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volume 5, issue 4, article 113, 2004.

Received 20 August, 2004;

accepted 01 September, 2004.

Communicated by:A. Lupa¸s

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Journal of Inequalities in Pure and Applied Mathematics

SIMULTANEOUS APPROXIMATION BY LUPA ¸S MODIFIED OPERATORS WITH WEIGHTED FUNCTION OF

SZASZ OPERATORS

NAOKANT DEO

Department of Applied Mathematics Delhi College of Engineering Bawana Road, Delhi - 110042, India.

EMail:dr_naokant_deo@yahoo.com

c

2000Victoria University ISSN (electronic): 1443-5756 151-04

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Simultaneous Approximation by Lupa ¸s Modified Operators with

Weighted Function of Szasz Operators

Naokant Deo

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J. Ineq. Pure and Appl. Math. 5(4) Art. 113, 2004

http://jipam.vu.edu.au

Abstract

In the present paper, we consider a new modification of the Lupa¸s operators with the weight function of Szasz operators and study simultaneous approximation. Here we obtain a Voronovskaja type asymptotic formula and an estimate of error in simultaneous approximation for these Lupa¸s-Szasz operators.

2000 Mathematics Subject Classification:41A28, 41A36.

Key words: Simultaneous approximation, Lupa¸s operators, Szasz operators.

1. Introduction

Lupa¸s proposed a family of linear positive operators mapping C[0,∞) into C[0,∞), the class of all bounded and continuous functions on[0,∞),namely,

(L_nf) (x) =

∞

X

k=0

n+k−1 k

x^k (1 +x)^n+kf

k n

,

wherex∈[0,∞).

Motivated by the integral modification of Bernstein polynomials by Derrien- nic [1], Sahai and Prasad [3] modified the operatorsLnfor functions integrable onC[0,∞)as

(Mnf) (x) = (n−1)

∞

X

k=0

Pn,k(x) Z ∞

0

Pn,k(y)f(y)dy, where

P_n,k(t) =

n+k−1 k

t^k (1 +t)^n+k.

Integral modification of Szasz-Mirakyan operators were studied by Gupta [2].

Now we consider another modification of Lupa¸s operators with the weight function of Szasz operators, which are defined as

(1.1) (Bnf) (x) =n

∞

X

k=0

Pn,k(x) Z ∞

0

Sn,k(y)f(y)dy

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where

Pn,k(x) =

n+k−1 k

x^k(1 +x)^−n−k and

Sn,k(y) = e^−ny(ny)^k k! .

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2. Basic Results

The following lemmas are useful for proving the main results.

Lemma 2.1. Letm ∈N⁰, n∈N, if we define

T_n,m(x) =n

∞

X

k=0

P_n+r,k(x) Z ∞

0

S_n,k+r(y)(y−x)^mdy

then

(i) T_n,0(x) = 1, T_n,1(x) = ^1+r(1+x)_n ,and

(2.1) Tn,2(x) = rx(1 +x) + 1 + [1 +r(1 +x)]²+nx(2 +x) n²

(ii) For allx≥0,

Tn,m(x) =O 1

n[^m+12 ]

.

(iii)

nT_n,m+1(x) =x(1+x)T_n,m+1⁽¹⁾ (x)+[m+1+r(1+x)]T_n,m(x)+mxT_n,m−1(x) wherem ≥2.

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Proof. The value ofT_n,0(x), T_n,1(x)easily follows from the definition, we give the proof of (iii) as follows.

x(x+ 1)T_n,m⁽¹⁾(x)

=n

∞

X

k=0

x(1 +x)P_n+r,k⁽¹⁾ (x) Z ∞

0

S_n,k+r(y)(y−x)^mdy

−mn

∞

X

k=0

x(1 +x)Pn+r,k(x) Z ∞

0

Sn,k+r(y)(y−x)^m−1dy.

Now using the identities

yS_n,k⁽¹⁾(y) = (k−ny)S_n,k(y), andx(1 +x)P_n,k⁽¹⁾(x) = (k−nx)P_n,k(x),we get

x(1 +x)T_n,m⁽¹⁾(x)

=n

∞

X

k=0

[k−(n+r)x]Pn+r,k(x) Z ∞

0

Sn,k+r(y)(y−x)^mdy

−mx(1 +x)Tn,m−1(x).

Therefore,

x(1 +x)[T_n,m⁽¹⁾(x) +mTn,m−1(x)]

=n

∞

X

k=0

P_n+r,k(x) Z ∞

0

[(k+r−ny) +n(y−x)−r(1 +x)]S_n,k+r(y)(y−x)^mdy

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=n

∞

X

k=0

P_n+r,k(x) Z ∞

0

yS_n,k+⁽¹⁾ _r(y)(y−x)^mdy+nT_n,m+1(x)−r(1 +x)T_n,m(x)

=n

∞

X

k=0

Pn+r,k(x) Z ∞

0

yS_n,k+r⁽¹⁾ (y)(y−x)^m+1dy

+nx

∞

X

k=0

P_n+r,k(x) Z ∞

0

S_n,k+r⁽¹⁾ (y)(y−x)^mdy +nT_n,m+1(x)−r(1 +x)T_n,m(x)

=−(m+ 1)T_n,m(x)−mxTn,m−1(x) +nT_n,m+1(x)−r(1 +x)T_n,m(x) This leads to proof of (iii).

Corollary 2.2. Let α and δ be positive numbers, then for every m ∈ N and x ∈ [0,∞), there exists a positive constant C_m,x depending on m andx such that

n

∞

X

k=0

P_n,k(x) Z

|t−x|≥δ

S_n,k(t)e^αtdt ≤C_m,xn^−m.

Lemma 2.3. Iff is differentiablertimes(r = 1,2,3, . . .)on[0,∞), then we have

(2.2) (B_n^(r)f)(x) = (n+r−1)!

n^r−1(n−1)!

∞

X

k=0

P_n+r,k(x) Z ∞

0

S_n,k+r(y)f^(r)(y)dy.

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Proof. Applying Leibniz’s theorem in (1.1) we have (B^(r)_n f)(x)

=n

r

X

i=0

∞

X

k=i

r i

(n+k+r−i−1)!

(n−1)!k! (−1)^r−ix^k−i(1 +x)^{−n−k−r+i}

× Z ∞

0

S_n,k(y)f(y)dy

=n

∞

X

k=0

(n+k+r−1)!

(n−1)!k! · x^k (1 +x)^n+k+r

Z ∞ 0

r

X

i=0

(−1)^r−ir i

S_n,k+i(y)f(y)dy

=n(n+r−1)!

(n−1)!

∞

X

k=0

P_n+r,k(x) Z ∞

0 r

X

i=0

(−1)^r−ir i

S_n,k+i(y)f(y)dy.

Again using Leibniz’s theorem, S_n,k+r^(r) (y) =

r

X

i=0

r i

(−1)ⁱn^re^−ny(ny)^k+i (k+i)!

=n^r

r

X

i=0

(−1)ⁱr i

S_n,k+i(y).

Hence

(B_n^(r)f)(x) = (n+r−1)!

n^r−1(n−1)!

∞

X

k=0

P_n+r,k(x) Z ∞

0

S^(r)

n,k+r(y)(−1)^rf(y)dy and integrating by partsrtimes, we get the required result.

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3. Main Results

Theorem 3.1. Let f be integrable in [0,∞), admitting a derivative of order (r+ 2) at a pointx∈[0,∞). Also supposef^(r)(x) =o(e^αx)asx→ ∞, then

n→∞lim n[(B_n^(r)f)(x)−f^(r)(x)] = [1 +r(1 +x)]f^(r+1)(x) +x(2 +x)f^(r+2)(x).

Proof. By Taylor’s formula, we get

(3.1) f^(r)(y)−f^(r)(x)

= (y−x)f^(r+1)(x) + (y−x)²

2 f^(r+2)(x) + (y−x)²

2 η(y, x), where

η(y, x) = f^(r)(y)−f^(r)(x)−(y−x)f^(r+1)(x)− ^(y−x)₂ ²f^(r+2)(x)

(y−x)² 2

if x6=y

= 0 if x=y.

Now, for arbitraryε >0,A>0∃ aδ >0s. t.

(3.2) |η(y, x)| ≤ε for |y−x|< δ, x≤A.

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Using (2.2) in (3.1) n^r(n−1)!

(n+r−1)!(B_n^(r)f)(x)−f^(r)(x)

=n

∞

X

k=0

P_n+r,k(x) Z ∞

0

S_n,k+r(y)f^(r)(y)dy−f^(r)(x)

=n

∞

X

k=0

P_n+r,k(x) Z ∞

0

S_n,k+r(y)

f^(r)(y)−f^(r)(x) dy

=Tn,1f^(r+1)(x) +Tn,2f^(r+2)(x) + En,r(x), where

E_n,r(x) = n 2

∞

X

k=0

P_n+r,k(x) Z ∞

0

S_n,k+r(y)(y−x)²η(y, x)dy.

In order to completely prove the theorem it is sufficient to show that nE_n,r(x)→0 as n→ ∞.

Now

nE_n,r(x) =R_n,r,1(x) +R_n,r,2(x), where

R_n,r,1(x) = n² 2

∞

X

k=0

P_n+r,k(x) Z

|y−x|<δ

S_n,k+r(y)(y−x)²η(x, y)dy

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and

R_n,r,2(x) = n² 2

∞

X

k=0

P_n+r,k(x) Z

|y−x|>δ

S_n,k+r(y)(y−x)²η(y, x)dy By (3.2) and (2.1)

|Rn,r,1(x)|< nε 2





n

∞

X

k=0

Pn+r,k(x) Z

|y−x|≤δ

Sn,k+r(y)(y−x)²dy



 (3.3) 

≤εx(2 +x) asn→ ∞.

Finally we estimateR_n,r,2(x). Using Corollary2.2we have R_n,r,2(x) = n²

2

∞

X

k=0

P_n+r,k(x) Z

|y−x|>δ

S_n,k+r(y)e^αydy (3.4)

= n

2M_m,xn^−m = 0 asn→ ∞.

Theorem 3.2. Letf ∈C^(r+1)[0, a]and letw(f^(r+1);·)be the modulus of conti- nuity off^(r+1), thenr= 0,1,2, . . .

(B_n^(r)f)(x)−f^(r)(x)

≤ [1 +r(1 +a)]

n

f^(r+1)(x)

+ 1 n²

q

Tn,2(a) + T_n,2(a) 2

w f^(r+1);n⁻²

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wherek·kis the sup norm[0, a].

Proof. We have by Taylor’s expansion f^(r)(y)−f^(r)(x)

= (y−x)f^(r+1)(x) + Z y

x

[f^(r+1)(t)−f^(r+1)(x)]dt

× n^r(n−1)!

(n+r−1)!(B_n^(r)f)(x)−f^(r)(x)

=n

∞

X

k=0

P_n+r,k(x) Z ∞

0

S_n,k+r(y)

f^(r)(y)−f^(r)(x) dy

=n

∞

X

k=0

P_n+r,k(x) Z ∞

0

S_n,k+r(y)

(y−x)f^(r+1)(x)

+ Z y

x

[f^(r+1)(t)−f^(r+1)(x)]dt

dy.

Also

f^(r+1)(t)−f^(r+1)(x) ≤

1 + |t−x|

δ

w(f^(r+1);δ) Hence

n^r(n−1)!

(n+r−1)!(B_n^(r)f)(x)−f^(r)(x)

≤ |Tn,1| ·

f^(r+1)(x) +

pTn,2

+ |T_n,2| 2δ

·w(f^(r+1);δ).

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By Schwarz’s inequality. Choosingδ = _n¹2 and using (i) and (2.1) we obtain the required result.

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References

[1] M.M. DERRIENNIC, Sur l’approximation de fonctions integrables sur [0,1] par des polynomes de Bernstein modifies, J. Approx, Theory, 31 (1981), 325–343.

[2] V. GUPTA, A note on modified Szasz operators, Bull. Inst. Math. Academia Sinica, 21(3) (1993), 275–278.

[3] A. SAHAIANDG. PRASAD, On simultaneous approximation by modified Lupa¸s operators, J. Approx., Theory, 45 (1985), 122–128.