Note on approximation properties of generalized Durrmeyer operators

Abstract

Purpose

To study the rate of convergence of q analogue of Durrmeyer operator generalization proposed by N. Deo.

Methods

We first estimate moments of q-Durrmeyer operators. We also study the rate of convergence our operators.

Results

We use Maple programming to draw the graphs for the approximation process for two operators. In all graphs, we observe that either classical operator has sharp convergence or both operators behave alike after a large number of iterations.

Conclusions

We conclude that the modified operator does not improve the approximation process.

Introduction

To approximate Lebesgue integrable functions on the interval [0,1], Durrmeyer introduced the integral modification of the well-known Bernstein polynomials. In 1981, Derriennic[1] first studied these operators in details. The Durrmeyer operators are defined as

$$D_n(f;x)=(n+1)\sum _{k=0}^n{p}_{n,k}\left(x\right)\underset{0}{\overset{1}{\int }}f\left(t\right)p_{n,k}\left(t\right)\mathit{\text{dt}}$$

(1)

where the Bernstein basis function is defined by$p_{n,k}\left(x\right)=\left(\begin{array}{c}n\\ k\end{array}\right)x^k{(1-x)}^{n-k}$. Recently, Deo et al.[2] introduced a new version of Durrmeyer operators as

$$M_n(f;x)=n{\left(1+\frac{1}{n}\right)}^2\sum _{k=0}^n{p}_{n,k}^{\ast }\left(x\right)\underset{0}{\overset{\frac{n}{n+1}}{\int }}f\left(t\right)p_{n,k}^{\ast }\left(t\right)\mathit{\text{dt}}$$

(2)

where

$$p_{n,k}^{\ast }\left(x\right)={\left(1+\frac{1}{n}\right)}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^k{\left(\frac{n}{n+1}-x\right)}^{n-k}$$

and$x\in [0,1-\frac{1}{n+1}]$. Several researchers have studied the q-analogue of Bernstein polynomials and established many interesting approximation properties. For important work in this direction, we refer the readers to[3–9], etc. In 2008, Gupta[10] introduced q-analogue of Durrmeyer operators as

$$D_{n,q}(f;x)=[n+1]\sum _{k=0}^n{q}^{-k}{p}_{n,k}(q;x)\underset{0}{\overset{1}{\int }}f\left(t\right)p_{n,k}(q;\mathit{\text{qt}})d_{q}t,$$

where

$$p_{\mathit{\text{nk}}}(q;x):=\left[\begin{array}{c}n\\ k\end{array}\right]x^k{(1-x)}_q^{n-k}.$$

In 2011, Gupta and Sharma[11] gave a recurrence formula and established better approximation for q-Durrmeyer operators. Before introducing the operators, we mention some basic definitions of q calculus (see[12]).

Let q > 0. For each nonnegative integer k, the q-integer [k] and the q-factorial [k]! are respectively defined by

$$\left[k\right]:=\left\{\begin{array}{l}(1-q^k)/(1-q),q\ne 1\\ k,q=1\end{array},\right.$$

and

$$\left[k\right]!:=\left\{\begin{array}{l}\left[k\right][k-1]···\left[1\right],k\ge 1\\ 1,k=0\end{array}\right..$$

For the integers n, k satisfying n ≥ k ≥ 0, the q-binomial coefficients are defined by

$$\left[\begin{array}{c}n\\ k\end{array}\right]:=\frac{\left[n\right]!}{\left[k\right]![n-k]!}.$$

We define

$${(a+b)}_q^n=\prod _{j=0}^{n-1}(a+q^{j}b)=(a+b)(a+\mathit{\text{qb}})\dots (a+q^{n-1}b).$$

The q-analogue of integration in the interval [0, a], discovered by Thomae[12], is defined by

$$\underset{0}{\overset{a}{\int }}f\left(t\right)d_{q}t:=a(1-q)\sum _{n=1}^{\infty }f\left(a{q}^n\right)q^{n}0<q<1.$$

We set

$$\begin{array}{ll}{p}_{n,k}^{\ast }(q;x)& =\frac{{[n+1]}^n}{{\left[n\right]}^n}\left[\begin{array}{c}n\\ k\end{array}\right]x^k{\left(\frac{\left[n\right]}{[n+1]}-x\right)}_q^{n-k}\\ =\left[\begin{array}{c}n\\ k\end{array}\right]{\left(\frac{[n+1]}{\left[n\right]}x\right)}^k{\left(1-\frac{[n+1]}{\left[n\right]}x\right)}_q^{n-k}.\end{array}$$

For$f\in C\left[0,\frac{\left[n\right]}{[n+1]}\right],$ we introduce the following q-Durrmeyer operators as

$$D_{n,q}^{\ast }(f;x)=\frac{{[n+1]}^2}{\left[n\right]}\sum _{k=0}^n{q}^{-k}{p}_{n,k}^{\ast }(q;x)\underset{0}{\overset{\frac{\left[n\right]}{[n+1]}}{\int }}f\left(t\right)p_{n,k}^{\ast }(q;\mathit{\text{qt}})d_q\mathrm{t.}$$

(3)

It can be easily verified that in case q = 1, the operators defined by Equation 3 reduce to generalized Durrmeyer operators (Equation 2).

Methods

Estimation of moments

Remark 1

For s = 0,1,… and by the definition of q-Beta function (see[12]), we have

$$\begin{array}{ll}\underset{0}{\overset{\frac{\left[n\right]}{[n+1]}}{\int }}{t}^s{p}_{n,k}^{\ast }(q;\mathit{\text{qt}})d_{q}t& =\frac{{[n+1]}^n}{{\left[n\right]}^n}\left[\begin{array}{c}n\\ k\end{array}\right]q^k\underset{0}{\overset{\frac{\left[n\right]}{[n+1]}}{\int }}{t}^{k+s}\\ \times {\left(\frac{\left[n\right]}{[n+1]}-\mathit{\text{qt}}\right)}_q^{n-k}{d}_{q}t\\ =\frac{{\left[n\right]}^{s+1}}{{[n+1]}^{s+1}}\left[\begin{array}{c}n\\ k\end{array}\right]q^k\\ \times \underset{0}{\overset{1}{\int }}{t}^{k+s}{\left(1-\mathit{\text{qt}}\right)}_q^{n-k}{d}_{q}t\\ =\frac{{\left[n\right]}^{s+1}}{{[n+1]}^{s+1}}\left[\begin{array}{c}n\\ k\end{array}\right]q^k\frac{[k+s]![n-k]!}{[s+n+1]!}\\ =\frac{{\left[n\right]}^{s+1}}{{[n+1]}^{s+1}}{q}^k\frac{\left[n\right]![k+s]!}{\left[k\right]![s+n+1]!}.\end{array}$$

Theorem 1

We have

$$\begin{array}{l}{D}_{n,q}^{\ast }(1;x)=1,\\ D_{n,q}^{\ast }(t;x)=\frac{\left[n\right]}{[n+1][n+2]}(1+\mathit{\text{qx}}[n+1\left]\right),\\ D_{n,q}^{\ast }(t^2;x)=\frac{(1+q){\left[n\right]}^2+q{(1+q)}^{2}x[n+1]{\left[n\right]}^2+q^4{[n+1]}^2{x}^2\left[n\right][n-1]}{{[n+1]}^2[n+2][n+3]}.\end{array}$$

Proof

Using Remark 1, the first identity can be proved easily.Consider

$$\begin{array}{ll}{D}_{n,q}^{\ast }(t;x)& =\frac{{[n+1]}^2}{\left[n\right]}\sum _{k=0}^n{q}^{-k}{p}_{n,k}^{\ast }(q;x)\\ \times \frac{{\left[n\right]}^2}{{[n+1]}^2}{q}^k\frac{\left[n\right]![k+1]!}{\left[k\right]![n+2]!}\\ =\frac{\left[n\right]}{[n+1]\left[n+2\right]}\sum _{k=0}^n\left[\begin{array}{c}n\\ k\end{array}\right]{\left(\frac{[n+1]}{\left[n\right]}x\right)}^k\\ \times {\left(1-\frac{[n+1]}{\left[n\right]}x\right)}_q^{n-k}[k+1]\\ =\frac{\left[n\right]}{[n+1]\left[n+2\right]}(1+q[n+1\left]x\right).\end{array}$$

Further,

$$\begin{array}{ll}{D}_{n,q}^{\ast }(t^2;x)& =\frac{{[n+1]}^2}{\left[n\right]}\sum _{k=0}^n{q}^{-k}{p}_{n,k}^{\ast }(q;x)\frac{{\left[n\right]}^3}{{[n+1]}^3}{q}^k\\ \times \frac{\left[n\right]![k+2]!}{\left[k\right]![n+3]!}\\ =\frac{{\left[n\right]}^2}{{[n+1]}^2[n+2][n+3]}\\ \times \sum _{k=0}^n{p}_{n,k}^{\ast }(q;x)[k+1][k+2]\\ =\frac{{\left[n\right]}^2}{{[n+1]}^2[n+2][n+3]}\\ \times \sum _{k=0}^n{p}_{n,k}^{\ast }(q;x)(1+q+q^2[k\left]\right)(1+q[k\left]\right)\\ =\frac{{\left[n\right]}^2}{{[n+1]}^2[n+2][n+3]}\\ \times \sum _{k=0}^n{p}_{n,k}^{\ast }(q;x)(1+q+q(1+2q+q^2\left)\right[k]\\ +q^4\left[k\right][k-1])\\ =\frac{1}{{[n+1]}^2[n+2][n+3]}\left((1+q){\left[n\right]}^2\right.\\ +q{(1+q)}^{2}x[n+1]{\left[n\right]}^2\\ \left.+q^4{[n+1]}^2{x}^2\left[n\right][n-1]\right).\end{array}$$

□

Remark 2

We have the following central moments:

$$\begin{array}{ll}{D}_{n,q}^{\ast }(t-x;x)& =\frac{\left[n\right]}{[n+1][n+2]}(1+\mathit{\text{qx}}[n+1\left]\right)-x,\\ D_n^{\ast }({(t-x)}^2;x)& =x^2\left(1+\frac{q^4\left[n\right][n-1]}{[n+2][n+3]}-\frac{2q\left[n\right]}{[n+2]}\right)\\ +x\left(\frac{q{(1+q)}^2{\left[n\right]}^2}{[n+1][n+2][n+3]}\right.\\ \left.-\frac{2\left[n\right]}{[n+1][n+2]}\right)\\ +\frac{(1+q){\left[n\right]}^2}{{[n+1]}^2[n+2][n+3]}.\end{array}$$

Remark 3

For the special case q = 1, we have the following moments:

$$\begin{array}{ll}{D}_n^{\ast }(1;x)& =1,\\ D_n^{\ast }(t;x)& =\frac{n}{(n+1)(n+2)}(1+x(n+1\left)\right),\\ D_n^{\ast }(t^2;x)& =\frac{1}{{(n+1)}^2(n+2)(n+3)}\\ \times \left(2n^2+4x(n+1)n^2+{(n+1)}^2{x}^2(n^2-n)\right).\end{array}$$

Remark 4

For the special case q = 1, we have the following central moments:

$$\begin{array}{ll}{D}_n^{\ast }(t-x;x)& =\frac{n}{(n+1)(n+2)}-\frac{2x}{n+2},\\ D_n^{\ast }({(t-x)}^2;x)& =\frac{1}{{(n+1)}^2(n+2)(n+3)}\\ +\frac{x^2(-2n^2+4n+6)+x(2n^2-6n)}{(n+1)(n+2)(n+3)}.\end{array}$$

Remark 5

It is observed that the operators$D_{n,q}^{\ast }(f;x)$ reproduce only the constant functions, not the linear ones.

Order of approximation

The modulus of continuity of f∈C[0,a], denoted by ω(f,δ), is defined by

$$\omega (f,\delta )=\underset{|x-y|\le \delta ;x,y\in [0,a]}{\text{sup}}\left|f\right(x)-f(y\left)\right|.$$

The modulus of continuity possesses the following property:

$$\omega (f,\mathrm{\lambda \delta })\le (1+\lambda )\omega (f,\delta ).$$

(4)

Theorem 2

Let (q _n ) _n be a sequence satisfying q _n → 1 as n → ∞. Then,

$$\left|D_{n,q_n}^{\ast }\right(f;x)-f(x\left)\right|\le 2\omega (f,\surd {\delta }_n)$$

for all $f\in C\left[0,\frac{\left[n\right]}{[n+1]}\right]$ , where

$${\delta }_n=D_{n,q_n}^{\ast }\left({(t-x)}^2;x\right).$$

Proof

$$\left|D_{n,q_n}^{\ast }\right(f;x)-f(x\left)\right|\le D_{n,q_n}^{\ast }\left(\right|f\left(t\right)-f\left(x\right)|;x)$$

Also, in view of Equation 4,

$$\left|f\right(t)-f(x\left)\right|\le \left(1+\frac{{(t-x)}^2}{{\delta }^2}\right)\omega (f,\delta ).$$

(5)

By using Equations 3 and 5, we get

$$\begin{array}{ll}\left|D_{n,q_n}^{\ast }\right(f;x)-f(x\left)\right|& \le \frac{{[n+1]}_{q_n}^2}{{\left[n\right]}_{q_n}}\sum _{k=0}^n{q}^{-k}{p}_{n,k}^{\ast }(q_n;x)\\ \times \underset{0}{\overset{\frac{{\left[n\right]}_{q_n}}{{[n+1]}_{q_n}}}{\int }}\left|f\right(t)-f(x\left)\right|p_{n,k}^{\ast }(q_n;\mathit{\text{qt}})d_{q}t\\ =\left(D_{n,q_n}^{\ast }\left(1;x\right)+\frac{1}{{\delta }^2}{D}_{n,q_n}^{\ast }\left({(t-x)}^2;x\right)\right)\\ \times \omega (f,\delta ).\end{array}$$

But, by Remark 2,

$$\begin{array}{ll}{D}_{n,q_n}^{\ast }\left({(t-x)}^2;x\right)& =x^2\left(1+\frac{q_n^4{\left[n\right]}_{q_n}{[n-1]}_{q_n}}{{[n+2]}_{q_n}{[n+3]}_{q_n}}-\frac{2q_n{\left[n\right]}_{q_n}}{{[n+2]}_{q_n}}\right)\\ +x\left(\frac{q_n{(1+q_n)}^2{\left[n\right]}_{q_n}^2}{{[n+1]}_{q_n}{[n+2]}_{q_n}{[n+3]}_{q_n}}\right.\\ \left.-\frac{2{\left[n\right]}_{q_n}}{{[n+1]}_{q_n}{[n+2]}_{q_n}}\right)\\ +\frac{(1+q_n){\left[n\right]}_{q_n}^2}{{[n+1]}_{q_n}^2{[n+2]}_{q_n}{[n+3]}_{q_n}}.\end{array}$$

Therefore, we get

$$\underset{n\to \infty }{\text{lim}}{D}_{n,q_n}^{\ast }\left({(t-x)}^2;x\right)=0$$

because${\left[n\right]}_{q_n}\to \infty$ as q _n → 1. So, letting${\delta }_n=D_{n,q_n}^{\ast }\left({(t-x)}^2;x\right)$ and taking δ = √ δ _n , we finally get

$$\left|D_{n,q_n}^{\ast }\right(f;x)-f|\le 2\omega (f,\surd {\delta }_n).$$

□

As usual, f ∈ Li p _M (α), M > 0 and 0 < α < 1 , if the inequality

$$\left|f\right(t)-f(x\left)\right|\le M|t-x{|}^{\alpha }$$

(6)

holds for all t,x∈[0,1].

In the following theorem, we will compute the rate of convergence by means of Lipschitz class.

Theorem 3

For all f ∈ Lip _M (α) and $x\in \left[0,\frac{\left[n\right]}{[n+1]}\right]$ , we have

$$\left|D_{n,q_n}^{\ast }\right(f;x)-f(x\left)\right|\le M{\delta }_n^{\alpha /2},$$

where${\delta }_n=D_{n,q_n}^{\ast }({(t-x)}^2;x)$.

Proof

Using inequality (6) and Hölder’s inequality with$p=\frac{2}{\alpha }$,$q=\frac{2}{2-\alpha }$, we get

$$\begin{array}{ll}\left|D_{n,q_n}^{\ast }\right(f;x)-f(x\left)\right|& \le D_{n,q_n}^{\ast }\left(\right|f\left(t\right)-f\left(x\right)|;x)\\ \le M{D}_{n,q_n}^{\ast }\left(\right|t-x{|}^{\alpha };x)\\ \le M{D}_{n,q_n}^{\ast }{\left(\right|t-x{|}^2;x)}^{\alpha /2},\end{array}$$

taking${\delta }_n=D_{n,q_n}^{\ast }({(t-x)}^2;x)$, we get

$$\left|D_{n,q_n}^{\ast }\right(f;x)-f(x\left)\right|\le M{\delta }_n^{\alpha /2}.$$

□

Results and discussion

In this section, we try to check whether the new generalization of operator introduced by Deo et al.[2] and given by Equation 2 has better approximation properties than the classical operator given by Equation 1. We use Maple programming to plot curves for f(x) = x (Figure1) and f(x) =x² (Figure2) for the two operators. In all graphs, we observe that either classical operator has sharp convergence or both operators behave alike after a large number of iterations. Moreover, in[2], the modification interval of approximation is$\left[0,1-\frac{1}{n+1}\right]$, which is smaller than that of the classical operator.

Figure 1

Curves for f(x) = x . Red line, original curve; yellow line, classical operator; black line, generalized operator.

Figure 2

Curves for f(x) = x². Red line, original curve; yellow line, classical operator; black line, generalized operator.

Conclusion

So, one can conclude that the modified operator (Equation 2) does not improve the approximation process.