On the convergence of a kind of q-gamma operators

Abstract

In this paper, we introduce a kind of q-gamma operators based on the concept of a q-integral. We estimate moments of these operators and establish direct and local approximation theorems of the operators. The estimates on the rate of convergence and weighted approximation of the operators are obtained, a Voronovskaya asymptotic formula is also presented.

MSC:41A10, 41A25, 41A36.

1 Introduction

In recent years, the applications of q-calculus in the approximation theory is one of the main areas of research. After q-Bernstein polynomials were introduced by Phillips [1] in 1997, many researchers have performed studies in this field; we mention some of them [1–4].

In 2007, Karsli [5] introduced and estimated the rate of convergence for functions with derivatives of bounded variation on $[0,\mathrm{\infty })$ of new gamma type operators as follows:

$$L_n(f;x)=\frac{(2n+3)!x^{n+3}}{n!(n+2)!}{\int }_0^{\mathrm{\infty }}\frac{t^n}{{(x+t)}^{2n+4}}f(t)dt,x>0.$$

(1)

In 2009, Karsli, Gupta and Izgi [6] gave an estimate of the rate of pointwise convergence of these operators (1) on a Lebesgue point of bounded variation function f defined on the interval $(0,\mathrm{\infty })$. In 2010, Karsli and Özarslan [7] obtained some direct local and global approximation results and gave a Voronoskaya-type theorem for the operators (1). As the application of q-calculus in approximation theory is an active field, it seems there are no papers mentioning the q analogue of these operators defined in (1). Inspired by Aral and Gupta [2], they defined a generalization of q-Baskakov type operators using q-Beta integral and obtained some important approximation properties, which motivates us to introduce the q analogue of this kind of gamma operators.

Before introducing the operators, we mention certain definitions based on q-integers; details can be found in [8, 9]. For any fixed real number $0<q\le 1$ and each nonnegative integer k, we denote q-integers by ${[k]}_q$, where

$${[k]}_q=\{\begin{array}{ll}\frac{1-q^k}{1-q},& q\ne 1,\\ k,& q=1.\end{array}$$

Also, q-factorial and q-binomial coefficients are defined as follows:

$$\begin{array}{c}{[k]}_q!=\{\begin{array}{ll}{[k]}_q{[k-1]}_q\cdots {[1]}_q,& k=1,2,\dots ,\\ 1,& k=0,\end{array}\hfill \\ {\left[\begin{array}{c}n\\ k\end{array}\right]}_q=\frac{{[n]}_q!}{{[k]}_q!{[n-k]}_q!}(n\ge k\ge 0).\hfill \end{array}$$

The q-improper integrals are defined as (see [10])

$${\int }_0^{\mathrm{\infty }/A}f(x)d_{q}x=(1-q)\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(\frac{q^n}{A}\right)\frac{q^n}{A},A>0,$$

provided the sums converge absolutely.

The q-beta integral is defined by

$$B_q(t;s)=K(A;t){\int }_0^{\mathrm{\infty }/A}\frac{x^{t-1}}{{(1+x)}_q^{t+s}}{d}_{q}x,$$

(2)

where $K(x;t)=\frac{1}{x+1}{x}^t{(1+\frac{1}{x})}_q^t{(1+x)}_q^{1-t}$ and ${(a+b)}_q^{\tau }={\prod }_{j=0}^{\tau -1}(a+q^{j}b)$, $\tau >0$.

In particular for any positive integer m, n,

$$K(x,n)=q^{\frac{n(n-1)}{2}},K(x,0)=1\text{and}{B}_q(m;n)=\frac{{\mathrm{\Gamma }}_q(m){\mathrm{\Gamma }}_q(n)}{{\mathrm{\Gamma }}_q(m+n)},$$

(3)

where ${\mathrm{\Gamma }}_q(t)$ is the q-gamma function satisfying the following functional equations:

$${\mathrm{\Gamma }}_q(t+1)={[t]}_q{\mathrm{\Gamma }}_q(t),{\mathrm{\Gamma }}_q(1)=1$$

(see [3]).

For $f\in C[0,\mathrm{\infty })$, $q\in (0,1)$ and $n\in \mathbb{N}$, we introduce a kind of q-gamma operators $G_{n,q}(f;x)$ as follows:

$$G_{n,q}(f;x)=\frac{{[2n+3]}_q!{(q^{n+\frac{3}{2}}x)}^{n+3}{q}^{\frac{n(n+1)}{2}}}{{[n]}_q!{[n+2]}_q!}{\int }_0^{\mathrm{\infty }/A}\frac{t^n}{{(q^{n+\frac{3}{2}}x+t)}_q^{2n+4}}f(t)d_{q}t.$$

(4)

Note that for $q\to 1^{-}$, $G_{n,1^{-}}(f;x)$ become the gamma operators defined in (1).

2 Some preliminary results

In order to obtain the approximation properties of the operators $G_{n,q}$, we need the following lemmas.

Lemma 1 For any $k\in \mathbb{N}$, $k\le n+2$ and $q\in (0,1)$, we have

$$G_{n,q}(t^k;x)=\frac{{[n+k]}_q!{[n-k+2]}_q!}{{[n]}_q!{[n+2]}_q!}{q}^{\frac{2k-k^2}{2}}{x}^k.$$

(5)

Proof Using the properties of q-beta integral, we have

$$\begin{array}{rcl}{G}_{n,q}(t^k;x)& =& \frac{{[2n+3]}_q!{(q^{n+\frac{3}{2}}x)}^{n+3}{q}^{\frac{n(n+1)}{2}}}{{[n]}_q!{[n+2]}_q!}{\int }_0^{\mathrm{\infty }/A}\frac{t^{n+k}}{{(q^{n+\frac{3}{2}}x+t)}_q^{2n+4}}{d}_{q}t\\ =& \frac{x^k{[2n+3]}_q!q^{\frac{n(n+1)}{2}}{q}^{k(n+\frac{3}{2})}}{{[n]}_q!{[n+2]}_q!}{\int }_0^{\mathrm{\infty }/A}\frac{{(\frac{t}{q^{n+\frac{3}{2}}x})}^{n+k}}{{(1+\frac{t}{q^{n+\frac{3}{2}}x})}_q^{2n+4}}{d}_q\left(\frac{t}{q^{n+\frac{3}{2}}x}\right)\\ =& \frac{x^k{[2n+3]}_q!q^{\frac{n(n+1)}{2}}{q}^{k(n+\frac{3}{2})}}{{[n]}_q!{[n+2]}_q!}\frac{B_q(n+k+1;n-k+3)}{K(A;n+k+1)}\\ =& \frac{{[n+k]}_q!{[n-k+2]}_q!}{{[n]}_q!{[n+2]}_q!}{q}^{\frac{2k-k^2}{2}}{x}^k.\end{array}$$

Lemma 1 is proved. □

Lemma 2 The following equalities hold:

$$G_{n,q}(1;x)=1,G_{n,q}(t;x)=\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}x,G_{n,q}(t^2;x)=x^2,$$

(6)

$$G_{n,q}(t^3;x)=\frac{{[n+3]}_q}{q^{3/2}{[n]}_q}{x}^3,G_{n,q}(t^4;x)=\frac{{[n+3]}_q{[n+4]}_q}{q^4{[n]}_q{[n-1]}_q}{x}^4,$$

(7)

$$G_{n,q}({(t-x)}^2;x)=2x^2(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}),$$

(8)

$$G_{n,q}({(t-x)}^4;x)=(1+\frac{{[n+3]}_q{[n+4]}_q}{q^4{[n]}_q{[n-1]}_q}-\frac{4{[n+3]}_q}{q^{3/2}{[n]}_q}-\frac{4\sqrt{q}{[n+1]}_q}{{[n+2]}_q})x^4.$$

(9)

Proof From Lemma 1, taking $k=0,1,2,3,4$, we get (6) and (7). Since $G_{n,q}({(t-x)}^2;x)=G_{n,q}(t^2;x)-2x{G}_{n,q}(t;x)+x^2$ and $G_{n,q}({(t-x)}^4;x)=G_{n,q}(t^4;x)-4x{G}_{n,q}(t^3;x)+6x^2{G}_{n,q}(t^2;x)-4x^3{G}_{n,q}(t;x)+x^4$, using (6), (7), we obtain (8) and (9) easily. □

Remark 1 Note that for $q\to 1^{-}$, from Lemma 2, we have

$$\begin{array}{c}{G}_{n,1^{-}}(1;x)=1,G_{n,1^{-}}(t;x)=\frac{n+1}{n+2}x,G_{n,1^{-}}(t^2;x)=x^2,\hfill \\ G_{n,1^{-}}({(t-x)}^2;x)=\frac{2}{n+2}{x}^2,\hfill \end{array}$$

which is the moments and central moments of the operators defined in (1).

3 Local approximation

In this section we establish direct and local approximation theorems in connection with the operators $G_{n,q}(f,x)$.

We denote the space of all real-valued continuous bounded functions f defined on the interval $[0,\mathrm{\infty })$ by $C_B[0,\mathrm{\infty })$. The norm $\parallel \cdot \parallel$ on the space $C_B[0,\mathrm{\infty })$ is given by $\parallel f\parallel =sup\{|f(x)|:x\in [0,\mathrm{\infty })\}$.

Further, let us consider Peetre’s K-functional

$$K_2(f,\delta )=\underset{g\in W^2}{inf}\{\parallel f-g\parallel +\delta \parallel g^{″}\parallel \},$$

where $\delta >0$ and $W^2=\{g\in C_B[0,\mathrm{\infty }):g^{\prime },g^{″}\in C_B[0,\mathrm{\infty })\}$.

For $f\in C_B[0,\mathrm{\infty })$, the modulus of continuity of second order is defined by

$${\omega }_2(f,\delta )=\underset{0<h\le \delta }{sup}\underset{x\in [0,\mathrm{\infty })}{sup}|f(x+2h)-2f(x+h)+f(x)|.$$

By [[11], p.177] there exists an absolute constant $C>0$ such that

$$K_2(f,\delta )\le C{\omega }_2(f,\sqrt{\delta }),\delta >0.$$

(10)

Our first result is a direct local approximation theorem for the operators $G_{n,q}(f,x)$.

Theorem 1 For $q\in (0,1)$, $x\in [0,\mathrm{\infty })$ and $f\in C_B[0,\mathrm{\infty })$, we have

$$|G_{n,q}(f;x)-f(x)|\le C{\omega }_2(f;\sqrt{{\alpha }_{n,q}(x)})+\omega (f;{\beta }_{n,q}(x)),$$

(11)

where C is a positive constant,

$${\alpha }_{n,q}(x)=(\frac{3}{4}-\frac{3\sqrt{q}{[n+1]}_q}{4{[n+2]}_q})x^2,{\beta }_{n,q}(x)=(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})x.$$

(12)

Proof

Let us define the auxiliary operators

$${\tilde{G}}_{n,q}(f;x)=G_{n,q}(f;x)-f\left(\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}x\right)+f(x),$$

(13)

$$x\in [0,\mathrm{\infty })$$

. The operators ${\tilde{G}}_{n,q}(f;x)$ are linear and preserve the linear functions

$${\tilde{G}}_{n,q}(t-x;x)=0$$

(14)

(see (6)).

Let $g\in C_B^2$. By Taylor’s expansion

$$g(t)=g(x)+g^{\prime }(x)(t-x)+{\int }_x^t(t-u)g^{″}(u)du,t\in [0,\mathrm{\infty }),$$

and (14), we get

$${\tilde{G}}_{n,q}(g;x)=g(x)+{\tilde{G}}_{n,q}({\int }_x^t(t-u)g^{″}(u)du;x).$$

Hence, by (13) and (8), we have

$$\begin{array}{r}|{\tilde{G}}_{n,q}(g;x)-g(x)|\\ \le \left|G_{n,q}({\int }_x^t(t-u)g^{″}(u)du;x)\right|+|{\int }_{\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}x}^x(u-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}x)g^{″}(u)du|\\ \le G_{n,q}\left(\right|{\int }_x^t(t-u)|g^{″}(u)|du|;x)+{\int }_{\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}x}^x|u-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}x||g^{″}(u)|du\\ \le [G_{n,q}({(t-x)}^2;x)+{(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})}^2{x}^2]\parallel g^{″}\parallel \\ =(3-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})x^2\parallel g^{″}\parallel \\ \le 3(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})x^2\parallel g^{″}\parallel .\end{array}$$

On the other hand, by (13), (4) and (6), we have

$$|{\tilde{G}}_{n,q}(f;x)|\le |G_{n,q}(f;x)|+2\parallel f\parallel \le \parallel f\parallel G_{n,q}(1;x)+2\parallel f\parallel \le 3\parallel f\parallel .$$

(15)

Now (13) and (15) imply

$$\begin{array}{r}|G_{n,q}(f;x)-f(x)|\\ \le |{\tilde{G}}_{n,q}(f-g;x)-(f-g)(x)|+|{\tilde{G}}_{n,q}(g;x)-g(x)|+|f\left(\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}x\right)-f(x)|\\ \le 4\parallel f-g\parallel +3(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})x^2\parallel g^{″}\parallel +\omega [f;(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})x].\end{array}$$

Hence taking infimum on the right-hand side over all $g\in W^2$, we get

$$|G_{n,q}(f;x)-f(x)|\le 4K_2[f;(\frac{3}{4}-\frac{3\sqrt{q}{[n+1]}_q}{4{[n+2]}_q})x^2]+\omega [f;(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})x].$$

By (10), for every $q\in (0,1)$, we have

$$|G_{n,q}(f;x)-f(x)|\le C{\omega }_2(f;\sqrt{{\alpha }_{n,q}(x)})+\omega (f;{\beta }_{n,q}(x)),$$

where ${\alpha }_{n,q}(x)$ and ${\beta }_{n,q}(x)$ are defined in (12). This completes the proof of Theorem 1. □

Remark 2 Let $q=\{q_n\}$ be a sequence satisfying $0<q_n<1$ and ${lim}_n{q}_n=1$, we have ${lim}_n{\alpha }_{n,q_n}=0$ and ${lim}_n{\beta }_{n,q_n}(x)=0$, these give us the pointwise rate of convergence of the operators $G_{n,q_n}(f;x)$ to $f(x)$.

4 Rate of convergence

Let $B_{x^2}[0,\mathrm{\infty })$ be the set of all functions f defined on $[0,\mathrm{\infty })$ satisfying the condition $|f(x)|\le M_f(1+x^2)$, where $M_f$ is a constant depending only on f. We denote the subspace of all continuous functions belonging to $B_{x^2}[0,\mathrm{\infty })$ by $C_{x^2}[0,\mathrm{\infty })$. Also, let $C_{x^2}^{\ast }[0,\mathrm{\infty })$ be the subspace of all functions $f\in C_{x^2}[0,\mathrm{\infty })$, for which ${lim}_{x\to \mathrm{\infty }}\frac{f(x)}{1+x^2}$ is finite. The norm on $C_{x^2}^{\ast }[0,\mathrm{\infty })$ is ${\parallel f\parallel }_{x^2}={sup}_{x\in [0,\mathrm{\infty })}\frac{|f(x)|}{1+x^2}$. We denote the usual modulus of continuity of f on the closed interval $[0,a]$ ($a>0$) by

$${\omega }_a(f,\delta )=\underset{|t-x|\le \delta }{sup}\underset{x,t\in [0,a]}{sup}|f(t)-f(x)|.$$

Obviously, for function $f\in C_{x^2}[0,\mathrm{\infty })$, the modulus of continuity ${\omega }_a(f,\delta )$ tends to zero.

Theorem 2 Let $f\in C_{x^2}[0,\mathrm{\infty })$, $q\in (0,1)$ and ${\omega }_{a+1}(f,\delta )$ be the modulus of continuity on the finite interval $[0,a+1]\subset [0,\mathrm{\infty })$, where $a>0$. Then we have

$$\begin{array}{r}{\parallel G_{n,q}(f)-f\parallel }_{C[0,a]}\\ \le 12M_f{a}^2(1+a^2)(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})+2{\omega }_{a+1}(f,\sqrt{2}a\sqrt{1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}}).\end{array}$$

(16)

Proof For $x\in [0,a]$ and $t>a+1$, we have

$$|f(t)-f(x)|\le M_f(2+x^2+t^2)\le M_f[2+3x^2+2{(t-x)}^2],$$

hence we obtain

$$|f(t)-f(x)|\le 6M_f(1+a^2){(t-x)}^2.$$

(17)

For $x\in [0,a]$ and $t\le a+1$, we have

$$|f(t)-f(x)|\le {\omega }_{a+1}(f,|t-x|)\le (1+\frac{|t-x|}{\delta }){\omega }_{a+1}(f,\delta ),\delta >0.$$

(18)

From (17) and (18), we get

$$|f(t)-f(x)|\le 6M_f(1+a^2){(t-x)}^2+(1+\frac{|t-x|}{\delta }){\omega }_{a+1}(f,\delta ).$$

(19)

For $x\in [0,a]$ and $t\ge 0$, by Schwarz’s inequality and Lemma 2, we have

$$\begin{array}{r}|G_{n,q}(f,x)-f(x)|\\ \le G_{n,q}(|f(t)-f(x)|,x)\\ \le 6M_f(1+a^2)G_{n,q}({(t-x)}^2,x)+{\omega }_{a+1}(f,\delta )(1+\frac{1}{\delta }\sqrt{G_{n,q}({(t-x)}^2,x)})\\ \le 12M_f{a}^2(1+a^2)(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})+{\omega }_{a+1}(f,\delta )(1+\frac{\sqrt{2}a}{\delta }\sqrt{1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}}).\end{array}$$

By taking $\delta =\sqrt{2}a\sqrt{1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}}$, we get the assertion of Theorem 2. □

5 Weighted approximation and Voronovskaya-type asymptotic formula

Now we will discuss the weighted approximation theorem.

Theorem 3 Let the sequence $q=\{q_n\}$ satisfy $0<q_n<1$ and $q_n\to 1$ as $n\to \mathrm{\infty }$, for $f\in C_{x^2}^{\ast }[0,\mathrm{\infty })$, we have

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel G_{n,q_n}(f)-f\parallel }_{x^2}=0.$$

(20)

Proof By using the Korovkin theorem in [12], we see that it is sufficient to verify the following three conditions.

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel G_{n,q_n}(t^v;x)-x^v\parallel }_{x^2},v=0,1,2.$$

(21)

Since $G_{n,q_n}(1;x)=1$ and $G_{n,q_n}(t^2;x)=x^2$, (20) holds true for $v=0$ and $v=2$.

Finally, for $v=1$, we have

$$\begin{array}{rcl}{\parallel G_{n,q_n}(t;x)-x\parallel }_{x^2}& =& \underset{x\in [0,\mathrm{\infty })}{sup}\frac{|G_{n,q_n}(t;x)-x|}{1+x^2}\\ =& (1-\frac{\sqrt{q_n}{[n+1]}_{q_n}}{{[n+2]}_{q_n}})\underset{x\in [0,\mathrm{\infty })}{sup}\frac{x}{1+x^2}\\ \le & 1-\frac{\sqrt{q_n}{[n+1]}_{q_n}}{{[n+2]}_{q_n}},\end{array}$$

since ${lim}_{n\to \mathrm{\infty }}{q}_n=1$, we get ${lim}_{n\to \mathrm{\infty }}\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q}=1$, so the condition of (21) holds for $v=1$ as $n\to \mathrm{\infty }$, then the proof of Theorem 3 is completed. □

Finally, we give a Voronovskaya-type asymptotic formula for $G_{n,q}(f;x)$ by means of the second and fourth central moments.

Theorem 4 Let $q:=\{q_n\}$ be a sequence satisfying $0<q_n<1$, ${lim}_n{q}_n=1$ and ${lim}_n{q}_n^n=1$. For $f\in C_{x^2}^2[0,\mathrm{\infty })$, ($f(x)$ is a twice differentiable function in $[0,\mathrm{\infty })$), the following equality holds

$$\underset{n\to \mathrm{\infty }}{lim}{[n+2]}_q(G_{n,q}(f;x)-f(x))=-f^{\prime }(x)x+f^{″}(x)x^2$$

(22)

for every $x\in [0,A]$, $A>0$.

Proof Let $x\in [0,\mathrm{\infty })$ be fixed. By the Taylor formula, we may write

$$f(t)=f(x)+f^{\prime }(x)(t-x)+\frac{1}{2}{f}^{″}(x){(t-x)}^2+r(t;x){(t-x)}^2,$$

(23)

where $r(t;x)$ is the Peano form of the remainder, $r(t;x)\in C_{x^2}[0,\mathrm{\infty })$ and using L’Hopital’s rule, we have

$$\begin{array}{rcl}\underset{t\to x}{lim}r(t;x)& =& \underset{t\to x}{lim}\frac{f(t)-f(x)-f^{\prime }(x)(t-x)-\frac{1}{2}{f}^{″}(x){(t-x)}^2}{{(t-x)}^2}\\ =& \underset{t\to x}{lim}\frac{f^{\prime }(t)-f^{\prime }(x)-f^{″}(x)(t-x)}{2(t-x)}=\underset{t\to x}{lim}\frac{f^{″}(t)-f^{″}(x)}{2}=0.\end{array}$$

Applying $G_{n,q}(f;x)$ to (23), we obtain

$$\begin{array}{rl}{[n+2]}_q(G_{n,q}(f;x)-f(x))=& f^{\prime }(x){[n+2]}_q{G}_{n,q}(t-x;x)+\frac{f^{″}(x)}{2}{[n+2]}_q{G}_{n,q}({(t-x)}^2;x)\\ +{[n+2]}_q{G}_{n,q}(r(t;x){(t-x)}^2;x).\end{array}$$

By the Cauchy-Schwarz inequality, we have

$$G_{n,q}(r(t;x){(t-x)}^2;x)\le \sqrt{G_{n,q}(r^2(t;x);x)}\sqrt{G_{n,q}({(t-x)}^4;x)}.$$

(24)

Since $r^2(x;x)=0$, then it follows from Theorem 3 that

$$\underset{n\to \mathrm{\infty }}{lim}{G}_{n,q}(r^2(t;x);x)=r^2(x;x)=0.$$

(25)

Now, from (24), (25) and Lemma 2, we get immediately

$$\underset{n\to \mathrm{\infty }}{lim}{[n+2]}_q{G}_{n,q}(r(t;x){(t-x)}^2;x)=0,\underset{n\to \mathrm{\infty }}{lim}{[n+2]}_q{G}_{n,q}(t-x;x)=-x,$$

and since $q^{n+1}={[n+2]}_q-{[n+1]}_q\le {[n+2]}_q-\sqrt{q}{[n+1]}_q\le {[n+2]}_q-q{[n+1]}_q=1$, we have

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}{[n+2]}_q{G}_{n,q}({(t-x)}^2;x)& =& \underset{n\to \mathrm{\infty }}{lim}{[n+2]}_q(1-\frac{\sqrt{q}{[n+1]}_q}{{[n+2]}_q})2x^2\\ =& \underset{n\to \mathrm{\infty }}{lim}({[n+2]}_q-\sqrt{q}{[n+1]}_q)2x^2=2x^2.\end{array}$$

Theorem 4 is proved. □