广义的单变量超几何级数定义如下,参见文献^{[2, 16]} \[_{1 + p}{F_q}\left[ {\begin{array}{*{20}{c}} {{a_0},{a_1}, \cdots ,{a_p}} \\ {{b_1}, \cdots ,{b_q}} \end{array}|z} \right] = \sum\limits_{k = 0}^\infty {\frac{{{{({a_0})}_k}{{({a_1})}_k} \cdots {{({a_p})}_k}}}{{k!{{({b_1})}_k} \cdots {{({b_q})}_k}}}} {z^k},\] 其中,当 $n=1,2,\cdots$时,升阶乘$(x)_n=x(x+1)\cdots(x+n-1)$,而当$n=0$时,$(x)_0=1$ .

由Srivastava和Daoust^[17]提出的双变量超几何级数定义为 \[\begin{gathered} F_{C:D;D'}^{A:B;B'}\left[ {\begin{array}{*{20}{c}} {[(a):\vartheta ,\varphi ]:[(b):\psi ];[(b'):\psi ']} \\ {[(c):\xi ,\eta ]:[(d):\zeta ];[(d'):\zeta ']} \end{array}|x,y} \right] \hfill \\ : = \sum\limits_{m,n = 0}^\infty {\frac{{\prod\limits_{j = 1}^A {{{({a_j})}_{m{\vartheta _j} + n{\varphi _j}}}} \prod\limits_{j = 1}^B {{{({b_j})}_{m{\psi _j}}}} \prod\limits_{j = 1}^{B'} {{{({b_{j'}})}_{n{\psi _{j'}}}}} }}{{\prod\limits_{j = 1}^C {{{({c_j})}_{m{\xi _j} + n{\eta _j}}}} \prod\limits_{j = 1}^D {{{({d_j})}_{m{\zeta _j}}}} \prod\limits_{j = 1}^{D'} {{{({d_{j'}})}_{n{\zeta _{j'}}}}} }}} \frac{{{x^m}{y^n}}}{{m!n!}}, \hfill \\ \end{gathered} \] 收敛条件为 $$ 1+\sum_{j=1}^C\xi_j+\sum_{j=1}^D\zeta_j-\sum_{j=1}^A\vartheta_j-\sum_{j=1}^B\psi_j\geq0, $$ $$ 1+\sum_{j=1}^C\eta_j+\sum_{j=1}^{D'}\zeta_j'-\sum_{j=1}^A\varphi_j-\sum_{j=1}^{B'}\psi_j'\geq0.$$ 当然,为了保证这个级数的收敛性,也可以适当的选取$|x|$和$|y|$的值,参见文献^[10].

当参数$\vartheta,\varphi,\psi,\psi',\xi,\eta,\zeta$和$\zeta'$ 全部取$1$时, 上述双变量超几何级数就是著名的Kampé de Fériet函数,参见文献^{[1, 14]} \[\begin{gathered} F_{C:D;D'}^{A:B;B'}\left[ {\begin{array}{*{20}{c}} {{\alpha _1}, \cdots ,{\alpha _A}:{a_1}, \cdots ,{a_B};{c_1}, \cdots ,{c_{B'}}} \\ {{\beta _1}, \cdots ,{\beta _C}:{b_1}, \cdots ,{b_D};{d_1}, \cdots ,{d_{D'}}} \end{array}|x,y} \right] \hfill \\ = \sum\limits_{m,n = 0}^\infty {\frac{{{{({\alpha _1}, \cdots ,{\alpha _A})}_{m + n}}{{({a_1}, \cdots ,{a_B})}_m}{{({c_1}, \cdots ,{c_{B'}})}_n}}}{{{{({\beta _1}, \cdots ,{\beta _C})}_{m + n}}{{({b_1}, \cdots ,{b_D})}_m}{{({d_1}, \cdots ,{d_{D'}})}_n}}}} \frac{{{x^m}{y^n}}}{{m!n!}}, \hfill \\ \end{gathered} \] 其中,$(a_1,a_2,\cdots,a_r)_n=(a_1)_n(a_2)_n\cdots(a_r)_n$.我们可以在文献中查阅到大量的Kampé de Fériet函数的简化和求和公式, 例如文献^{[3, 4, 6, 7, 11, 12, 13, 15]}.

在2005年,应用三个二次变换公式,Chen-Srivastava^[5] 推导出一些双变量超几何级数的简化公式. 最近,使用迭代技术, Srivastava等^[18]获得一系列Srivastava-Daoust双变量超几何级数的简化公式. 在这些工作的基础上,本文将建立两个一般的双变量超几何级数的变换公式, 以此推导一批类型为$F_{1:0;\mu}^{1:1;\lambda}$和$F_{2:0;\mu}^{2:1;\lambda}$ 的超几何级数的简化公式. 本文中,由于参数 $\psi,\psi',\zeta$和$\zeta'$都为$1$,我们将在$F_{C:D;D'}^{A:B;B'}$的表示式中略去它们.

定理 2.1  (变换公式) 对任意复数序列$\{{\Omega}(j)\}$,下列变换公式成立 \[\sum\limits_{i,j = 0}^\infty \Omega (j)\frac{{{{(\alpha )}_{i + 2j}}{{(\beta )}_i}{{(\frac{{\alpha + 1}}{2} - \beta )}_j}}}{{{{(1 + \alpha - \beta )}_{i + 2j}}}}\frac{{{x^i}{y^j}}}{{i!j!}}\tag{2.1a}\] \[\begin{array}{*{20}{l}} { = {{(1 - x)}^{ - \alpha }}\sum\limits_{n = 0}^\infty {\frac{{{{(\frac{\alpha }{2})}_n}{{(\frac{{\alpha + 1}}{2} - \beta )}_n}}}{{n!{{(1 + \alpha - \beta )}_n}}}} {{\left( {\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right)}^n}} \\ { \times \sum\limits_{j = 0}^n \Omega (j)\frac{{{{(\frac{{\alpha + 1}}{2})}_j}{{( - n)}_j}}}{{j!{{(1 + \alpha - \beta + n)}_j}}}{{\left( {\frac{y}{x}} \right)}^j},{\text{ }}} \tag{2.1b}\end{array}\] 这里,假设所有的级数都是收敛的.

证  将式(2.1a)重新表示为 \[\sum\limits_{j = 0}^\infty \Omega (j){\frac{{{{(\alpha )}_{2j}}{{(\frac{{\alpha + 1}}{2} - \beta )}_j}{y^j}}}{{j!{{(1 + \alpha - \beta )}_{2j}}}}_2}{F_1}\left[ {\begin{array}{*{20}{c}} {\alpha + 2j,\beta } \\ {1 + \alpha - \beta + 2j} \end{array}|x} \right].\tag{2.2}\] 将式(2.2)中的$_2F_1$ -级数应用如下变换公式,参见文献^[9]和[2,2.3节 (2)式]) \[_2{F_1}\left[ {\begin{array}{*{20}{c}} {a,b} \\ {1 + a - b} \end{array}|z} \right] = (1 - z)_2^{ - a}{F_1}\left[ {\begin{array}{*{20}{c}} {\frac{a}{2},\frac{1}{2} + \frac{a}{2} - b} \\ {1 + a - b} \end{array}|\frac{{ - 4z}}{{{{(1 - z)}^2}}}} \right],\tag{2.3}\] 我们可以计算式(2.1a)为 \[{(1 - x)^{ - \alpha }}\sum\limits_{i,j = 0}^\infty \Omega (j)\frac{{{{(\frac{\alpha }{2})}_{i + j}}{{(\frac{{\alpha + 1}}{2} - \beta )}_{i + j}}{{(\frac{{\alpha + 1}}{2})}_j}}}{{{{(1 + \alpha - \beta )}_{i + 2j}}}}\frac{{{{\left( {\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right)}^i}{{\left( {\frac{{4y}}{{{{(1 - x)}^2}}}} \right)}^j}}}{{i!j!}}.\] 将上式中的下标$i$置换为$n-j$并做一些简化计算,可得公式(2.1b).

推论 2.1  (简化公式)\[\begin{gathered} F_{1:0;0}^{1:1;1}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2]:\beta ;\frac{{\alpha + 1}}{2} - \beta } \\ {[1 + \alpha - \beta :1,2]: - ; - } \end{array}|x,x} \right] \hfill \\ = (1 - x)_3^{ - \alpha }{F_2}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{1 + \alpha - 2\beta }}{4},\frac{{3 + \alpha - 2\beta }}{4}} \\ {\frac{{1 + \alpha - \beta }}{2},\frac{{\alpha - \beta }}{2} + 1} \end{array}|\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right]. \hfill \\ \end{gathered} \]

证  我们列出Vandermonde-Chu恒等式如下,参见文献[2,2.3节]和[16,(1.7.7)式] \[_2{F_1}\left[ {\begin{array}{*{20}{c}} { - n,a} \\ b \end{array}|1} \right] = \frac{{{{(b - a)}_n}}}{{{{(b)}_n}}}.\tag{2.4}\] 在定理2.1中,令Ω(j)= 1并且y = x,利用Vandermonde-Chu恒等式(2.4)计算公式(2.1b)中的内和为 $${{{{({{\alpha + 1} \over 2} - \beta + n)}_n}} \over {{{(1 + \alpha - \beta + n)}_n}}}.$$ 经过化简,可得结论.

推论 2.2  (简化公式) \[\begin{gathered} F_{1:0;0}^{1:1;1}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2]:\beta ;\alpha - \beta } \\ {[1 + \alpha - \beta :1,2]: - ; - } \end{array}|x,x} \right] \hfill \\ = (1 - x)_2^{ - \alpha }{F_1}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{1}{2} - \frac{\beta }{2}} \\ {1 + \frac{{\alpha - \beta }}{2}} \end{array}|\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right]. \hfill \\ \end{gathered} \]

证  在定理2.1中,令$y=x$, \begin{eqnarray*} {\Omega}(j)=\frac{(\alpha-\beta)_j}{(\frac{\alpha+1}{2}-\beta)_j} , \end{eqnarray*} 利用如下Dixon定理,参见文献[16,Ⅲ. 9] \[_3{F_2}\left[ {\begin{array}{*{20}{c}} {a,b, - n} \\ {1 + a - b,1 + a + n} \end{array}|1} \right] = \frac{{{{(1 + a)}_n}{{(1 + \frac{a}{2} - b)}_n}}}{{{{(1 + \frac{a}{2})}_n}{{(1 + a - b)}_n}}},\tag{2.5}\] 计算(2.1b)式的内和,化简后,即得结论.

推论 2.3  (求和公式) \[F_{1:0;1}^{1:1;2}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2]:{\text{ }}\beta ;{\text{ }}\alpha - \beta ,1 + \frac{{\alpha - \beta }}{2}} \\ {[1 + \alpha - \beta :1,2]:{\text{ }} - ;{\text{ }}\frac{{\alpha - \beta }}{2}} \end{array}|x, - x} \right] = {(1 + x)^{ - \alpha }}.\]

证  在定理2.1中,令$y=-x$, \begin{eqnarray*} {\Omega}(j)=\frac{(\alpha-\beta)_j(1+\frac{\alpha-\beta}{2})_j} {(\frac{\alpha-\beta}{2})_j(\frac{\alpha+1}{2}-\beta)_j} , \end{eqnarray*} 公式(2.1b)中的内和可由下列公式计算,参见文献[16,Ⅲ. 11] \[_4{F_3}\left[ {\begin{array}{*{20}{c}} {a,1 + \frac{a}{2},b, - n} \\ {\frac{a}{2},1 + a - b,1 + a + n} \end{array}| - 1} \right] = \frac{{{{(1 + a)}_n}}}{{{{(1 + a - b)}_n}}},\tag{2.6}\] 经过化简,我们能够得到上述求和公式.

值得注意的是,这个求和公式的右端不依赖于参数$\beta$.

推论 2.4  (简化公式) \[\begin{gathered} F_{1:0;2}^{1:1;3}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2]:\beta ;\alpha - \beta ,1 + \frac{{\alpha - \beta }}{2},\gamma } \\ {[1 + \alpha - \beta :1,2]: - ;\frac{{\alpha - \beta }}{2},1 + \alpha - \beta - \gamma } \end{array}|x,x} \right] \hfill \\ = (1 - x)_2^{ - \alpha }{F_1}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{1 + \alpha }}{2} - \beta - \gamma } \\ {1 + \alpha - \beta - \gamma } \end{array}|\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right]. \hfill \\ \end{gathered} \]

证  在定理2.1中,令$y=x$, \begin{eqnarray*} {\Omega}(j)=\frac{(\alpha-\beta)_j(1+\frac{\alpha-\beta}{2})_j(\gamma)_j} {(\frac{\alpha-\beta}{2})_j(\frac{\alpha+1}{2}-\beta)_j(1+\alpha-\beta-\gamma)_j} , \end{eqnarray*} 公式(2.1b)中的内和可由下列公式计算,参见文献[16,Ⅲ. 13] \[_5{F_4}\left[ {\begin{array}{*{20}{c}} {a,1 + \frac{a}{2},b,c, - n} \\ {\frac{a}{2},1 + a - b,1 + a - c,1 + a + n} \end{array}|1} \right] = \frac{{{{(1 + a)}_n}{{(1 + a - b - c)}_n}}}{{{{(1 + a - b)}_n}{{(1 + a - c)}_n}}},\tag{2.7}\] 推论得证.

推论 2.5  (简化公式) \[\begin{gathered} F_{1:0;1}^{1:1;2}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2]:\beta ;\frac{{1 + \alpha }}{2} - \beta ,1 + \frac{{1 + \alpha }}{4}} \\ {[1 + \alpha - \beta :1,2]: - ;\frac{{1 + \alpha }}{4}} \end{array}|x,x} \right] \hfill \\ = (1 - x)_3^{ - \alpha }{F_2}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{\alpha - 1 - 2\beta }}{4},\frac{{\alpha + 1 - 2\beta }}{4}} \\ {\frac{{1 + \alpha - \beta }}{2},\frac{{\alpha - \beta + 2}}{2}} \end{array}|\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right]. \hfill \\ \end{gathered} \]

证  在定理2.1中,令$y=x$,${\Omega}(j)=(1+\frac{1+\alpha}{4})_j/(\frac{1+\alpha}{4})_j$, 利用下列公式,参见文献[16,Ⅲ. 15]和[2,4.5节,(1.1)式] \[_3{F_2}\left[ {\begin{array}{*{20}{c}} {a,1 + \frac{a}{2}, - n} \\ {\frac{a}{2},b} \end{array}|1} \right] = \frac{{(b - a - 1 - n){{(b - a)}_{n - 1}}}}{{{{(b)}_n}}},\tag{2.8} \] 化简式子(2.1b),推论得证.

定理 3.1  (变换公式) 对任意复数序列$\{{\Omega}(j)\}$,下列变换公式成立 \[\sum\limits_{i,j = 0}^\infty \Omega (j)\frac{{{{(\alpha )}_{i + 2j}}{{(\beta )}_{i + j}}{{(\gamma )}_i}{{(1 + \alpha - \beta - \gamma )}_j}}}{{{{(1 + \alpha - \gamma )}_{i + 2j}}{{(1 + \alpha - \beta )}_{i + j}}}}\frac{{{x^i}{y^j}}}{{i!j!}}\tag{3.1a}\] \[\begin{array}{*{20}{l}} { = {{(1 - x)}^{ - \alpha }}\sum\limits_{n = 0}^\infty {\frac{{{{(\alpha )}_{2n}}{{(1 + \alpha - \beta - \gamma )}_n}}}{{n!{{(1 + \alpha - \beta )}_n}{{(1 + \alpha - \gamma )}_n}}}} {{\left( {\frac{{ - x}}{{{{(1 - x)}^2}}}} \right)}^n}onumber} \\ {[4mm] \times \sum\limits_{j = 0}^n {\frac{{\Omega (j){{(\beta )}_j}{{( - n)}_j}}}{{j!{{(1 + \alpha - \gamma + n)}_j}}}} {{\left( {\frac{y}{x}} \right)}^j},} \tag{3.1b}\end{array}\] 这里假设所有级数均收敛.

证  利用Whipple二次变换公式(参见文献[8,p190,4.5(1)式]) \[_3{F_2}\left[ {\begin{array}{*{20}{c}} {\alpha ,\beta ,\gamma } \\ {1 + \alpha - \beta ,1 + \alpha - \gamma } \end{array}{\text{|}}z} \right] = (1 - z)_3^{ - \alpha }{F_2}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{\alpha + 1}}{2},1 + \alpha - \beta - \gamma } \\ {1 + \alpha - \beta ,1 + \alpha - \gamma } \end{array}{\text{|}}\frac{{ - 4z}}{{{{(1 - z)}^2}}}} \right],\] 计算公式(3.1a)如下 \[\begin{gathered} ({\text{3}}.{\text{1a}}) = \sum\limits_{j = 0}^\infty \Omega (j)\frac{{{{(\alpha )}_{2j}}{{(\beta )}_j}{{(1 + \alpha - \beta - \gamma )}_j}{y^j}}}{{j!{{(1 + \alpha - \beta )}_j}{{(1 + \alpha - \gamma )}_{2j}}}}{ \times _3}{F_2}\left[ {\begin{array}{*{20}{c}} {\alpha + 2j,\beta + j,\gamma } \\ {1 + \alpha - \beta + j,1 + \alpha - \gamma + 2j} \end{array}{\text{|}}x} \right] \hfill \\ = \sum\limits_{j = 0}^\infty \Omega (j)\frac{{{{(\alpha )}_{2j}}{{(\beta )}_j}{{(1 + \alpha - \beta - \gamma )}_j}{y^j}}}{{j!{{(1 + \alpha - \beta )}_j}{{(1 + \alpha - \gamma )}_{2j}}}}{(1 - x)^{ - \alpha - 2j}} \hfill \\ { \times _3}{F_2}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2} + j,\frac{{\alpha + 1}}{2} + j,1 + \alpha - \beta - \gamma + j} \\ {1 + \alpha - \beta + j,1 + \alpha - \gamma + 2j} \end{array}|\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right] \hfill \\ = {(1 - x)^{ - \alpha }}\sum\limits_{i,j = 0}^\infty \Omega (j)\frac{{{{(\alpha )}_{2i + 2j}}{{(1 + \alpha - \beta - \gamma )}_{i + j}}{{(\beta )}_j}}}{{{{(1 + \alpha - \beta )}_{i + j}}{{(1 + \alpha - \gamma )}_{i + 2j}}}}\frac{{{{\left( {\frac{{ - x}}{{{{(1 - x)}^2}}}} \right)}^i}{{\left( {\frac{y}{{{{(1 - x)}^2}}}} \right)}^j}}}{{i!j!}}. \hfill \\ \end{gathered} \] 置换下标$i$为$n-j$,化简后得到公式(3.1b).

推论 3.1  (简化公式) \[F_{2:0;0}^{2:1;1}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2],[\beta :1,1]:\gamma ;1 + \alpha - \beta - \gamma } \\ {[1 + \alpha - \gamma :1,2],[1 + \alpha - \beta :1,1]: - ; - } \end{array}{\text{|}}x,x} \right]\] \[ = (1 - x)_4^{ - \alpha }{F_3}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{1 + \alpha }}{2},\frac{{1 + \alpha - \beta - \gamma }}{2},\frac{{\alpha - \beta - \gamma }}{2} + 1} \\ {1 + \alpha - \beta ,\frac{{1 + \alpha - \gamma }}{2},\frac{{\alpha - \gamma }}{2} + 1} \end{array}|\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right],\tag{3.2}\] \[F_{2:0;1}^{2:1;2}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2],[\beta :1,1]:{\text{ }}\gamma ;{\text{ }}1 + \alpha - \beta - \gamma ,\alpha - \gamma } \\ {[1 + \alpha - \gamma :1,2],[1 + \alpha - \beta :1,1]:{\text{ }} - ;{\text{ }}1 + \alpha - \beta - \gamma } \end{array}|x,x} \right]\] \[ = (1 - x)_3^{ - \alpha }{F_2}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{1 + \alpha }}{2},1 + \frac{{\alpha - \gamma }}{2} - \beta } \\ {1 + \alpha - \beta ,1 + \frac{{\alpha - \gamma }}{2}} \end{array}|\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right],\tag{3.3}\] \[F_{2:0;1}^{2:1;2}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2],[\beta :1,1]:\gamma ;\alpha - \gamma ,1 + \frac{{\alpha - \gamma }}{2}} \\ {[1 + \alpha - \gamma :1,2],[1 + \alpha - \beta :1,1]: - ;\frac{{\alpha - \gamma }}{2}} \end{array}{\text{|}}x, - x} \right]\] \[ = (1 - x)_2^{ - \alpha }{F_1}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{1 + \alpha }}{2}} \\ {1 + \alpha - \beta } \end{array}{\text{|}}\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right],\tag{3.4}\] \[F_{2:0;2}^{2:1;3}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2],[\beta :1,1]:\gamma ;\alpha - \gamma ,1 + \frac{{\alpha - \gamma }}{2},\delta } \\ {[1 + \alpha - \gamma :1,2],[1 + \alpha - \beta :1,1]: - ;\frac{{\alpha - \gamma }}{2},1 + \alpha - \gamma - \delta } \end{array}|x,x} \right]\] \[ = (1 - x)_3^{ - \alpha }{F_2}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{1 + \alpha }}{2},1 + \alpha - \beta - \gamma - \delta } \\ {1 + \alpha - \beta ,1 + \alpha - \gamma - \delta } \end{array}{\text{|}}\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right],\tag{3.5}\] \[F_{2:0;1}^{2:1;2}\left[ {\begin{array}{*{20}{c}} {[\alpha :1,2],[\beta :1,1]:\gamma ;1 + \alpha - \beta - \gamma ,1 + \frac{\beta }{2}} \\ {[1 + \alpha - \gamma :1,2],[1 + \alpha - \beta :1,1]: - ;\frac{\beta }{2}} \end{array}|x,x} \right]\] \[ = (1 - x)_4^{ - \alpha }{F_3}\left[ {\begin{array}{*{20}{c}} {\frac{\alpha }{2},\frac{{1 + \alpha }}{2},\frac{{\alpha - \beta - \gamma }}{2},\frac{{1 + \alpha - \beta - \gamma }}{2}} \\ {1 + \alpha - \beta ,\frac{{1 + \alpha - \gamma }}{2},\frac{{\alpha - \gamma }}{2} + 1} \end{array}|\frac{{ - 4x}}{{{{(1 - x)}^2}}}} \right].\tag{3.6}\]

证  在定理3.1中,令${\Omega}(j)=1$,$y=x$, 利用Vandermonde-Chu恒等式(2.4)计算(3.1b)式的内和,易得简化公式(3.2).

在定理3.1中,令${\Omega}(j)=(\alpha-\gamma)_j/(1+\alpha-\beta-\gamma)_j$,$y=x$, (3.1b)式中的内和由公式(2.5)化简,即得简化公式(3.3).

在定理3.1中,令 \begin{eqnarray*} {\Omega}(j)=\frac{(\alpha-\gamma)_j(1+\frac{\alpha-\gamma}{2})_j} {(\frac{\alpha-\gamma}{2})_j(1+\alpha-\beta-\gamma)_j} \end{eqnarray*} 并且令$y=-x$,将公式(2.6)应用到计算(3.1b)式的内和,可得简化公式(3.4).

在定理3.1中,令 \begin{eqnarray*} {\Omega}(j)=\frac{(\alpha-\gamma)_j(1+\frac{\alpha-\gamma}{2})_j(\delta)_j} {(\frac{\alpha-\gamma}{2})_j(1+\alpha-\beta-\gamma)_j(1+\alpha-\gamma-\delta)_j}, \end{eqnarray*} 并且令$y=x$,利用公式(2.7)计算(3.1b)式的内和,即得简化公式(3.5).

在定理3.1中,令${\Omega}(j)=(1+\frac{\beta}{2})_j/(\frac{\beta}{2})_j$以及$y=x$, 利用公式(2.8)计算(3.1b)式的内和,可得简化公式(3.6).