Approximation by Modified Jain-Baskakov Operators Vishnu Narayan Mishraa,b,1 , Preeti Sharmaa , Marius Birouc a Department
of Applied Mathematics & Humanities, Sardar Vallabhbhai National Institute of Technology, Ichchhanath Mahadev Dumas Road, Surat -395 007 (Gujarat), India b L. 1627 Awadh Puri Colony Beniganj, Phase -III, Opposite - Industrial Training Institute (I.T.I.), Ayodhya Main Road Faizabad-224 001, (Uttar Pradesh) c Mathematics Department, Technical University of Cluj Napoca, Cluj Napoca, Romania
arXiv:1508.05309v2 [math.FA] 9 Sep 2015
Dedicated to the 66-th Happy Anniversary of our great Master and friend academician Prof. Dr. Heiner Gonska
Abstract In the present paper we discuss the approximation properties of Jain-Baskakov operators with parameter c. The present paper deals with the modified forms of the Baskakov basis functions. We establish some direct results, which include the asymptotic formula and error estimation in terms of the modulus of continuity and weighted approximation. Keywords: Jain operators, Baskakov operators, King-type operators, Weighted approximation. 2000 MSC: 41A25, 41A36.
1. Introduction In 1972, G.C. Jain [7] introduced and studied the following class of positive linear operators Pn[β] (f, x) =
∞ X
ωβ (v, nx)f
v=0
v n
,
(1.1)
where ωβ (v, nx) = nx(nx + vβ)v−1
e−(nx+vβ) , v!
(1.2)
and β ∈ [0, 1), f ∈ C(R+ ). In the particular case β = 0, Pn0 , n ∈ N, turns into well-known Szasz-Mirakjan operators [β]
[12]. A Kantorovich-type extension of Pn was given in [14]. In the last few decades several authors contributed to the approximation theory, fractional differential and integral equations have recently been applied in various areas of engineering, mathematics, physics and bio-engineering and other applied sciences (see [2], [6], [13], [15], [16], [17], [18] and references therein). Very recently Patel and Mishra [10] introduced modified Jain-Baskakov operators as follows β (f, x) Dn,c
=
∞
(n − c) X ωβ (v, nx) c v=1
Z
∞
pn,v−1,c (t)f (t)dt + e−nx f (0),
(1.3)
0
Email addresses:
[email protected] ( Vishnu Narayan Mishra),
[email protected] (Preeti Sharma),
[email protected] ( Marius Birou) 1 Corresponding author
Preprint submitted to Results in Mathematics
September 10, 2015
Γ(n/c + v − 1) (ct)v−1 · and ωβ (v, nx) are as defined in (1.2). Γ(v)Γ(n/c) (1 + ct)n/c+v−1 As a special case, i.e., c = 1, the operators (1.3) reduce to the Jain-Baskakov operators which is defined in [10]. where pn,v−1,c (t) = c
The operators given by (1.3) have different approximation properties than the operators defined for c = 1. The present paper deals with the study of some direct results in ordinary approximation for the modified JainBaskakov operators with parameter c. We study weighted approximation theorems, rate of convergence of these operators.
2. Auxiliary Results [β]
Lemma 1. ([7], [3]) For Pn (tm , x), m = 0, ..., 4, we have Pn[β] (1, x) = 1,
Pn[β] (t, x) =
Pn[β] (t3 , x) =
Pn[β] (t4 , x)
=
x , 1−β
Pn[β] (t2 , x) =
x x2 + , 2 (1 − β) n(1 − β)3
x3 3x2 (6β 4 − 6β 3 − 2β − 1)x + − , (1 − β)3 n(1 − β)4 n2 (1 − β)5
x4 6x3 (36β 4 − 72β 3 + 36β 2 − 8β − 7)x2 + − (1 − β)4 n(1 − β)5 n2 (1 − β)6 5 4 3 2 (105β − 14β − 2β + 12β + 8β + 1)x . + n3 (1 − β)7
Lemma 2. For n > 5c > 0 , we have β β Dn,c (1, x) = 1, Dn,c (t, x) =
nx , (n − 2c)(1 − β)
" # x 2 − 2β + β 2 x2 n2 , + = (n − 2c)(n − 3c) (1 − β)2 n(1 − β)3 1 n2 x2 (−(1 − β)xn + 3(β 2 − 2β + 2)) β , + o Dn,c (t3 , x) = (1 − β)4 (n − 2c)(n − 3c)(n − 4c) n n3 x3 ((1 − β)xn + 6(β 2 − 2β + 2)) 1 β 4 Dn,c (t , x) = . +o 5 (1 − β) (n − 2c)(n − 3c)(n − 4c)(n − 5c) n β Dn,c (t2 , x)
Proof. For n > (j + 1)c, we have Z ∞ tj pn,v−1,c (t)dt 0
It follows that β Dn,c (1, x) =
β Dn,c (t, x)
cΓ(n/c + v − 1) Γ(v + j)Γ(n/c − j − 1) · Γ(v)Γ(n/c) cj+1 Γ(n/c + v − 1) (v + j − 1)...v . = c (n − c)...(n − (j + 1)c)
=
∞ c n−c X + e−nx = Pn[β] (1, x), ωβ (v, nx) · c n=1 n−c
= =
∞ n−c X cv ωβ (v, nx) · c n=1 (n − c)(n − 2c)
∞ n v n X P [β] (t, x), ωβ (v, nx) = n − 2c n=1 n n − 2c n
2
β Dn,c (t2 , x)
= = =
β Dn,c (t3 , x) =
= =
β Dn,c (t4 , x) =
= =
∞ cv(v + 1) n−c X ωβ (v, nx) · c n=1 (n − c)(n − 2c)(n − 3c)
∞ X n2 v2 + v ωβ (v, nx) (n − 2c)(n − 3c) n=1 n2 n2 1 [β] [β] 2 Pn (t , x) + Pn (t, x) , (n − 2c)(n − 3c) n
∞ n−c X cv(v + 1)(v + 2) ωβ (v, nx) · c n=1 (n − c)(n − 2c)(n − 3c)(n − 4c)
∞ X v 3 + 3v 2 + 2v n3 ωβ (v, nx) (n − 2c)(n − 3c)(n − 4c) n=1 n3 n3 3 2 Pn[β] (t3 , x) + Pn[β] (t2 , x) + 2 Pn[β] (t, x) , (n − 2c)(n − 3c)(n − 4c) n n ∞ n−c X cv(v + 1)(v + 2)(v + 3) ωβ (v, nx) · c n=1 (n − c)(n − 2c)(n − 3c)(n − 4c)(n − 5c)
∞ X n4 v 4 + 6v 3 + 11v 2 + 6v ωβ (v, nx) (n − 2c)(n − 3c)(n − 4c)(n − 5c) n=1 n4
n4 × (n − 2c)(n − 3c)(n − 4c)(n − 5c) 6 11 6 × Pn[β] (t4 , x) + Pn[β] (t3 , x) + 2 Pn[β] (t2 , x) + 3 Pn[β] (t, x) . n n n
Using Lemma 1 we get the conclusion. Lemma 3. For n > 5c > 0, we have µβ1,n,c
=
β Dn,c (t − x, x) =
µβ2,n,c
=
β Dn,c ((t − x)2 , x)
= µβ4,n,c
= =
x(nβ + 2c(1 − β)) , (n − 2c)(1 − β)
x2 n2 β 2 + n c + 4cβ − 5cβ 2 + 6c2 − 12c2 β + 6c2 β 2 nx 2 − 2β + β 2 + , (n − 2c)(n − 3c)(1 − β)2 (n − 2c)(n − 3c)(1 − β)3
β Dn,c ((t − x)4 , x)
(1 − β)n3 β 2 x4 (2c(3 + 4β − 7β 2 ) + β 2 n) + 6n3 β 2 x2 (β 2 − 2β + 2) +o (n − 2c)(n − 3c)(n − 4c)(n − 5c)(1 − β)5
1 . n
Proof. The proof follows from Lemma 2.
3. Direct theorem Let the space CB [0, ∞) of all continuous and bounded functions be endowed with the norm kf k = sup{|f (x)| : x ∈ [0, ∞)}. Further let us consider the following K-functional: K2 (f, δ) = inf 2 {kf − gk + δkg ′′ k}, g∈W
3
(3.1)
where δ > 0 and W 2 = {g ∈ CB [0, ∞) : g ′ , g ′′ ∈ CB [0, ∞)}. By the methods as given in [11], there exists an absolute constant M > 0 such that
where
√ K2 (f, δ) ≤ M ω2 (f, δ), √ ω2 (f, δ) =
sup
√ 0 3c > 0, we have β |Dn,c (f, x)
− f (x)|
≤
ω
f, µβ1,n,c (x)
+ M ω2 f,
r
2
µβ1,n,c (x)
+
µβ2,n,c (x)
!
,
where M is a positive constant. Proof. We are introducing the auxiliary operators as follows (2c − 2cβ + nβ)x β β ˆ + f (x). Dn,c (f, x) = Dn,c (f, x) − f x + (n − 2c)(1 − β) 2 Let g ∈ W∞ and x, t ∈ [0, ∞). By Taylor’s expansion we have ′
g(t) = g(x) + (t − x)g (x) + Applying
ˆβ , D n,c
Z
t
x
(t − u)g ′′ (u)du.
we get β β β ˆn,c ˆ n,c ˆ n,c (g, x) − g(x) = g ′ (x)D ((t − x), x) + D D
Z
t x
(t − u)g ′′ (u)du, x .
Now, β ˆ n,c |D (g, x) − g(x)|
Z t β ′′ ˆ Dn,c (t − u)g (u)du , x x β Dn,c (t − x)2 , x kg ′′ k Z x+ (2c−2cβ+nβ)x (n−2c)(1−β) (2c − 2cβ + nβ)x x+ + − u g ′′ (u)du (n − 2c)(1 − β) " x 2 # (2c − 2cβ + nβ)x β kg ′′ k. µ2,n,c (x) + (n − 2c)(1 − β)
≤ ≤
≤ Since, β |Dn,c (f, x)|
≤ (n − c)/c
β |Dn,c (f, x) − f (x)|
∞ X v=1
≤
≤
wβ (v, nx)
Z
∞ 0
pn,v−1,c (t)|f (t)|dt + e−nx |f (0)| ≤ kf k,
ˆ β (f − g, x) − (f − g)(x)| + |D ˆ β (g, x) − g(x)| |D n,c n,c (2c − 2cβ + nβ)x + x + − f (x) (n − 2c)(1 − β) " 2 # (2c − 2cβ + nβ)x β kg ′′ k 2kf − gk + µ2,n,c (x) + (n − 2c)(1 − β) (2c − 2cβ + nβ)x +ω f, . (n − 2c)(1 − β) 4
Taking infimum overall g ∈ W 2 , we get β |Dn,c (f, x)
− f (x)| ≤
(2c − 2cβ + nβ)x + K2 (n − 2c)(1 − β) (2c − 2cβ + nβ)x +ω f, . (n − 2c)(1 − β) f, µβ2,n,c (x)
2 !
In view of (3.2), we have β |Dn,c (f, x) − f (x)|
≤
s
(2c − 2cβ + nβ)x M ω2 f, µβ2,n,c (x) + (n − 2c)(1 − β) (2c − 2cβ + nβ)x , +ω f, (n − 2c)(1 − β)
2
which proves the Theorem. 3.1. Rate of Convergence Let Bρ0 (R+ ) be the set of all functions f defined on R+ satisfying the condition |f (x)| ≤ Mf (1 + x2 ), where Mf
is a constant depending only on f . Now, Cρ0 (R+ ) can be represented as {f ∈ Bρ0 (R+ ) : f is continuous and
f (x) ρ0 (x)
is convergent as x → ∞}. For any positive a, by
ωa (f, δ) = sup
sup |f (t) − f (x)|
|t−x|≤δ x,t∈[0,a]
we denote the usual modulus of continuity of f on the closed interval [0, a]. We know that for a function f ∈ Cρ0 (R+ ), the modulus of the continuity ωa (f, δ) tends to zero. Now, we give a rate of convergence theorem for the Modified Jain-Baskakov operators. Theorem 2. Let f ∈ Cρ0 (R+ ) and ωa+1 (f, δ) be its modulus of continuity on the finite interval [0, a + 1] ⊂ R+ , where a > 0. Then for n > 3c, β (f ) kDn,c
− f kC[0,a] ≤ 6Mf (1 + a
2
)µβ2,n,c (x)
q β + 2ωa+1 f, µ2,n,c (x) .
Proof. For x ∈ [0, a] and t > a + 1, since t − x > 1, we have |f (t) − f (x)| ≤
Mf (2 + x2 + t2 )
≤
Mf 2 + 3x2 + 2(t − x)2
≤
6Mf (1 + a2 )(t − x)2 .
(3.3)
For x ∈ [0, a] and t ≤ a + 1, we have |t − x| |f (x) − f (t)| ≤ ωa+1 (f, |t − x|) ≤ 1 + ωa+1 (f, δ) δ
(3.4)
with δ > 0. From (3.3) and (3.4) we can write |t − x| ωa+1 (f, δ), |f (t) − f (x)| ≤ 6Mf (1 + a )(t − x) + 1 + δ 2
2
5
(3.5)
for x ∈ [0, a] and t ≥ 0. Thus β |Dn,c (f, x) − f (x)|
≤
β Dn,c (|f (t) − f (x)|, x)
≤
β 6Mf (1 + a2 )Dn,c ((t − x)2 , x) 1 β 2 1/2 +ωa+1 (f, δ) 1 + Dn,c ((t − x) , x) . δ
Hence, by Schwarz’s inequality and Lemma 3, for x ∈ [0, a] β |Dn,c (f, x)
By taking δ =
− f (x)| ≤ 6Mf (1 + a
2
)µβ2,n,c (x)
1 + ωa+1 (f, δ) 1 + δ
q β µ2,n,c (x) .
q µβ2,n,c (x), we get
q β kDn,c (f ) − f kC[0,a] ≤ 6Mf (1 + a2 )µβ2,n,c (x) + 2ωa+1 f, µβ2,n,c (x) .
This completes the proof of Theorem. 3.2. Weighted approximation Now we shall discuss the weighted approximation theorem, where the approximation formula holds true on the interval R+ . Theorem 3. If f ∈ Cρ0 (R+ ), lim βn = 0 and n > 3c. Then, n→∞
βn (f ) − f kρ0 = 0. lim kDn,c
n→∞
Proof. Using the theorem in [4] we see that it is sufficient to verify the following three conditions βn r (t , x) − xr kρ0 = 0, r = 0, 1, 2. lim kDn,c
(3.6)
n→∞
βn (1, x) = 1, the first condition of (3.6) is satisfied for r = 0. Since Dn,c
Now, for n > 2c βn (t, x) − xkρ0 kDn,c
βn |Dn,c (t, x) − x| 1 + x2 x∈[0,∞) n ≤ − 1 sup (n − 2c)(1 − βn )
=
sup
x∈[0,∞)
≤
x 1 + x2
2c nβn + (n − 2c) (n − 2c)(1 − βn )
βn (t, x) − xkρ0 = 0 as n → ∞ with βn → 0. which implies kDn,c
Similarly, we can write for n > 3c βn 2 (t , x) − x2 kρ0 kDn,c
βn 2 |Dn,c (t , x) − x2 | 1 + x2 x∈[0,∞) x2 n2 sup − 1 ≤ 2 (n − 2c)(n − 3c)(1 − βn )2 x∈[0,∞) 1 + x n 2 − 2βn + βn2 x sup + (n − 2c)(n − 3c)(1 − βn )3 x∈[0,∞) 1 + x2
=
sup
6
which implies that βn 2 (t , x) − x2 kρ0 = 0 with βn → 0. lim kDn,c
n→∞
Thus the proof is completed. We give the following theorem to approximate all function in Cρ0 (R+ ). This type of results are given in [5] for locally integrable functions. Theorem 4. Let βn → 0 as n → ∞. For each f ∈ Cρ0 (R+ ) and λ > 0, we have lim
n→∞
βn (f, x) − f (x)| |Dn,c = 0. (1 + x2 )1+λ x∈[0,∞)
sup
Proof. For any fixed x0 > 0, βn |Dn,c (f, x) − f (x)| (1 + x2 )1+λ x∈[0,∞)
sup
≤
sup x≤x0
βn βn |Dn,c |Dn,c (f, x) − f (x)| (f, x) − f (x)| + sup 2 1+λ (1 + x ) (1 + x2 )1+λ x≥x0
βn (f ) − f kC[0,x0 ] ≤ kDn,c
+kf kρ0 sup
x≥x0
βn |Dn,c (1 + t2 , x)| |f (x)| + sup . 2 1+λ 2 1+λ (1 + x ) x≥x0 (1 + x )
The first term of the above inequality tends to zero from Theorem 2. By Lemma 2 for any fixed x0 > 0 it is easily βn (1 + t2 , x)| |Dn,c seen that sup → 0 as n → ∞ with βn → 0. We can choose x0 > 0 so large that the last part of (1 + x2 )1+λ x≥x0 the above inequality can be made small enough. Thus the proof is completed. Theorem 5. Let b > 0 and βn ∈ (0, 1) such that nβn → l ∈ R as n → ∞. Then for every f ∈ C[0, b], twice differentiable at a fixed point x ∈ (0, b), we have βn (f, x) − f (x)) = x(l + 2c)f ′ (x) + lim n(Dn,c
n→∞
x(2 + xc) ′′ f (x). 2
. Proof. From Lemma 3 we have n = x(l + 2c), lim nµβ1,n,c
n→∞
n lim nµβ2,n,c = x(2 + xc),
n→∞
n lim nµβ4,n,c = 0.
n→∞
Using Theorem 2.1 from [1] we get the conclusion.
4. King’s Approach To make the convergence faster in 2003, King [8] proposed an approach to modify the classical Bernstein polynomial, so that the sequence preserve test functions e0 and e2 . After this approach many researcher contributed in this direction. Motivated by such type of operators, we generalized King-type approach of the operators (1.3) as
7
follows: For β ∈ [0, 1), x ∈ [0, ∞) and f ∈ Cρ0 (R+ ), we introduced King-type positive linear operators as ∗β Dn,c (f, x)
where rn (x) =
=
∞
(n − c) X ωβ (v, nrn (x)) c v=1
Z
∞
pn,v−1 (t)f (t)dt + e−nrn (x) f (0),
(4.1)
0
(n − 2c)(1 − β)x and ωβ (v, nx) and pn,v−1 (t) as same as defined in (1.2) and (1.3). n
Lemma 4. The following holds ∗β Dn,c (1, x) = 1,
∗β Dn,c (t, x) = x,
(2 − 2β + β 2 ) n − 2c 2 x, n > 3c, x + (n − 3c) (n − 3c)(1 − β)2 1 x2 n((1 − β 2 )(n − 4c)x + 3(β 2 − 2β + 2)) ∗β 3 , n > 4c, + o Dn,c (t , x) = (1 − β)2 (n − 3c)(n − 4c) n n2 x3 ((1 − β)2 (n − 6c)x + 6(β 2 − 2β + 2)) 1 ∗β 4 Dn,c (t , x) = , n > 5c. + o (1 − β)2 (n − 3c)(n − 4c)(n − 5c) n ∗β 2 Dn,c (t , x) =
∗β ∗β The proof follows from Lemma 2 and linearity of Dn,c . We observe that the operators Dn,c are preserves linear ∗β functions that means, Dn,c (a + bt, x) = a + bx for constants a, b.
Lemma 5. The following holds ∗β µ∗β 1,n,c = Dn,c (t − x, x) = 0,
c (2 − 2β + β 2 ) x, n > 3c, x2 + n − 3c (n − 3c)(1 − β)2 1 ∗β , n > 5c. = Dn,c ((t − x)4 , x) = o n
∗β 2 µ∗β 2,n,c = Dn,c ((t − x) , x) =
µ∗β 4,n,c The proof follows from Lemma 4.
∗βn Theorem 6. Let Dn,c , n > 3c, be defined as in (4.1), where lim βn = 0. For any compact B ⊂ R+ and for each n→∞
f ∈ Cρ0 (R+ ) one has
∗βn (f, x) = f (x), uniformly in x ∈ B. lim Dn,c
n→∞
The proof follows from Lemma 4 and Korovkin theorem [9]. Theorem 7. For f ∈ CB [0, ∞), 0 < β < 1 and n > 3, we have q ∗β |Dn,c (f, x) − f (x)| ≤ C1 ω2 f, µ ˆβn,2,c , where C1 is a positive constant. 2 Proof. Let g ∈ W∞ and x ∈ [0, ∞). Using Taylor’s expansion Z t ′ g(t) = g(x) + g (x)(t − x) + (t − u)g ′′ (u)du, t ∈ [0, ∞) x
and Lemma 4, we have
8
∗β Dn,c (g, x) − g(x) = Dn∗β
Z t Notice that, (t − u)g ′′ (u)du ≤ (t − x)2 kg ′′ k. Thus
Z
t
x
(t − u)g ′′ (u)du .
x
′′ ∗β ∗β |Dn,c (g, x) − g(x)| ≤ Dn,c ((t − x)2 , x)kg ′′ (x)k = µ∗β 2,n,c kg k.
∗β Since |Dn,c (f, x)| ≤ kf k,
∗β |Dn,c (f, x) − f (x)|
≤
∗β ∗β |Dn,c (f − g, x) − (f − g)(x)| + |Dn,c (g, x) − g(x)|
≤
′′ 2kf − gk + µ∗β 2,n,c kg k.
2 Finally taking the infimum on right side over all g ∈ W∞ and using (3.2), we get the desired result. ∗β Now, we establish the Voronovskaja type asymptotic formula for the operators Dn,c :
Theorem 8. Let b > 0 and βn ∈ (0, 1) such that nβn → 0 as → ∞. Then for every f ∈ C[0, b], twice differentiable at a fixed point x ∈ (0, b), we have
Proof. From Lemma 5 we have
f ′′ (x) ∗βn (f, x) − f (x) = x(2 + xc) lim n Dn,c . n→∞ 2 n lim nµ∗β 1,n,c = 0,
n→∞
n lim nµ∗β 2,n,c = x(2 + xc),
n→∞
n lim nµ∗β 4,n,c = 0.
n→∞
Using Theorem 2.1 from [1] we get the conclusion. References:
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