# 超几何函数

（重定向自超几何级数

## 超几何级数

${\displaystyle c}$ 不是0,-1,-2...时，对于|z| < 1，超几何函数可用如下幂级数定义

${\displaystyle \,_{2}F_{1}(a,b;c;z)=\sum _{n=0}^{\infty }{a^{(n)}b^{(n)} \over c^{(n)}}\,{z^{n} \over n!}}$

${\displaystyle q^{(n)}=\left\{{\begin{array}{ll}1&{\mbox{if }}n=0\\q(q+1)\cdots (q+n-1)&{\mbox{if }}n>0\end{array}}\right.}$

ab0或负整数时级数只有有限项。

## 特殊情形

${\displaystyle \ln(1+z)=z\,_{2}F_{1}(1,1;2;-z)}$ .
${\displaystyle (1-z)^{-a}=\,_{2}F_{1}(a,1;1;z)}$
${\displaystyle \arcsin z=z\,_{2}F_{1}\left({\tfrac {1}{2}},{\tfrac {1}{2}};{\tfrac {3}{2}};z^{2}\right)}$

${\displaystyle M(a,c,z)=\lim _{b\rightarrow \infty }{}_{2}F_{1}(a,b;c;b^{-1}z)}$

${\displaystyle {}_{2}F_{1}(a,1-a;c;z)=\Gamma (c)z^{\tfrac {1-c}{2}}(1-z)^{\tfrac {c-1}{2}}P_{-a}^{1-c}(1-2z)}$

n

${\displaystyle {}_{2}F_{1}(-n,\alpha +1+\beta +n;\alpha +1;x)={\frac {n!}{(\alpha +1)_{n}}}P_{n}^{(\alpha ,\beta )}(1-2x)}$  其它特殊情形还包括Krawtchouk多项式, Meixner多项式, Meixner–Pollaczek多项式

${\displaystyle \tau ={\rm {i}}{\frac {{}_{2}F_{1}({\frac {1}{2}},{\frac {1}{2}};1;1-z)}{{}_{2}F_{1}({\frac {1}{2}},{\frac {1}{2}};1;z)}}}$

${\displaystyle z=\kappa ^{2}(\tau )={\frac {\theta _{2}(\tau )^{4}}{\theta _{3}(\tau )^{4}}}}$

τ的椭圆模函数.

${\displaystyle B_{x}(p,q)={\frac {x^{p}}{p}}{}_{2}F_{1}(p,1-q;p+1;x)}$

${\displaystyle K(k)={\tfrac {\pi }{2}}\,_{2}F_{1}\left({\tfrac {1}{2}},{\tfrac {1}{2}};1;k^{2}\right)}$
${\displaystyle E(k)={\tfrac {\pi }{2}}\,_{2}F_{1}\left(-{\tfrac {1}{2}},{\tfrac {1}{2}};1;k^{2}\right)}$

## 超几何方程

${\displaystyle z\left(z{\frac {\rm {d}}{{\rm {d}}z}}+a\right)\left(z{\frac {\rm {d}}{{\rm {d}}z}}+b\right)w=z{\frac {\rm {d}}{{\rm {d}}z}}\left(z{\frac {\rm {d}}{{\rm {d}}z}}+c-1\right)w,\quad w(z)={}_{2}F_{1}(a,b;c;z)}$ .

${\displaystyle z(1-z){\frac {\mathrm {d} ^{2}w}{\mathrm {d} z^{2}}}+\left[c-(a+b+1)z\right]{\frac {\mathrm {d} w}{\mathrm {d} z}}-abw=0.}$

### 正则奇点 0 附近的解

${\displaystyle \rho (\rho -1)+c\rho =0}$

c不是整数时，超几何方程在 0 附近的两个线性无关的正则特解为：

${\displaystyle \,_{2}F_{1}(a,b;c;z){\text{ and }}z^{1-c}\,_{2}F_{1}(1+a-c,1+b-c;2-c;z)}$

c 为 1 时，方程只有一个正则解。当 c 为其余整数时，另一个线性无关的正则特解涉及对数项。

${\displaystyle \,_{2}F_{1}(a,b;c;z){\text{ and }}\,G_{2,2}^{2,0}(1-a,1-b;0,c-1;z),{\text{ if }}c\in \mathbb {Z} ^{+}}$
${\displaystyle \,z^{1-c}\,_{2}F_{1}(1+a-c,1+b-c;2-c;z){\text{ and }}\,G_{2,2}^{2,0}(1-a,1-b;0,1-c;z),{\text{ if }}c\in \mathbb {Z} _{0}^{-}}$

### 正则奇点 1 附近的解

${\displaystyle t(1-t){\frac {d^{2}w}{dt^{2}}}+\left[1+a+b-c-(a+b+1)t\right]{\frac {dw}{dt}}-abw=0.}$

a+b-c 不是整数时，两个线性无关的正则特解为：

${\displaystyle \,_{2}F_{1}(a,b;1+a+b-c;1-z){\text{ and }}(1-z)^{c-a-b}\,_{2}F_{1}(c-b,c-a;1-a-b+c;1-z)}$

### 正则奇点 ∞ 附近的解

a-b 不是整数时，两个线性无关的正则特解为：

${\displaystyle z^{-a}\,_{2}F_{1}\left(a,1+a-c;1+a-b;z^{-1}\right){\text{ and }}z^{-b}\,_{2}F_{1}\left(b,1+b-c;1+b-a;z^{-1}\right).}$

### 李代数参数与连接关系

${\displaystyle F_{\alpha ,\beta ,\mu }(z)={}_{2}F_{1}(a,b;c;z)}$
${\displaystyle \alpha =c-1,\beta =a+b-c,\mu =b-a}$
${\displaystyle a={\frac {1+\alpha +\beta -\mu }{2}},b={\frac {1+\alpha +\beta +\mu }{2}},c=1+\alpha }$

${\displaystyle {\text{At }}0:F_{\alpha ,\beta ,\mu }(z){\text{ and }}z^{-\alpha }F_{-\alpha ,\beta ,-\mu }(z)}$
${\displaystyle {\text{At }}1:F_{\beta ,\alpha ,\mu }(1-z){\text{ and }}(1-z)^{-\beta }F_{-\beta ,\alpha ,-\mu }(1-z)}$
${\displaystyle {\text{At }}\infty :(-z)^{\frac {-1-\alpha -\beta +\mu }{2}}F_{-\mu ,\beta ,-\alpha }(z^{-1}){\text{ and }}(-z)^{\frac {-1-\alpha -\beta -\mu }{2}}F_{\mu ,\beta ,\alpha }(z^{-1})}$

${\displaystyle G(m;n,p)={\frac {\pi }{\sin m\pi \Gamma (n)\Gamma (p)}}=G(m;p,n)}$
${\displaystyle \mathbf {F} _{\alpha ,\beta ,\mu }(z)={\frac {1}{\Gamma (1+\alpha )}}F_{\alpha ,\beta ,\mu }(z)}$

${\displaystyle \mathbf {F} _{\beta ,\alpha ,\mu }(1-z)=G(-\alpha ;a-\alpha ,b-\alpha )\mathbf {F} _{\alpha ,\beta ,\mu }(z)+G(\alpha ;a,b)z^{-\alpha }\mathbf {F} _{-\alpha ,\beta ,-\mu }(z),}$
${\displaystyle {\begin{array}{rcl}(1-z)^{-\beta }\mathbf {F} _{-\beta ,\alpha ,-\mu }(1-z)&=&G(-\alpha ;1-a,1-b)\mathbf {F} _{\alpha ,\beta ,\mu }(z)+G(\alpha ;b-\beta ,a-\beta )z^{-\alpha }\mathbf {F} _{-\alpha ,\beta ,-\mu }(z)\\&=&G(-\alpha ;1-a,1-b)\mathbf {F} _{\alpha ,\beta ,\mu }(z)+G(\alpha ;1-(a-\alpha ),1-(b-\alpha ))z^{-\alpha }\mathbf {F} _{-\alpha ,\beta ,-\mu }(z);\end{array}}}$
${\displaystyle (-z)^{-a}\mathbf {F} _{-\mu ,\beta ,-\alpha }(z^{-1})=G(-\alpha ;1-b,a-\alpha )\mathbf {F} _{\alpha ,\beta ,\mu }(z)+G(\alpha ;a,a-\beta )z^{-\alpha }\mathbf {F} _{-\alpha ,\beta ,-\mu }(z),}$
${\displaystyle {\begin{array}{rcl}(-z)^{-b}\mathbf {F} _{\mu ,\beta ,\alpha }(z^{-1})&=&G(-\alpha ;1-a,b-\alpha )\mathbf {F} _{\alpha ,\beta ,\mu }(z)+G(\alpha ;b,b-\beta )z^{-\alpha }\mathbf {F} _{-\alpha ,\beta ,-\mu }(z)\\&=&G(-\alpha ;1-a,1-(a-\beta ))\mathbf {F} _{\alpha ,\beta ,\mu }(z)+G(\alpha ;b,1-(a-\alpha ))z^{-\alpha }\mathbf {F} _{-\alpha ,\beta ,-\mu }(z).\end{array}}}$

## 积分表示

${\displaystyle \mathrm {B} (a,c-a){}_{2}F_{1}(a,b;c;z)=\int _{1}^{\infty }t^{b-c}(t-1)^{c-a-1}(t-z)^{-b}\mathrm {d} t,\Re (c)>\Re (a)>0,|\arg(1-z)|<\pi }$

### 证明

${\displaystyle p(a,b,c;t,z)=t^{b-c}(t-1)^{c-a-1}(t-z)^{-b-2},\quad w(a,b,c;t,z)=(t-z)^{2}p(a,b,c;t,z);}$

${\displaystyle {\frac {\partial w}{\partial z}}=b(t-z)p(a,b,c;t,z),\quad {\frac {\partial ^{2}w}{\partial z^{2}}}=b(b+1)p(a,b,c;t,z)}$
${\displaystyle {\begin{array}{cl}&z(1-z){\frac {\partial ^{2}w}{\partial z^{2}}}+\left[c-(a+b+1)z\right]{\frac {\partial w}{\partial z}}-abw\\=&bp(a,b,c;t,z)\left\{z(1-z)(b+1)+[c-(a+b+1)z](t-z)-a(t-z)^{2}\right\}\\=&bp(a,b,c;t,z)\left\{-at^{2}+[c-(b-a+1)z]t+(b-c+1)z\right\}\\=&bp(a,b,c;t,z)\left\{(b-c+1)(t-1)(t-z)+(c-a)t(t-z)+(-b-1)t(t-1)\right\}\\=&b{\frac {\partial }{\partial t}}[t(t-1)(t-z)p(a,b,c;t,z)],\end{array}}}$

## 变换公式

### 分式线性变换

#### Pfaff 变换

Pfaff 变换将正则奇点 1 和 ∞ 交换（也就是将李代数参数中的 βμ 对换）：

${\displaystyle {}_{2}F_{1}(a,b;c;z)=(1-z)^{-b}\,{}_{2}F_{1}(c-a,b;c;{\tfrac {z}{z-1}}),\quad |\arg(1-z)|<\pi }$

a,b 的对称性自然有：

${\displaystyle {}_{2}F_{1}(a,b;c;z)=(1-z)^{-a}\,{}_{2}F_{1}(a,c-b;c;{\tfrac {z}{z-1}}),\quad |\arg(1-z)|<\pi }$
##### 证明

Pfaff 变换可以根据超几何方程得到。事实上，令

${\displaystyle u={\tfrac {z}{z-1}}=1+{\tfrac {1}{z-1}}}$

${\displaystyle z={\tfrac {u}{u-1}},\quad (1-z)^{a}=(1-u)^{-a},\quad {\tfrac {\mathrm {d} u}{\mathrm {d} z}}=-(1-u)^{2},\quad {\tfrac {\mathrm {d} ^{2}u}{\mathrm {d} z^{2}}}=-(1-u)^{3}}$
${\displaystyle {\begin{array}{cl}&z(1-z){\tfrac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}[(1-z)^{-b}w]+\left[c-(a+b+1)z\right]{\tfrac {\mathrm {d} }{\mathrm {d} z}}[(1-z)^{-b}w]-ab(1-z)^{-b}w\\=&(1-z)^{-b-1}\left\{z[b(b+1)+2b(1-z){\tfrac {\mathrm {d} }{\mathrm {d} z}}+(1-z)^{2}{\tfrac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}]+[c-(a+b+1)z][b+(1-z){\tfrac {\mathrm {d} }{\mathrm {d} z}}]-ab(1-z)\right\}w\\=&(1-z)^{-b-1}\left\{z(1-z)^{2}{\tfrac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}+(1-z)[c-(a-b+1)z]{\tfrac {\mathrm {d} }{\mathrm {d} z}}+b(c-a)\right\}w\\=&(1-u)^{b+1}\left\{-u(1-u){\tfrac {\mathrm {d} ^{2}}{\mathrm {d} u^{2}}}+2u{\tfrac {\mathrm {d} }{\mathrm {d} u}}-(1-u)[c+(a-b+1)(1-u)^{-1}u]{\tfrac {\mathrm {d} }{\mathrm {d} u}}+b(c-a)\right\}w\\=&-(1-u)^{b+1}\left\{u(1-u){\tfrac {\mathrm {d} ^{2}}{\mathrm {d} u^{2}}}+[c-(c-a+b+1)u]{\tfrac {\mathrm {d} }{\mathrm {d} u}}-b(c-a)\right\}w\end{array}}}$

${\displaystyle w={}_{2}F_{1}(c-a,b;c;u)}$

w(u) 满足的超几何方程知等号右边为 0，再考虑函数 (1-z)-bw(z)z=0 附近的性质即可得到 Pfaff 变换的公式。

#### Euler 变换

Pfaff 变换可以导出 Euler 变换，它将李代数参数 β 变成 -β

${\displaystyle {\begin{array}{rcl}{}_{2}F_{1}(a,b;c;z)&=&(1-z)^{-b}\,{}_{2}F_{1}(c-a,b;c;{\frac {z}{z-1}})\\&=&(1-z)^{-b}\left(1-{\frac {z}{z-1}}\right)^{a-c}\,{}_{2}F_{1}\left(c-a,c-b;c;{\frac {\frac {z}{z-1}}{{\frac {z}{z-1}}-1}}\right)\\&=&(1-z)^{c-a-b}\,{}_{2}F_{1}(c-a,c-b;c;z),\quad |\arg(1-z)|<\pi \end{array}}}$

Pfaff 变换和 Euler 变换都是分式线性变换的例子，这得名于等式两边的超几何函数的宗量的联系，参见莫比乌斯变换

${\displaystyle F_{\alpha ,\beta ,\mu }{\xrightarrow {\text{Pfaff}}}F_{\alpha ,\mu ,\beta }\equiv F_{\alpha ,\mu ,-\beta }{\xrightarrow {\text{Pfaff}}}F_{\alpha ,-\beta ,\mu }}$

### 二次变换

${\displaystyle {}_{2}F_{1}(a,b;2a;z)=(1-z)^{-{\tfrac {b}{2}}}\,_{2}F_{1}(a-{\tfrac {b}{2}},{\tfrac {b}{2}};a+{\tfrac {1}{2}};{\tfrac {z^{2}}{4z-4}}),\quad |\arg(1-z)|<\pi }$

#### 证明

${\displaystyle {\begin{array}{cl}&z(1-z){\tfrac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}[(1-z)^{-{\tfrac {b}{2}}}w]+\left[c-(a+b+1)z\right]{\tfrac {\mathrm {d} }{\mathrm {d} z}}[(1-z)^{-{\tfrac {b}{2}}}w]-ab(1-z)^{-{\tfrac {b}{2}}}w\\=&(1-z)^{-{\tfrac {b}{2}}-1}\left\{z[{\tfrac {b}{2}}({\tfrac {b}{2}}+1)+b(1-z){\tfrac {\mathrm {d} }{\mathrm {d} z}}+(1-z)^{2}{\tfrac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}]+[c-(a+b+1)z][{\tfrac {b}{2}}+(1-z){\tfrac {\mathrm {d} }{\mathrm {d} z}}]-ab(1-z)\right\}w\\=&(1-z)^{-{\tfrac {b}{2}}-1}\left\{z(1-z)^{2}{\tfrac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}+(1-z)[c-(a+1)z]{\tfrac {\mathrm {d} }{\mathrm {d} z}}+{\tfrac {b}{4}}[2(c-2a)+(2a-b)z]\right\}w\\\end{array}}}$

${\displaystyle c=2a,\quad u={\tfrac {z^{2}}{4z-4}}={\tfrac {1}{4}}(z+1-{\tfrac {1}{1-z}})}$

${\displaystyle 1-u={\tfrac {(z-2)^{2}}{4(1-z)}},\quad {\tfrac {\mathrm {d} u}{\mathrm {d} z}}={\tfrac {z(z-2)}{4(1-z)^{2}}},\quad {\tfrac {\mathrm {d} ^{2}u}{\mathrm {d} z^{2}}}=-{\tfrac {1}{2(1-z)^{3}}}}$
${\displaystyle {\begin{array}{cl}&z(1-z)^{2}{\tfrac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}+(1-z)[c-(a+1)z]{\tfrac {\mathrm {d} }{\mathrm {d} z}}+{\tfrac {b}{4}}[2(c-2a)+(2a-b)z]\\=&z(1-z)^{2}{\tfrac {\mathrm {d} ^{2}}{\mathrm {d} z^{2}}}+(1-z)[2a-(a+1)z]{\tfrac {\mathrm {d} }{\mathrm {d} z}}+{\tfrac {b}{2}}(a-{\tfrac {b}{2}})z\\=&{\tfrac {z^{3}(z-2)^{2}}{16(1-z)^{2}}}{\tfrac {\mathrm {d} ^{2}}{\mathrm {d} u^{2}}}-{\tfrac {z}{2(1-z)}}{\tfrac {\mathrm {d} }{\mathrm {d} u}}+{\tfrac {z(z-2)(2a-az-z)}{4(1-z)}}{\tfrac {\mathrm {d} }{\mathrm {d} u}}+{\tfrac {b}{2}}(a-{\tfrac {b}{2}})z\\=&-z\left\{u(1-u){\tfrac {\mathrm {d} ^{2}}{\mathrm {d} u^{2}}}+[a+{\tfrac {1}{2}}-(a+1)u]{\tfrac {\mathrm {d} }{\mathrm {d} u}}-{\tfrac {b}{2}}(a-{\tfrac {b}{2}})\right\}\end{array}}}$

${\displaystyle w=\,{}_{2}F_{1}(a-{\tfrac {b}{2}},{\tfrac {b}{2}};a+{\tfrac {1}{2}};u)}$

#### 其它例子

${\displaystyle F_{\alpha ,\beta ,\mu }(z)=f(z)F_{\alpha ',\beta ',\mu '}(g(z)),\quad P(z)}$

${\displaystyle \alpha ,\mu ,\mu }$  ${\displaystyle {\tfrac {\alpha }{2}},\mu ,{\tfrac {1}{2}}}$  ${\displaystyle (1-{\tfrac {1}{2}}z)^{-b}}$  ${\displaystyle \left({\tfrac {z}{2-z}}\right)^{2}}$  ${\displaystyle |\arg(1-z)|<\pi }$
${\displaystyle \mu ,\beta ,\mu }$  ${\displaystyle \mu ,{\tfrac {\beta }{2}},{\tfrac {1}{2}}}$  ${\displaystyle (1+z)^{-b}}$  ${\displaystyle {\tfrac {4z}{(1+z)^{2}}}}$  ${\displaystyle |z|<1}$
${\displaystyle \alpha ,\alpha ,\mu }$  ${\displaystyle \alpha ,{\tfrac {\mu }{2}},{\tfrac {1}{2}}}$  ${\displaystyle (1-2z)^{-b}}$  ${\displaystyle {\tfrac {4z(z-1)}{(1-2z)^{2}}}}$  ${\displaystyle \Re z<{\tfrac {1}{2}}}$

${\displaystyle {\tfrac {2\Gamma ({\tfrac {1}{2}})\Gamma ({\tfrac {a+b+1}{2}})}{\Gamma ({\tfrac {a+1}{2}})\Gamma ({\tfrac {b+1}{2}})}}F_{-{\tfrac {1}{2}},\beta ,{\tfrac {\mu }{2}}}(z)=F_{\beta ,\beta ,\mu }\left({\tfrac {1}{2}}-{\tfrac {1}{2}}{\sqrt {z}}\right)+F_{\beta ,\beta ,\mu }\left({\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {z}}\right),\quad |\arg z|<\pi ,|\arg(1-z)|<\pi }$

## 特殊值

### z=0

${\displaystyle {}_{2}F_{1}(a,b;c;0)=1}$

### z=1

${\displaystyle {}_{2}F_{1}(a,b;c;1)={\tfrac {\mathrm {B} (a,c-a-b)}{\mathrm {B} (a,c-a)}}={\tfrac {\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}},\quad \Re (c)>\Re (a+b)}$

### z=-1

${\displaystyle {}_{2}F_{1}(a,b;1+a-b;-1)={\frac {\Gamma (1+a-b)\Gamma (1+{\tfrac {1}{2}}a)}{\Gamma (1+a)\Gamma (1+{\tfrac {1}{2}}a-b)}}}$

### z=1/2

${\displaystyle _{2}F_{1}\left(a,b;{\tfrac {1}{2}}\left(1+a+b\right);{\tfrac {1}{2}}\right)={\frac {\Gamma ({\tfrac {1}{2}})\Gamma ({\tfrac {1}{2}}\left(1+a+b\right))}{\Gamma ({\tfrac {1}{2}}\left(1+a)\right)\Gamma ({\tfrac {1}{2}}\left(1+b\right))}}.}$
${\displaystyle _{2}F_{1}\left(a,1-a;c;{\tfrac {1}{2}}\right)={\frac {\Gamma ({\tfrac {1}{2}}c)\Gamma ({\tfrac {1}{2}}\left(1+c\right))}{\Gamma ({\tfrac {1}{2}}\left(c+a\right))\Gamma ({\tfrac {1}{2}}\left(1+c-a\right))}}.}$

## 参考文献

• ^ Dereziński, Jan. Hypergeometric type functions and their symmetries. arXiv:1305.3113.