Note On Jakimovski-Leviatan Operators Preserving e–x

J. P. King [7] introduced a modification of the well known Bernstein polynomials which preserve constant and the x² test function. This modification provide better approximation over the usual Bernstein polynomials. In addition, a modification of Szászirakyan operators is presented that reproduces the functions 1 and e^2ax, a > 0 fixed and also are proved uniform convergence, order of approximation via a certain weighted modulus of continuity, and a quantitative Voronovskaya-type theorem by T. Acar [13]. Many different applications of similar type of operators have studied in [7]-[9].

In [1], Jakimovski and Leviatan constructed a new type of operators P_n by using Appell polynomials given as below; $\begin{matrix} g (u) = \sum_{n = 0}^{\infty} a_{n} u^{n}, g (1) \neq 1 \end{matrix}$ $\begin{array}{} \displaystyle g(u)=\sum_{n=0}^\infty a_{n}u^n, g(1)\neq 1 \end{array} $ be an analytic function in the disk $\begin{matrix} p_{k} (x) = \sum_{i = 0}^{k} a_{i} \frac{x^{k - i}}{(k - i)!}, \end{matrix}$ $\begin{array}{} \displaystyle p_{k}(x)=\sum_{i=0}^k a_{i}\frac{x^{k-i}}{(k-i)!}, \end{array} $ k ∈ ℕ be the Appell polynomials defined by the identity

$\begin{matrix} g (u) e^{u x} = \sum_{k = 0}^{\infty} p_{k} (x) u^{k} . \end{matrix}$ $$\begin{array}{} \displaystyle g(u)e^{ux}=\sum_{k=0}^\infty p_{k}(x)u^k. \end{array}$$(1)

Let E[0, ∞) indicate the class of functions of exponential type on [0, ∞) which satisfy the property |f(x)| ≤ β e^αx for some finite constants α, β > 0.

In [1], the authors considered the operator P_n:E → C[0, ∞)

$\begin{matrix} P_{n} (f; x) = \frac{e^{- n x}}{g (1)} \sum_{k = 0}^{\infty} p_{k} (n x) f (\frac{k}{n}) . \end{matrix}$ $$\begin{array}{} \displaystyle P_{n}(f;x)=\frac{e^{-nx}}{g(1)}\sum_{k=0}^{\infty}p_{k}\left(nx\right)f\left(\frac{k}{n}\right). \end{array}$$(2)

Remark 1

If g(1) = 1 in (1), we obtain $\begin{matrix} p_{k} (x) = \frac{x^{k}}{k!} \end{matrix}$ $\begin{array}{} \displaystyle p_{k}(x)=\frac{x^k}{k!} \end{array}$ and we get classical Szász-Mirakjan operator which is given by

$\begin{matrix} S_{n} (f; x) = e^{- n x} \sum_{k = 0}^{\infty} \frac{{(n x)}^{k}}{k!} f (\frac{k}{n}) . \end{matrix}$ $$\begin{array}{} \displaystyle S_{n}(f;x)=e^{-nx}\sum_{k=0}^{\infty}\frac{\left(nx\right)^k}{k!}f\left(\frac{k}{n}\right). \end{array}$$

B. Wood in [6] proved that the operators P_n are positive if and only if $\begin{matrix} \frac{a_{n}}{g (1)} \geq 0 \end{matrix}$ $\begin{array}{} \displaystyle \frac{a_{n}}{g(1)}\geq 0 \end{array} $ for n ∈ ℕ. In [5], Ciupa studied the rate of convergence of these operators. The convergence of these operators in a weighted space of functions on a positive semi-axis and estimate the approximation by using a new type of weighted modulus of continuity introduced by A.D. Gadjiev and A. Aral in [8] were studied in [3].

2

Main Results

In this section, we consider the following modified form of generalization of Jakimovski-Leviatan operators

$\begin{matrix} E_{n} (f; x) = \frac{e^{- n a_{n} (x)}}{g (1)} \sum_{k = 0}^{\infty} a_{k} \frac{{(n a_{n} (x))}^{k}}{k!} f (\frac{k}{n}) \end{matrix}$ $$\begin{array}{} \displaystyle E_{n}(f;x)=\frac{e^{-na_{n}(x)}}{g(1)}\sum_{k=0}^{\infty}a_{k}\frac{\left(na_{n}(x)\right)^k}{k!}f\left(\frac{k}{n}\right) \end{array}$$(3)

x ≥ 0, n ∈ ℕ such that the conditions

$\begin{matrix} E_{n} (e^{- t}; x) = e^{- x} \end{matrix}$ $$\begin{array}{} \displaystyle E_{n}(e^{-t};x)=e^{-x} \end{array}$$(4)

are satisfied for all x and all n. Therefore by simple computation, we get

$\begin{matrix} e^{- x} = e^{n a_{n} (x) [e^{- 1 / n} - 1]} . \end{matrix}$ $$\begin{array}{} \displaystyle e^{-x}=e^{na_{n}(x)\left[e^{-1/n}-1\right]}. \end{array} $$

This implies

$\begin{matrix} a_{n} (x) = \frac{- x}{n (e^{- 1 / n} - 1)} \end{matrix}$ $$\begin{array}{} \displaystyle a_{n}(x)=\frac{-x}{n(e^{-1/n}-1)} \end{array} $$(5)

in other words we can write as

$\begin{matrix} a_{n} (x) = \frac{x e^{1 / n}}{n (e^{1 / n} - 1)} . \end{matrix}$ $$\begin{array}{} \displaystyle a_{n}(x)=\frac{xe^{1/n}}{n(e^{1/n}-1)}. \end{array} $$

Thus the operator (3) can be rewritten the following form:

$\begin{matrix} E_{n} (f; x) = \frac{e^{\frac{x e^{1 / n}}{(1 - e^{1 / n})}}}{g (1)} \sum_{k = 0}^{\infty} a_{k} \frac{(- 1)^{k}}{k!} {(\frac{x e^{1 / n}}{(e^{1 / n} - 1)})}^{k} f (\frac{k}{n}) \end{matrix}$ $$\begin{array}{} \displaystyle E_{n}(f;x)=\frac{e^{\frac{xe^{1/n}}{(1-e^{1/n})}}}{g(1)}\sum_{k=0}^{\infty}a_{k}\frac{(-1)^k}{k!}\left(\frac{xe^{1/n}}{(e^{1/n}-1)}\right)^kf\left(\frac{k}{n}\right) \end{array} $$(6)

Lemma 1

The moment for the operator (3) may be given as

$\begin{matrix} E_{n} (e^{A t}; x) = e^{n a_{n} (x) [e^{A / n} - 1]} . \end{matrix}$ $$\begin{array}{} \displaystyle E_{n}\left(e^{At};x\right)=e^{na_{n}(x)\left[e^{A/n}-1\right]}. \end{array} $$(7)

Proof

From (5), we have

$\begin{matrix} \begin{aligned} E_{n} (e^{A t}; x) = & \frac{e^{- n a_{n} (x)}}{g (1)} \sum_{k = 0}^{\infty} a_{k} \frac{{(n a_{n} (x))}^{k}}{k!} e^{A \frac{k}{n}} \\ = & \frac{1}{g (1)} \sum_{k = 0}^{\infty} a_{k} \frac{e^{- n a_{n} (x)} {(n a_{n} (x) e^{A / n})}^{k}}{k!} \\ = & e^{n a_{n} (x) [e^{A / n} - 1]} \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle E_{n}\left(e^{At};x\right)=&\frac{e^{-na_{n}(x)}}{g(1)}\sum_{k=0}^{\infty}a_{k}\frac{\left(na_{n}(x)\right)^k}{k!}e^{A\frac{k}{n}}\\ \displaystyle=&\frac{1}{g(1)}\sum_{k=0}^{\infty}a_{k}\frac{e^{-na_{n}(x)}\left(na_{n}(x)e^{A/n}\right)^k}{k!}\\ \displaystyle=&e^{na_{n}(x)\left[e^{A/n}-1\right]} \end{split} \end{array}$$(8)

□

Lemma 2

If the operator E_n is defined by (3), then with e_r(t) = t^r, r = 0, 1, 2... the moments as follows;

$\begin{matrix} \begin{aligned} E_{n} (e_{0}; x) = & 1 \\ E_{n} (e_{1}; x) = & a_{n} (x) \\ E_{n} (e_{2}; x) = & a_{n}^{2} (x) + a_{n} (x) / n \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle E_{n}\left(e_{0};x\right)=&1\\ \displaystyle E_{n}\left(e_{1};x\right)=&a_{n}(x)\\ \displaystyle E_{n}\left(e_{2};x\right)=&a_{n}^2(x)+a_{n}(x)/n\\ \end{split} \end{array} $$(9)

Lemma 3

By Lemma (2) the central moments for E_n operator

$\begin{matrix} E_{n} (ϕ_{x}^{m} (t); x) = E_{n} ((t - x)^{m}; x), m = 0, 1, 2 \end{matrix}$ $$\begin{array}{} \displaystyle E_{n}\left(\phi_{x}^m(t);x\right)=E_{n}\left((t-x)^m;x\right), m=0,1,2 \end{array} $$(10)

are given by

$\begin{matrix} \begin{aligned} E_{n} (ϕ_{x}^{0} (t); x) = & 1 \\ E_{n} (ϕ_{x}^{1} (t); x) = & a_{n} (x) - x \\ E_{n} (ϕ_{x}^{2} (t); x) = & (a_{n} (x) - x)^{2} + a_{n} (x) / n . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle E_{n}\left(\phi_{x}^0(t);x\right)=&1\\ \displaystyle E_{n}\left(\phi_{x}^1(t);x\right)=&a_{n}(x)-x\\ \displaystyle E_{n}\left(\phi_{x}^2(t);x\right)=&(a_{n}(x)-x)^2+a_{n}(x)/n. \end{split} \end{array}$$(11)

Furthermore,

$\begin{matrix} lim_{n \to \infty} n (- x n (e^{- 1 / n} - 1) - x) = \frac{x}{2}, \end{matrix}$ $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}n\left(-x{n(e^{-1/n}-1)}-x\right)=\frac{x}{2}, \end{array}$$(12)

$\begin{matrix} lim_{n \to \infty} n ((a_{n} (x) - x)^{2} + \frac{a_{n} (x)}{n}) = x . \end{matrix}$ $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}n\left((a_{n}(x)-x)^2+\frac{a_{n}(x)}{n}\right)=x. \end{array}$$(13)

3

A Quantitative Result

In this section, we represent the rate of uniform convergence for the E_n operators. In [12], the uniform convergence estimate for any sequence of positive linear operators were established by Boyanov and Veselinov. In [11], Holhoş presented the following theorem:

Theorem 1

([11]) If a sequence of linear positive operators L_n:C^*[0, ∞) → C^*[0, ∞) satisfy the equalities

$\begin{matrix} \begin{aligned} {∥L_{n} e_{0} - 1∥}_{[0, \infty)} = & α_{n} \\ {∥L_{n} (e^{- t}) - e^{- x}∥}_{[0, \infty)} = & β_{n} \\ {∥L_{n} (e^{- 2 t}) - e^{- 2 x}∥}_{[0, \infty)} = & γ_{n} \end{aligned} \end{matrix}$ $$\begin{array}{c} \begin{split} \displaystyle \left\|L_{n}e_{0}-1\right\|_{[0,\infty)}=&\alpha_{n}\\ \displaystyle\left\|L_{n}(e^{-t})-e^{-x}\right\|_{[0,\infty)}=&\beta_{n}\\ \displaystyle\left\|L_{n}(e^{-2t})-e^{-2x}\right\|_{[0,\infty)}=&\gamma_{n}\\ \end{split} \end{array} $$(14)

then, for f ∈ C^*[0, ∞), we have

$\begin{matrix} {∥L_{n} f - f∥}_{[0, \infty)} \leq α_{n} {∥f∥}_{[0, \infty)} + (2 + α_{n}) ω^{*} (f, \sqrt{α_{n} + 2 β_{n} + γ_{n}}), \end{matrix}$ $$\begin{array}{} \displaystyle \left\|L_{n}f-f\right\|_{[0,\infty)}\leq \alpha_{n}\left\|f\right\|_{[0,\infty)}+(2+\alpha_{n})\omega^*(f,\sqrt{\alpha_{n}+2\beta_{n}+\gamma_{n}}), \end{array}$$(15)

where the modulus of continuity is defined as:

$\begin{matrix} ω^{*} (f, δ) = sup_{\binom{|e^{- x} - e^{- t}| \leq δ}{x, t > 0}} |f (t) - f (x)| . \end{matrix}$ $$\begin{array}{} \displaystyle \omega^*(f,\delta)=\sup_{\left|e^{-x}-e^{-t}\right|\leq \delta \atop x,t \gt0}\left|f(t)-f(x)\right|. \end{array}$$(16)

Now, we will proof the quantitative estimate for E_n operators defined by (3) which preserve e^–x function:

Theorem 2

For f ∈ C^*[0, ∞), we get

$\begin{matrix} {∥E_{n} f - f∥}_{[0, \infty)} \leq 2 ω^{*} (f; \sqrt{2 β_{n} + γ_{n}}), \end{matrix}$ $$\begin{array}{} \displaystyle \left\|E_{n}f-f\right\|_{[0,\infty)}\leq 2\omega^*\left(f;\sqrt{2\beta_{n}+\gamma_{n}}\right), \end{array} $$(17)

where

$\begin{matrix} \begin{aligned} β_{n} = & {∥E_{n} (e^{- t}) - e^{- x}∥}_{[0, \infty)}, \\ γ_{n} = & {∥E_{n} (e^{- 2 t}) - e^{- 2 x}∥}_{[0, \infty)} . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle \beta_{n}=& \left\|E_{n}(e^{-t})-e^{-x}\right\|_{[0,\infty)},\\ \displaystyle\gamma_{n}=& \left\|E_{n}(e^{-2t})-e^{-2x}\right\|_{[0,\infty)}. \end{split} \end{array}$$

Here, β_n and γ_n tend to zero as n goes to infinity, therefore E_n converges uniformly to f.

Proof

By using the equalities (3) and (5), we get

$\begin{matrix} \begin{aligned} E_{n} (e^{- λ t}; x) = & \frac{e^{- n a_{n} (x)}}{g (1)} \sum_{k = 0}^{\infty} a_{k} \frac{{(n a_{n} (x))}^{k}}{k!} e^{- λ \frac{k}{n}} \\ = & e^{- n a_{n} (x)} e^{n a_{n} (x) e^{- λ / n}} \\ = & e^{n a_{n} (x) (e^{- λ / n} - 1)} \\ = & e^{\frac{x}{e^{λ / n}} (\frac{e^{λ / n} - 1}{e^{- 1 / n} - 1})} . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle E_{n}\left(e^{-\lambda t};x\right)=&\frac{e^{-na_{n}(x)}}{g(1)}\sum_{k=0}^{\infty}a_{k}\frac{\left(na_{n}(x)\right)^k}{k!}e^{-\lambda \frac{k}{n}}\\ \displaystyle=&e^{-na_{n}(x)}e^{na_{n}(x)e^{-\lambda /n}}\\ \displaystyle=&e^{na_{n}(x)\left(e^{-\lambda /n}-1\right)}\\ \displaystyle=&e^{\frac{x}{e^{\lambda/n}}\left(\frac{e^{\lambda /n}-1}{e^{-1/n}-1}\right)}. \end{split} \end{array} $$

Firstly, let take λ = 1. Using the inequality

$\begin{matrix} \frac{u - v}{\ln u - \ln v} < \frac{u + v}{2} \end{matrix}$ $$\begin{array}{} \displaystyle \frac{u-v}{\ln u-\ln v}\lt\frac{u+v}{2} \end{array}$$

for 0 < v < u, we get

$\begin{matrix} e^{- x u_{n}} - e^{- x} < \frac{1 - u_{n}}{2} (x e^{- x u_{n}} + x e^{- x}) . \end{matrix}$ $$\begin{array}{} \displaystyle e^{-xu_{n}}-e^{-x}\lt\frac{1-u_{n}}{2}\left(xe^{-xu_{n}}+xe^{-x}\right). \end{array}$$

On the other hand, since

$\begin{matrix} max_{x > 0} x e^{- b x} = \frac{1}{e b} \end{matrix}$ $$\begin{array}{} \displaystyle \max_{x\gt0}xe^{-bx}=\frac{1}{eb} \end{array}$$

for every b > 0, we can write as

$\begin{matrix} \begin{aligned} e^{- x u_{n}} - e^{- x} < & \frac{1 - u_{n}}{2} (\frac{1}{e u_{n}} + \frac{1}{e}) \\ = & \frac{1 - u_{n}^{2}}{2 e u_{n}} \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle e^{-xu_{n}}-e^{-x}\lt&\frac{1-u_{n}}{2}\left(\frac{1}{eu_{n}}+\frac{1}{e}\right)\\ =&\frac{1-u_{n}^2}{2eu_{n}} \end{split} \end{array} $$

where $\begin{matrix} u_{n} = - \frac{1}{e^{1 / n}} (\frac{e^{1 / n} - 1}{e^{- 1 / n} - 1}) . \end{matrix}$ $\begin{array}{} \displaystyle u_{n}=-\frac{1}{e^{1/n}}\left(\frac{e^{1/n}-1}{e^{-1/n}-1}\right). \end{array} $ Thus

$\begin{matrix} {∥E_{n} (e^{- t}; x) - e^{- x}∥}_{[0, \infty)} = β_{n} < \frac{1 - u_{n}^{2}}{2 e u_{n}} \to 0, \end{matrix}$ $$\begin{array}{} \displaystyle \left\|E_{n}(e^{-t};x)-e^{-x}\right\|_{[0,\infty)}=\beta_{n}\lt \frac{1-u_{n}^2}{2eu_{n}}\rightarrow 0, \end{array}$$

as n → ∞.

For λ = 2, we have

$\begin{matrix} \begin{aligned} e^{- x v_{n}} - e^{- 2 x} < & \frac{2 - v_{n}}{2} (x e^{- x v_{n}} + x e^{- 2 x}) \\ < & \frac{2 - v_{n}}{2} (\frac{1}{e v_{n}} + \frac{1}{2 e}) \\ = & \frac{4 - v_{n}^{2}}{4 e v_{n}} \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle e^{-xv_{n}}-e^{-2x}\lt&\frac{2-v_{n}}{2}\left(xe^{-xv_{n}}+xe^{-2x}\right)\\ \displaystyle\lt&\frac{2-v_{n}}{2}\left(\frac{1}{ev_{n}}+\frac{1}{2e}\right)\\ \displaystyle=&\frac{4-v_{n}^2}{4ev_{n}} \end{split} \end{array} $$

where $\begin{matrix} v_{n} = - \frac{1}{e^{2 / n}} (\frac{e^{2 / n} - 1}{e^{- 2 / n} - 1}) . \end{matrix}$ $\begin{array}{} \displaystyle v_{n}=-\frac{1}{e^{2/n}}\left(\frac{e^{2/n}-1}{e^{-2/n}-1}\right). \end{array}$ Therefore

$\begin{matrix} {∥E_{n} (e^{- 2 t}; x) - e^{- 2 x}∥}_{[0, \infty)} = γ_{n} < \frac{4 - v_{n}^{2}}{4 e v_{n}} \to 0, \end{matrix}$ $$\begin{array}{} \displaystyle \left\|E_{n}(e^{-2t};x)-e^{-2x}\right\|_{[0,\infty)}=\gamma_{n}\lt \frac{4-v_{n}^2}{4ev_{n}}\rightarrow 0, \end{array}$$

as n → ∞. Hence the proof is complete.□

4

A Quantitative Voronovskaya type theorem

Now, we will proof a Quantitative Voronovskaya type theorem for E_n operators.

Theorem 3

Let f', f˝ ∈ C^*[0, ∞). Then, we have

$\begin{matrix} |n [E_{n} (f; x) - f (x)] - \frac{x}{2} f^{'} (x) - \frac{x}{2} f^{″} (x)| \leq |r_{n} (x)| |f^{'} (x)| + |q_{n} (x)| |f^{″} (x)| + 2 (2 g_{n} (x) + x + s_{n} (x)) ω^{*} (f^{″}; 1 / \sqrt{n}) \end{matrix}$ $$\begin{array}{} \displaystyle \left|n[E_{n}(f;x)-f(x)]-\frac{x}{2}f'(x)-\frac{x}{2}f''(x)\right| \leq \left|r_{n}(x)\right|\left|f'(x)\right|+\left|q_{n}(x)\right|\left|f''(x)\right|+2\left(2g_{n}(x)+x+s_{n}(x)\right)\omega^*(f'';1/ \sqrt{n}) \end{array}$$

where

$\begin{matrix} \begin{aligned} r_{n} (x) = & n E_{n} (ϕ_{x}^{1} (t); x) - \frac{x}{2} \\ q_{n} (x) = & \frac{1}{2} (n E_{n} (ϕ_{x}^{2} (t); x) - x) \\ s_{n} (x) = & \sqrt{n^{2} E_{n} ((e^{- x} - e^{- t})^{4}; x)} \sqrt{n^{2} E_{n} ((t - x)^{4}; x)} . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle r_{n}(x)=&nE_{n}(\phi_{x}^1(t);x)-\frac{x}{2} \\ q_{n}(x)=&\frac{1}{2}\left(nE_{n}(\phi_{x}^2(t);x)-x\right) \\ \displaystyle s_{n}(x)=&\sqrt{n^2E_{n}\left((e^{-x}-e^{-t})^4;x\right)}\sqrt{n^2E_{n}\left((t-x)^4;x\right)}. \end{split} \end{array}$$

Proof

By the Taylor’s expansion of f, we have

$\begin{matrix} f (t) = f (x) + f^{'} (x) (t - x) + \frac{f^{″} (x)}{2} (t - x)^{2} + h (t, x) (t - x)^{2}, \end{matrix}$ $$\begin{array}{} \displaystyle f(t)=f(x)+f'(x)(t-x)+\frac{f''(x)}{2}(t-x)^2+h(t,x)(t-x)^2, \end{array} $$(18)

where

$\begin{matrix} h (t, x) = \frac{f^{″} (η) - f^{″} (x)}{2} \end{matrix}$ $$\begin{array}{} \displaystyle h(t,x)=\frac{f''(\eta)-f''(x)}{2} \end{array} $$

is a continous function and η is a number between x and t. Applying the E_n operator to both sides of equality (18), we get

$\begin{matrix} |E_{n} (f; x) - f (x) - f^{'} (x) E_{n} (ϕ_{x}^{1} (t); x) - \frac{f^{″} (x)}{2} E_{n} (ϕ_{x}^{2} (t); x)| \leq |E_{n} (h (t, x) ϕ_{x}^{2} (t); x)| . \end{matrix}$ $$\begin{array}{} \displaystyle \left|E_{n}(f;x)-f(x)-f'(x)E_{n}(\phi_{x}^1(t);x)-\frac{f''(x)}{2}E_{n}(\phi_{x}^2(t);x)\right| \leq \left|E_{n}(h(t,x)\phi_{x}^2(t);x)\right|. \end{array} $$

$\begin{matrix} \begin{aligned} |n [E_{n} (f; x) - f (x)] - \frac{x}{2} f^{'} (x) - \frac{x}{2} f^{″} (x)| \leq & |n E_{n} (ϕ_{x}^{1} (t); x) - \frac{x}{2}| |f^{'} (x)| \\ + & \frac{1}{2} |n E_{n} (ϕ_{x}^{2} (t); x) - x| |f^{″} (x)| + |n E_{n} (h (t, x) ϕ_{x}^{2} (t); x)| . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle \left|n[E_{n}(f;x)-f(x)]-\frac{x}{2}f'(x)-\frac{x}{2}f''(x)\right|\leq& \left|nE_{n}(\phi_{x}^1(t);x)-\frac{x}{2}\right|\left|f'(x)\right|\\ \displaystyle+&\frac{1}{2}\left|nE_{n}(\phi_{x}^2(t);x)-x\right|\left|f''(x)\right|+\left|nE_{n}(h(t,x)\phi_{x}^2(t);x)\right|. \end{split} \end{array}$$

Take $\begin{matrix} r_{n} (x) = n E_{n} (ϕ_{x}^{1} (t); x) - \frac{x}{2} \end{matrix}$ $\begin{array}{} \displaystyle r_{n}(x)=nE_{n}(\phi_{x}^1(t);x)-\frac{x}{2} \end{array} $ and $\begin{matrix} q_{n} (x) = \frac{1}{2} (n E_{n} (ϕ_{x}^{2} (t); x) - x) . \end{matrix}$ $\begin{array}{} \displaystyle q_{n}(x)=\frac{1}{2}\left(nE_{n}(\phi_{x}^2(t);x)-x\right). \end{array}$ Thus, we get

$\begin{matrix} |n [E_{n} (f; x) - f (x)] - \frac{x}{2} f^{'} (x) - \frac{x}{2} f^{″} (x)| \leq |r_{n} (x)| |f^{'} (x)| + |q_{n} (x)| |f^{″} (x)| + |n E_{n} (h (t, x) ϕ_{x}^{2} (t); x)| \end{matrix}$ $$\begin{array}{} \displaystyle \left|n[E_{n}(f;x)-f(x)]-\frac{x}{2}f'(x)-\frac{x}{2}f''(x)\right| \leq \left|r_{n}(x)\right|\left|f'(x)\right|+\left|q_{n}(x)\right|\left|f''(x)\right|+\left|nE_{n}(h(t,x)\phi_{x}^2(t);x)\right| \end{array}$$(19)

In order to complete the proof of the theorem, we must find the last term of the inequality (19). Here, we also know that the equalities given in (12) and (13) r_n(x)→ 0, q_n(x)→ 0 as n → ∞ at any point x ∈ [0, ∞).

Using the property

$\begin{matrix} |f (t) - f (x)| \leq (1 + \frac{(e^{- t} - e^{- x})^{2}}{δ^{2}}) ω^{*} (f; δ), δ > 0 \end{matrix}$ $$\begin{array}{} \displaystyle \left|f(t)-f(x)\right|\leq\left(1+\frac{(e^{-t}-e^{-x})^2}{\delta^2}\right)\omega^*(f;\delta), \delta\gt0 \end{array} $$

we have

$\begin{matrix} |h (t, x)| \leq (1 + \frac{(e^{- t} - e^{- x})^{2}}{δ^{2}}) ω^{*} (f^{″}; δ) . \end{matrix}$ $$\begin{array}{} \displaystyle \left|h(t,x)\right|\leq\left(1+\frac{(e^{-t}-e^{-x})^2}{\delta^2}\right)\omega^*(f'';\delta). \end{array}$$

If |e^–x–e^–t| ≤ δ, then |h(t,x)| ≤ 2ω^*(f˝;δ). In case |e^–x–e^–t|> δ, then we get |h(t, x)| ≤ $\begin{matrix} 2 \frac{(e^{- t} - e^{- x})^{2}}{δ^{2}} ω^{*} (f^{″}; δ) . \end{matrix}$ $\begin{array}{} \displaystyle 2\frac{(e^{-t}-e^{-x})^2}{\delta^2}\omega^*(f'';\delta). \end{array}$ Hence

$\begin{matrix} |h (t, x)| \leq 2 (1 + \frac{(e^{- t} - e^{- x})^{2}}{δ^{2}} ω^{*} (f^{″}; δ)) . \end{matrix}$ $$\begin{array}{} \displaystyle \left|h(t,x)\right|\leq2\left(1+\frac{(e^{-t}-e^{-x})^2}{\delta^2}\omega^*(f'';\delta)\right). \end{array}$$

By using this inequality and applying Cauchy-Schwarz inequality, we get

$\begin{matrix} \begin{aligned} n . E_{n} (|h (t, x)| ϕ_{x}^{2} (t); x) \leq & 2 n ω^{*} (f^{″}; δ) E_{n} (ϕ_{x}^{2} (t); x) \\ + & \frac{2 n}{δ^{2}} ω^{*} (f^{″}; δ) \sqrt{E_{n} ((e^{- x} - e^{- t})^{4}; x)} \sqrt{E_{n} ((t - x)^{4}; x)} . \end{aligned} \end{matrix}$ $$\begin{array}{} \begin{split} \displaystyle n.E_{n}\left(\left|h(t,x)\right|\phi_{x}^2(t);x\right)\leq& 2n\omega^*(f'';\delta)E_{n}\left(\phi_{x}^2(t);x\right)\\ \displaystyle+&\frac{2n}{\delta^2}\omega^*(f'';\delta)\sqrt{E_{n}\left((e^{-x}-e^{-t})^4;x\right)}\sqrt{E_{n}\left((t-x)^4;x\right)}. \end{split} \end{array}$$

Choosing $\begin{matrix} δ = 1 / \sqrt{n} \end{matrix}$ $\begin{array}{} \displaystyle \delta=1/\sqrt{n} \end{array}$ and letting

$\begin{matrix} s_{n} (x) = \sqrt{n^{2} E_{n} ((e^{- x} - e^{- t})^{4}; x)} \sqrt{n^{2} E_{n} ((t - x)^{4}; x)}, \end{matrix}$ $$\begin{array}{} \displaystyle s_{n}(x)=\sqrt{n^2E_{n}\left((e^{-x}-e^{-t})^4;x\right)}\sqrt{n^2E_{n}\left((t-x)^4;x\right)}, \end{array} $$

we obtain our result which was claim in the theorem.□

Corollary 1

Let f, f˝ ∈ C^*[0, ∞). Then we get

$\begin{matrix} lim_{n \to \infty} n [E_{n} (f; x) - f (x)] = \frac{x}{2} [f^{'} (x) + f^{″} (x)] \end{matrix}$ $$\begin{array}{} \displaystyle \lim_{n\rightarrow \infty}n[E_{n}(f;x)-f(x)]=\frac{x}{2}[f'(x)+f''(x)] \end{array}$$

for any x ∈ [0, ∞).

5

Conclusion

In this paper, it is studied the theoretical aspects of Jakimovski-Leviatan operators which reproduce constant and e^–x functions. A theorem for determining uniform convergence order of a quantitative estimate for the modified operators are presented. We also prove a quantitative Voronovskya type theorem. For the following studies, the convergence of the operators by illustrative graphics in Maple to certain functions are investigated.

Idioma:: Inglés

Calendario de la edición:: 1 veces al año
Temas de la revista:: Ciencias de la vida, Ciencias de la vida, otros, Matemáticas, Matemáticas aplicadas, Matemáticas generales, Física, Física, otros

RSS Feed de revista

Note On Jakimovski-Leviatan Operators Preserving e–x

Ecem Acar

Aydın İzgi

Sevilay Kirci Serenbay

Publicado en línea: 26 dic 2019

Páginas: 543 - 550

Recibido: 26 abr 2019

Aceptado: 22 jul 2019

DOI: https://doi.org/10.2478/AMNS.2019.2.00051

Palabras claveJakimovski-Leviatan operators, Exponential functions, Quantitative asymptotic formula, Uniform convergence, Voronovskya type theorem

© 2019 Ecem Acar et al., published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 Public License.

Note On Jakimovski-Leviatan Operators Preserving e^–x

Palabras clave
Jakimovski-Leviatan operators, Exponential functions, Quantitative asymptotic formula, Uniform convergence, Voronovskya type theorem