_{n}\right)_{n \geq 0}(sn)n0 is a Sheffer set, s n = S 1 p n s n = S 1 p n s_(n)=S^(-1)p_(n)s_{n}=S^{-1} p_{n}sn=S1pn. These operators generalize the binomial operators of T. Popoviciu.

MSC 2000. 41A36, 05A40.

1. INTRODUCTION

Let P P PPP be the linear space of all polynomials with real coefficients, and P n P n P_(n)P_{n}Pn the linear space of all polynomials of degree at most n n nnn.
We will consider some linear operators defined on P P PPP. We will denote by I I III the identity and by D D DDD the derivative. The shift operator E a : P P E a : P P E^(a):P rarr PE^{a}: P \rightarrow PEa:PP is defined by E a p ( x ) = p ( x + a ) E a p ( x ) = p ( x + a ) E^(a)p(x)=p(x+a)E^{a} p(x)=p(x+a)Eap(x)=p(x+a).
A linear operator T T TTT which commutes with all shift operators is called a shift invariant operator. In symbols, E a T = T E a E a T = T E a E^(a)T=TE^(a)E^{a} T=T E^{a}EaT=TEa, for all real a a aaa.
Let us remind that if T 1 T 1 T_(1)T_{1}T1 and T 2 T 2 T_(2)T_{2}T2 are shift invariant operators, then T 1 T 2 = T 2 T 1 T 1 T 2 = T 2 T 1 T_(1)T_(2)=T_(2)T_(1)T_{1} T_{2}= T_{2} T_{1}T1T2=T2T1.
Definition 1. A shift invariant operator for which Q x = Q x = Qx=Q x=Qx= const 0 0 !=0\neq 00 is called a delta operator.
By a polynomial sequence we shall denote a sequence of polynomials p n ( x ) p n ( x ) p_(n)(x)p_{n}(x)pn(x), n = 0 , 1 , 2 , n = 0 , 1 , 2 , n=0,1,2,dotsn=0,1,2, \ldotsn=0,1,2, where p n ( x ) p n ( x ) p_(n)(x)p_{n}(x)pn(x) is of degree exactly n n nnn for all n n nnn.
A sequence of binomial type is a polynomial sequence ( p n ) n 0 p n n 0 (p_(n))_(n >= 0)\left(p_{n}\right)_{n \geq 0}(pn)n0 with p 0 ( x ) = 1 p 0 ( x ) = 1 p_(0)(x)=1p_{0}(x)=1p0(x)=1 and satisfying the identities
p n ( x + y ) = k = 0 n ( n k ) p k ( x ) p n k ( y ) , p n ( x + y ) = k = 0 n ( n k ) p k ( x ) p n k ( y ) , p_(n)(x+y)=sum_(k=0)^(n)((n)/(k))p_(k)(x)p_(n-k)(y),p_{n}(x+y)=\sum_{k=0}^{n}\binom{n}{k} p_{k}(x) p_{n-k}(y),pn(x+y)=k=0n(nk)pk(x)pnk(y),
for all x , y x , y x,yx, yx,y and n = 0 , 1 , 2 , n = 0 , 1 , 2 , n=0,1,2,dotsn=0,1,2, \ldotsn=0,1,2,.
Definition 2. Let Q Q QQQ be a delta operator and ( p n ( x ) ) n 0 p n ( x ) n 0 (p_(n)(x))_(n >= 0)\left(p_{n}(x)\right)_{n \geq 0}(pn(x))n0 a polynomial sequence. If
i) p 0 ( x ) = 1 p 0 ( x ) = 1 quadp_(0)(x)=1\quad p_{0}(x)=1p0(x)=1,
ii) p n ( 0 ) = 0 , n = 1 , 2 , p n ( 0 ) = 0 , n = 1 , 2 , quadp_(n)(0)=0,n=1,2,dots\quad p_{n}(0)=0, n=1,2, \ldotspn(0)=0,n=1,2,,
iii) Q p n = n p n 1 , n = 1 , 2 , Q p n = n p n 1 , n = 1 , 2 , quad Qp_(n)=np_(n-1),n=1,2,dots\quad Q p_{n}=n p_{n-1}, n=1,2, \ldotsQpn=npn1,n=1,2,,
then ( p n ) p n (p_(n))\left(p_{n}\right)(pn) is called the sequence of basic polynomials for Q Q QQQ.
Proposition 1. 8].
i) Every delta operator has a unique sequence of basic polynomials.
ii) If p n ( x ) p n ( x ) p_(n)(x)p_{n}(x)pn(x) is a basic sequence for some delta operator Q Q QQQ, then it is binomial.
iii) If p n ( x ) p n ( x ) p_(n)(x)p_{n}(x)pn(x) is a binomial sequence, then it is a basic sequence for some delta operator Q Q QQQ.
Let X X XXX be the multiplication operator defined as ( X p ) ( x ) = x p ( x ) ( X p ) ( x ) = x p ( x ) (Xp)(x)=xp(x)(X p)(x)=x p(x)(Xp)(x)=xp(x) for every polynomial p p ppp.
For any operator T T TTT defined on P P PPP, the operator T = T X X T T = T X X T T^(')=TX-XTT^{\prime}=T X-X TT=TXXT is called the Pincherle derivative of the operator T T TTT.
Proposition 2. 8].
i) If T T TTT is a shift invariant operator, then its Pincherle derivative is also a shift invariant operator.
ii) If Q Q QQQ is a delta operator, then its Pincherle derivative Q Q Q^(')Q^{\prime}Q is an invertible operator.
Proposition 3. [8], [11]. If ( p n ( x ) ) n 0 p n ( x ) n 0 (p_(n)(x))_(n >= 0)\left(p_{n}(x)\right)_{n \geq 0}(pn(x))n0 is a sequence of basic polynomials for the delta operator Q Q QQQ then
i) p n ( x ) = X ( Q ) 1 p n 1 ( x ) , n = 1 , 2 , p n ( x ) = X Q 1 p n 1 ( x ) , n = 1 , 2 , quadp_(n)(x)=X(Q^('))^(-1)p_(n-1)(x),n=1,2,dots\quad p_{n}(x)=X\left(Q^{\prime}\right)^{-1} p_{n-1}(x), n=1,2, \ldotspn(x)=X(Q)1pn1(x),n=1,2,,
ii) p n ( x ) = x k = 0 n 1 ( n 1 k ) p n 1 k ( x ) p k + 1 ( 0 ) , n = 1 , 2 , p n ( x ) = x k = 0 n 1 ( n 1 k ) p n 1 k ( x ) p k + 1 ( 0 ) , n = 1 , 2 , quadp_(n)(x)=xsum_(k=0)^(n-1)((n-1)/(k))p_(n-1-k)(x)p_(k+1)^(')(0),n=1,2,dots\quad p_{n}(x)=x \sum_{k=0}^{n-1}\binom{n-1}{k} p_{n-1-k}(x) p_{k+1}^{\prime}(0), n=1,2, \ldotspn(x)=xk=0n1(n1k)pn1k(x)pk+1(0),n=1,2,
Definition 3. A polynomial sequence ( s n ( x ) ) n 0 s n ( x ) n 0 (s_(n)(x))_(n >= 0)\left(s_{n}(x)\right)_{n \geq 0}(sn(x))n0 is called a Sheffer set relative to the delta operator Q Q QQQ if:
i) s 0 ( x ) = s 0 ( x ) = quads_(0)(x)=\quad s_{0}(x)=s0(x)= const 0 0 !=0\neq 00
ii) Q s n = n s n 1 , n = 1 , 2 , Q s n = n s n 1 , n = 1 , 2 , Qs_(n)=ns_(n-1),n=1,2,dotsQ s_{n}=n s_{n-1}, n=1,2, \ldotsQsn=nsn1,n=1,2,
An Appel set is a Sheffer set relative to the derivative D D DDD.
Proposition 4. 11. Let Q Q QQQ be a delta operator with basic polynomial set ( p n ( x ) ) n 0 p n ( x ) n 0 (p_(n)(x))_(n >= 0)\left(p_{n}(x)\right)_{n \geq 0}(pn(x))n0 and ( s n ( x ) ) n 0 s n ( x ) n 0 (s_(n)(x))_(n >= 0)\left(s_{n}(x)\right)_{n \geq 0}(sn(x))n0 a polynomial sequence. The next statements are equivalent:
i) s n ( x ) s n ( x ) s_(n)(x)s_{n}(x)sn(x) is a Sheffer set relative to Q Q QQQ.
ii) There exists an invertible shift invariant operator S S SSS such that s n ( x ) = S 1 p n ( x ) s n ( x ) = S 1 p n ( x ) s_(n)(x)=S^(-1)p_(n)(x)s_{n}(x)= S^{-1} p_{n}(x)sn(x)=S1pn(x).
iii) For all x , y R x , y R x,y inRx, y \in \mathbb{R}x,yR and n = 0 , 1 , 2 , n = 0 , 1 , 2 , n=0,1,2,dotsn=0,1,2, \ldotsn=0,1,2,, the following identity holds:
s n ( x + y ) = k = 0 n ( n k ) p k ( x ) s n k ( y ) s n ( x + y ) = k = 0 n ( n k ) p k ( x ) s n k ( y ) s_(n)(x+y)=sum_(k=0)^(n)((n)/(k))p_(k)(x)s_(n-k)(y)s_{n}(x+y)=\sum_{k=0}^{n}\binom{n}{k} p_{k}(x) s_{n-k}(y)sn(x+y)=k=0n(nk)pk(x)snk(y)
From the previous Proposition it results that the pair ( Q , S Q , S Q,SQ, SQ,S ) gives us a unique Sheffer set.

2. THE OPERATORS CONSTRUCTED BY MEANS OF SHEFFER POLYNOMIALS AND THEIR CONVERGENCE

In 1931 in [9] Tiberiu Popoviciu has used binomial sequences in order to construct some operators of the form
(1) ( L n f ) ( x ) = 1 p n ( 1 ) k = 0 n ( n k ) p k ( x ) p n k ( 1 x ) f ( k n ) (1) L n f ( x ) = 1 p n ( 1 ) k = 0 n ( n k ) p k ( x ) p n k ( 1 x ) f k n {:(1)(L_(n)f)(x)=(1)/(p_(n)(1))sum_(k=0)^(n)((n)/(k))p_(k)(x)p_(n-k)(1-x)f((k)/(n)):}\begin{equation*} \left(L_{n} f\right)(x)=\frac{1}{p_{n}(1)} \sum_{k=0}^{n}\binom{n}{k} p_{k}(x) p_{n-k}(1-x) f\left(\frac{k}{n}\right) \tag{1} \end{equation*}(1)(Lnf)(x)=1pn(1)k=0n(nk)pk(x)pnk(1x)f(kn)
where f C [ 0 , 1 ] f C [ 0 , 1 ] f in C[0,1]f \in C[0,1]fC[0,1] and x [ 0 , 1 ] x [ 0 , 1 ] x in[0,1]x \in[0,1]x[0,1]. These operators are called binomial operators.
Such operators and their generalizations have been studied by the Romanian mathematicians as: D. D. Stancu, A. Lupaş, L. Lupaş, G. Moldovan, C. Manole, O. Agratini, A. Vernescu, and others.
Let Q Q QQQ be a delta operator and S S SSS an invertible shift invariant operator. Let ( p n ( x ) ) n 0 p n ( x ) n 0 (p_(n)(x))_(n >= 0)\left(p_{n}(x)\right)_{n \geq 0}(pn(x))n0 be the sequence of basic polynomials for Q Q QQQ, and ( s n ( x ) ) n 0 s n ( x ) n 0 (s_(n)(x))_(n >= 0)\left(s_{n}(x)\right)_{n \geq 0}(sn(x))n0 a Sheffer set relative to Q , s n = S 1 p n Q , s n = S 1 p n Q,s_(n)=S^(-1)p_(n)Q, s_{n}=S^{-1} p_{n}Q,sn=S1pn with s n ( 1 ) 0 s n ( 1 ) 0 s_(n)(1)!=0s_{n}(1) \neq 0sn(1)0 for any positive integer n n nnn.
In this note we want to study the operators L n Q , S : C [ 0 , 1 ] C [ 0 , 1 ] L n Q , S : C [ 0 , 1 ] C [ 0 , 1 ] L_(n)^(Q,S):C[0,1]rarr C[0,1]L_{n}^{Q, S}: C[0,1] \rightarrow C[0,1]LnQ,S:C[0,1]C[0,1],
(2) ( L n Q , S f ) ( x ) = 1 s n ( 1 ) k = 0 n ( n k ) p k ( x ) s n k ( 1 x ) f ( k n ) (2) L n Q , S f ( x ) = 1 s n ( 1 ) k = 0 n ( n k ) p k ( x ) s n k ( 1 x ) f k n {:(2)(L_(n)^(Q,S)f)(x)=(1)/(s_(n)(1))sum_(k=0)^(n)((n)/(k))p_(k)(x)s_(n-k)(1-x)f((k)/(n)):}\begin{equation*} \left(L_{n}^{Q, S} f\right)(x)=\frac{1}{s_{n}(1)} \sum_{k=0}^{n}\binom{n}{k} p_{k}(x) s_{n-k}(1-x) f\left(\frac{k}{n}\right) \tag{2} \end{equation*}(2)(LnQ,Sf)(x)=1sn(1)k=0n(nk)pk(x)snk(1x)f(kn)
Because p k ( 0 ) = δ k , 0 p k ( 0 ) = δ k , 0 p_(k)(0)=delta_(k,0)p_{k}(0)=\delta_{k, 0}pk(0)=δk,0 (from the definition of basic polynomials), we have ( L n Q , S f ) ( 0 ) = f ( 0 ) L n Q , S f ( 0 ) = f ( 0 ) (L_(n)^(Q,S)f)(0)=f(0)\left(L_{n}^{Q, S} f\right)(0)=f(0)(LnQ,Sf)(0)=f(0).
In order to evaluate expression ( L n Q , S e m ) ( x ) L n Q , S e m ( x ) (L_(n)^(Q,S)e_(m))(x)\left(L_{n}^{Q, S} e_{m}\right)(x)(LnQ,Sem)(x), where e m ( x ) = x m e m ( x ) = x m e_(m)(x)=x^(m)e_{m}(x)=x^{m}em(x)=xm we shall make use of C. Manole's method for binomial operators (see [5]) which we have adapted to our purposes.
Let us introduce the polynomials
(3) S m ( x , y , n ) = k = 0 n ( n k ) p k ( x ) s n k ( y ) ( k n ) m (3) S m ( x , y , n ) = k = 0 n ( n k ) p k ( x ) s n k ( y ) k n m {:(3)S_(m)(x","y","n)=sum_(k=0)^(n)((n)/(k))p_(k)(x)s_(n-k)(y)((k)/(n))^(m):}\begin{equation*} S_{m}(x, y, n)=\sum_{k=0}^{n}\binom{n}{k} p_{k}(x) s_{n-k}(y)\left(\frac{k}{n}\right)^{m} \tag{3} \end{equation*}(3)Sm(x,y,n)=k=0n(nk)pk(x)snk(y)(kn)m
From Proposition 4 iii) we have S 0 ( x , y , n ) = s n ( x + y ) S 0 ( x , y , n ) = s n ( x + y ) S_(0)(x,y,n)=s_(n)(x+y)S_{0}(x, y, n)=s_{n}(x+y)S0(x,y,n)=sn(x+y).
In the following we consider that x x xxx is the variable. Let us denote θ = X ( Q ) 1 θ = X Q 1 theta=X(Q^('))^(-1)\theta= X\left(Q^{\prime}\right)^{-1}θ=X(Q)1.
From Proposition 3 i) it results that θ p k ( x ) = p k + 1 ( x ) θ p k ( x ) = p k + 1 ( x ) thetap_(k)(x)=p_(k+1)(x)\theta p_{k}(x)=p_{k+1}(x)θpk(x)=pk+1(x) and consequently the linear operator θ θ theta\thetaθ is called the shift operator for the sequence ( p n ) n 0 p n n 0 (p_(n))_(n >= 0)\left(p_{n}\right)_{n \geq 0}(pn)n0 (see [10]). Therefore θ Q p k ( x ) = θ ( k p k 1 ( x ) ) = k p k ( x ) θ Q p k ( x ) = θ k p k 1 ( x ) = k p k ( x ) theta Qp_(k)(x)=theta(kp_(k-1)(x))=kp_(k)(x)\theta Q p_{k}(x)=\theta\left(k p_{k-1}(x)\right)=k p_{k}(x)θQpk(x)=θ(kpk1(x))=kpk(x); consequently k k kkk is an eigenvalue for the operator θ Q θ Q theta Q\theta QθQ, with its eigenvector p k ( x ) p k ( x ) p_(k)(x)p_{k}(x)pk(x). We have
(4) ( θ Q ) m = k m p k ( x ) (4) ( θ Q ) m = k m p k ( x ) {:(4)(theta Q)^(m)=k^(m)p_(k)(x):}\begin{equation*} (\theta Q)^{m}=k^{m} p_{k}(x) \tag{4} \end{equation*}(4)(θQ)m=kmpk(x)
for every positive integer m m mmm, and then
S m ( x , y , n ) = 1 n m ( θ Q ) m k = 0 n ( n k ) p k ( x ) s n k ( y ) = 1 n m ( θ Q ) m S 0 ( x , y , n ) = 1 n m ( θ Q ) m s n ( x + y ) S m ( x , y , n ) = 1 n m ( θ Q ) m k = 0 n ( n k ) p k ( x ) s n k ( y ) = 1 n m ( θ Q ) m S 0 ( x , y , n ) = 1 n m ( θ Q ) m s n ( x + y ) {:[S_(m)(x","y","n)=(1)/(n^(m))(theta Q)^(m)sum_(k=0)^(n)((n)/(k))p_(k)(x)s_(n-k)(y)],[=(1)/(n^(m))(theta Q)^(m)S_(0)(x","y","n)=(1)/(n^(m))(theta Q)^(m)s_(n)(x+y)]:}\begin{aligned} S_{m}(x, y, n) & =\frac{1}{n^{m}}(\theta Q)^{m} \sum_{k=0}^{n}\binom{n}{k} p_{k}(x) s_{n-k}(y) \\ & =\frac{1}{n^{m}}(\theta Q)^{m} S_{0}(x, y, n)=\frac{1}{n^{m}}(\theta Q)^{m} s_{n}(x+y) \end{aligned}Sm(x,y,n)=1nm(θQ)mk=0n(nk)pk(x)snk(y)=1nm(θQ)mS0(x,y,n)=1nm(θQ)msn(x+y)
In this way we obtain
(5) S m ( x , y , n ) = 1 n m ( θ Q ) m E y s n ( x ) (5) S m ( x , y , n ) = 1 n m ( θ Q ) m E y s n ( x ) {:(5)S_(m)(x","y","n)=(1)/(n^(m))(theta Q)^(m)E^(y)s_(n)(x):}\begin{equation*} S_{m}(x, y, n)=\frac{1}{n^{m}}(\theta Q)^{m} E^{y} s_{n}(x) \tag{5} \end{equation*}(5)Sm(x,y,n)=1nm(θQ)mEysn(x)
Using the operational formula (see for instance [10])
( θ Q ) m = k = 0 n S ( m , k ) θ k Q k ( θ Q ) m = k = 0 n S ( m , k ) θ k Q k (theta Q)^(m)=sum_(k=0)^(n)S(m,k)theta^(k)Q^(k)(\theta Q)^{m}=\sum_{k=0}^{n} S(m, k) \theta^{k} Q^{k}(θQ)m=k=0nS(m,k)θkQk
where S ( m , k ) = [ 0 , 1 , , k ; e m ] S ( m , k ) = 0 , 1 , , k ; e m S(m,k)=[0,1,dots,k;e_(m)]S(m, k)=\left[0,1, \ldots, k ; e_{m}\right]S(m,k)=[0,1,,k;em] are the Stirling numbers of the second kind, relation (5) becomes:
(6) S m ( x , y , n ) = 1 n m k = 0 n S ( m , k ) θ k Q k E y s n ( x ) (6) S m ( x , y , n ) = 1 n m k = 0 n S ( m , k ) θ k Q k E y s n ( x ) {:(6)S_(m)(x","y","n)=(1)/(n^(m))sum_(k=0)^(n)S(m","k)theta^(k)Q^(k)E^(y)s_(n)(x):}\begin{equation*} S_{m}(x, y, n)=\frac{1}{n^{m}} \sum_{k=0}^{n} S(m, k) \theta^{k} Q^{k} E^{y} s_{n}(x) \tag{6} \end{equation*}(6)Sm(x,y,n)=1nmk=0nS(m,k)θkQkEysn(x)
Because Q Q QQQ is shift invariant and Q k s n ( x ) = n ( n 1 ) ( n k + 1 ) s n k ( x ) = n [ k ] s n k ( x ) Q k s n ( x ) = n ( n 1 ) ( n k + 1 ) s n k ( x ) = n [ k ] s n k ( x ) Q^(k)s_(n)(x)=n(n-1)dots(n-k+1)s_(n-k)(x)=n^([k])s_(n-k)(x)Q^{k} s_{n}(x)=n(n-1) \ldots(n-k+1) s_{n-k}(x)= n^{[k]} s_{n-k}(x)Qksn(x)=n(n1)(nk+1)snk(x)=n[k]snk(x) we obtain
(7) S m ( x , y , n ) = 1 n m k = 0 n ( n k ) k ! S ( m , k ) θ k E y s n k ( x ) , m N (7) S m ( x , y , n ) = 1 n m k = 0 n ( n k ) k ! S ( m , k ) θ k E y s n k ( x ) , m N {:(7)S_(m)(x","y","n)=(1)/(n^(m))sum_(k=0)^(n)((n)/(k))k!S(m","k)theta^(k)E^(y)s_(n-k)(x)","AA m inN^(**):}\begin{equation*} S_{m}(x, y, n)=\frac{1}{n^{m}} \sum_{k=0}^{n}\binom{n}{k} k!S(m, k) \theta^{k} E^{y} s_{n-k}(x), \forall m \in \mathbb{N}^{*} \tag{7} \end{equation*}(7)Sm(x,y,n)=1nmk=0n(nk)k!S(m,k)θkEysnk(x),mN
Theorem 1. If L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S is the linear operator defined by (2) then
( L n Q , S e 0 ) ( x ) = e 0 ( x ) (8) ( L n Q , S e 1 ) ( x ) = a n e 1 ( x ) ( L n Q , S e 2 ) ( x ) = b n x 2 + x ( a n b n c n ) , L n Q , S e 0 ( x ) = e 0 ( x ) (8) L n Q , S e 1 ( x ) = a n e 1 ( x ) L n Q , S e 2 ( x ) = b n x 2 + x a n b n c n , {:[(L_(n)^(Q,S)e_(0))(x)=e_(0)(x)],[(8)(L_(n)^(Q,S)e_(1))(x)=a_(n)e_(1)(x)],[(L_(n)^(Q,S)e_(2))(x)=b_(n)x^(2)+x(a_(n)-b_(n)-c_(n))","]:}\begin{align*} & \left(L_{n}^{Q, S} e_{0}\right)(x)=e_{0}(x) \\ & \left(L_{n}^{Q, S} e_{1}\right)(x)=a_{n} e_{1}(x) \tag{8}\\ & \left(L_{n}^{Q, S} e_{2}\right)(x)=b_{n} x^{2}+x\left(a_{n}-b_{n}-c_{n}\right), \end{align*}(LnQ,Se0)(x)=e0(x)(8)(LnQ,Se1)(x)=ane1(x)(LnQ,Se2)(x)=bnx2+x(anbncn),
where
a n = [ ( Q ) 1 s n 1 ] ( 1 ) s n ( 1 ) (9) b n = n 1 n [ ( Q ) 2 s n 2 ] ( 1 ) s n ( 1 ) c n = n 1 n [ ( Q ) 2 ( S 1 ) S s n 2 ] ( 1 ) s n ( 1 ) a n = Q 1 s n 1 ( 1 ) s n ( 1 ) (9) b n = n 1 n Q 2 s n 2 ( 1 ) s n ( 1 ) c n = n 1 n Q 2 S 1 S s n 2 ( 1 ) s n ( 1 ) {:[a_(n)=([(Q^('))^(-1)s_(n-1)](1))/(s_(n)(1))],[(9)b_(n)=(n-1)/(n)([(Q^('))^(-2)s_(n-2)](1))/(s_(n)(1))],[c_(n)=(n-1)/(n)([(Q^('))^(-2)(S^(-1))^(')Ss_(n-2)](1))/(s_(n)(1))]:}\begin{align*} & a_{n}=\frac{\left[\left(Q^{\prime}\right)^{-1} s_{n-1}\right](1)}{s_{n}(1)} \\ & b_{n}=\frac{n-1}{n} \frac{\left[\left(Q^{\prime}\right)^{-2} s_{n-2}\right](1)}{s_{n}(1)} \tag{9}\\ & c_{n}=\frac{n-1}{n} \frac{\left[\left(Q^{\prime}\right)^{-2}\left(S^{-1}\right)^{\prime} S s_{n-2}\right](1)}{s_{n}(1)} \end{align*}an=[(Q)1sn1](1)sn(1)(9)bn=n1n[(Q)2sn2](1)sn(1)cn=n1n[(Q)2(S1)Ssn2](1)sn(1)
Proof. Using the notation (3) we can write
(10) ( L n Q , S e m ) ( x ) = S m ( x , 1 x , n ) / s n ( 1 ) (10) L n Q , S e m ( x ) = S m ( x , 1 x , n ) / s n ( 1 ) {:(10)(L_(n)^(Q,S)e_(m))(x)=S_(m)(x","1-x","n)//s_(n)(1):}\begin{equation*} \left(L_{n}^{Q, S} e_{m}\right)(x)=S_{m}(x, 1-x, n) / s_{n}(1) \tag{10} \end{equation*}(10)(LnQ,Sem)(x)=Sm(x,1x,n)/sn(1)
Because S 0 ( x , 1 x , n ) = s n ( 1 ) S 0 ( x , 1 x , n ) = s n ( 1 ) S_(0)(x,1-x,n)=s_(n)(1)S_{0}(x, 1-x, n)=s_{n}(1)S0(x,1x,n)=sn(1) we have ( L n Q , S e 0 ) ( x ) = e 0 ( x ) L n Q , S e 0 ( x ) = e 0 ( x ) (L_(n)^(Q,S)e_(0))(x)=e_(0)(x)\left(L_{n}^{Q, S} e_{0}\right)(x)=e_{0}(x)(LnQ,Se0)(x)=e0(x).
As we have
(11) θ E y s n 1 ( x ) = X ( Q ) 1 E y s n 1 ( x ) = X E y ( Q ) 1 s n 1 ( x ) (11) θ E y s n 1 ( x ) = X Q 1 E y s n 1 ( x ) = X E y Q 1 s n 1 ( x ) {:(11)thetaE^(y)s_(n-1)(x)=X(Q^('))^(-1)E^(y)s_(n-1)(x)=XE^(y)(Q^('))^(-1)s_(n-1)(x):}\begin{equation*} \theta E^{y} s_{n-1}(x)=X\left(Q^{\prime}\right)^{-1} E^{y} s_{n-1}(x)=X E^{y}\left(Q^{\prime}\right)^{-1} s_{n-1}(x) \tag{11} \end{equation*}(11)θEysn1(x)=X(Q)1Eysn1(x)=XEy(Q)1sn1(x)
we obtain from (7): S 1 ( x , 1 x , n ) = x [ ( Q ) 1 s n 1 ] ( 1 ) S 1 ( x , 1 x , n ) = x Q 1 s n 1 ( 1 ) S_(1)(x,1-x,n)=x[(Q^('))^(-1)s_(n-1)](1)S_{1}(x, 1-x, n)=x\left[\left(Q^{\prime}\right)^{-1} s_{n-1}\right](1)S1(x,1x,n)=x[(Q)1sn1](1); consequently we get:
( L n Q , S e 1 ) ( x ) = [ ( Q ) 1 s n 1 ] ( 1 ) s n ( 1 ) x . L n Q , S e 1 ( x ) = Q 1 s n 1 ( 1 ) s n ( 1 ) x . (L_(n)^(Q,S)e_(1))(x)=([(Q^('))^(-1)s_(n-1)](1))/(s_(n)(1))x.\left(L_{n}^{Q, S} e_{1}\right)(x)=\frac{\left[\left(Q^{\prime}\right)^{-1} s_{n-1}\right](1)}{s_{n}(1)} x .(LnQ,Se1)(x)=[(Q)1sn1](1)sn(1)x.
Using the Pincherle derivative of the shift operator E y E y E^(y)E^{y}Ey
(12) ( E y ) = y E y = E y X X E y (12) E y = y E y = E y X X E y {:(12)(E^(y))^(')=yE^(y)=E^(y)X-XE^(y):}\begin{equation*} \left(E^{y}\right)^{\prime}=y E^{y}=E^{y} X-X E^{y} \tag{12} \end{equation*}(12)(Ey)=yEy=EyXXEy
we can write
θ E y s n k ( x ) = E y X ( Q ) 1 s n k ( x ) y E y ( Q ) 1 s n k ( x ) θ E y s n k ( x ) = E y X Q 1 s n k ( x ) y E y Q 1 s n k ( x ) thetaE^(y)s_(n-k)(x)=E^(y)X(Q^('))^(-1)s_(n-k)(x)-yE^(y)(Q^('))^(-1)s_(n-k)(x)\theta E^{y} s_{n-k}(x)=E^{y} X\left(Q^{\prime}\right)^{-1} s_{n-k}(x)-y E^{y}\left(Q^{\prime}\right)^{-1} s_{n-k}(x)θEysnk(x)=EyX(Q)1snk(x)yEy(Q)1snk(x)
Then
(13) θ 2 E y s n k ( x ) = X E y ( Q ) 1 E y X ( Q ) 1 s n k ( x ) y X E y ( Q ) 2 s n k ( x ) (13) θ 2 E y s n k ( x ) = X E y Q 1 E y X Q 1 s n k ( x ) y X E y Q 2 s n k ( x ) {:(13)theta^(2)E^(y)s_(n-k)(x)=XE^(y)(Q^('))^(-1)E^(y)X(Q^('))^(-1)s_(n-k)(x)-yXE^(y)(Q^('))^(-2)s_(n-k)(x):}\begin{equation*} \theta^{2} E^{y} s_{n-k}(x)=X E^{y}\left(Q^{\prime}\right)^{-1} E^{y} X\left(Q^{\prime}\right)^{-1} s_{n-k}(x)-y X E^{y}\left(Q^{\prime}\right)^{-2} s_{n-k}(x) \tag{13} \end{equation*}(13)θ2Eysnk(x)=XEy(Q)1EyX(Q)1snk(x)yXEy(Q)2snk(x)
Because s n k = S 1 p n k , X S 1 = S 1 X ( S 1 ) s n k = S 1 p n k , X S 1 = S 1 X S 1 s_(n-k)=S^(-1)p_(n-k),XS^(-1)=S^(-1)X-(S^(-1))^(')s_{n-k}=S^{-1} p_{n-k}, X S^{-1}=S^{-1} X-\left(S^{-1}\right)^{\prime}snk=S1pnk,XS1=S1X(S1) (from the definition of Pincherle derivative) and ( Q ) 1 p n k ( x ) = p n k + 1 ( x ) / x Q 1 p n k ( x ) = p n k + 1 ( x ) / x (Q^('))^(-1)p_(n-k)(x)=p_(n-k+1)(x)//x\left(Q^{\prime}\right)^{-1} p_{n-k}(x)=p_{n-k+1}(x) / x(Q)1pnk(x)=pnk+1(x)/x (from Proposition 3 i), we obtain
(14) θ 2 E y s n k ( x ) = X E y ( Q ) 1 s n k + 1 ( x ) X E y ( Q ) 2 ( S 1 ) S s n k ( x ) y X E y ( Q ) 2 s n k ( x ) (14) θ 2 E y s n k ( x ) = X E y Q 1 s n k + 1 ( x ) X E y Q 2 S 1 S s n k ( x ) y X E y Q 2 s n k ( x ) {:[(14)theta^(2)E^(y)s_(n-k)(x)=XE^(y)(Q^('))^(-1)s_(n-k+1)(x)-XE^(y)(Q^('))^(-2)(S^(-1))^(')Ss_(n-k)(x)-],[-yXE^(y)(Q^('))^(-2)s_(n-k)(x)]:}\begin{align*} \theta^{2} E^{y} s_{n-k}(x)= & X E^{y}\left(Q^{\prime}\right)^{-1} s_{n-k+1}(x)-X E^{y}\left(Q^{\prime}\right)^{-2}\left(S^{-1}\right)^{\prime} S s_{n-k}(x)- \tag{14}\\ & -y X E^{y}\left(Q^{\prime}\right)^{-2} s_{n-k}(x) \end{align*}(14)θ2Eysnk(x)=XEy(Q)1snk+1(x)XEy(Q)2(S1)Ssnk(x)yXEy(Q)2snk(x)
Replacing (11) and (14) in (7) we can write
S 2 ( x , y , n ) = x E y ( Q ) 1 s n 1 ( x ) n 1 n [ x E y ( Q ) 2 ( S 1 ) S s n 2 ( x ) y x E y ( Q ) 2 s n 2 ( x ) ] S 2 ( x , y , n ) = x E y Q 1 s n 1 ( x ) n 1 n x E y Q 2 S 1 S s n 2 ( x ) y x E y Q 2 s n 2 ( x ) {:[S_(2)(x","y","n)=xE^(y)(Q^('))^(-1)s_(n-1)(x)-(n-1)/(n)[xE^(y)(Q^('))^(-2)(S^(-1))^(')Ss_(n-2)(x)-:}],[{:-yxE^(y)(Q^('))^(-2)s_(n-2)(x)]]:}\begin{aligned} S_{2}(x, y, n)= & x E^{y}\left(Q^{\prime}\right)^{-1} s_{n-1}(x)-\frac{n-1}{n}\left[x E^{y}\left(Q^{\prime}\right)^{-2}\left(S^{-1}\right)^{\prime} S s_{n-2}(x)-\right. \\ & \left.-y x E^{y}\left(Q^{\prime}\right)^{-2} s_{n-2}(x)\right] \end{aligned}S2(x,y,n)=xEy(Q)1sn1(x)n1n[xEy(Q)2(S1)Ssn2(x)yxEy(Q)2sn2(x)]
From (10) and the previous relation one obtains expression L n Q , S e 2 L n Q , S e 2 L_(n)^(Q,S)e_(2)L_{n}^{Q, S} e_{2}LnQ,Se2 from theorem's conclusion.
Lemma 1. Let Q Q QQQ be a delta operator and S S SSS an invertible shift invariant operator. Let ( p n ( x ) ) n 0 p n ( x ) n 0 (p_(n)(x))_(n >= 0)\left(p_{n}(x)\right)_{n \geq 0}(pn(x))n0 be the sequence of basic polynomials for Q Q QQQ and ( s n ( x ) ) n 0 s n ( x ) n 0 (s_(n)(x))_(n >= 0)\left(s_{n}(x)\right)_{n \geq 0}(sn(x))n0 a Sheffer set relative to Q , s n = S 1 p n Q , s n = S 1 p n Q,s_(n)=S^(-1)p_(n)Q, s_{n}=S^{-1} p_{n}Q,sn=S1pn with s n ( 1 ) 0 s n ( 1 ) 0 s_(n)(1)!=0s_{n}(1) \neq 0sn(1)0 for any positive integer n n nnn. If p k ( 0 ) 0 p k ( 0 ) 0 p_(k)^(')(0) >= 0p_{k}^{\prime}(0) \geq 0pk(0)0 and s k ( 0 ) 0 s k ( 0 ) 0 s_(k)(0) >= 0s_{k}(0) \geq 0sk(0)0 for n = 0 , 1 , 2 , n = 0 , 1 , 2 , n=0,1,2,dotsn=0,1,2, \ldotsn=0,1,2, then the operator L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S defined by (2) is positive.
Proof. If p k ( 0 ) 0 p k ( 0 ) 0 p_(k)^(')(0) >= 0p_{k}^{\prime}(0) \geq 0pk(0)0 using Proposition 3 ii), it is easy to prove by induction that p k ( x ) 0 , k N p k ( x ) 0 , k N p_(k)(x) >= 0,AA k inNp_{k}(x) \geq 0, \forall k \in \mathbb{N}pk(x)0,kN and x [ 0 , 1 ] x [ 0 , 1 ] AA x in[0,1]\forall x \in[0,1]x[0,1].
If we consider x = 0 x = 0 x=0x=0x=0 in Proposition 4 iii) we obtain
s n ( x ) = k = 0 n ( n k ) p k ( x ) s n k ( 0 ) ; s n ( x ) = k = 0 n ( n k ) p k ( x ) s n k ( 0 ) ; s_(n)(x)=sum_(k=0)^(n)((n)/(k))p_(k)(x)s_(n-k)(0);s_{n}(x)=\sum_{k=0}^{n}\binom{n}{k} p_{k}(x) s_{n-k}(0) ;sn(x)=k=0n(nk)pk(x)snk(0);
accordingly, for s k ( 0 ) 0 s k ( 0 ) 0 s_(k)(0) >= 0s_{k}(0) \geq 0sk(0)0 and p k ( x ) 0 , k N , x [ 0 , 1 ] p k ( x ) 0 , k N , x [ 0 , 1 ] p_(k)(x) >= 0,AA k inN,AA x in[0,1]p_{k}(x) \geq 0, \forall k \in \mathbb{N}, \forall x \in[0,1]pk(x)0,kN,x[0,1], we have s k ( x ) 0 , k N s k ( x ) 0 , k N s_(k)(x) >= 0,AA k inNs_{k}(x) \geq 0, \forall k \in \mathbb{N}sk(x)0,kN and x [ 0 , 1 ] x [ 0 , 1 ] AA x in[0,1]\forall x \in[0,1]x[0,1]. Therefore the operator L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S is positive.
Lemma 2. If the operator L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S is positive, then a n [ 0 , 1 ] , b n 1 a n [ 0 , 1 ] , b n 1 a_(n)in[0,1],b_(n) <= 1a_{n} \in[0,1], b_{n} \leq 1an[0,1],bn1 and 0 c n min { 1 b n 2 , a n a n 2 } , n N 0 c n min 1 b n 2 , a n a n 2 , n N 0 <= c_(n) <= min{(1-b_(n))/(2),a_(n)-a_(n)^(2)},AA n inN0 \leq c_{n} \leq \min \left\{\frac{1-b_{n}}{2}, a_{n}-a_{n}^{2}\right\}, \forall n \in \mathbb{N}0cnmin{1bn2,anan2},nN, where a n , b n a n , b n a_(n),b_(n)a_{n}, b_{n}an,bn and c n c n c_(n)c_{n}cn are defined by (9).
Proof. Since 0 e 1 ( t ) 1 , t [ 0 , 1 ] 0 e 1 ( t ) 1 , t [ 0 , 1 ] 0 <= e_(1)(t) <= 1,AA t in[0,1]0 \leq e_{1}(t) \leq 1, \forall t \in[0,1]0e1(t)1,t[0,1] and the operator L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S is positive, we have 0 ( L n Q , S e 1 ) ( x ) 1 , x [ 0 , 1 ] 0 L n Q , S e 1 ( x ) 1 , x [ 0 , 1 ] 0 <= (L_(n)^(Q,S)e_(1))(x) <= 1,AA x in[0,1]0 \leq\left(L_{n}^{Q, S} e_{1}\right)(x) \leq 1, \forall x \in[0,1]0(LnQ,Se1)(x)1,x[0,1], and as ( L n Q , S e 1 ) ( x ) = a n x L n Q , S e 1 ( x ) = a n x (L_(n)^(Q,S)e_(1))(x)=a_(n)x\left(L_{n}^{Q, S} e_{1}\right)(x)=a_{n} x(LnQ,Se1)(x)=anx, we get a n [ 0 , 1 ] a n [ 0 , 1 ] a_(n)in[0,1]a_{n} \in[0,1]an[0,1].
From t ( 1 t ) 0 t ( 1 t ) 0 t(1-t) >= 0t(1-t) \geq 0t(1t)0 it results that ( L n Q , S e 1 ) ( x ) ( L n Q , S e 2 ) ( x ) 0 L n Q , S e 1 ( x ) L n Q , S e 2 ( x ) 0 (L_(n)^(Q,S)e_(1))(x)-(L_(n)^(Q,S)e_(2))(x) >= 0\left(L_{n}^{Q, S} e_{1}\right)(x)-\left(L_{n}^{Q, S} e_{2}\right)(x) \geq 0(LnQ,Se1)(x)(LnQ,Se2)(x)0, which leads to x ( 1 x ) b n + x c n 0 , x [ 0 , 1 ] x ( 1 x ) b n + x c n 0 , x [ 0 , 1 ] x(1-x)b_(n)+xc_(n) >= 0,AA x in[0,1]x(1-x) b_{n}+x c_{n} \geq 0, \forall x \in[0,1]x(1x)bn+xcn0,x[0,1] and choosing x = 1 x = 1 x=1x=1x=1, we get c n 0 c n 0 c_(n) >= 0c_{n} \geq 0cn0.
Since t 2 t + 1 / 4 0 t 2 t + 1 / 4 0 t^(2)-t+1//4 >= 0t^{2}-t+1 / 4 \geq 0t2t+1/40, we obtain ( L n Q , S e 2 ) ( x ) ( L n Q , S e 1 ) ( x ) + ( L n Q , S e 0 ) ( x ) / 4 0 , x [ 0 , 1 ] L n Q , S e 2 ( x ) L n Q , S e 1 ( x ) + L n Q , S e 0 ( x ) / 4 0 , x [ 0 , 1 ] (L_(n)^(Q,S)e_(2))(x)-(L_(n)^(Q,S)e_(1))(x)+(L_(n)^(Q,S)e_(0))(x)//4 >= 0,AA x in[0,1]\left(L_{n}^{Q, S} e_{2}\right)(x)-\left(L_{n}^{Q, S} e_{1}\right)(x)+\left(L_{n}^{Q, S} e_{0}\right)(x) / 4 \geq 0, \forall x \in[0,1](LnQ,Se2)(x)(LnQ,Se1)(x)+(LnQ,Se0)(x)/40,x[0,1], relation equivalent to x 2 b n x b n x c n + 1 / 4 0 , x [ 0 , 1 ] x 2 b n x b n x c n + 1 / 4 0 , x [ 0 , 1 ] x^(2)b_(n)-xb_(n)-xc_(n)+1//4 >= 0,AA x in[0,1]x^{2} b_{n}-x b_{n}-x c_{n}+1 / 4 \geq 0, \forall x \in[0,1]x2bnxbnxcn+1/40,x[0,1]. If we consider x = 1 / 2 x = 1 / 2 x=1//2x=1 / 2x=1/2, it results that c n ( 1 b n ) / 2 c n 1 b n / 2 c_(n) <= (1-b_(n))//2c_{n} \leq\left(1-b_{n}\right) / 2cn(1bn)/2 and because c n 0 c n 0 c_(n) >= 0c_{n} \geq 0cn0 we get b n 1 b n 1 b_(n) <= 1b_{n} \leq 1bn1.
Finally, from the Schwarz's inequality,
[ ( L n Q , S e 1 ) ( x ) ] 2 ( L n Q , S e 2 ) ( x ) ( L n Q , S e 0 ) ( x ) , L n Q , S e 1 ( x ) 2 L n Q , S e 2 ( x ) L n Q , S e 0 ( x ) , [(L_(n)^(Q,S)e_(1))(x)]^(2) <= (L_(n)^(Q,S)e_(2))(x)(L_(n)^(Q,S)e_(0))(x),\left[\left(L_{n}^{Q, S} e_{1}\right)(x)\right]^{2} \leq\left(L_{n}^{Q, S} e_{2}\right)(x)\left(L_{n}^{Q, S} e_{0}\right)(x),[(LnQ,Se1)(x)]2(LnQ,Se2)(x)(LnQ,Se0)(x),
we have a n 2 x 2 b n x 2 + x ( a n b n c n ) , x [ 0 , 1 ] a n 2 x 2 b n x 2 + x a n b n c n , x [ 0 , 1 ] a_(n)^(2)x^(2) <= b_(n)x^(2)+x(a_(n)-b_(n)-c_(n)),AA x in[0,1]a_{n}^{2} x^{2} \leq b_{n} x^{2}+x\left(a_{n}-b_{n}-c_{n}\right), \forall x \in[0,1]an2x2bnx2+x(anbncn),x[0,1]. For x = 1 x = 1 x=1x=1x=1 that implies c n a n a n 2 c n a n a n 2 c_(n) <= a_(n)-a_(n)^(2)c_{n} \leq a_{n}-a_{n}^{2}cnanan2.
Theorem 2. Let Q Q QQQ be a delta operator and S S SSS an invertible shift invariant operator. Let ( p n ( x ) ) n 0 p n ( x ) n 0 (p_(n)(x))_(n >= 0)\left(p_{n}(x)\right)_{n \geq 0}(pn(x))n0 be the sequence of basic polynomials for Q Q QQQ, with p n ( 0 ) 0 , n N p n ( 0 ) 0 , n N p_(n)^(')(0) >= 0,AA n inNp_{n}^{\prime}(0) \geq 0, \forall n \in \mathbb{N}pn(0)0,nN, and ( s n ( x ) ) n 0 s n ( x ) n 0 (s_(n)(x))_(n >= 0)\left(s_{n}(x)\right)_{n \geq 0}(sn(x))n0 a Sheffer set relative to Q , s n = S 1 p n Q , s n = S 1 p n Q,s_(n)=S^(-1)p_(n)Q, s_{n}=S^{-1} p_{n}Q,sn=S1pn with s n ( 1 ) 0 s n ( 1 ) 0 s_(n)(1)!=0s_{n}(1) \neq 0sn(1)0 and s n ( 0 ) 0 , n N s n ( 0 ) 0 , n N s_(n)(0) >= 0,AA n inNs_{n}(0) \geq 0, \forall n \in \mathbb{N}sn(0)0,nN. If f C [ 0 , 1 ] f C [ 0 , 1 ] f in C[0,1]f \in C[0,1]fC[0,1] and lim n a n = lim n b n = 1 lim n a n = lim n b n = 1 lim_(n rarr oo)a_(n)=lim_(n rarr oo)b_(n)=1\lim _{n \rightarrow \infty} a_{n}= \lim _{n \rightarrow \infty} b_{n}=1limnan=limnbn=1, where a n a n a_(n)a_{n}an and b n b n b_(n)b_{n}bn are defined by (9), then the operator L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S converges to the function f f fff, uniformly on the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1].
Proof. If lim n a n = 1 lim n a n = 1 lim_(n rarr oo)a_(n)=1\lim _{n \rightarrow \infty} a_{n}=1limnan=1 then lim n ( L n Q , S e 1 ) ( x ) = e 1 ( x ) lim n L n Q , S e 1 ( x ) = e 1 ( x ) lim_(n rarr oo)(L_(n)^(Q,S)e_(1))(x)=e_(1)(x)\lim _{n \rightarrow \infty}\left(L_{n}^{Q, S} e_{1}\right)(x)=e_{1}(x)limn(LnQ,Se1)(x)=e1(x). From Lemma 2 , c n a n a n 2 2 , c n a n a n 2 2,c_(n) <= a_(n)-a_(n)^(2)2, c_{n} \leq a_{n}-a_{n}^{2}2,cnanan2 so we have lim n c n = 0 lim n c n = 0 lim_(n rarr oo)c_(n)=0\lim _{n \rightarrow \infty} c_{n}=0limncn=0, and as lim n b n = 1 lim n b n = 1 lim_(n rarr oo)b_(n)=1\lim _{n \rightarrow \infty} b_{n}=1limnbn=1, we get lim n ( L n Q , S e 2 ) ( x ) = e 2 ( x ) lim n L n Q , S e 2 ( x ) = e 2 ( x ) lim_(n rarr oo)(L_(n)^(Q,S)e_(2))(x)=e_(2)(x)\lim _{n \rightarrow \infty}\left(L_{n}^{Q, S} e_{2}\right)(x)=e_{2}(x)limn(LnQ,Se2)(x)=e2(x). Therefore lim n ( L n Q , S e i ) ( x ) = e i ( x ) lim n L n Q , S e i ( x ) = e i ( x ) lim_(n rarr oo)(L_(n)^(Q,S)e_(i))(x)=e_(i)(x)\lim _{n \rightarrow \infty}\left(L_{n}^{Q, S} e_{i}\right)(x)=e_{i}(x)limn(LnQ,Sei)(x)=ei(x) for i = 0 , 1 , 2 i = 0 , 1 , 2 i=0,1,2i= 0,1,2i=0,1,2 so we can use the convergence criterion of Bohman-Korokvin.

3. REPRESENTATIONS OF THE OPERATOR L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S

Theorem 3. The operator L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S can be represented in the form
(15) ( L n Q , S f ) ( x ) = k = 0 n k ! n k ( n k ) [ 0 , 1 n , , k n ; f ] d k , n ( x ) (15) L n Q , S f ( x ) = k = 0 n k ! n k ( n k ) 0 , 1 n , , k n ; f d k , n ( x ) {:(15)(L_(n)^(Q,S)f)(x)=sum_(k=0)^(n)(k!)/(n^(k))((n)/(k))[0,(1)/(n),dots,(k)/(n);f]d_(k,n)(x):}\begin{equation*} \left(L_{n}^{Q, S} f\right)(x)=\sum_{k=0}^{n} \frac{k!}{n^{k}}\binom{n}{k}\left[0, \frac{1}{n}, \ldots, \frac{k}{n} ; f\right] d_{k, n}(x) \tag{15} \end{equation*}(15)(LnQ,Sf)(x)=k=0nk!nk(nk)[0,1n,,kn;f]dk,n(x)
where
d k , n ( x ) = 1 s n ( 1 ) ( θ k E 1 x s n k ) ( x ) d k , n ( x ) = 1 s n ( 1 ) θ k E 1 x s n k ( x ) d_(k,n)(x)=(1)/(s_(n)(1))(theta^(k)E^(1-x)s_(n-k))(x)d_{k, n}(x)=\frac{1}{s_{n}(1)}\left(\theta^{k} E^{1-x} s_{n-k}\right)(x)dk,n(x)=1sn(1)(θkE1xsnk)(x)
Moreover L n Q , S ( P m ) P m , m N L n Q , S P m P m , m N L_(n)^(Q,S)(P_(m))subeP_(m),AA m inNL_{n}^{Q, S}\left(P_{m}\right) \subseteq P_{m}, \forall m \in \mathbb{N}LnQ,S(Pm)Pm,mN.
Proof. From the Newton interpolation formula we have
f ( k n ) = j = 0 k j ! n j ( k j ) [ 0 , 1 n , , j n ; f ] f k n = j = 0 k j ! n j ( k j ) 0 , 1 n , , j n ; f f((k)/(n))=sum_(j=0)^(k)(j!)/(n^(j))((k)/(j))[0,(1)/(n),dots,(j)/(n);f]f\left(\frac{k}{n}\right)=\sum_{j=0}^{k} \frac{j!}{n^{j}}\binom{k}{j}\left[0, \frac{1}{n}, \ldots, \frac{j}{n} ; f\right]f(kn)=j=0kj!nj(kj)[0,1n,,jn;f]
If we denote w k , n ( x , y ) = ( n k ) p k ( x ) s n k ( y ) w k , n ( x , y ) = ( n k ) p k ( x ) s n k ( y ) w_(k,n)(x,y)=((n)/(k))p_(k)(x)s_(n-k)(y)w_{k, n}(x, y)=\binom{n}{k} p_{k}(x) s_{n-k}(y)wk,n(x,y)=(nk)pk(x)snk(y) then
k = 0 n w k , n ( x , y ) f ( k n ) = k = 0 n k ! n k [ 0 , 1 n , , k n ; f ] j = k n ( j k ) w j , n ( x , y ) k = 0 n w k , n ( x , y ) f k n = k = 0 n k ! n k 0 , 1 n , , k n ; f j = k n ( j k ) w j , n ( x , y ) sum_(k=0)^(n)w_(k,n)(x,y)f((k)/(n))=sum_(k=0)^(n)(k!)/(n^(k))[0,(1)/(n),dots,(k)/(n);f]sum_(j=k)^(n)((j)/(k))w_(j,n)(x,y)\sum_{k=0}^{n} w_{k, n}(x, y) f\left(\frac{k}{n}\right)=\sum_{k=0}^{n} \frac{k!}{n^{k}}\left[0, \frac{1}{n}, \ldots, \frac{k}{n} ; f\right] \sum_{j=k}^{n}\binom{j}{k} w_{j, n}(x, y)k=0nwk,n(x,y)f(kn)=k=0nk!nk[0,1n,,kn;f]j=kn(jk)wj,n(x,y)
But
j = k n ( j k ) w j , n ( x , y ) = ( n k ) j = k n ( n k j k ) p j ( x ) s n j ( y ) = ( n k ) j = 0 n k ( n k j ) p j + k ( x ) s n k j ( y ) = ( n k ) θ k j = 0 n k ( n k j ) p j ( x ) s n k j ( y ) = ( n k ) θ k E y s n k ( x ) j = k n ( j k ) w j , n ( x , y ) = ( n k ) j = k n ( n k j k ) p j ( x ) s n j ( y ) = ( n k ) j = 0 n k ( n k j ) p j + k ( x ) s n k j ( y ) = ( n k ) θ k j = 0 n k ( n k j ) p j ( x ) s n k j ( y ) = ( n k ) θ k E y s n k ( x ) {:[sum_(j=k)^(n)((j)/(k))w_(j,n)(x","y)=((n)/(k))sum_(j=k)^(n)((n-k)/(j-k))p_(j)(x)s_(n-j)(y)],[=((n)/(k))sum_(j=0)^(n-k)((n-k)/(j))p_(j+k)(x)s_(n-k-j)(y)],[=((n)/(k))theta^(k)sum_(j=0)^(n-k)((n-k)/(j))p_(j)(x)s_(n-k-j)(y)],[=((n)/(k))theta^(k)E^(y)s_(n-k)(x)]:}\begin{aligned} \sum_{j=k}^{n}\binom{j}{k} w_{j, n}(x, y) & =\binom{n}{k} \sum_{j=k}^{n}\binom{n-k}{j-k} p_{j}(x) s_{n-j}(y) \\ & =\binom{n}{k} \sum_{j=0}^{n-k}\binom{n-k}{j} p_{j+k}(x) s_{n-k-j}(y) \\ & =\binom{n}{k} \theta^{k} \sum_{j=0}^{n-k}\binom{n-k}{j} p_{j}(x) s_{n-k-j}(y) \\ & =\binom{n}{k} \theta^{k} E^{y} s_{n-k}(x) \end{aligned}j=kn(jk)wj,n(x,y)=(nk)j=kn(nkjk)pj(x)snj(y)=(nk)j=0nk(nkj)pj+k(x)snkj(y)=(nk)θkj=0nk(nkj)pj(x)snkj(y)=(nk)θkEysnk(x)
Since ( L n Q , S f ) ( x ) = 1 s n ( 1 ) k = 0 n w k , n ( x , 1 x ) f ( k n ) L n Q , S f ( x ) = 1 s n ( 1 ) k = 0 n w k , n ( x , 1 x ) f k n (L_(n)^(Q,S)f)(x)=(1)/(s_(n)(1))sum_(k=0)^(n)w_(k,n)(x,1-x)f((k)/(n))\left(L_{n}^{Q, S} f\right)(x)=\frac{1}{s_{n}(1)} \sum_{k=0}^{n} w_{k, n}(x, 1-x) f\left(\frac{k}{n}\right)(LnQ,Sf)(x)=1sn(1)k=0nwk,n(x,1x)f(kn) we obtain (15).
In order to show that L n Q , S ( P m ) P m L n Q , S P m P m L_(n)^(Q,S)(P_(m))subeP_(m)L_{n}^{Q, S}\left(P_{m}\right) \subseteq P_{m}LnQ,S(Pm)Pm we shall prove that deg ( d k , n ( x ) ) = k deg d k , n ( x ) = k deg(d_(k,n)(x))=k\operatorname{deg}\left(d_{k, n}(x)\right)=kdeg(dk,n(x))=k.
We remind that if ( p n p n p_(n)p_{n}pn ) is a basic sequence for Q = q ( D ) Q = q ( D ) Q=q(D)Q=q(D)Q=q(D) and h ( t ) h ( t ) h(t)h(t)h(t) is the compositional inverse of q ( t ) q ( t ) q(t)q(t)q(t), then the generating function for ( p n ) p n (p_(n))\left(p_{n}\right)(pn) is
(16) k = 0 p k ( x ) t k k ! = e x h ( t ) (16) k = 0 p k ( x ) t k k ! = e x h ( t ) {:(16)sum_(k=0)^(oo)p_(k)(x)(t^(k))/(k!)=e^(xh(t)):}\begin{equation*} \sum_{k=0}^{\infty} p_{k}(x) \frac{t^{k}}{k!}=e^{x h(t)} \tag{16} \end{equation*}(16)k=0pk(x)tkk!=exh(t)
and if s n = S 1 p n s n = S 1 p n s_(n)=S^(-1)p_(n)s_{n}=S^{-1} p_{n}sn=S1pn, with S = s ( D ) S = s ( D ) S=s(D)S=s(D)S=s(D) then
(17) k = 0 s k ( x ) t k k ! = 1 s ( h ( t ) ) e x h ( t ) (17) k = 0 s k ( x ) t k k ! = 1 s ( h ( t ) ) e x h ( t ) {:(17)sum_(k=0)^(oo)s_(k)(x)(t^(k))/(k!)=(1)/(s(h(t)))e^(xh(t)):}\begin{equation*} \sum_{k=0}^{\infty} s_{k}(x) \frac{t^{k}}{k!}=\frac{1}{s(h(t))} e^{x h(t)} \tag{17} \end{equation*}(17)k=0sk(x)tkk!=1s(h(t))exh(t)
If we differentiate the relation (16) m m mmm times with respect to t t ttt, we get
(18) k = 0 p k + m ( x ) t k k ! = d m d x m ( e x h ( t ) ) = ( x h 1 ( t ) + x 2 h 1 ( t ) + + x m h m ( t ) ) e x h ( t ) k = 0 p k + m ( x ) t k k ! = d m d x m e x h ( t ) = x h 1 ( t ) + x 2 h 1 ( t ) + + x m h m ( t ) e x h ( t ) sum_(k=0)^(oo)p_(k+m)(x)(t^(k))/(k!)=(d^(m))/(dx^(m))(e^(xh(t)))=(xh_(1)(t)+x^(2)h_(1)(t)+cdots+x^(m)h_(m)(t))e^(xh(t))\sum_{k=0}^{\infty} p_{k+m}(x) \frac{t^{k}}{k!}=\frac{d^{m}}{d x^{m}}\left(e^{x h(t)}\right)=\left(x h_{1}(t)+x^{2} h_{1}(t)+\cdots+x^{m} h_{m}(t)\right) e^{x h(t)}k=0pk+m(x)tkk!=dmdxm(exh(t))=(xh1(t)+x2h1(t)++xmhm(t))exh(t),
where every h i ( t ) h i ( t ) h_(i)(t)h_{i}(t)hi(t) is a product of derivatives of h ( t ) h ( t ) h(t)h(t)h(t).
Let us denote r ( k , m , x ) = j = 0 k ( k j ) p j + m ( x ) s k j ( 1 x ) r ( k , m , x ) = j = 0 k ( k j ) p j + m ( x ) s k j ( 1 x ) r(k,m,x)=sum_(j=0)^(k)((k)/(j))p_(j+m)(x)s_(k-j)(1-x)r(k, m, x)=\sum_{j=0}^{k}\binom{k}{j} p_{j+m}(x) s_{k-j}(1-x)r(k,m,x)=j=0k(kj)pj+m(x)skj(1x). Expanding 1 s ( h ( t ) ) h i ( t ) e h ( t ) = k 0 α i k t k k ! 1 s ( h ( t ) ) h i ( t ) e h ( t ) = k 0 α i k t k k ! (1)/(s(h(t)))h_(i)(t)e^(h(t))=sum_(k >= 0)alpha_(ik)(t^(k))/(k!)\frac{1}{s(h(t))} h_{i}(t) e^{h(t)}=\sum_{k \geq 0} \alpha_{i k} \frac{t^{k}}{k!}1s(h(t))hi(t)eh(t)=k0αiktkk!, from (17) and (18) we get r ( k , m , x ) = x α 1 k + x 2 α 2 k + + x m α m k r ( k , m , x ) = x α 1 k + x 2 α 2 k + + x m α m k r(k,m,x)=xalpha_(1k)+x^(2)alpha_(2k)+cdots+x^(m)alpha_(mk)r(k, m, x)=x \alpha_{1 k}+ x^{2} \alpha_{2 k}+\cdots+x^{m} \alpha_{m k}r(k,m,x)=xα1k+x2α2k++xmαmk.
Because d k , n ( x ) = r ( n k , k , x ) / s n ( 1 ) d k , n ( x ) = r ( n k , k , x ) / s n ( 1 ) d_(k,n)(x)=r(n-k,k,x)//s_(n)(1)d_{k, n}(x)=r(n-k, k, x) / s_{n}(1)dk,n(x)=r(nk,k,x)/sn(1) we obtain deg ( d k , n ( x ) ) = k deg d k , n ( x ) = k deg(d_(k,n)(x))=k\operatorname{deg}\left(d_{k, n}(x)\right)=kdeg(dk,n(x))=k.
Suppose that p P m p P m p inP_(m)p \in P_{m}pPm. Then [ 0 , 1 n , , k n ; p ] = 0 0 , 1 n , , k n ; p = 0 [0,(1)/(n),dots,(k)/(n);p]=0\left[0, \frac{1}{n}, \ldots, \frac{k}{n} ; p\right]=0[0,1n,,kn;p]=0 for k m + 1 k m + 1 k >= m+1k \geq m+1km+1, and using (15) we get L n Q , S ( P m ) P m L n Q , S P m P m L_(n)^(Q,S)(P_(m))subeP_(m)L_{n}^{Q, S}\left(P_{m}\right) \subseteq P_{m}LnQ,S(Pm)Pm.
Remark 1. For Q = D Q = D Q=DQ=DQ=D (it means that s n s n s_(n)s_{n}sn is an Appell set A n A n A_(n)A_{n}An ) we have θ = X θ = X theta=X\theta= Xθ=X, therefore in this case d k , n = A n k ( 1 ) A n ( 1 ) x k d k , n = A n k ( 1 ) A n ( 1 ) x k d_(k,n)=(A_(n-k)(1))/(A_(n)(1))x^(k)d_{k, n}=\frac{A_{n-k}(1)}{A_{n}(1)} x^{k}dk,n=Ank(1)An(1)xk. This representation for operators constructed with Appell sequences was given by C. Manole in [5].
Theorem 4. Suppose that all the assumptions of Theorem 2 are true, then there exists θ 1 n , θ 2 n , θ 3 n [ 0 , 1 ] θ 1 n , θ 2 n , θ 3 n [ 0 , 1 ] theta_(1n),theta_(2n),theta_(3n)in[0,1]\theta_{1 n}, \theta_{2 n}, \theta_{3 n} \in[0,1]θ1n,θ2n,θ3n[0,1] such that x [ 0 , 1 ] x [ 0 , 1 ] AA x in[0,1]\forall x \in[0,1]x[0,1] and f C [ 0 , 1 ] f C [ 0 , 1 ] AA f in C[0,1]\forall f \in C[0,1]fC[0,1] we have
( L n Q , S f ) ( x ) = f ( a n x ) + α ( x , n ) [ θ 1 n , θ 2 n , θ 3 n ; f ] L n Q , S f ( x ) = f a n x + α ( x , n ) θ 1 n , θ 2 n , θ 3 n ; f (L_(n)^(Q,S)f)(x)=f(a_(n)x)+alpha(x,n)[theta_(1n),theta_(2n),theta_(3n);f]\left(L_{n}^{Q, S} f\right)(x)=f\left(a_{n} x\right)+\alpha(x, n)\left[\theta_{1 n}, \theta_{2 n}, \theta_{3 n} ; f\right](LnQ,Sf)(x)=f(anx)+α(x,n)[θ1n,θ2n,θ3n;f]
where α ( x , n ) = x 2 ( b n a n 2 ) + x ( a n b n c n ) α ( x , n ) = x 2 b n a n 2 + x a n b n c n alpha(x,n)=x^(2)(b_(n)-a_(n)^(2))+x(a_(n)-b_(n)-c_(n))\alpha(x, n)=x^{2}\left(b_{n}-a_{n}^{2}\right)+x\left(a_{n}-b_{n}-c_{n}\right)α(x,n)=x2(bnan2)+x(anbncn).
Proof. First we shall prove that f ( a n x ) ( L n Q , S f ) ( x ) f a n x L n Q , S f ( x ) f(a_(n)x) <= (L_(n)^(Q,S)f)(x)f\left(a_{n} x\right) \leq\left(L_{n}^{Q, S} f\right)(x)f(anx)(LnQ,Sf)(x) for every convex function f f fff.
Let us denote c k = 1 s n ( 1 ) ( n k ) p k ( x ) s n k ( 1 x ) c k = 1 s n ( 1 ) ( n k ) p k ( x ) s n k ( 1 x ) c_(k)=(1)/(s_(n)(1))((n)/(k))p_(k)(x)s_(n-k)(1-x)c_{k}=\frac{1}{s_{n}(1)}\binom{n}{k} p_{k}(x) s_{n-k}(1-x)ck=1sn(1)(nk)pk(x)snk(1x) and x k = k n , k = 0 , 1 , , n x k = k n , k = 0 , 1 , , n x_(k)=(k)/(n),k=0,1,dots,nx_{k}=\frac{k}{n}, k=0,1, \ldots, nxk=kn,k=0,1,,n.
We have c k 0 , k = 0 n c k = 1 c k 0 , k = 0 n c k = 1 c_(k) >= 0,sum_(k=0)^(n)c_(k)=1c_{k} \geq 0, \sum_{k=0}^{n} c_{k}=1ck0,k=0nck=1 and x k > 0 , k N x k > 0 , k N x_(k) > 0,AA k inNx_{k}>0, \forall k \in \mathbb{N}xk>0,kN. If f f fff is a convex function then f ( k = 0 n c k x k ) k = 0 n c k f ( x k ) f k = 0 n c k x k k = 0 n c k f x k f(sum_(k=0)^(n)c_(k)x_(k)) <= sum_(k=0)^(n)c_(k)f(x_(k))f\left(\sum_{k=0}^{n} c_{k} x_{k}\right) \leq \sum_{k=0}^{n} c_{k} f\left(x_{k}\right)f(k=0nckxk)k=0nckf(xk); but
k = 0 n c k x k = ( L n Q , S e 1 ) ( x ) = a n x and k = 0 n c k f ( x k ) = ( L n Q , S f ) ( x ) k = 0 n c k x k = L n Q , S e 1 ( x ) = a n x  and  k = 0 n c k f x k = L n Q , S f ( x ) sum_(k=0)^(n)c_(k)x_(k)=(L_(n)^(Q,S)e_(1))(x)=a_(n)x quad" and "quadsum_(k=0)^(n)c_(k)f(x_(k))=(L_(n)^(Q,S)f)(x)\sum_{k=0}^{n} c_{k} x_{k}=\left(L_{n}^{Q, S} e_{1}\right)(x)=a_{n} x \quad \text { and } \quad \sum_{k=0}^{n} c_{k} f\left(x_{k}\right)=\left(L_{n}^{Q, S} f\right)(x)k=0nckxk=(LnQ,Se1)(x)=anx and k=0nckf(xk)=(LnQ,Sf)(x)
therefore we get f ( a n x ) ( L n Q , S f ) ( x ) f a n x L n Q , S f ( x ) f(a_(n)x) <= (L_(n)^(Q,S)f)(x)f\left(a_{n} x\right) \leq\left(L_{n}^{Q, S} f\right)(x)f(anx)(LnQ,Sf)(x).
If we consider the formula
f ( a n x ) = ( L n Q , S f ) ( x ) + ( R n f ) ( x ) f a n x = L n Q , S f ( x ) + R n f ( x ) f(a_(n)x)=(L_(n)^(Q,S)f)(x)+(R_(n)f)(x)f\left(a_{n} x\right)=\left(L_{n}^{Q, S} f\right)(x)+\left(R_{n} f\right)(x)f(anx)=(LnQ,Sf)(x)+(Rnf)(x)
we have ( R n f ) 0 R n f 0 (R_(n)f) <= 0\left(R_{n} f\right) \leq 0(Rnf)0 for every convex function f f fff.
Since ( R n e i ) ( x ) = 0 R n e i ( x ) = 0 (R_(n)e_(i))(x)=0\left(R_{n} e_{i}\right)(x)=0(Rnei)(x)=0 for i = 0 , 1 i = 0 , 1 i=0,1i=0,1i=0,1, the degree of exactness of the previous formula is one and then there exist θ 1 n , θ 2 n , θ 3 n [ 0 , 1 ] θ 1 n , θ 2 n , θ 3 n [ 0 , 1 ] theta_(1n),theta_(2n),theta_(3n)in[0,1]\theta_{1 n}, \theta_{2 n}, \theta_{3 n} \in[0,1]θ1n,θ2n,θ3n[0,1] such that the remainder can be represented in the following form
( R n f ) ( x ) = ( R n e 2 ) ( x ) [ θ 1 n , θ 2 n , θ 3 n ; f ] R n f ( x ) = R n e 2 ( x ) θ 1 n , θ 2 n , θ 3 n ; f (R_(n)f)(x)=(R_(n)e_(2))(x)[theta_(1n),theta_(2n),theta_(3n);f]\left(R_{n} f\right)(x)=\left(R_{n} e_{2}\right)(x)\left[\theta_{1 n}, \theta_{2 n}, \theta_{3 n} ; f\right](Rnf)(x)=(Rne2)(x)[θ1n,θ2n,θ3n;f]
where ( R n e 2 ) ( x ) = x 2 ( a n 2 b n ) + x ( b n + c n a n ) R n e 2 ( x ) = x 2 a n 2 b n + x b n + c n a n (R_(n)e_(2))(x)=x^(2)(a_(n)^(2)-b_(n))+x(b_(n)+c_(n)-a_(n))\left(R_{n} e_{2}\right)(x)=x^{2}\left(a_{n}^{2}-b_{n}\right)+x\left(b_{n}+c_{n}-a_{n}\right)(Rne2)(x)=x2(an2bn)+x(bn+cnan), so we obtain the conclusion.

4. EXAMPLES

  1. If S = I S = I S=IS=IS=I then s n = p n s n = p n s_(n)=p_(n)s_{n}=p_{n}sn=pn and in this case the operator defined by (2) becomes the binomial operator (1) introduced by Tiberiu Popoviciu in 9 .
    1.1. For Q = D Q = D Q=DQ=DQ=D the basic sequence is p n ( x ) = x n p n ( x ) = x n p_(n)(x)=x^(n)p_{n}(x)=x^{n}pn(x)=xn and L n D , I L n D , I L_(n)^(D,I)L_{n}^{D, I}LnD,I is the Bernstein operator B n B n B_(n)B_{n}Bn.
    1.2. If Q Q QQQ is Abel operator A = E β D A = E β D A=E^(-beta)DA=E^{-\beta} DA=EβD we have p n ( x ) = x ( x + n β ) n 1 p n ( x ) = x ( x + n β ) n 1 p_(n)(x)=x(x+n beta)^(n-1)p_{n}(x)=x(x+n \beta)^{n-1}pn(x)=x(x+nβ)n1 and L n A , I L n A , I L_(n)^(A,I)L_{n}^{A, I}LnA,I is the second operator introduced by Cheney and Sharma in [1,
( L n A , I f ) ( x ) = = 1 ( 1 + n β ) n 1 k = 0 n ( n k ) x ( x + k β ) k 1 ( 1 x ) ( 1 x + ( n k ) β ) n k 1 f ( k n ) L n A , I f ( x ) = = 1 ( 1 + n β ) n 1 k = 0 n ( n k ) x ( x + k β ) k 1 ( 1 x ) ( 1 x + ( n k ) β ) n k 1 f k n {:[(L_(n)^(A,I)f)(x)=],[=(1)/((1+n beta)^(n-1))sum_(k=0)^(n)((n)/(k))x(x+k beta)^(k-1)(1-x)(1-x+(n-k)beta)^(n-k-1)f((k)/(n))]:}\begin{aligned} & \left(L_{n}^{A, I} f\right)(x)= \\ & =\frac{1}{(1+n \beta)^{n-1}} \sum_{k=0}^{n}\binom{n}{k} x(x+k \beta)^{k-1}(1-x)(1-x+(n-k) \beta)^{n-k-1} f\left(\frac{k}{n}\right) \end{aligned}(LnA,If)(x)==1(1+nβ)n1k=0n(nk)x(x+kβ)k1(1x)(1x+(nk)β)nk1f(kn)
1.3. For Laguerre delta operator L = D D + I L = D D + I L=(D)/(D+I)L=\frac{D}{D+I}L=DD+I the basic sequence is l n ( x ) = k = 0 n ( n k ) ( n 1 ) ! ( k 1 ) ! x k l n ( x ) = k = 0 n ( n k ) ( n 1 ) ! ( k 1 ) ! x k l_(n)(x)=sum_(k=0)^(n)((n)/(k))((n-1)!)/((k-1)!)x^(k)l_{n}(x)= \sum_{k=0}^{n}\binom{n}{k} \frac{(n-1)!}{(k-1)!} x^{k}ln(x)=k=0n(nk)(n1)!(k1)!xk and the coresponding binomial operator has been considered by T. Popoviciu.
1.4. The delta operator Q = 1 α α = 1 α ( I E α ) Q = 1 α α = 1 α I E α Q=(1)/(alpha)grad_(alpha)=(1)/(alpha)(I-E^(-alpha))Q=\frac{1}{\alpha} \nabla_{\alpha}=\frac{1}{\alpha}\left(I-E^{-\alpha}\right)Q=1αα=1α(IEα) has the basic sequence p n ( x ) = x [ n , α ] = x ( x + α ) ( x + ( n 1 ) α ) p n ( x ) = x [ n , α ] = x ( x + α ) ( x + ( n 1 ) α ) p_(n)(x)=x^([n,-alpha])=x(x+alpha)dots(x+(n-1)alpha)p_{n}(x)=x^{[n,-\alpha]}=x(x+\alpha) \ldots(x+(n-1) \alpha)pn(x)=x[n,α]=x(x+α)(x+(n1)α) and in this case we obtain the operator
( S n f ) ( x ) = 1 1 [ n , α ] k = 0 n ( n k ) x [ k , α ] ( 1 x ) [ n k , α ] f ( k n ) S n f ( x ) = 1 1 [ n , α ] k = 0 n ( n k ) x [ k , α ] ( 1 x ) [ n k , α ] f k n (S_(n)f)(x)=(1)/(1^([n,-alpha]))sum_(k=0)^(n)((n)/(k))x^([k,-alpha])(1-x)^([n-k,-alpha])f((k)/(n))\left(S_{n} f\right)(x)=\frac{1}{1^{[n,-\alpha]}} \sum_{k=0}^{n}\binom{n}{k} x^{[k,-\alpha]}(1-x)^{[n-k,-\alpha]} f\left(\frac{k}{n}\right)(Snf)(x)=11[n,α]k=0n(nk)x[k,α](1x)[nk,α]f(kn)
which has been introduced and investigated in detail by D. D. Stancu in [14], [16] and other papers.
1.5. The exponential polynomials t n ( x ) = k = 0 n S ( n , k ) x k = e x k = 0 k n x k k ! t n ( x ) = k = 0 n S ( n , k ) x k = e x k = 0 k n x k k ! t_(n)(x)=sum_(k=0)^(n)S(n,k)x^(k)=e^(-x)sum_(k=0)^(oo)(k^(n)x^(k))/(k!)t_{n}(x)=\sum_{k=0}^{n} S(n, k) x^{k}= e^{-x} \sum_{k=0}^{\infty} \frac{k^{n} x^{k}}{k!}tn(x)=k=0nS(n,k)xk=exk=0knxkk!, where S ( n , k ) S ( n , k ) S(n,k)S(n, k)S(n,k) denote the Stirling numbers of the second kind, are basic polynomials for the delta operator T = ln ( I + D ) T = ln ( I + D ) T=ln(I+D)T=\ln (I+D)T=ln(I+D). The approximation operator construct by means of the exponential polynomials
( L n T f ) ( x ) = 1 t n ( 1 ) k = 0 n ( n k ) t k ( x ) t n k ( 1 x ) f ( k n ) L n T f ( x ) = 1 t n ( 1 ) k = 0 n ( n k ) t k ( x ) t n k ( 1 x ) f k n (L_(n)^(T)f)(x)=(1)/(t_(n)(1))sum_(k=0)^(n)((n)/(k))t_(k)(x)t_(n-k)(1-x)f((k)/(n))\left(L_{n}^{T} f\right)(x)=\frac{1}{t_{n}(1)} \sum_{k=0}^{n}\binom{n}{k} t_{k}(x) t_{n-k}(1-x) f\left(\frac{k}{n}\right)(LnTf)(x)=1tn(1)k=0n(nk)tk(x)tnk(1x)f(kn)
was studied by C. Manole in [5].
1.6. If we take the delta operator Q = G = 1 α E β α = 1 α ( E β E α β Q = G = 1 α E β α = 1 α E β E α β Q=G=(1)/(alpha)E^(-beta)grad_(alpha)=(1)/(alpha)(E^(-beta)-:}E^(-alpha-beta)Q=G=\frac{1}{\alpha} E^{-\beta} \nabla_{\alpha}=\frac{1}{\alpha}\left(E^{-\beta}-\right. E^{-\alpha-\beta}Q=G=1αEβα=1α(EβEαβ ) its basic sequence is p n ( x ) = x ( x + α + n β ) [ n 1 , α ] p n ( x ) = x ( x + α + n β ) [ n 1 , α ] p_(n)(x)=x(x+alpha+n beta)^([n-1,-alpha])p_{n}(x)=x(x+\alpha+n \beta)^{[n-1,-\alpha]}pn(x)=x(x+α+nβ)[n1,α] and the operator
( L n G f ) ( x ) = 1 ( 1 + n β ) [ n , α ] k = 0 n ( n k ) x ( x + α + k β ) [ k 1 , α ] ( 1 x ) ( 1 x + ( n k ) β ) [ n k , α ] f ( k n ) L n G f ( x ) = 1 ( 1 + n β ) [ n , α ] k = 0 n ( n k ) x ( x + α + k β ) [ k 1 , α ] ( 1 x ) ( 1 x + ( n k ) β ) [ n k , α ] f k n {:[(L_(n)^(G)f)(x)=(1)/((1+n beta)^([n,-alpha]))],[*sum_(k=0)^(n)((n)/(k))x(x+alpha+k beta)^([k-1,-alpha])(1-x)(1-x+(n-k)beta)^([n-k,-alpha])f((k)/(n))]:}\begin{aligned} & \left(L_{n}^{G} f\right)(x)=\frac{1}{(1+n \beta)^{[n,-\alpha]}} \\ & \cdot \sum_{k=0}^{n}\binom{n}{k} x(x+\alpha+k \beta)^{[k-1,-\alpha]}(1-x)(1-x+(n-k) \beta)^{[n-k,-\alpha]} f\left(\frac{k}{n}\right) \end{aligned}(LnGf)(x)=1(1+nβ)[n,α]k=0n(nk)x(x+α+kβ)[k1,α](1x)(1x+(nk)β)[nk,α]f(kn)
was investigated by D. D. Stancu, G. Moldovan. In [18] D. D. Stancu and M. R. Occorsio have studied this operator with the nodes k + γ n + δ , 0 γ δ k + γ n + δ , 0 γ δ (k+gamma)/(n+delta),0 <= gamma <= delta\frac{k+\gamma}{n+\delta}, 0 \leq \gamma \leq \deltak+γn+δ,0γδ.
2. If Q = D Q = D Q=DQ=DQ=D and S S SSS is an invertible shift invariant operator then p n ( x ) = x n p n ( x ) = x n p_(n)(x)=x^(n)p_{n}(x)=x^{n}pn(x)=xn and s n = A n = S 1 x n s n = A n = S 1 x n s_(n)=A_(n)=S^(-1)x^(n)s_{n}=A_{n}=S^{-1} x^{n}sn=An=S1xn is an Appell set. The operator of the form
( L n D , S f ) ( x ) = 1 A n ( 1 ) k = 0 n ( n k ) x k A n k ( 1 x ) f ( k n ) L n D , S f ( x ) = 1 A n ( 1 ) k = 0 n ( n k ) x k A n k ( 1 x ) f k n (L_(n)^(D,S)f)(x)=(1)/(A_(n)(1))sum_(k=0)^(n)((n)/(k))x^(k)A_(n-k)(1-x)f((k)/(n))\left(L_{n}^{D, S} f\right)(x)=\frac{1}{A_{n}(1)} \sum_{k=0}^{n}\binom{n}{k} x^{k} A_{n-k}(1-x) f\left(\frac{k}{n}\right)(LnD,Sf)(x)=1An(1)k=0n(nk)xkAnk(1x)f(kn)
was introduced and investigated by C. Manole in [5].
2.1. If S = ( I + D ) 1 S = ( I + D ) 1 S=(I+D)^(-1)S=(I+D)^{-1}S=(I+D)1 the coresponding Appell set is A n ( x ) = x n + n x n 1 A n ( x ) = x n + n x n 1 A_(n)(x)=x^(n)+nx^(n-1)A_{n}(x)=x^{n}+n x^{n-1}An(x)=xn+nxn1 and then
( L n D , ( I + D ) 1 f ) ( x ) = 1 n + 1 k = 0 n ( n k ) x k ( 1 x ) n k ( n k + 1 x ) f ( k n ) L n D , ( I + D ) 1 f ( x ) = 1 n + 1 k = 0 n ( n k ) x k ( 1 x ) n k ( n k + 1 x ) f k n (L_(n)^(D,(I+D)^(-1))f)(x)=(1)/(n+1)sum_(k=0)^(n)((n)/(k))x^(k)(1-x)^(n-k)(n-k+1-x)f((k)/(n))\left(L_{n}^{D,(I+D)^{-1}} f\right)(x)=\frac{1}{n+1} \sum_{k=0}^{n}\binom{n}{k} x^{k}(1-x)^{n-k}(n-k+1-x) f\left(\frac{k}{n}\right)(LnD,(I+D)1f)(x)=1n+1k=0n(nk)xk(1x)nk(nk+1x)f(kn)
  1. If we take Q = A = E β D Q = A = E β D Q=A=E^(-beta)DQ=A=E^{-\beta} DQ=A=EβD and S = E β Q = I β D S = E β Q = I β D S=E^(beta)Q^(')=I-beta DS=E^{\beta} Q^{\prime}=I-\beta DS=EβQ=IβD then p n ( x ) = x ( x + n β ) n 1 p n ( x ) = x ( x + n β ) n 1 p_(n)(x)=x(x+n beta)^(n-1)p_{n}(x)= x(x+n \beta)^{n-1}pn(x)=x(x+nβ)n1 is the basic sequence for Q Q QQQ and s n ( x ) = ( x + n β ) n s n ( x ) = ( x + n β ) n s_(n)(x)=(x+n beta)^(n)s_{n}(x)=(x+n \beta)^{n}sn(x)=(x+nβ)n a Sheffer set for Q Q QQQ we obtain the first operator introduced by Cheney and Sharma in [1]:
( L n A , I β D f ) ( x ) = 1 ( 1 + n β ) n k = 0 n ( n k ) x ( x + k β ) k 1 ( 1 x + ( n k ) β ) n k f ( k n ) L n A , I β D f ( x ) = 1 ( 1 + n β ) n k = 0 n ( n k ) x ( x + k β ) k 1 ( 1 x + ( n k ) β ) n k f k n (L_(n)^(A,I-beta D)f)(x)=(1)/((1+n beta)^(n))sum_(k=0)^(n)((n)/(k))x(x+k beta)^(k-1)(1-x+(n-k)beta)^(n-k)f((k)/(n))\left(L_{n}^{A, I-\beta D} f\right)(x)=\frac{1}{(1+n \beta)^{n}} \sum_{k=0}^{n}\binom{n}{k} x(x+k \beta)^{k-1}(1-x+(n-k) \beta)^{n-k} f\left(\frac{k}{n}\right)(LnA,IβDf)(x)=1(1+nβ)nk=0n(nk)x(x+kβ)k1(1x+(nk)β)nkf(kn)
  1. For Q = 1 α E β α = 1 α ( E β E α β ) Q = 1 α E β α = 1 α E β E α β Q=(1)/(alpha)E^(-beta)grad_(alpha)=(1)/(alpha)(E^(-beta)-E^(-alpha-beta))Q=\frac{1}{\alpha} E^{-\beta} \nabla_{\alpha}=\frac{1}{\alpha}\left(E^{-\beta}-E^{-\alpha-\beta}\right)Q=1αEβα=1α(EβEαβ) and S = E α + β Q = 1 α ( ( α + β ) I β E α ) S = E α + β Q = 1 α ( ( α + β ) I β E α S=E^(alpha+beta)Q^(')=(1)/(alpha)((alpha+beta)I-{: betaE^(alpha))S=E^{\alpha+\beta} Q^{\prime}=\frac{1}{\alpha}((\alpha+\beta) I- \left.\beta E^{\alpha}\right)S=Eα+βQ=1α((α+β)IβEα) we have p n ( x ) = x ( x + α + n β ) [ n 1 , α ] p n ( x ) = x ( x + α + n β ) [ n 1 , α ] p_(n)(x)=x(x+alpha+n beta)^([n-1,-alpha])p_{n}(x)=x(x+\alpha+n \beta)^{[n-1,-\alpha]}pn(x)=x(x+α+nβ)[n1,α] and s n ( x ) = ( x + n β ) [ n , α ] s n ( x ) = ( x + n β ) [ n , α ] s_(n)(x)=(x+n beta)^([n,-alpha])s_{n}(x)=(x+n \beta)^{[n,-\alpha]}sn(x)=(x+nβ)[n,α] therefore the operator L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S in this case is
( L n [ α , β ] f ) ( x ) = = 1 ( 1 + n β ) [ n , α ] k = 0 n ( n k ) x ( x + α + k β ) [ k 1 , α ] ( 1 x + ( n k ) β ) [ n k , α ] f ( k n ) L n [ α , β ] f ( x ) = = 1 ( 1 + n β ) [ n , α ] k = 0 n ( n k ) x ( x + α + k β ) [ k 1 , α ] ( 1 x + ( n k ) β ) [ n k , α ] f k n {:[(L_(n)^([alpha,beta])f)(x)=],[=(1)/((1+n beta)^([n,-alpha]))sum_(k=0)^(n)((n)/(k))x(x+alpha+k beta)^([k-1,-alpha])(1-x+(n-k)beta)^([n-k,-alpha])f((k)/(n))]:}\begin{aligned} & \left(L_{n}^{[\alpha, \beta]} f\right)(x)= \\ & =\frac{1}{(1+n \beta)^{[n,-\alpha]}} \sum_{k=0}^{n}\binom{n}{k} x(x+\alpha+k \beta)^{[k-1,-\alpha]}(1-x+(n-k) \beta)^{[n-k,-\alpha]} f\left(\frac{k}{n}\right) \end{aligned}(Ln[α,β]f)(x)==1(1+nβ)[n,α]k=0n(nk)x(x+α+kβ)[k1,α](1x+(nk)β)[nk,α]f(kn)
If we replace x x xxx with s ( x ) s ( x ) s(x)s(x)s(x) we obtain a operator which has been studied by G . Moldovan in [6]. He has found the value of this operator for the monomials e i e i e_(i)e_{i}ei for i = 1 , 2 i = 1 , 2 i=1,2i=1,2i=1,2 using some generalized identities of Vandermonde type.
We want to find the sequences a n , b n , c n a n , b n , c n a_(n),b_(n),c_(n)a_{n}, b_{n}, c_{n}an,bn,cn, which appears in ( L n [ α , β ] e i ) ( x ) L n [ α , β ] e i ( x ) (L_(n)^([alpha,beta])e_(i))(x)\left(L_{n}^{[\alpha, \beta]} e_{i}\right)(x)(Ln[α,β]ei)(x), using relations (9).
The Pincherle derivative of Q Q QQQ is
Q = β α E β + ( 1 + β α ) E α β = E α β ( I β α Δ α ) Q = β α E β + 1 + β α E α β = E α β I β α Δ α Q^(')=-(beta )/(alpha)E^(-beta)+(1+(beta )/(alpha))E^(-alpha-beta)=E^(-alpha-beta)(I-(beta )/(alpha)Delta_(alpha))Q^{\prime}=-\frac{\beta}{\alpha} E^{-\beta}+\left(1+\frac{\beta}{\alpha}\right) E^{-\alpha-\beta}=E^{-\alpha-\beta}\left(I-\frac{\beta}{\alpha} \Delta_{\alpha}\right)Q=βαEβ+(1+βα)Eαβ=Eαβ(IβαΔα)
so
( Q ) 1 = E α + β k 0 β k ( Δ α α ) k Q 1 = E α + β k 0 β k Δ α α k (Q^('))^(-1)=E^(alpha+beta)sum_(k >= 0)beta^(k)((Delta_(alpha))/(alpha))^(k)\left(Q^{\prime}\right)^{-1}=E^{\alpha+\beta} \sum_{k \geq 0} \beta^{k}\left(\frac{\Delta_{\alpha}}{\alpha}\right)^{k}(Q)1=Eα+βk0βk(Δαα)k
Since s n 1 ( x ) = ( x + ( n 1 ) β ) [ n 1 , α ] = E ( n 2 ) α + ( n 1 ) β x [ n 1 , α ] s n 1 ( x ) = ( x + ( n 1 ) β ) [ n 1 , α ] = E ( n 2 ) α + ( n 1 ) β x [ n 1 , α ] s_(n-1)(x)=(x+(n-1)beta)^([n-1,-alpha])=E^((n-2)alpha+(n-1)beta)x^([n-1,alpha])s_{n-1}(x)=(x+(n-1) \beta)^{[n-1,-\alpha]}=E^{(n-2) \alpha+(n-1) \beta} x^{[n-1, \alpha]}sn1(x)=(x+(n1)β)[n1,α]=E(n2)α+(n1)βx[n1,α] and x [ n , α ] = x ( x 1 ) ( x ( n 1 ) α ) x [ n , α ] = x ( x 1 ) ( x ( n 1 ) α ) x^([n,alpha])=x(x-1)dots(x-(n-1)alpha)x^{[n, \alpha]}= x(x-1) \ldots(x-(n-1) \alpha)x[n,α]=x(x1)(x(n1)α) is the basic sequence for the delta operator Δ α α Δ α α (Delta_(alpha))/(alpha)\frac{\Delta_{\alpha}}{\alpha}Δαα, we have ( Δ α α ) k s n 1 ( x ) = E ( n 2 ) α + ( n 1 ) β ( Δ α α ) k x [ n 1 , α ] = ( n 1 ) [ k ] E ( n 2 ) α + ( n 1 ) β Δ α α k s n 1 ( x ) = E ( n 2 ) α + ( n 1 ) β Δ α α k x [ n 1 , α ] = ( n 1 ) [ k ] E ( n 2 ) α + ( n 1 ) β ((Delta_(alpha))/(alpha))^(k)s_(n-1)(x)=E^((n-2)alpha+(n-1)beta)((Delta_(alpha))/(alpha))^(k)x^([n-1,alpha])=(n-1)^([k])E^((n-2)alpha+(n-1)beta)\left(\frac{\Delta_{\alpha}}{\alpha}\right)^{k} s_{n-1}(x)=E^{(n-2) \alpha+(n-1) \beta}\left(\frac{\Delta_{\alpha}}{\alpha}\right)^{k} x^{[n-1, \alpha]}=(n-1)^{[k]} E^{(n-2) \alpha+(n-1) \beta}(Δαα)ksn1(x)=E(n2)α+(n1)β(Δαα)kx[n1,α]=(n1)[k]E(n2)α+(n1)β. x [ n 1 k , α ] x [ n 1 k , α ] *x^([n-1-k,alpha])\cdot x^{[n-1-k, \alpha]}x[n1k,α]. Because a n = [ ( Q ) 1 s n 1 ] ( 1 ) s n ( 1 ) a n = Q 1 s n 1 ( 1 ) s n ( 1 ) a_(n)=([(Q^('))^(-1)s_(n-1)](1))/(s_(n)(1))a_{n}=\frac{\left[\left(Q^{\prime}\right)^{-1} s_{n-1}\right](1)}{s_{n}(1)}an=[(Q)1sn1](1)sn(1) we get
a n = k = 0 n 1 ( n 1 k ) k ! β k ( 1 + n β ) [ k + 1 , α ] a n = k = 0 n 1 ( n 1 k ) k ! β k ( 1 + n β ) [ k + 1 , α ] a_(n)=sum_(k=0)^(n-1)((n-1)/(k))(k!beta^(k))/((1+n beta)^([k+1,-alpha]))a_{n}=\sum_{k=0}^{n-1}\binom{n-1}{k} \frac{k!\beta^{k}}{(1+n \beta)^{[k+1,-\alpha]}}an=k=0n1(n1k)k!βk(1+nβ)[k+1,α]
Since b n = n 1 n [ ( Q ) 2 s n 2 ] ( 1 ) s n ( 1 ) b n = n 1 n Q 2 s n 2 ( 1 ) s n ( 1 ) b_(n)=(n-1)/(n)([(Q^('))^(-2)s_(n-2)](1))/(s_(n)(1))b_{n}=\frac{n-1}{n} \frac{\left[\left(Q^{\prime}\right)^{-2} s_{n-2}\right](1)}{s_{n}(1)}bn=n1n[(Q)2sn2](1)sn(1) and
( Q ) 2 = E 2 α + 2 β ( I β α Δ α ) 2 = E 2 α + 2 β k 0 ( k + 1 ) β k ( Δ α α ) k Q 2 = E 2 α + 2 β I β α Δ α 2 = E 2 α + 2 β k 0 ( k + 1 ) β k Δ α α k (Q^('))^(-2)=E^(2alpha+2beta)(I-(beta )/(alpha)Delta_(alpha))^(-2)=E^(2alpha+2beta)sum_(k >= 0)(k+1)beta^(k)((Delta_(alpha))/(alpha))^(k)\left(Q^{\prime}\right)^{-2}=E^{2 \alpha+2 \beta}\left(I-\frac{\beta}{\alpha} \Delta_{\alpha}\right)^{-2}=E^{2 \alpha+2 \beta} \sum_{k \geq 0}(k+1) \beta^{k}\left(\frac{\Delta_{\alpha}}{\alpha}\right)^{k}(Q)2=E2α+2β(IβαΔα)2=E2α+2βk0(k+1)βk(Δαα)k
we obtain
b n = n 1 n k = 0 n 2 ( n 2 k ) ( k + 1 ) ! β k ( 1 + n β ) [ k + 2 , α ] b n = n 1 n k = 0 n 2 ( n 2 k ) ( k + 1 ) ! β k ( 1 + n β ) [ k + 2 , α ] b_(n)=(n-1)/(n)sum_(k=0)^(n-2)((n-2)/(k))((k+1)!beta^(k))/((1+n beta)^([k+2,-alpha]))b_{n}=\frac{n-1}{n} \sum_{k=0}^{n-2}\binom{n-2}{k} \frac{(k+1)!\beta^{k}}{(1+n \beta)^{[k+2,-\alpha]}}bn=n1nk=0n2(n2k)(k+1)!βk(1+nβ)[k+2,α]
The Pincherle derivative of S 1 S 1 S^(-1)S^{-1}S1 may be written in the form
( S 1 ) = ( α + β ) E α β ( Q ) 1 E α β ( Q ) 2 Q = E α β ( Q ) 2 ( ( α + β ) Q + Q ) . S 1 = ( α + β ) E α β Q 1 E α β Q 2 Q = E α β Q 2 ( α + β ) Q + Q . {:[(S^(-1))^(')=-(alpha+beta)E^(-alpha-beta)(Q^('))^(-1)-E^(-alpha-beta)(Q^('))^(-2)Q^('')],[=-E^(-alpha-beta)(Q^('))^(-2)((alpha+beta)Q^(')+Q^('')).]:}\begin{aligned} \left(S^{-1}\right)^{\prime} & =-(\alpha+\beta) E^{-\alpha-\beta}\left(Q^{\prime}\right)^{-1}-E^{-\alpha-\beta}\left(Q^{\prime}\right)^{-2} Q^{\prime \prime} \\ & =-E^{-\alpha-\beta}\left(Q^{\prime}\right)^{-2}\left((\alpha+\beta) Q^{\prime}+Q^{\prime \prime}\right) . \end{aligned}(S1)=(α+β)Eαβ(Q)1Eαβ(Q)2Q=Eαβ(Q)2((α+β)Q+Q).
Because Q = 1 α ( β 2 E β ( α + β ) 2 E α β ) Q = 1 α β 2 E β ( α + β ) 2 E α β Q^('')=(1)/(alpha)(beta^(2)E^(-beta)-(alpha+beta)^(2)E^(-alpha-beta))Q^{\prime \prime}=\frac{1}{\alpha}\left(\beta^{2} E^{-\beta}-(\alpha+\beta)^{2} E^{-\alpha-\beta}\right)Q=1α(β2Eβ(α+β)2Eαβ) and ( α + β ) Q + Q = β E β ( α + β ) Q + Q = β E β (alpha+beta)Q^(')+Q^('')=-betaE^(-beta)(\alpha+\beta) Q^{\prime}+Q^{\prime \prime}=-\beta E^{-\beta}(α+β)Q+Q=βEβ this implies
( Q ) 2 ( S 1 ) S = β ( Q ) 3 E β = β E 3 α + 2 β ( I β α Δ α ) 3 = E 3 α + 2 β k 0 ( k + 1 ) ( k + 2 ) 2 β k + 1 ( 1 α Δ α ) k . Q 2 S 1 S = β Q 3 E β = β E 3 α + 2 β I β α Δ α 3 = E 3 α + 2 β k 0 ( k + 1 ) ( k + 2 ) 2 β k + 1 1 α Δ α k . {:[(Q^('))^(-2)(S^(-1))^(')S=beta(Q^('))^(-3)E^(-beta)],[=betaE^(3alpha+2beta)(I-(beta )/(alpha)Delta_(alpha))^(-3)],[=E^(3alpha+2beta)sum_(k >= 0)((k+1)(k+2))/(2)beta^(k+1)((1)/(alpha)Delta_(alpha))^(k).]:}\begin{aligned} \left(Q^{\prime}\right)^{-2}\left(S^{-1}\right)^{\prime} S & =\beta\left(Q^{\prime}\right)^{-3} E^{-\beta} \\ & =\beta E^{3 \alpha+2 \beta}\left(I-\frac{\beta}{\alpha} \Delta_{\alpha}\right)^{-3} \\ & =E^{3 \alpha+2 \beta} \sum_{k \geq 0} \frac{(k+1)(k+2)}{2} \beta^{k+1}\left(\frac{1}{\alpha} \Delta_{\alpha}\right)^{k} . \end{aligned}(Q)2(S1)S=β(Q)3Eβ=βE3α+2β(IβαΔα)3=E3α+2βk0(k+1)(k+2)2βk+1(1αΔα)k.
From (9) and the previous relation we get
c n = n 1 2 n ( ( 1 + n α + n β ) k = 0 n 3 ( n 2 k ) ( k + 2 ) ! β k + 1 ( 1 + n β ) [ k + 3 , α ] + n ! β n 1 ( 1 + n β ) [ n , α ] ) c n = n 1 2 n ( 1 + n α + n β ) k = 0 n 3 ( n 2 k ) ( k + 2 ) ! β k + 1 ( 1 + n β ) [ k + 3 , α ] + n ! β n 1 ( 1 + n β ) [ n , α ] c_(n)=(n-1)/(2n)((1+n alpha+n beta)sum_(k=0)^(n-3)((n-2)/(k))((k+2)!beta^(k+1))/((1+n beta)^([k+3,-alpha]))+(n!beta^(n-1))/((1+n beta)^([n,-alpha])))c_{n}=\frac{n-1}{2 n}\left((1+n \alpha+n \beta) \sum_{k=0}^{n-3}\binom{n-2}{k} \frac{(k+2)!\beta^{k+1}}{(1+n \beta)^{[k+3,-\alpha]}}+\frac{n!\beta^{n-1}}{(1+n \beta)^{[n,-\alpha]}}\right)cn=n12n((1+nα+nβ)k=0n3(n2k)(k+2)!βk+1(1+nβ)[k+3,α]+n!βn1(1+nβ)[n,α])

5. EVALUATION OF THE ORDERS OF APPROXIMATION

Now we establish some estimates of the order of approximation of a function f C [ 0 , 1 ] f C [ 0 , 1 ] f in C[0,1]f \in C[0,1]fC[0,1] by means of the operator L n Q , S L n Q , S L_(n)^(Q,S)L_{n}^{Q, S}LnQ,S, defined by (2).
According to a result of O. Shisha and B. Mond [13], we can write
| f ( x ) ( L n Q , S f ) ( x ) | [ 1 + 1 δ 2 L n Q , S ( ( t x ) 2 ; x ) ] ω 1 ( f ; δ ) , δ R + f ( x ) L n Q , S f ( x ) 1 + 1 δ 2 L n Q , S ( t x ) 2 ; x ω 1 ( f ; δ ) , δ R + |f(x)-(L_(n)^(Q,S)f)(x)| <= [1+(1)/(delta^(2))L_(n)^(Q,S)((t-x)^(2);x)]omega_(1)(f;delta),quad delta inR^(+)\left|f(x)-\left(L_{n}^{Q, S} f\right)(x)\right| \leq\left[1+\frac{1}{\delta^{2}} L_{n}^{Q, S}\left((t-x)^{2} ; x\right)\right] \omega_{1}(f ; \delta), \quad \delta \in \mathbb{R}^{+}|f(x)(LnQ,Sf)(x)|[1+1δ2LnQ,S((tx)2;x)]ω1(f;δ),δR+
Using the relations (8) we have
L n Q , S ( ( t x ) 2 ; x ) = x 2 ( b n 2 a n + 1 ) + x ( a n b n c n ) L n Q , S ( t x ) 2 ; x = x 2 b n 2 a n + 1 + x a n b n c n L_(n)^(Q,S)((t-x)^(2);x)=x^(2)(b_(n)-2a_(n)+1)+x(a_(n)-b_(n)-c_(n))L_{n}^{Q, S}\left((t-x)^{2} ; x\right)=x^{2}\left(b_{n}-2 a_{n}+1\right)+x\left(a_{n}-b_{n}-c_{n}\right)LnQ,S((tx)2;x)=x2(bn2an+1)+x(anbncn)
so we get
| f ( x ) ( L n Q , S f ) ( x ) | [ 1 + 1 δ 2 [ x 2 ( b n 2 a n + 1 ) + x ( a n b n c n ) ] ] ω 1 ( f ; δ ) f ( x ) L n Q , S f ( x ) 1 + 1 δ 2 x 2 b n 2 a n + 1 + x a n b n c n ω 1 ( f ; δ ) |f(x)-(L_(n)^(Q,S)f)(x)| <= [1+(1)/(delta^(2))[x^(2)(b_(n)-2a_(n)+1)+x(a_(n)-b_(n)-c_(n))]]omega_(1)(f;delta)\left|f(x)-\left(L_{n}^{Q, S} f\right)(x)\right| \leq\left[1+\frac{1}{\delta^{2}}\left[x^{2}\left(b_{n}-2 a_{n}+1\right)+x\left(a_{n}-b_{n}-c_{n}\right)\right]\right] \omega_{1}(f ; \delta)|f(x)(LnQ,Sf)(x)|[1+1δ2[x2(bn2an+1)+x(anbncn)]]ω1(f;δ).
One observes that if b n 2 a n + 1 < 0 b n 2 a n + 1 < 0 b_(n)-2a_(n)+1 < 0b_{n}-2 a_{n}+1<0bn2an+1<0 then x 2 ( b n 2 a n + 1 ) + x ( a n b n c n ) ( a n b n c n ) 2 4 ( 2 a n b n 1 ) , x [ 0 , 1 ] x 2 b n 2 a n + 1 + x a n b n c n a n b n c n 2 4 2 a n b n 1 , x [ 0 , 1 ] x^(2)(b_(n)-2a_(n)+1)+x(a_(n)-b_(n)-c_(n)) <= ((a_(n)-b_(n)-c_(n))^(2))/(4(2a_(n)-b_(n)-1)),AA x in[0,1]x^{2}\left(b_{n}-2 a_{n}+1\right)+x\left(a_{n}-b_{n}-c_{n}\right) \leq \frac{\left(a_{n}-b_{n}-c_{n}\right)^{2}}{4\left(2 a_{n}-b_{n}-1\right)}, \forall x \in[0,1]x2(bn2an+1)+x(anbncn)(anbncn)24(2anbn1),x[0,1].
By choosing δ = 1 n δ = 1 n delta=(1)/(sqrtn)\delta=\frac{1}{\sqrt{n}}δ=1n we can state
Theorem 5. If f C [ 0 , 1 ] f C [ 0 , 1 ] f in C[0,1]f \in C[0,1]fC[0,1] and k N k N EE k inN\exists k \in \mathbb{N}kN such as b n 2 a n + 1 < 0 , n k b n 2 a n + 1 < 0 , n k b_(n)-2a_(n)+1 < 0,AA n >= kb_{n}-2 a_{n}+1<0, \forall n \geq kbn2an+1<0,nk, then we can give the following estimation of the order of approximation, by means of the first modulus of continuity
f L n Q , S f ( 1 + n 4 ( a n b n c n ) 2 ( 2 a n b n 1 ) ) ω 1 ( f ; 1 n ) , n k f L n Q , S f 1 + n 4 a n b n c n 2 2 a n b n 1 ω 1 f ; 1 n , n k ||f-L_(n)^(Q,S)f|| <= (1+(n)/(4)((a_(n)-b_(n)-c_(n))^(2))/((2a_(n)-b_(n)-1)))omega_(1)(f;(1)/(sqrtn)),quad n >= k\left\|f-L_{n}^{Q, S} f\right\| \leq\left(1+\frac{n}{4} \frac{\left(a_{n}-b_{n}-c_{n}\right)^{2}}{\left(2 a_{n}-b_{n}-1\right)}\right) \omega_{1}\left(f ; \frac{1}{\sqrt{n}}\right), \quad n \geq kfLnQ,Sf(1+n4(anbncn)2(2anbn1))ω1(f;1n),nk
where a n , b n , c n a n , b n , c n a_(n),b_(n),c_(n)a_{n}, b_{n}, c_{n}an,bn,cn are defined by (9).
In the case of binomial operators of positive type defined by (1), since S = I = O S = I = O S^(')=I^(')=OS^{\prime}=I^{\prime}=OS=I=O we have
(19) a n = 1 , c n = 0 , b n = n 1 n [ ( Q ) 2 p n 2 ] ( 1 ) p n ( 1 ) (19) a n = 1 , c n = 0 , b n = n 1 n Q 2 p n 2 ( 1 ) p n ( 1 ) {:(19)a_(n)=1","quadc_(n)=0","quadb_(n)=(n-1)/(n)([(Q^('))^(-2)p_(n-2)](1))/(p_(n)(1)):}\begin{equation*} a_{n}=1, \quad c_{n}=0, \quad b_{n}=\frac{n-1}{n} \frac{\left[\left(Q^{\prime}\right)^{-2} p_{n-2}\right](1)}{p_{n}(1)} \tag{19} \end{equation*}(19)an=1,cn=0,bn=n1n[(Q)2pn2](1)pn(1)
Then b n 2 a n + 1 = b n 1 < 0 , n N b n 2 a n + 1 = b n 1 < 0 , n N b_(n)-2a_(n)+1=b_(n)-1 < 0,AA n inNb_{n}-2 a_{n}+1=b_{n}-1<0, \forall n \in \mathbb{N}bn2an+1=bn1<0,nN therefore the previous inequality reduces to
f L n f ( 5 4 + n 4 d n ) ω 1 ( f ; 1 n ) f L n f 5 4 + n 4 d n ω 1 f ; 1 n ||f-L_(n)f|| <= ((5)/(4)+(n)/(4)d_(n))omega_(1)(f;(1)/(sqrtn))\left\|f-L_{n} f\right\| \leq\left(\frac{5}{4}+\frac{n}{4} d_{n}\right) \omega_{1}\left(f ; \frac{1}{\sqrt{n}}\right)fLnf(54+n4dn)ω1(f;1n)
where
(20) d n = n 1 n b n = n 1 n ( 1 [ ( Q ) 2 p n 2 ] ( 1 ) p n ( 1 ) ) (20) d n = n 1 n b n = n 1 n 1 Q 2 p n 2 ( 1 ) p n ( 1 ) {:(20)d_(n)=(n-1)/(n)-b_(n)=(n-1)/(n)(1-([(Q^('))^(-2)p_(n-2)](1))/(p_(n)(1))):}\begin{equation*} d_{n}=\frac{n-1}{n}-b_{n}=\frac{n-1}{n}\left(1-\frac{\left[\left(Q^{\prime}\right)^{-2} p_{n-2}\right](1)}{p_{n}(1)}\right) \tag{20} \end{equation*}(20)dn=n1nbn=n1n(1[(Q)2pn2](1)pn(1))
We mention that this inequality was established by D. D. Stancu in [18].
In order to find an evaluation of the order of approximation using both moduli of smoothness ω 1 ω 1 omega_(1)\omega_{1}ω1 and ω 2 ω 2 omega_(2)\omega_{2}ω2 we can use a result of H. H. Gonska and R. K. Kovacheva included in the following
Lemma 3. [2]. If I = [ a , b ] I = [ a , b ] I=[a,b]I=[a, b]I=[a,b] is a compact interval of the real axis and I 1 = [ a 1 , b 1 ] I 1 = a 1 , b 1 I_(1)=[a_(1),b_(1)]I_{1}= \left[a_{1}, b_{1}\right]I1=[a1,b1] is a subinterval of it, and if we assume that L : C ( I ) C ( I 1 ) L : C ( I ) C I 1 L:C(I)rarr C(I_(1))L: C(I) \rightarrow C\left(I_{1}\right)L:C(I)C(I1) is a positive operator, such that L e 0 = e 0 L e 0 = e 0 Le_(0)=e_(0)L e_{0}=e_{0}Le0=e0 and 0 δ 1 2 ( b a ) 0 δ 1 2 ( b a ) 0 <= delta <= (1)/(2)(b-a)0 \leq \delta \leq \frac{1}{2}(b-a)0δ12(ba), then we have
| f ( x ) L ( f ( t ) ; x ) | 2 δ | L ( t x ; x ) | ω 1 ( f ; δ ) + + 3 2 [ 1 + 1 δ | L ( t x ; x ) | + 1 2 δ 2 L ( ( t x ) 2 ; x ) ] ω 2 ( f ; δ ) | f ( x ) L ( f ( t ) ; x ) | 2 δ | L ( t x ; x ) | ω 1 ( f ; δ ) + + 3 2 1 + 1 δ | L ( t x ; x ) | + 1 2 δ 2 L ( t x ) 2 ; x ω 2 ( f ; δ ) {:[|f(x)-L(f(t);x)| <= (2)/(delta)|L(t-x;x)|omega_(1)(f;delta)+],[+(3)/(2)[1+(1)/(delta)|L(t-x;x)|+(1)/(2delta^(2))L((t-x)^(2);x)]omega_(2)(f;delta)]:}\begin{aligned} |f(x)-L(f(t) ; x)| \leq & \frac{2}{\delta}|L(t-x ; x)| \omega_{1}(f ; \delta)+ \\ & +\frac{3}{2}\left[1+\frac{1}{\delta}|L(t-x ; x)|+\frac{1}{2 \delta^{2}} L\left((t-x)^{2} ; x\right)\right] \omega_{2}(f ; \delta) \end{aligned}|f(x)L(f(t);x)|2δ|L(tx;x)|ω1(f;δ)++32[1+1δ|L(tx;x)|+12δ2L((tx)2;x)]ω2(f;δ)
Using the relations (8) we obtain the inequality
| f ( x ) ( L n Q , S f ) ( x ) | 2 δ | ( a n 1 ) x | ω 1 ( f ; δ ) + + 3 2 [ 1 + 1 δ | ( a n 1 ) x | + 1 2 δ 2 [ x 2 ( b n 2 a n + 1 ) + x ( a n b n c n ) ] ] ω 2 ( f ; δ ) f ( x ) L n Q , S f ( x ) 2 δ a n 1 x ω 1 ( f ; δ ) + + 3 2 1 + 1 δ a n 1 x + 1 2 δ 2 x 2 b n 2 a n + 1 + x a n b n c n ω 2 ( f ; δ ) {:[|f(x)-(L_(n)^(Q,S)f)(x)| <= (2)/(delta)|(a_(n)-1)x|omega_(1)(f;delta)+],[+(3)/(2)[1+(1)/(delta)|(a_(n)-1)x|+(1)/(2delta^(2))[x^(2)(b_(n)-2a_(n)+1)+x(a_(n)-b_(n)-c_(n))]]omega_(2)(f;delta)]:}\begin{aligned} & \left|f(x)-\left(L_{n}^{Q, S} f\right)(x)\right| \leq \frac{2}{\delta}\left|\left(a_{n}-1\right) x\right| \omega_{1}(f ; \delta)+ \\ & +\frac{3}{2}\left[1+\frac{1}{\delta}\left|\left(a_{n}-1\right) x\right|+\frac{1}{2 \delta^{2}}\left[x^{2}\left(b_{n}-2 a_{n}+1\right)+x\left(a_{n}-b_{n}-c_{n}\right)\right]\right] \omega_{2}(f ; \delta) \end{aligned}|f(x)(LnQ,Sf)(x)|2δ|(an1)x|ω1(f;δ)++32[1+1δ|(an1)x|+12δ2[x2(bn2an+1)+x(anbncn)]]ω2(f;δ)
If b n 2 a n + 1 < 0 b n 2 a n + 1 < 0 b_(n)-2a_(n)+1 < 0b_{n}-2 a_{n}+1<0bn2an+1<0 the previous inequality implies
f L n Q , S f 2 δ ( 1 a n ) ω 1 ( f ; δ ) + 3 2 [ 1 + 1 δ ( 1 a n ) + 1 8 δ 2 ( a n b n c n ) 2 ( 2 a n b n 1 ) ] ω 2 ( f ; δ ) . f L n Q , S f 2 δ 1 a n ω 1 ( f ; δ ) + 3 2 1 + 1 δ 1 a n + 1 8 δ 2 a n b n c n 2 2 a n b n 1 ω 2 ( f ; δ ) . ||f-L_(n)^(Q,S)f|| <= (2)/(delta)(1-a_(n))omega_(1)(f;delta)+(3)/(2)[1+(1)/(delta)(1-a_(n))+(1)/(8delta^(2))((a_(n)-b_(n)-c_(n))^(2))/((2a_(n)-b_(n)-1))]omega_(2)(f;delta).\left\|f-L_{n}^{Q, S} f\right\| \leq \frac{2}{\delta}\left(1-a_{n}\right) \omega_{1}(f ; \delta)+\frac{3}{2}\left[1+\frac{1}{\delta}\left(1-a_{n}\right)+\frac{1}{8 \delta^{2}} \frac{\left(a_{n}-b_{n}-c_{n}\right)^{2}}{\left(2 a_{n}-b_{n}-1\right)}\right] \omega_{2}(f ; \delta) .fLnQ,Sf2δ(1an)ω1(f;δ)+32[1+1δ(1an)+18δ2(anbncn)2(2anbn1)]ω2(f;δ).
By choosing δ = 1 n δ = 1 n delta=(1)/(sqrtn)\delta=\frac{1}{\sqrt{n}}δ=1n we get
f L n Q , S f 2 n ( 1 a n ) ω 1 ( f ; 1 n ) + + 3 2 [ 1 + n ( 1 a n ) + n 8 ( a n b n c n ) 2 ( 2 a n b n 1 ) ] ω 2 ( f ; 1 n ) . f L n Q , S f 2 n 1 a n ω 1 f ; 1 n + + 3 2 1 + n 1 a n + n 8 a n b n c n 2 2 a n b n 1 ω 2 f ; 1 n . {:[||f-L_(n)^(Q,S)f|| <= 2sqrtn(1-a_(n))omega_(1)(f;(1)/(sqrtn))+],[+(3)/(2)[1+sqrtn(1-a_(n))+(n)/(8)((a_(n)-b_(n)-c_(n))^(2))/((2a_(n)-b_(n)-1))]omega_(2)(f;(1)/(sqrtn)).]:}\begin{aligned} \left\|f-L_{n}^{Q, S} f\right\| \leq & 2 \sqrt{n}\left(1-a_{n}\right) \omega_{1}\left(f ; \frac{1}{\sqrt{n}}\right)+ \\ & +\frac{3}{2}\left[1+\sqrt{n}\left(1-a_{n}\right)+\frac{n}{8} \frac{\left(a_{n}-b_{n}-c_{n}\right)^{2}}{\left(2 a_{n}-b_{n}-1\right)}\right] \omega_{2}\left(f ; \frac{1}{\sqrt{n}}\right) . \end{aligned}fLnQ,Sf2n(1an)ω1(f;1n)++32[1+n(1an)+n8(anbncn)2(2anbn1)]ω2(f;1n).
If we consider the binomial operator introduced by Tiberiu Popoviciu, using (19) and the previous relation, we arrive at an inequality which has found by D. D. Stancu (see [18])
f L n f 3 16 ( 9 + n d n ) ω 2 ( f ; 1 n ) , f L n f 3 16 9 + n d n ω 2 f ; 1 n , ||f-L_(n)f|| <= (3)/(16)(9+nd_(n))omega_(2)(f;(1)/(sqrtn)),\left\|f-L_{n} f\right\| \leq \frac{3}{16}\left(9+n d_{n}\right) \omega_{2}\left(f ; \frac{1}{\sqrt{n}}\right),fLnf316(9+ndn)ω2(f;1n),
where d n d n d_(n)d_{n}dn is defined by (20).

REFERENCES

[1] Cheney, E. W. and Sharma, A., On a generalization of Bernstein polynomials, Riv. Mat. Univ. Parma, 5, pp. 77-82, 1964.
[2] Gonska, H. H. and Kovacheva, R. K., The second order modulus revisited: remarks, applications, problems, Conferenze del Seminario di Matematica Univ. Bari, 257, pp. 132, 1994.
[3] Lupaş, L. and Lupaş, A., Polynomials of binomial type and approximation operators, Studia Univ. Babes-Bolyai, Mathematica, 32, pp. 61-69, 1987.
[4] Lupaş, A., Approximation operators of binomial type, Proc. IDoMAT 98, International Series of Numerical Mathematics, ISNM 132, Birkhäuser Verlag, Basel, pp. 175-198, 1999.
[5] Manole, C., Developments in series of generalized Appell polynomials, with applications to the approximation of functions, Ph.D. Thesis, Cluj-Napoca, Romania, 1984 (in Romanian).
[6] Moldovan, G., Generalizations of the S. N. Bernstein operators, Ph.D. Thesis, ClujNapoca, Romania, 1971 (in Romanian).
[7] Moldovan, G., Discrete convolutions and positive operators I, Annales Univ. Sci. Budapest R. Eötvös, 15, pp. 31-34, 1972.
[8] Mullin, R. and Rota, G. C., On the foundations of combinatorial theory III, theory of binomial enumeration, in: B. Harris, ed., Graph Theory and Its Applications, Academic Press, New York, pp. 167-213, 1970.
[9] Popoviciu, T., Remarques sur les poynômes binomiaux, Bul. Soc. Ştiinte Cluj, 6, pp 146-148, 1931.
[10] Roman, S., Operational formulas, Linear and Multilinear Algebra, 12, pp. 1-20, 1982.
[11] Rota, G. C., Kahaner, D. and Odlyzko, A., Finite operator calculus, J. Math. Anal. Appl., 42, pp. 685-760, 1973.
[12] Sablonnière, P., Positive Bernstein-Sheffer operators, J. Approx. Theory, 83, pp. 330-341, 1995.
[13] Shisha, O. and Mond, B., The degree of convergence of linear positive operators, Proc. Nat. Acad. Sci. U.S.A., 60, pp. 1196-1200, 1968.
[14] Stancu, D. D., Approximation of functions by a new class of linear positive operators, Rev. Roum. Math. Pures Appl., 13, pp. 1173-1194, 1968.
[15] Stancu, D. D., On a generalization of the Bernstein polynomials, Studia Univ. BabeşBolyai, Cluj, 14, pp. 31-45, 1969.
[16] Stancu, D. D., Approximation properties of a class of linear positive operators, Studia Univ. Babeş-Bolyai, Cluj, 15, pp. 31-38, 1970.
[17] Stancu, D. D., Approximation of functions by means of some new classes of positive linear operators, Numerische Methoden der Approximationstheorie, Proc. Conf. Oberwolfach 1971 ISNM 16, Birkhäuser-Verlag, Basel, pp. 187-203, 1972.
[18] Stancu, D. D. and Occorsio, M.R., On approximation by binomial operators of Tiberiu Popoviciu type, Rev. Anal. Numér. Théor. Approx., 27 no.1, pp. 167-181, 1998. 줄
[19] Stancu, D. D. and Cismaşiu, C., On an approximating linear positive operator of Cheney-Sharma, Rev. Anal. Numér. Théor. Approx., 26, pp. 221-227, 1997.«
Received September 12, 2000.

  1. "T. Popoviciu" Institute of Numerical Analysis, P.O. Box 68-1, 3400 Cluj-Napoca, Romania, e-mail: craciun@ictp-acad.math.ubbcluj.ro.