General Mathematics Vol. 18, No. 2 (2010), 185–199
On the degree of approximation by new Durrmeyer
type operators 1
Naokant Deo, Suresh P. Singh
Abstract
In this paper, we define a new kind of positive linear operators and
study basic properties as well as Voronovskaya type results. In the last
section of this paper we establish the error estimation for simultaneous
approximation in terms of higher order modulus of continuity by using
the technique of linear approximating method viz Steklov mean.
2000 Mathematics Subject Classification: 41A30, 41A36.
Key words and phrases: Positive linear operators, Voronovskaya type
results.
1
Introduction
In the year 1957, Baskakov [1] introduced the following operators
µ ¶
∞
X
k
bn,k (x)f
(1)
Bn (f, x) =
,
n
k=0
1
Received 13 September, 2009
Accepted for publication (in revised form) 24 November, 2009
185
186
N. Deo, S. P. Singh
where
µ
bn,k (x) =
n+k−1
k
¶
xk
,
(1 + x)n+k
x ∈ [0, ∞).
Important modifications had been studied by Sahai & Prasad [14] and
Heilmann [9] on Baskakov operators after these milestone modifications, various researchers have given different type modification of Baskakov operators
and studied several good results. Now we are giving another modification of
Baskakov operators:
(2)
Vn (f, x) =
∞
X
µ
pn,k (x)f
k=0
where
µ
pn,k (x) =
n
n+1
¶n+1


¶
k
,
n+1

n+k
k
xk
³
1−
1
n+1
+x
´n+k+1
and Durrmeyer variants of these operators are:
Z ∞
∞
X
pn,k (x)
pn,k (t)f (t)dt.
(3)
Dn (f, x) = (n + 1)
0
k=0
Let Cγ [0, ∞) = {f ∈ C [0, ∞) : |f (x)| ≤ M tγ , for some γ > 0} we define
the norm k.k on Cγ [0, ∞) by
kf kγ = sup |f (t)| t−γ
0≤t<∞
We note that the order of approximation by these operators (3) is at best
¡
¢
O n−1 , howsoever smooth the function may be. Thus to improve the order of approximation, we consider May [13] type linear combination of the
operators (3) as described below:
For d0 , d1 , d2 , ..., dk arbitrary but fixed distinct positive integers, the lin¡ ¡
¢ ¢
ear combination Dn f, d0 , d1 , d2 , ..., dk , x of Ddj n (f, x), j = 0, 1, 2, ..., k are
defined as:
Dn (f, (d0 , d1 , d2 , ..., dk ) , x) =
k
X
j=0
C(j, k)Ddj n (f, x),
On the degree of approximation by new Durrmeyer type operators
where
C(j, k) =
k
Y
i=0
i6=j
dj
f or k 6= 0
dj − di
and
187
C(0, 0) = 1.
Very recently Deo et al. [4] have studied new Bernstein type operators
and established a Voronovskaya type asymptotic formula and an estimate of
error in terms of modulus of continuity in simultaneous approximation for the
linear combinations. In [5], Deo and Singh have given some theorems on the
approximation of the r-th derivative of a function f by the same operators.
Deo [3] has studied Voronovskaya type result for Lupaş type operators and
he [2] has also given iterative combinations of Baskakov operator.
In the present paper, we study some ordinary approximation results including Voronovskaya type results. At the end of this paper we obtain an
estimate of error in terms of higher order modulus of continuity in simultaneous approximation for the linear combination of the operators (3).
2
Properties and Basic Results
In this section we write some basic results to prove our theorem.
Lemma 1 For n ≥ 1 one obtains,
Vn (1, x) = 1
µ
¶
1
Vn (t, x) = 1 +
x
n
µ
¶
¡2 ¢
3
2
x
Vn t , x = 1 + + 2 x2 +
n n
n
Lemma 2 For m ∈ N 0 (the set of non-negative integers), the m-th order
moment of the operator is defined as
Un,m (x) =
∞
X
k=0
µ
pn,k (x)
¶m
k
−x .
n+1
188
N. Deo, S. P. Singh
Consequently, Un,0 (x) = 1 and Un,1 (x) = x/n. There holds the recurrence
relation
µ
nUn,m+1 (x) = x 1 −
¶h
i
1
0
+ x Un,m (x) + mUn,m−1 (x) + xUn,m (x)
n+1
Proof. It is easily observed that
µ
¶
·
¸
1
nk
(4)
x 1−
+ x p0n,k (x) =
− (n + 1)x pn,k (x).
n+1
n+1
Hence the result. Thus
(i) Un,m (x) is a polynomial in x of degree ≤ m;
³
´
m+1
(ii) For every x ∈ [0, ∞) , Un,m (x) = O n−[ 2 ] , where bαc denotes the
integral part of α.
Lemma 3 Let the m-th order moment be defined by
Z ∞
∞
X
pn,k (x)
pn,k (t)(t − x)m dt
Tn,m (x) = (n + 1)
k=0
0
then
(5)
(6)
Tn,0 (x) = 1, Tn,1 (x) =
n(1 + 2x) + 2x
, n > 1,
n2 − 1
¡
¢
2 n2 + 4x2 + 4nx + 3n2 x + 7nx2 + 2n2 x2 − n3 x − n3 x2
Tn,2 (x) =
(n + 1)(n2 − 1)(n − 2)
and
(7)
µ
¶
1
(n − m − 1)Tn,m+1 (x) = (m + 1) 1 −
+ 2x Tn,m (x)
n+1
µ
¶
£ 0
¤
1
+x 1−
+ x Tn,m
(x) + 2mTn,m−1 (x) .
n+1
Further, for all x ∈ [0, ∞)
(8)
µ h (m+1) i ¶
−
2
Tn,m (x) = O n
.
On the degree of approximation by new Durrmeyer type operators
189
Proof. We can easily obtain (5) and (6) by using the definition of Tn,m (x).
For the proof of (7), we proceed as follows. First
µ
¶
1
0
x 1−
+ x Tn,m
(x)
n+1
µ
¶
Z ∞
∞
X
1
0
=x 1−
+ x (n + 1)
pn,k (x)
pn,k (t)(t − x)m dt
n+1
0
k=0
µ
¶
1
− mx 1 −
+ x Tn,m−1 (x).
n+1
Now, using inequality (4) two times, then we get
µ
¶
£ 0
¤
1
x 1−
+ x Tn,m
(x) + mTn,m−1 (x)
n+1
¸
Z ∞
∞ ·
X
nk
− (n + 1)x pn,k (x)
= (n + 1)
pn,k (t)(t − x)m dt
n+1
0
k=0
·
¸
Z
∞
∞
X
nk
= (n+1)
pn,k (x)
−(n+1)t pn,k (t)(t−x)m dt+(n+1)Tn,m+1 (x)
n+1
0
k=0
¶
Z
∞
∞ µ
X
1
pn,k (x)
t 1−
+t p0n,k (t)(t−x)m dt+(n+1)Tn,m+1 (x)
= (n+1)
n+1
0
k=0
¶
µ
¶¸
Z ∞ ·µ
∞
X
1
1
2
pn,k (x)
1−
= (n+1)
+2x (t−x)+(t−x) +x 1−
+x
n+1
n+1
0
k=0
.p0n,k (t)(t − x)m dt + (n + 1)Tn,m+1 (x)
µ
¶
1
= −(m + 1) 1 −
+ 2x Tn,m (x) + (n − m − 1)Tn,m+1 (x)
n+1
¶
µ
1
+ x Tn,m−1 (x).
− mx 1 −
n+1
This leads to (7). The proof of (8) easily follow from (5) and (7).
Lemma 4 There exists the polynomials qi,j,r (x) independent of n and k such
that
µ
r
x 1−
¶r r
X
1
d
+x
pn,k (x) =
(n+1)i {k − (n + 1)x}j φi,j,r (x)pn,k (x).
r
n+1
dx
2i+j≤r
i,j≥0
190
N. Deo, S. P. Singh
The proof of this lemma proceeds exactly on the lines of that of a results by
Lorentz [12, p. 26].
Lemma 5 Let f be r times differentiable on [0, ∞) such that f (r−1) = O (tα ),
for some α > 0 as t → ∞ then for r = 1, 2, 3, ... and n > α + r, we have
∞
Dn(r) (f, x)
(9)
(n + 1)(n − r)!(n + r)! X
=
pn+r,k (x)
(n!)2
k=0
Z
∞
pn−r,k+r (t)f (t)dt
0
Proof. We have by Leibnitz theorem
Dn(r) (f, x)
µ
= (n+1)
Z
0
=
¶n+1 X
r X
∞
i=0 k=i



r−i
r
 (−1)
i
(n+k+r−i)
³
n!(k−i)!
xk−i
´n+k+r+1−i
1
1− n+1
+x
∞
.
=
n
n+1
pn,k (t)f (t)dt
(n+1)(n+r)!
n!
µ
(n + 1)(n + r)!
n!
n+1
n
µ
¶r X
r
∞ X
k=0 i=0
n+1
n
¶r X
∞



r
i
Z
 (−1)r−i pn+r,k (x)
Z
pn+r,k (x)
k=0
0
r
∞X
∞
pn,k+i (t)f (t)dt
0

(−1)r−i 

r
i
i=0
 pn,k+i (t)f (t)dt
Again applying Leibnitz theorem
(r)
pn−r,k+r (t) =
¶
r µ
X
n+1 r
i=0
n
 
r
n!
(−1)i   pn,k+i (t)
(n − r)!
i
∞
Dn(r) (f, x)
(n + 1)(n − r)!(n + r)! X
pn+r,k (x)
=
(n!)2
k=0
Z
0
∞
(r)
(−1)r pn−r,k+r (t)f (t)dt.
Further integrating by parts r times, we get the required result.
On the degree of approximation by new Durrmeyer type operators
191
Lemma 6 Let f ∈ Cγ [0, ∞) , if f (2k+r+2) exists at a point x ∈ Cγ [0, ∞),
then
h
i 2k+r+2
X
¡ ¡
¢ ¢
lim nk+1 Dn(r) f d0 , d1 , d2 , ..., dk , x − f (r) (x) =
Q(i, k, r, x)f (i) (x),
n→∞
i=r
where Q(i, k, r, x) are certain polynomials in x.
The proof of the above Lemma follows easily along the lines of [8, 11].
3
Voronovskaya Type Results
Theorem 1 If a function f is such that its first and second order derivatives
are bounded in [0, ∞), then
(10)
lim (n + 1) {Dn (f, x) − f (x)} = f 0 (x)(1 + 2x) − x(1 + x)f 00 (x)
n→∞
Proof. Using Taylor’s theorem we write that
(11)
f (t) − f (x) = (t − x)f 0 (x) +
(t − x)2
(t − x)2 00
f (x) +
η(t, x),
2!
2!
where η(t, x) is a bounded function ∀t, x and lim η(t, x) = 0. Now applying (3)
t→x
and (11), we get
Dn (f, x) − f (x) = f 0 (x)Dn (t − x, x) +
¡
¢
f 00 (x)
Dn (t − x)2 , x + I1
2
where
¡
¢
1
I1 = Dn (t − x)2 η(t, x), x .
2
Using (5) and (6), we get
f 00 (x)
Dn (f, x) − f (x) = f 0 (x)Tn,1 (x) +
Tn,2 (x) + I1
2
½
¾
n(1 + 2x) + 2x
= f 0 (x)
n2 − 1
¾
½ 2
n + 4x2 + 4nx + 3n2 x + 7nx2 + 2n2 x2 − n3 x − n3 x2
00
+ I1 ,
+ f (x)
(n + 1)(n2 − 1)(n − 2)
192
N. Deo, S. P. Singh
therefore
½
0
n(1 + 2x) + 2x
n−1
¾
(n + 1) {Dn (f, x) − f (x)} = f (x)
(¡
¢)
2 + nx(4 + 7x) + n2 (1 + 3x + 2x2 ) − n3 x(1 + x)
4x
+ f 00 (x)
+ (n + 1)I1 .
(n2 − 1)(n − 2)
Now, we have to show that as n → ∞, the value of (n + 1)I1 → 0. Let ε > 0
be given since η(t, x) → 0 as t → 0, then there exists δ > 0 such that when
|t − x| < δ we have |η(t, x)| < ε and when |t − x| ≥ δ, we write
|η(t, x)| ≤ M < M
(t − x)2
.
δ2
Thus, for all t, x ∈ [0, ∞)
|η(t, x)| ≤ ε + M
(t − x)2
δ2
µ
µ
¶ ¶
M (t − x)2
2
(n + 1)I1 ≤ (n + 1)Dn (t − x) ε +
,x
δ2
¡
¢
¡
¢ M
≤ ε(n + 1)Dn (t − x)2 , x + 2 (n + 1)Dn (t − x)4 , x
δ
Using (6) and (8), we see that,
(n + 1)I1 → 0 as n → ∞.
This leads to (10).
Corollary 1 We can also get the following result:
(12)
lim (n − 1) {Dn (f, x) − f (x)} = f 0 (x)(1 + 2x) − x(1 + x)f 00 (x)
n→∞
2 [0, ∞) then we have
Theorem 2 If g ∈ CB
(13)
|Dn (g, x) − g(x)| ≤ λn (x) kgkC 2
B
where
λn (x) =
n(1 + 2x) + 2x
n2 − 1
On the degree of approximation by new Durrmeyer type operators
193
Proof. We write that
(14)
1
g(t)g(x) = (t − x)g 0 (x) + (t − x)2 g 00 (ξ)
2
where t ≤ ξ ≤ x. Now applying (3) on (13)
|Dn (g, x) − g(x)|
¯
° °
1 ¯ ¯¯
≤ °g 0 ° |Dn ((t − x), x)| + ¯g 00 ¯ ¯Dn ((t − x)2 , x)¯
2
°
n(1 + 2x) + 2x °
0
°g °
≤
n2 − 1
½ 2
¾
°
n + 4x2 + 4nx + 3n2 x + 7nx2 + 2n2 x2 − n3 x − n3 x2 °
°g 00 °
+
(n + 1)(n2 − 1)(n − 2)
©° ° ° °ª
≤ λn (x) °g 0 ° + °g 00 ° ≤ λn (x) kgkC 2
B
Theorem 3 For f ∈ CB [0, ∞), we obtain
( Ã p
!
)
µ
¶
λn (x)
λn (x)
(15) |Dn (f, x) − f (x)| ≤ A ω2 f,
+ min 1,
kf kCB ,
2
2
where constant A depends on f and λn (x).
2 [0, ∞) we write
Proof. for f ∈ CB [0, ∞) and g ∈ CB
Dn (f, x) − f (x) = Dn (f, x) − Dn (g, x) + Dn (g, x) − g(x) + g(x) − f (x)
by using (13) and Peetre K−functions, we get
|Dn (f, x) − f (x)| = |Dn (f, x) − Dn (g, x)| + |Dn (g, x) − g(x)| + |g(x) − f (x)|
≤ kDn f k kf − gk + λn (x) kgkC 2 + kf − gk
B
≤ 2 kf − gk + λn (x) kgkC 2
B
½
¾
½
¾
1
1
≤ 2 kf − gk + λn (x) kgkC 2 ≤ 2K f ; λn (x)
B
2
2
½ µ
¶
µ
¶
¾
p
1
1
≤ 2A ω2 f,
λn (x) + min 1, λn (x) kf kCB .
2
2
This complete the proof.
194
4
N. Deo, S. P. Singh
Rate of Convergence
Definition 1 Let us suppose that 0 < a < a1 < b3 < b1 < b < ∞, for sufficiently small δ > 0, the (2k +2)-th order Steklov mean f2k+2,δ (t) corresponding
to f (t) ∈ Cγ [0, ∞) is defined by
Z
(16) f2k+2,δ (t) = δ
−(2k+2)
δ/2
Z
Z
δ/2
δ/2
...
−δ/2
−δ/2
−δ/2
n
o 2k+2
Y
2k+2
dti ,
f (x) − ∆η f (t)
i=1
where
2k+2
X
1
η=
ti and t ∈ [a, b] .
2k + 2
i=1
It is easily checked (see e.g. [6, 10]) that
(i) f2k+2,δ has continuous derivatives up to order (2k + 2) on [a, b];
° (r) °
(ii) °f2k+2,δ °C[a1 ,b1 ] ≤ M1 δ −r ωr (f, δ, a, b), r = 1, 2, ..., (2k + 2);
°
°
(iii) °f − f2k+2,δ °C[a1 ,b1 ] ≤ M2 ω2k+2 (f, δ, a, b);
°
°
° °
(iv) °f2k+2,δ °C[a1 ,b1 ] ≤ M3 °f °γ ,
0
where Mi s, i = 1, 2, 3, are certain unrelated constants independent of f and δ.
Theorem 4 For f (r) ∈ Cγ [0, ∞) and 0 < a < a1 < b1 < b < ∞. Then for n
sufficiently large
°
°
° (r)
°
°Dn (f, (d0 , d1 , d2 , ..., dk ), .) − f (r) °
C[a1 ,b1 ]
n
³
´
o
≤ max C1 ω2k+2 f (r) , n−1/2 , a, b , C2 n−(k+1) kf kγ ,
where constant C1 = C1 (k, r) and C2 = C2 (k, r, f ).
On the degree of approximation by new Durrmeyer type operators
195
Proof. By linearity property
°
°
° (r)
°
°Dn (f, ((d0 , d1 , d2 , ..., dk )) , .)°
C[a1 ,b1 ]
°
°
° (r)
°
≤ °Dn ((f − f2k+2,δ ) , (d0 , d1 , d2 , ..., dk ) , .)°
C[a1 ,b1 ]
°
°
° (r)
°
(r)
+ °Dn (f2k+2,δ , (d0 , d1 , d2 , ..., dk ) , .) − f2k+2,δ °
C[a1 ,b1 ]
°
°
° (r)
°
(r)
+ °f − f2k+2,δ °
C[a1 ,b1 ]
= E1 + E2 + E3 , say.
¡
¢
(r)
Since, f2k+2,δ (t) = f (r) 2k+2,δ (t), by property (iii) of Steklov mean, we have
³ (r)
´
E3 ≤ C1 ω2k+2 f , δ, a, b
Next by Lemma 6, we get
E2 ≤ C2 n
−(k+1)
2k+r+2
X °
°
° (j) °
°f2k+2,δ °
j=r
C[a,b]
.
By applying the interpolation property due to Goldberg and Meir [7] for each
j = r, r + 1, ..., 2k + r + 2, we have
½
°
°
°
°
° (j) °
° (2k+r+2) °
≤ C3 kf2k+2,δ kC[a,b] + °f2k+2,δ °
°f2k+2,δ °
C[a,b]
C[a,b]
¾
.
Therefore, by applying properties (ii) and (iv) of Steklov mean, we get
n
³
´o
E2 ≤ C4 n−(k+1) kf kγ + δ −(2k+2) ω2k+2 f (r) , δ .
Finally, we shall estimate E1 , choosing a∗ , b∗ satisfying the condition 0 < a <
a∗ < a1 < b1 < b∗ < b < ∞. Also let ψ(t) denotes the characteristic function
of the interval [a∗ , b∗ ], then
°
°
°
°
E1 ≤ °Dn(r) (ψ(t) (f (t) − f2k+2,δ (t)) (d0 , d1 , d2 , ..., dk ) , .)°
C[a1 ,b1 ]
°
°
° (r)
°
+ °Dn ((1 − ψ(t)) (f (t) − f2k+2,δ (t)) (d0 , d1 , d2 , ..., dk ) , .)°
C[a1 ,b1 ]
= E4 + E5 , say.
196
N. Deo, S. P. Singh
We may note here that to estimate E4 and E5 , it is enough to consider their
expressions without the linear combinations. By Lemma 5, we have
∞
(n + 1)(n − r)!(n + r)! X
pn+r,k (x)
(n!)2
k=0
Z ∞
³
´
(r)
.
pn−r,k+r (t)ψ(t) f (r) (t) − f2k+2,δ (t) dt.
Dn(r) (ψ(t) (f (t) − f2k+2,δ (t)) , x) =
0
Hence
°
°
° (r)
°
°Dn (ψ(t) (f (t) − f2k+2,δ (t)) , k, .)°
[a1 ,b1 ]
°
°
°
°
(r)
≤ C5 °f (r) − f2k+2,δ °
[a∗ ,b∗ ]
Now for x ∈ [a1 , b1 ] and t ∈ [0, ∞)/[a∗ , b∗ ], we can choose a δ1 > 0 satisfying
|t − x| ≥ δ1 . Therefore, by Lemma 4 and Schwarz inequality, we have
¯
¯
¯
¯
I ≡ ¯Dn(r) ((1 − ψ(t)) (f (t) − f2k+2,δ (t)) , x)¯
≤ (n + 1)
Z
X
ni
2i+j≤r
i,j≥0
∞
X
|φ (x)|
´r
³ i,j,r
pn,k (x)|k − (n + 1)x|j
1
r
x 1 − n+1 + x k=0
∞
.
pn,k (t) (1 − ψ(t)) |f (t) − f2k+2,δ (t)|dt
0
X
≤ C6 kf kγ (n + 1)
∞
X
ni
k=0
2i+j≤r
i,j≥0
X
≤ C6 δ1−2s kf kγ (n + 1)
µZ
.
0
ni
¶1/2 µZ
pn,k (t)dt
≤ C6 δ1−2s kf kγ
(
. (n + 1)
X
(
ni
0
k=0
pn,k (x)|k − (n + 1)x|j
¶1/2
pn,k (t)(t − x) dt
4s
)1/2
pn,k (x) {k − (n + 1)x}2j
k=0
Z
pn,k (x)
∞
∞
X
2i+j≤r
i,j≥0
∞
X
∞
X
0
∞
|t−x|≥δ1
k=0
2i+j≤r
i,j≥0
∞
Z
pn,k (x)|k − (n + 1)x|j
)1/2
pn,k (t)(t − x)4s dt
pn,k (t)dt
On the degree of approximation by new Durrmeyer type operators
197
Hence by Lemma 2 and Lemma 3, we have
I ≤ C7 kf kγ
X
δ1−2m O
³
´
i+ 2j −s)
(
n
≤ C7 n−q kf kγ
2i+j≤r
i,j≥0
where q = (s − r/2). Now choose s > 0 such that q ≥ k + 1. Now we obtain
I ≤ C7 n−(k+1) kf kγ .
Therefore by property (iii) of Steklov mean, we get
°
°
°
°
(r)
E1 ≤ C8 °f (r) − f2k+2,δ ° ∗ ∗ + C7 n−(k+1) kf kγ
C[a ,b ]
³
´
≤ C9 ω2k+2 f (r) , δ, a, b + C7 n−(k+1) kf kγ .
Choosing δ = n−1/2 , the theorem follows.
References
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On the degree of approximation by new Durrmeyer type operators
Naokant Deo
Department of Applied Mathematics,
Delhi Technological University(Formerly Delhi College of Engineering),
Bawana Road, Delhi-110042, India.
e-mail: dr naokant deo@yahoo.com
Suresh P. Singh
Department of Mathematics,
G. G. University,
Bilaspur (C. G.)-495009, India.
e-mail: drspsingh1@rediffmail.com
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