阿佩爾函數

阿佩爾函數是法國數學家（Paul Apell)在1880年為推廣高斯超幾何函數而創建的一組雙變數函數，定義如下

$$F_1(a,b_1,b_2,c;x,y) = \sum_{m,n=0}^\infty \frac{(a)_{m+n} (b_1)_m (b_2)_n} {(c)_{m+n} \,m! \,n!} \,x^m y^n ~,$$

$$F_2(a,b_1,b_2,c_1,c_2;x,y) = \sum_{m,n=0}^\infty \frac{(a)_{m+n} (b_1)_m (b_2)_n} {(c_1)_m (c_2)_n \,m! \,n!} \,x^m y^n ~,$$

$$F_3(a_1,a_2,b_1,b_2,c;x,y) = \sum_{m,n=0}^\infty \frac{(a_1)_m (a_2)_n (b_1)_m (b_2)_n} {(c)_{m+n} \,m! \,n!} \,x^m y^n ~,$$

$$F_4(a,b,c_1,c_2;x,y) = \sum_{m,n=0}^\infty \frac{(a)_{m+n} (b)_{m+n}} {(c_1)_m (c_2)_n \,m! \,n!} \,x^m y^n ~.$$

其中的符號$:(a)_{m+n}$是階乘冪

阿佩爾函數是嫪麗切拉函數和Kampé_de_Fériet函數的特例。

## 歸遞關係

$$(a-b_1-b_2) F_1(a,b_1,b_2,c; x,y) - a \,F_1(a+1,b_1,b_2,c; x,y) + b_1 F_1(a,b_1+1,b_2,c; x,y) + b_2 F_1(a,b_1,b_2+1,c; x,y) = 0 ~,$$

$$c \,F_1(a,b_1,b_2,c; x,y) - (c-a) F_1(a,b_1,b_2,c+1; x,y) - a \,F_1(a+1,b_1,b_2,c+1; x,y) = 0 ~,$$

$$c \,F_1(a,b_1,b_2,c; x,y) + c(x-1) F_1(a,b_1+1,b_2,c; x,y) - (c-a)x \,F_1(a,b_1+1,b_2,c+1; x,y) = 0 ~,$$

$$c \,F_1(a,b_1,b_2,c; x,y) + c(y-1) F_1(a,b_1,b_2+1,c; x,y) - (c-a)y \,F_1(a,b_1,b_2+1,c+1; x,y) = 0 ~.$$

其它式子

$(y-x) F_1(a, b_1+1, b_2+1,c,x,y) = y \, F_1(a,b_1,b_2+1,c,x,y) - x \, F_1(a,b_1+1,b_2,c,x,y)$

可從這四個關係導出。

$$c \,F_3(a_1,a_2,b_1,b_2,c; x,y) + (a_1+a_2-c) F_3(a_1,a_2,b_1,b_2,c+1; x,y) - a_1 F_3(a_1+1,a_2,b_1,b_2,c+1; x,y) - a_2 F_3(a_1,a_2+1,b_1,b_2,c+1; x,y) = 0 ~,$$

$$c \,F_3(a_1,a_2,b_1,b_2,c; x,y) - c \,F_3(a_1+1,a_2,b_1,b_2,c; x,y) + b_1 x \,F_3(a_1+1,a_2,b_1+1,b_2,c+1; x,y) = 0 ~,$$

$$c \,F_3(a_1,a_2,b_1,b_2,c; x,y) - c \,F_3(a_1,a_2+1,b_1,b_2,c; x,y) + b_2 y \,F_3(a_1,a_2+1,b_1,b_2+1,c+1; x,y) = 0 ~,$$

$$c \,F_3(a_1,a_2,b_1,b_2,c; x,y) - c \,F_3(a_1,a_2,b_1+1,b_2,c; x,y) + a_1 x \,F_3(a_1+1,a_2,b_1+1,b_2,c+1; x,y) = 0 ~,$$

$$c \,F_3(a_1,a_2,b_1,b_2,c; x,y) - c \,F_3(a_1,a_2,b_1,b_2+1,c; x,y) + a_2 y \,F_3(a_1,a_2+1,b_1,b_2+1,c+1; x,y) = 0 ~.$$

## 導數與微分方程

$$\frac {\partial} {\partial x} F_1(a,b_1,b_2,c; x,y) = \frac {a b_1} {c} F_1(a+1,b_1+1,b_2,c+1; x,y) ~,$$

$$\frac {\partial} {\partial y} F_1(a,b_1,b_2,c; x,y) = \frac {a b_2} {c} F_1(a+1,b_1,b_2+1,c+1; x,y) ~.$$

$$\left( x(1-x) \frac {\partial^2} {\partial x^2} + y(1-x) \frac {\partial^2} {\partial x \partial y} + [c - (a+b_1+1) x] \frac {\partial} {\partial x} - b_1 y \frac {\partial} {\partial y} - a b_1 \right) F_1(x,y) = 0 ~,$$

$$\left( y(1-y) \frac {\partial^2} {\partial y^2} + x(1-y) \frac {\partial^2} {\partial x \partial y} + [c - (a+b_2+1) y] \frac {\partial} {\partial y} - b_2 x \frac {\partial} {\partial x} - a b_2 \right) F_1(x,y) = 0 ~.$$

$$\frac {\partial} {\partial x} F_3(a_1,a_2,b_1,b_2,c; x,y) = \frac {a_1 b_1} {c} F_3(a_1+1,a_2,b_1+1,b_2,c+1; x,y) ~,$$

$$\frac {\partial} {\partial y} F_3(a_1,a_2,b_1,b_2,c; x,y) = \frac {a_2 b_2} {c} F_3(a_1,a_2+1,b_1,b_2+1,c+1; x,y) ~.$$

$$\left( x(1-x) \frac {\partial^2} {\partial x^2} + y \frac {\partial^2} {\partial x \partial y} + [c - (a_1+b_1+1) x] \frac {\partial} {\partial x} - a_1 b_1 \right) F_3(x,y) = 0 ~,$$

$$\left( y(1-y) \frac {\partial^2} {\partial y^2} + x \frac {\partial^2} {\partial x \partial y} + [c - (a_2+b_2+1) y] \frac {\partial} {\partial y} - a_2 b_2 \right) F_3(x,y) = 0 ~.$$

## 積分關係

$$F_1(a,b_1,b_2,c; x,y) = \frac{\Gamma(c)} {\Gamma(a) \Gamma(c-a)} \int_0^1 t^{a-1} (1-t)^{c-a-1} (1-xt)^{-b_1} (1-yt)^{-b_2} \,\mathrm{d}t, \quad \real \,c > \real \,a > 0 ~.$$

## 特例

$$F(\phi,k) = \int_0^\phi \frac{\mathrm{d} \theta} {\sqrt{1 - k^2 \sin^2 \theta}} = \sin \phi \,F_1(\tfrac 1 2, \tfrac 1 2, \tfrac 1 2, \tfrac 3 2; \sin^2 \phi, k^2 \sin^2 \phi), \quad |\real \,\phi| < \frac \pi 2 ~,$$

$$E(\phi, k) = \int_0^\phi \sqrt{1 - k^2 \sin^2 \theta} \,\mathrm{d} \theta = \sin \phi \,F_1(\tfrac 1 2, \tfrac 1 2, -\tfrac 1 2, \tfrac 3 2; \sin^2 \phi, k^2 \sin^2 \phi), \quad |\real \,\phi| < \frac \pi 2 ~,$$

$$\Pi(n,k) = \int_0^{\pi/2} \frac{\mathrm{d} \theta} {(1 - n \sin^2 \theta) \sqrt{1 - k^2 \sin^2 \theta}} = \frac {\pi} {2} \,F_1(\tfrac 1 2, 1, \tfrac 1 2, 1; n,k^2) ~.$$