The proof of this theorem is similar to those of Goldman et al. [
1] and [
22]. Let
\(\digamma_{k}^{p,q} ( x,t ) \) and
\(\mathbf{F}_{k}^{p,q} ( x,t,c ) \) be generating functions of
\((p,q)\)-Bernstein polynomials. So we have the following formal definitions, respectively:
$$ \digamma_{k}^{p,q} ( x,t ) =\sum _{n=k}^{\infty }B_{k,n}(x;p,q)\frac{t^{n}}{ [ n ] _{p,q}!} $$
(3.4)
and
$$ \mathbf{F}_{k}^{p,q} ( x,t,c ) =\sum _{n=k}^{\infty }B_{k,n}(x;p,q)\frac{t^{n}}{ ( ( 1,c ) ; ( p,q ) ) _{n}}. $$
(3.5)
By substituting the right-hand side of (
3.2) into (
3.5), we have
$$\begin{aligned} &\mathbf{F}_{k}^{p,q} ( x,t,c ) \\ &\quad=\sum _{n=k}^{\infty }\frac{p^{\binom{k}{2}-\binom{n}{2}} ( ( 1,x ) ; ( p,q ) ) _{k} ( ( p^{k},q^{k+1} ) ; ( p,q ) ) _{n-k}x^{k}}{ ( ( p,q ) ; ( p,q ) ) _{n-k}}\frac{t^{n}}{ ( ( 1,c ) ; ( p,q ) ) _{n}} \\ &\quad=\frac{x^{k}t^{k}p^{\binom{k}{2}}}{ ( ( 1,c ) ; ( p,q ) ) _{k}}\sum_{n=k}^{\infty} \frac{p^{-\binom{n}{2}} ( ( 1,x ) ; ( p,q ) ) _{k} ( ( p^{k},q^{k+1} ) ; ( p,q ) ) _{n-k}}{ ( ( p,q ) ; ( p,q ) ) _{n-k} ( ( p^{k},cq^{k} ) ; ( p,q ) ) _{n-k}}t^{n-k} \\ &\quad={}_{2}\Phi_{1} \bigl( ( 1,x ), \bigl( p^{k+1},q^{k+1} \bigr) ; \bigl( p^{k},cq^{k} \bigr), ( p,q ),t \bigr) \frac {x^{k}t^{k}p^{\binom{k}{2}-\binom{n}{2}}}{ ( ( 1,c ) ; ( p,q ) ) _{k}}. \end{aligned}$$
(3.6)
From (
3.4) and (
3.5), we have the following equality:
$$\begin{aligned} \digamma_{k}^{p,q} ( x,t ) &=\sum _{n=k}^{\infty }B_{k,n}(x;p,q)\frac{t^{n}}{ [ n ] _{p,q}!} \\ &=\sum_{n=k}^{\infty}B_{k,n}(x;p,q) \frac{ ( p-q ) ^{n}t^{n}}{ ( ( p,q ) ; ( p,q ) ) _{n}} \\ &=\mathbf{F}_{k}^{p,q} \bigl( ( 1,x ), ( p-q ) t, ( p,q ) \bigr). \end{aligned}$$
(3.7)
From the right-hand side of (
3.7), we get
$$\begin{aligned} &\mathbf{F}_{k}^{p,q} \bigl( ( 1,x ), ( p-q ) t, ( p,q ) \bigr) \\ &\quad=\sum_{n=k}^{\infty}\frac{p^{\binom{k}{2}-\binom{n}{2}} ( ( 1,x ) ; ( p,q ) ) _{k} ( ( p^{k},q^{k+1} ) ; ( p,q ) ) _{n-k}x^{k}}{ ( ( p,q ) ; ( p,q ) ) _{n-k} ( ( p,q ) ; ( p,q ) ) _{n-k}} \frac{ ( p-q ) ^{n}t^{n}}{ ( ( p,q ) ; ( p,q ) ) _{n}} \\ &\quad=\frac{p^{\binom{k}{2}} ( x ( p-q ) t ) ^{k}}{ ( ( p,q ) ; ( p,q ) ) _{k}}\sum_{n=k}^{\infty } \frac{ ( ( 1,x ) ; ( p,q ) ) _{n-k} ( ( p^{k},q^{k+1} ) ; ( p,q ) ) _{n-k}t^{n-k}}{ ( ( p,q ) ; ( p,q ) ) _{n-k} ( ( p^{k+1},q^{k+1} ) ; ( p,q ) ) _{n-k}} \\ &\quad={}_{1}\Phi_{0} \bigl( ( 1,x ),-; ( p,q ), ( p-q ) t \bigr) \frac{p^{\binom{k}{2}-\binom{n}{2}} ( x ( p-q ) t ) ^{k}}{ ( ( p,q ) ; ( p,q ) ) _{k}} \end{aligned}$$
by using the definition of
\((p,q)\)-hypergeometric functions
$$\begin{aligned} &=\frac{p^{\binom{k}{2}-\binom{n}{2}} ( x ( p-q ) t ) ^{k}}{ ( ( p,q ) ; ( p,q ) ) _{k}}\frac { ( ( p,x ( p-q ) t ), ( p,q ) ) _{\infty}}{ ( ( p, ( p-q ) t ), ( p,q ) ) _{\infty}} \\ &=p^{\binom{k}{2}-\binom{n}{2}}\frac{ ( xt ) ^{k}}{ [ k ] _{p,q}!}\frac{ ( ( p,x ( p-q ) t ), ( p,q ) ) _{\infty}}{ ( ( p, ( p-q ) t ) , ( p,q ) ) _{\infty}} \\ &=p^{\binom{k}{2}-\binom{n}{2}}\frac{ ( xt ) ^{k}}{ [ k ] _{p,q}!}\frac{e_{p,q}( ( p-q ) t)}{e_{p,q}( ( p-q ) xt)}. \end{aligned}$$
From (
2.12), we obtain that
$$ \digamma_{k}^{p,q} ( x,t ) =p^{\binom{k}{2}-\binom {n}{2}} \frac{ ( xt ) ^{k}}{ [ k ] _{p,q}!}\frac{\varepsilon _{p,q}(t)}{\varepsilon_{p,q}(xt)}. $$
Therefore, we arrive at the desired result. □