阿佩尔函数

${\displaystyle (a-b_{1}-b_{2})F_{1}(a,b_{1},b_{2},c;x,y)-a\,F_{1}(a+1,b_{1},b_{2},c;x,y)+b_{1}F_{1}(a,b_{1}+1,b_{2},c;x,y)+b_{2}F_{1}(a,b_{1},b_{2}+1,c;x,y)=0~,}$

${\displaystyle c\,F_{1}(a,b_{1},b_{2},c;x,y)-(c-a)F_{1}(a,b_{1},b_{2},c+1;x,y)-a\,F_{1}(a+1,b_{1},b_{2},c+1;x,y)=0~,}$

${\displaystyle c\,F_{1}(a,b_{1},b_{2},c;x,y)+c(x-1)F_{1}(a,b_{1}+1,b_{2},c;x,y)-(c-a)x\,F_{1}(a,b_{1}+1,b_{2},c+1;x,y)=0~,}$

${\displaystyle c\,F_{1}(a,b_{1},b_{2},c;x,y)+c(y-1)F_{1}(a,b_{1},b_{2}+1,c;x,y)-(c-a)y\,F_{1}(a,b_{1},b_{2}+1,c+1;x,y)=0~.}$

其它式子[1]可從這四個關係導出。

${\displaystyle c\,F_{3}(a_{1},a_{2},b_{1},b_{2},c;x,y)+(a_{1}+a_{2}-c)F_{3}(a_{1},a_{2},b_{1},b_{2},c+1;x,y)-a_{1}F_{3}(a_{1}+1,a_{2},b_{1},b_{2},c+1;x,y)-a_{2}F_{3}(a_{1},a_{2}+1,b_{1},b_{2},c+1;x,y)=0~,}$

${\displaystyle c\,F_{3}(a_{1},a_{2},b_{1},b_{2},c;x,y)-c\,F_{3}(a_{1}+1,a_{2},b_{1},b_{2},c;x,y)+b_{1}x\,F_{3}(a_{1}+1,a_{2},b_{1}+1,b_{2},c+1;x,y)=0~,}$

${\displaystyle c\,F_{3}(a_{1},a_{2},b_{1},b_{2},c;x,y)-c\,F_{3}(a_{1},a_{2}+1,b_{1},b_{2},c;x,y)+b_{2}y\,F_{3}(a_{1},a_{2}+1,b_{1},b_{2}+1,c+1;x,y)=0~,}$

${\displaystyle c\,F_{3}(a_{1},a_{2},b_{1},b_{2},c;x,y)-c\,F_{3}(a_{1},a_{2},b_{1}+1,b_{2},c;x,y)+a_{1}x\,F_{3}(a_{1}+1,a_{2},b_{1}+1,b_{2},c+1;x,y)=0~,}$

${\displaystyle c\,F_{3}(a_{1},a_{2},b_{1},b_{2},c;x,y)-c\,F_{3}(a_{1},a_{2},b_{1},b_{2}+1,c;x,y)+a_{2}y\,F_{3}(a_{1},a_{2}+1,b_{1},b_{2}+1,c+1;x,y)=0~.}$

${\displaystyle {\frac {\partial }{\partial x}}F_{1}(a,b_{1},b_{2},c;x,y)={\frac {ab_{1}}{c}}F_{1}(a+1,b_{1}+1,b_{2},c+1;x,y)~,}$

${\displaystyle {\frac {\partial }{\partial y}}F_{1}(a,b_{1},b_{2},c;x,y)={\frac {ab_{2}}{c}}F_{1}(a+1,b_{1},b_{2}+1,c+1;x,y)~.}$

${\displaystyle \left(x(1-x){\frac {\partial ^{2}}{\partial x^{2}}}+y(1-x){\frac {\partial ^{2}}{\partial x\partial y}}+[c-(a+b_{1}+1)x]{\frac {\partial }{\partial x}}-b_{1}y{\frac {\partial }{\partial y}}-ab_{1}\right)F_{1}(x,y)=0~,}$

${\displaystyle \left(y(1-y){\frac {\partial ^{2}}{\partial y^{2}}}+x(1-y){\frac {\partial ^{2}}{\partial x\partial y}}+[c-(a+b_{2}+1)y]{\frac {\partial }{\partial y}}-b_{2}x{\frac {\partial }{\partial x}}-ab_{2}\right)F_{1}(x,y)=0~.}$

${\displaystyle {\frac {\partial }{\partial x}}F_{3}(a_{1},a_{2},b_{1},b_{2},c;x,y)={\frac {a_{1}b_{1}}{c}}F_{3}(a_{1}+1,a_{2},b_{1}+1,b_{2},c+1;x,y)~,}$

${\displaystyle {\frac {\partial }{\partial y}}F_{3}(a_{1},a_{2},b_{1},b_{2},c;x,y)={\frac {a_{2}b_{2}}{c}}F_{3}(a_{1},a_{2}+1,b_{1},b_{2}+1,c+1;x,y)~.}$

${\displaystyle \left(x(1-x){\frac {\partial ^{2}}{\partial x^{2}}}+y{\frac {\partial ^{2}}{\partial x\partial y}}+[c-(a_{1}+b_{1}+1)x]{\frac {\partial }{\partial x}}-a_{1}b_{1}\right)F_{3}(x,y)=0~,}$

${\displaystyle \left(y(1-y){\frac {\partial ^{2}}{\partial y^{2}}}+x{\frac {\partial ^{2}}{\partial x\partial y}}+[c-(a_{2}+b_{2}+1)y]{\frac {\partial }{\partial y}}-a_{2}b_{2}\right)F_{3}(x,y)=0~.}$

${\displaystyle F_{1}(a,b_{1},b_{2},c;x,y)={\frac {\Gamma (c)}{\Gamma (a)\Gamma (c-a)}}\int _{0}^{1}t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b_{1}}(1-yt)^{-b_{2}}\,\mathrm {d} t,\quad \Re \,c>\Re \,a>0~.}$

${\displaystyle F(\phi ,k)=\int _{0}^{\phi }{\frac {\mathrm {d} \theta }{\sqrt {1-k^{2}\sin ^{2}\theta }}}=\sin \phi \,F_{1}({\tfrac {1}{2}},{\tfrac {1}{2}},{\tfrac {1}{2}},{\tfrac {3}{2}};\sin ^{2}\phi ,k^{2}\sin ^{2}\phi ),\quad |\Re \,\phi |<{\frac {\pi }{2}}~,}$

${\displaystyle E(\phi ,k)=\int _{0}^{\phi }{\sqrt {1-k^{2}\sin ^{2}\theta }}\,\mathrm {d} \theta =\sin \phi \,F_{1}({\tfrac {1}{2}},{\tfrac {1}{2}},-{\tfrac {1}{2}},{\tfrac {3}{2}};\sin ^{2}\phi ,k^{2}\sin ^{2}\phi ),\quad |\Re \,\phi |<{\frac {\pi }{2}}~,}$

${\displaystyle \Pi (n,k)=\int _{0}^{\pi /2}{\frac {\mathrm {d} \theta }{(1-n\sin ^{2}\theta ){\sqrt {1-k^{2}\sin ^{2}\theta }}}}={\frac {\pi }{2}}\,F_{1}({\tfrac {1}{2}},1,{\tfrac {1}{2}},1;n,k^{2})~.}$

Q阿佩尔函数

1. 例如：${\displaystyle (y-x)F_{1}(a,b_{1}+1,b_{2}+1,c,x,y)=y\,F_{1}(a,b_{1},b_{2}+1,c,x,y)-x\,F_{1}(a,b_{1}+1,b_{2},c,x,y)}$

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