超几何变换

先来看一些封闭形式,比如和式 km(nk)\displaystyle \sum_{k \leqslant m} \binom{n}{k} 对应于

(nm)F(1,mnm+1| 1),nm0\displaystyle \binom{n}{m} F\left(\begin{gathered} 1, -m \\ n-m+1 \end{gathered} \middle| \ -1\right), \quad n \geqslant m \geqslant 0

将上式进行化简,等价于 (nm)k(mk)(nm+kk)=kn!m!(nm)!m!(nm)!(mk)!(nm+k)!\displaystyle \binom{n}{m} \sum_k \frac{ \binom{m}{k} }{ \binom{n-m+k}{k} } = \sum_k \frac{n!}{m!(n-m)!} \cdot \frac{m!(n-m)!}{(m-k)!(n-m+k)!}

进一步化简, k(nmk)=km(nk)\displaystyle \sum_k \binom{n}{m-k} = \sum_{k'\leqslant m} \binom{n}{k'},证明完毕

反射定律

1(1z)aF(a,bc| z1z)=F(a,cbc| z)\displaystyle \frac{1}{(1-z)^a} F\left(\begin{gathered} a, b \\ c \end{gathered} \middle| \ \frac{-z}{1-z}\right) = F\left(\begin{gathered} a, c-b \\ c \end{gathered} \middle| \ z \right)

根据二项展开,不难写出左边等于

k0akbkck(z)kk!m0(k+a+m1m)zm\displaystyle \sum_{k \geqslant 0}\frac{a^{\overline{k}} b^{\overline{k}}}{c^{\overline{k}}} \cdot \frac{(-z)^k}{k!} \cdot \sum_{m \geqslant 0} \binom{k+a+m-1}{m}z^m
k+m=nk+m = n,求和可以写成,n0znk0akbkckk!(1)k(n+a1nk)\displaystyle \sum_{n \geqslant 0} z^n \sum_{k \geqslant 0} \frac{a^{\overline{k}} b^{\overline{k}}}{c^{\overline{k}} k!} (-1)^k \binom{n+a-1}{n-k}
znz^n 的系数展开
(a1)!a(a+1)(a+k1)b(b+1)(b+k1)(a1)!c(c+1)(c+k1)k!(1)k(n+a1)!(nk)!(a+k1)!\displaystyle \frac{(a-1)!a(a+1)\cdots (a+k-1) \cdot b(b+1)\cdots (b+k-1)}{(a-1)!c(c+1)\cdots (c+k-1) \cdot k!} \cdot (-1)^k \frac{(n+a-1)!}{(n-k)! (a+k-1)!}
(a+k1)!(a+k-1)! 可以约去,将 (1)k(-1)^k 开始的后半部分做一个变形

(1)k(n+a1)!(nk+1)(nk+2)n(nk)!(nk+1)(nk+2)n\displaystyle (-1)^k \frac{(n+a-1)! \cdot (n-k+1)(n-k+2) \cdots n}{(n-k)! (n-k+1)(n-k+2)\cdots n}
综上所述,可以将原式化简成
bkckk!(n+a1)!(n)kn!(a1)!=(n+a1n)F(a,b,nc,a| 1)\displaystyle \frac{b^{\overline{k}} }{c^{\overline{k}} k!} \cdot \frac{(n+a-1)! \cdot (-n)^{\overline{k}}}{n! (a-1)!} = \binom{n+a-1}{n} F\left(\begin{gathered} a, b, -n \\ c, a \end{gathered} \middle| \ 1 \right)

根据范德蒙卷积,多项式的超几何表示这一节内容的 (3)\textbf{(3)},原式为 ann!(cb)ncn\displaystyle \frac{a^{\overline{n}}}{n!} \frac{(c-b)^{\overline{n}}}{c^{\overline{n}}}

库默尔公式和反射定律

c=1+bac = 1+b-a 代入反射定律中,可以得到

2aF(a,1a1+ba| 12)=F(a,b1+ba| 1)=(b/2)!b!(ba)b/2\displaystyle 2^{-a} F\left(\begin{gathered} a, 1-a \\ 1+b-a \end{gathered} \middle| \ \frac{1}{2}\right) = F\left(\begin{gathered} a, b \\ 1+b-a \end{gathered} \middle| \ -1 \right) = \frac{(b/2)!}{b!} (b-a)^{\underline{b/2}}

a=na= -n 就可以得到一个新的恒等式

k0(n)k(1+n)k(1+b+n)k2kk!=k(nk)(12)k(n+kk)/(n+b+kk)\displaystyle \sum_{k\geqslant 0} \frac{(-n)^{\overline{k}} (1+n)^{\overline{k}} }{(1+b+n)^{\overline{k}}} \frac{2^{-k}}{k!} = \sum_k \binom{n}{k} \left(\frac{-1}{2} \right)^k \binom{n+k}{k}\bigg / \binom{n+b+k}{k}

=2n(b/2)!(b+n)!b!(b/2+n)!\displaystyle \quad \quad = 2^{-n}\frac{(b/2)!(b+n)!}{b!(b/2+n)!}

其他一些恒等式

km(m+kk)2k=2m\displaystyle \sum_{k \leqslant m} \binom{m+k}{k}2^{-k} = 2^m,令 k=mkk' = m - k 代入,可以得到

k0(2mkmk)2k=2m\displaystyle \sum_{k \geqslant 0} \binom{2m-k}{m-k}2^k = 2^m,写成超几何形式,令 tk+1tk=2(km)(k+1)(k2m)(k+1)\displaystyle \frac{t_{k+1}}{t_k} = \frac{2(k-m)(k+1)}{(k-2m)(k+1)}

(2mm)F(1,m2m| 2)=22m\displaystyle \binom{2m}{m} F\left(\begin{gathered} 1, -m \\ -2m \end{gathered} \middle| \ 2\right) = 2^{2m}

但这个形式并不是标准形式,因为下参数 2m-2m 被禁止的

(2mm)F(1,m2m| 2)=22m\displaystyle \binom{2m}{m} F\left(\begin{gathered} 1, -m \\ -2m \end{gathered} \middle| \ 2\right) = 2^{2m} 表示 k0(2mkmk)2k\displaystyle \sum_{k \geqslant 0}\binom{2m-k}{m-k} 2^k
我们想要令 k>mk > m 的项都为 00(二项式下指标 <0< 0 时值都为 00

注意到上参数,(m)k(-m)^{\overline{k}} 形式是 ((m+k1)(m+k))(\cdots (-m+k-1)(-m+k) \cdots),在 k>mk > m 的时候,一定包含 00
所以对于 k>mk > m 的项,FF 展开分子为 00

为了使极限有良好定义,对超几何函数 FF下参数取极限,具体来说

(2mm)limϵF(1,m2m+ϵ| 2)=22m,m0\displaystyle \binom{2m}{m} \lim_{\epsilon \to \infty} F\left(\begin{gathered} 1, -m \\ -2m+\epsilon \end{gathered} \middle| \ 2\right) = 2^{2m}, \quad m \geqslant 0

这样,分子在 k>mk > m 时候为 00,分母 (2m)k(-2m)^{\overline{k}}k>2mk > 2m 时才为 00,这样以来,极限刚好给出了良好的定义

(1)\textbf{(1)}

(2mm)limεF(1,m2m+ε| 2)=22m,m0\displaystyle \binom{2m}{m} \lim_{\varepsilon \to \infty} F\left(\begin{gathered} 1, -m \\ -2m+\varepsilon \end{gathered} \middle| \ 2\right) = 2^{2m}, \quad m \geqslant 0

根据 (1)\textbf{(1)} 可以推出其他恒等式,在反射定律中,令 z=2,a=m,b=1,c=2m+εz = 2, a = -m, b = 1, c = -2m + \varepsilon

(1)mlimε0F(1,m2m+ε| 2)=limε0F(m,2m1+ε2m+ε| 2)\displaystyle (-1)^m \lim_{\varepsilon \to 0} F\left(\begin{gathered} 1, -m \\ -2m+\varepsilon \end{gathered} \middle| \ 2\right) = \lim_{\varepsilon \to 0} F\left(\begin{gathered} -m, -2m-1+\varepsilon \\ -2m+\varepsilon \end{gathered} \middle| \ 2\right)

=limε0k0(m)k(2m1+ε)k(2m+ε)k2kk!=km(mk)(2m+1)k(2m)k(2)k\displaystyle \quad \quad = \lim_{\varepsilon \to 0} \sum_{k \geqslant 0} \frac{(-m)^{\overline{k}} (-2m-1+\varepsilon)^{\overline{k}} }{(-2m + \varepsilon)^{\overline{k}} } \cdot \frac{2^k}{k!} = \sum_{k \leqslant m} \binom{m}{k} \frac{(2m+1)^{\underline{k}}}{(2m)^{\underline{k}}} (-2)^k

这就给出了一个公式

km(mk)2m+12m+1k(2)k=(1)m22m/(2mm)\displaystyle \sum_{k \leqslant m} \binom{m}{k} \frac{2m+1}{2m+1-k}(-2)^k = (-1)^m 2^{2m} \bigg / \binom{2m}{m},这里用了公式 (1)\textbf{(1)}
=1/(1/2m),m0\displaystyle \quad \quad = 1 \bigg / \binom{-1/2}{m}, \quad m \geqslant 0

(1)m22mm!m!(2m)(2m1)21=(1)mm!m!m(m12)(m22)112\displaystyle (-1)^m 2^{2m} \frac{m!m!}{(2m)(2m-1)\cdots 2 \cdot 1} = (-1)^m \cdot \frac{m!m!}{m(m-\frac{1}{2}) (m - \frac{2}{2}) \cdots 1 \cdot \frac{1}{2} }

=m!m!(m(m1)1)((12+m)(32+m)(12))(1)m\displaystyle \quad \quad = \frac{m!m!}{ (m(m-1)\cdots 1) \cdot \left( (-\frac{1}{2}+m) (-\frac{3}{2}+m) \cdots (\frac{1}{2}) \right)} \cdot (-1)^m

=1/(1/2m),m0\displaystyle \quad \quad = 1 \bigg / \binom{-1/2}{m}, \quad m \geqslant 0

微分法

二项级数的部分和恒等式 描述如下

km(m+rk)xkymk=km(rk)(x)k(x+y)mk\displaystyle \sum_{k \leqslant m} \binom{m+r}{k}x^k y^{m - k} = \sum_{k \leqslant m} \binom{-r}{k} (-x)^k (x+y)^{m-k}

两边同时对 yy 微分 nn 次,然后令 mnkm-n-k 代替 kk,很容易得出

k0(m+rmnk)(n+kn)xmnkyk=k0(rmnk)(n+kn)(x)mnk(x+y)k\displaystyle \sum_{k \geqslant 0} \binom{m+r}{m-n-k}\binom{n+k}{n} x^{m-n-k}y^k = \sum_{k \geqslant 0} \binom{-r}{m-n-k} \binom{n+k}{n}(-x)^{m-n-k}(x+y)^k

展开写成超几何函数形式,左边记 tk+1tk\displaystyle \frac{t_{k+1}}{t_k},右边记 tk+1tk\displaystyle \frac{t'_{k+1}}{t'_{k}}

左边超几何表示 t0F(nm,n+1n+r+1| y)t_0 \displaystyle F\left(\begin{gathered} n-m, n+1 \\ n+r+1 \end{gathered} \middle| \ -y\right)

右边超几何表示 t0F(nm,n+1nmr+1| 1+y)t_0' \displaystyle F\left(\begin{gathered} n-m, n+1 \\ n-m-r+1 \end{gathered} \middle| \ 1+y\right)

这里令 {nm=an+1=nn+r+1=cnmr+1=acn+1\displaystyle \begin{cases}n-m = a \\ n+1 = -n' \\ n+r+1 = c \\ n-m-r+1 = a-c-n'+1\end{cases}

另外 t0t0=(ca1+n)!(c1)!(c1+n)!(ca1)!\displaystyle \frac{t_0'}{t_0} = \frac{(c-a-1+n')! \cdot (c-1)!}{(c-1+n')! \cdot (c-a-1)!},实际上这里做了替换

{ca=m+r+1r1=c+n1\displaystyle \begin{cases} c-a = m+r+1 \\ r-1=c+n'-1 \end{cases}

于是我们有 t0t0=(ca)ncn=(ac)n(c)n\displaystyle \frac{t_0'}{t_0} = \frac{(c-a)^{\overline{n'}}}{c^{\overline{n'}}} = \frac{(a-c)^{\underline{n'}}}{(-c)^{\underline{n'}}}

可以推出超几何变换

F(a,nc| z)=(ac)n(c)nF(a,n1n+ac| 1z)\displaystyle F\left(\begin{gathered} a, -n \\ c \end{gathered} \middle| \ z\right) = \frac{(a-c)^{\underline{n}}}{(-c)^{\underline{n}}} F\left(\begin{gathered} a, -n \\ 1-n+a-c \end{gathered} \middle| \ 1-z\right)

超几何级数的微分

ddzF(a1,,amb1,,bn| z)=k1a1kamkzk1b1kbnk(k1)!\displaystyle \frac{d}{dz} F\left(\begin{gathered} a_1, \cdots, a_m \\ b_1, \cdots, b_n \end{gathered} \middle| \ z\right) = \sum_{k \geqslant 1} \frac{a_1^{\overline{k}} \cdots a_m^{\overline{k}} z^{k-1} }{ b_1^{\overline{k}} \cdots b_n^{\overline{k}} (k-1)!}

作代换,令 k1=kk-1 = k',从而有

=k+11a1k+1amk+1zkb1k+1bnk+1(k)!=k0a1(a1+1)kam(am+1)kzkb1(b1+1)kbn(bn+1)k(k)!\displaystyle \quad \quad = \sum_{k+1 \geqslant 1} \frac{a_1^{\overline{k+1}} \cdots a_m^{\overline{k+1}} z^{k} }{ b_1^{\overline{k+1}} \cdots b_n^{\overline{k+1}} (k)!} = \sum_{k \geqslant 0} \frac{a_1 (a_1+1)^{\overline{k}} \cdots a_m (a_m+1)^{\overline{k}} z^{k} }{ b_1(b_1+1)^{\overline{k}} \cdots b_n(b_n+1)^{\overline{k}} (k)!}

=a1amb1bnF(a1+1,,am+1b1+1,,bn+1| z)\displaystyle \quad \quad = \frac{a_1\cdots a_m}{b_1 \cdots b_n} F\left(\begin{gathered} a_1+1, \cdots, a_m+1 \\ b_1+1, \cdots, b_n+1 \end{gathered} \middle| \ z\right)

ddzF(a1,,amb1,,bn| z)=a1amb1bnF(a1+1,,am+1b1+1,,bn+1| z)\displaystyle \frac{d}{dz} F\left(\begin{gathered} a_1, \cdots, a_m \\ b_1, \cdots, b_n \end{gathered} \middle| \ z\right) = \frac{a_1\cdots a_m}{b_1 \cdots b_n} F\left(\begin{gathered} a_1+1, \cdots, a_m+1 \\ b_1+1, \cdots, b_n+1 \end{gathered} \middle| \ z\right)

微分算子

记算子 ϑ=zddz\displaystyle \vartheta = z \frac{d}{dz}

容易推出

ddzF(a1,,amb1,,bn| z)=zk1a1kamkzk1b1kbnk(k1)!=k0ka1kamkzkb1kbnk(k)!\displaystyle \frac{d}{dz} F\left(\begin{gathered} a_1, \cdots, a_m \\ b_1, \cdots, b_n \end{gathered} \middle| \ z\right) = z \sum_{k \geqslant 1} \frac{a_1^{\overline{k}} \cdots a_m^{\overline{k}} z^{k-1} }{ b_1^{\overline{k}} \cdots b_n^{\overline{k}} (k-1)!} = \sum_{k \geqslant 0} \frac{ k a_1^{\overline{k}} \cdots a_m^{\overline{k}} z^{k} }{ b_1^{\overline{k}} \cdots b_n^{\overline{k}} (k)!}

(ϑ+a1)F(a1,,amb1,,bn| z)=k0a1(a1+1)ka2kamkzkb1kbnkk!=a1F(a1+1,a2,,amb1,,bn| z)\displaystyle (\vartheta + a_1)F\left(\begin{gathered} a_1, \cdots, a_m \\ b_1, \cdots, b_n \end{gathered} \middle| \ z\right) = \sum_{k \geqslant 0} \frac{a_1 (a_1+1)^{\overline{k}} a_2^{\overline{k}} \cdots a_m^{\overline{k}} z^k }{b_1^{\overline{k}} \cdots b_n^{\overline{k}} k! } = a_1 \cdot F\left(\begin{gathered} a_1+1, a_2, \cdots, a_m \\ b_1, \cdots, b_n \end{gathered} \middle| \ z\right)

(ϑ+b11)F(a1,,amb1,,bn| z)=k0(k+b11)a1ka2kamkzkb1kbnkk!=k0(b11)a1ka2kamkzk(b11)kb2kbnkk!\displaystyle (\vartheta + b_1-1)F\left(\begin{gathered} a_1, \cdots, a_m \\ b_1, \cdots, b_n \end{gathered} \middle| \ z\right) = \sum_{k \geqslant 0} \frac{(k+b_1-1) a_1^{\overline{k}} a_2^{\overline{k}} \cdots a_m^{\overline{k}} z^k }{b_1^{\overline{k}} \cdots b_n^{\overline{k}} k! } = \sum_{k \geqslant 0} \frac{(b_1-1) a_1^{\overline{k}} a_2^{\overline{k}} \cdots a_m^{\overline{k}} z^k }{(b_1-1)^{\overline{k}} b_2^{\overline{k}} \cdots b_n^{\overline{k}} k! }

=(b11)F(a1,,amb11,b2,,bn| z)\displaystyle \quad \quad = (b_1-1) F\left(\begin{gathered} a_1, \cdots, a_m \\ b_1-1, b_2, \cdots, b_n \end{gathered} \middle| \ z\right)

从而有超几何函数微分方程如下

(ϑ+a1)(ϑ+a2)(ϑ+am)F=(a1am)F(a1+1,,am+1b1,,bn| z)\displaystyle (\vartheta + a_1)(\vartheta + a_2) \cdots (\vartheta + a_m) F = (a_1\cdots a_m) F\left(\begin{gathered} a_1+1, \cdots, a_m+1 \\ b_1, \cdots, b_n \end{gathered} \middle| \ z\right)

(ϑ+b11)(ϑ+bn1)F=(b11)(bn1)F(a1,,amb11,,bn1| z)\displaystyle (\vartheta + b_1-1) \cdots (\vartheta + b_n - 1) F = (b_1-1)\cdots (b_n-1) F\left(\begin{gathered} a_1, \cdots, a_m \\ b_1-1, \cdots, b_n-1 \end{gathered} \middle| \ z\right)

记算子 D=ddz\displaystyle \bm{D} = \frac{d}{dz},从而

D(ϑ+b11)(ϑ+bn1)F=(ϑ+a1)(ϑ+a2)(ϑ+am)F(1)\displaystyle \bm{D} (\vartheta + b_1-1) \cdots (\vartheta + b_n - 1) F = (\vartheta + a_1)(\vartheta + a_2) \cdots (\vartheta + a_m) F \quad \quad \bold{(1)}

反过来,我们可以从微分方程回到幂级数,不妨设 F(z)=k0tkzk\displaystyle F(z) = \sum_{k \geqslant 0} t_kz^k 是满足上式子的幂级数

左边,观察 [zk+1]\displaystyle [z^{k+1}] 系数,对于 (ϑ+bn1)(\vartheta + b_n - 1)
其系数为 Q(z)=(k+1)tk+1zk+1+(bn1)tk+1zk+1=(k+bn)tk+1zk+1Q(z) = (k+1)t_{k+1}z^{k+1} + (b_n - 1)t_{k+1}z^{k+1} = (k+b_n) t_{k+1}z^{k+1}

D(ϑ+b11)(ϑ+bn1)F(z)=[zk](k+1)(k+b1)(k+bn)tk+1\displaystyle \bm{D} (\vartheta + b_1-1) \cdots (\vartheta + b_n - 1) F(z) = [z^k](k+1)(k+b_1)\cdots (k+b_n) t_{k+1}
注意这里对 zz 求导,自然会有 (k+1)(k+1) 这一项

同理,右边为 [zk](k+a1)(k+a2)(k+an)tk\displaystyle [z^k](k+a_1)(k+a_2)\cdots (k+a_n) t_k

由此 tk+1tk=(k+a1)(k+a2)(k+am)(k+b1)(k+bn)(k+1)\displaystyle \frac{t_{k+1}}{t_k} = \frac{(k+a_1)(k+a_2) \cdots (k+a_m)}{(k+b_1)\cdots (k+b_n) (k+1)},恰好对于超几何函数表达式

另外,更为普遍的是 22 个上参数的超几何表达式

z(1z)F(z)+(cz(a+b+1))F(z)abF(z)=0(2)\displaystyle z(1-z)F''(z) + (c - z(a+b+1)) F'(z) - ab F(z) = 0 \quad \textbf{(2)}

F(z)=F(a,bc| z)F(z) = \displaystyle F\left(\begin{gathered} a, b \\ c \end{gathered} \middle| \ z\right),根据上式,D(ϑ+c1)F=(ϑ+a)(ϑ+b)F\displaystyle \bm{D}(\vartheta + c - 1)F = (\vartheta + a)(\vartheta + b) F

其中 (ϑ+c1)F=zF(z)+(c1)F(z)(\vartheta + c - 1)F = \displaystyle zF'(z) + (c-1)F(z),对其求导
左边 F(z)+zF(z)+(c1)F(z)F'(z) + zF''(z) + (c-1)F'(z),右边 zF(z)+z2F(z)+bzF(z)+azF(z)+abF(z)\displaystyle zF'(z) + z^2 F''(z) + bz F'(z) + az F'(z) + ab F(z)

z(1z)F(z)+(cz(a+b+1))F(z)abF(z)=0\displaystyle z(1-z)F''(z) + (c - z(a+b+1)) F'(z) - ab F(z) = 0

如果幂级数 F(z)=k0tkzk\displaystyle F(z) = \sum_{k \geqslant 0} t_k z^k 满足 (1)\textbf{(1)},那么我们可以有

tk+1tk=(k+a1)(k+am)(k+b1)(k+bn)(k+1)\displaystyle \frac{t_{k+1}}{t_k} = \frac{(k+a_1)\cdots (k+a_m)}{(k+b_1)\cdots (k+b_n) (k+1)}

从而 F(z)F(z) 就是 t0F(a1,,am; b1,,bn; z)t_0 F(a_1, \cdots, a_m; \ b_1, \cdots, b_n; \ z)

微分法应用

算子 ϑ=zddz\displaystyle \vartheta = z \frac{\text{d}}{dz},所以对于 (1)\bold{(1)} 而言

右边形式为 αkzkF(k)(z)\displaystyle \alpha_k z^k F^{(k)}(z),左边有形式 βkzk1F(k)(z)\displaystyle \beta_k z^{k-1} F^{(k)}(z)

所以微分方程总是可以写成

zn1(βnzαn)F(n)(z)++(β1zα1)F(z)α0F(z)=0(3)\displaystyle z^{n-1}(\beta_n - z\alpha_n)F^{(n)}(z) + \cdots + (\beta_1 - z \alpha_1)F'(z) - \alpha_0 F(z) = 0 \quad \textbf{(3)}

写成全 ϑ\vartheta 形式,可以用 zz 乘以 (1)\textbf{(1)} 的两边

ϑ(ϑ+b11)(ϑ+bn1)F=z(ϑ+a1)(ϑ+am)F\vartheta(\vartheta + b_1 - 1) \cdots (\vartheta + b_n - 1) F = z(\vartheta + a_1) \cdots (\vartheta + a_m)F

ϑ=(ϑ+11)\vartheta = (\vartheta + 1 - 1) 对应超几何表达式中的 (k+1)(k+1)(ϑ+bj1)(\vartheta + b_j - 1) 对应因子 (k+bj)(k+b_j),对应到超几何函数中的 bjkb_j^{\overline{k}}

右边的 zz 对应 zkz^k(ϑ+aj)(\vartheta + a_j) 对应 ajka_j^{\overline{k}}

高斯恒等式

高斯恒等式重要的理论依据是基于 22 个上参数的恒等式的超几何表达,即上式 (2)\textbf{(2)}

F(2a,2ba+b+12| x)=F(a,ba+b+12| 4t(1t))\displaystyle F\left(\begin{gathered} 2a, 2b \\ a+b+\frac{1}{2} \end{gathered} \middle| \ x\right) = F\left(\begin{gathered} a, b \\ a+b+\frac{1}{2} \end{gathered} \middle| \ 4t(1-t) \right)

先来看右边,令 x=4t(1t)x = 4t(1-t),那么右边形式满足微分方程

x(1x)d2Pdx2+(a+b+12(a+b+1)x)dPdxabP=0(2)\displaystyle x(1-x)\frac{d^2 P}{dx^2} + \left(a+b+\frac{1}{2} - (a+b+1)x \right) \frac{d P}{dx} - ab P = 0 \quad \textbf{(2)}

作变换,令 x=4t(1t)x = 4t(1-t),那么有 dxdt=48t,dtdx=14(12t)\displaystyle \frac{dx}{dt} = 4-8t, \quad \frac{dt}{dx} = \frac{1}{4(1-2t)}

dPdx=dPdtdtdx=148tdPdt\displaystyle \frac{dP}{dx} = \frac{dP}{dt} \cdot \frac{dt}{dx} = \frac{1}{4-8t} \cdot \frac{dP}{dt}

d2Pdx2=ddx(148t)dPdt+148tddx(dPdt)\displaystyle \frac{d^2 P}{dx^2} = \frac{d}{dx} \left(\frac{1}{4-8t} \right) \cdot \frac{dP}{dt} + \frac{1}{4-8t} \cdot \frac{d}{dx} \left(\frac{dP}{dt} \right)

=ddt(148t)dtdxdPdt+1(48t)2d2Pdt2\quad \quad = \displaystyle \frac{d}{dt} \left(\frac{1}{4-8t} \right) \cdot \frac{dt}{dx} \cdot \frac{dP}{dt} + \frac{1}{(4-8t)^2} \frac{d^2 P}{dt^2}

=116(12t)2d2Pdt2+18(12t)3dPdt\quad \quad = \displaystyle \frac{1}{16(1-2t)^2} \frac{d^2 P}{dt^2} + \frac{1}{8(1-2t)^3} \frac{dP}{dt}

原式划归如下
14t(1t)d2Pdt2+t(1t)2(12t)dPdt+(a+b+12)(14t(1t))14(12t)2dPdt2t(1t)4(12t)dPdtabP=0\displaystyle \frac{1}{4}t(1-t) \frac{d^2 P}{dt^2} + \frac{t(1-t)}{2(1-2t)} \frac{dP}{dt} + (a+b+\frac{1}{2})(1-4t(1-t)) \frac{1}{4(1-2t)^2} \frac{dP}{dt} - \frac{2t(1-t)}{4(1-2t)} \frac{dP}{dt} - abP = 0

t(1t)d2Pdt2+(a+b+12(2a+2b+1)t)dPdt4abP=0\displaystyle t(1-t) \frac{d^2 P}{dt^2} + \left(a+b+\frac{1}{2} - (2a + 2b + 1)t \right) \frac{dP}{dt} - 4abP = 0

将高斯恒等式左边代入 (2)\textbf{(2)},恰好和上式一致,问题证毕

超几何恒等式关于二项式系数的推论

先来看一个关于复数 1z!\displaystyle \frac{1}{z!} 的定义

1z!=limn(n+zn)nz(4)\displaystyle \frac{1}{z!} = \lim_{n \to \infty} \binom{n+z}{n} n^{-z} \quad \quad \textbf{(4)}

证明其实很简单,因为 limn(n+m)mnm=1\displaystyle \lim_{n \to \infty} \frac{(n+m)^{\overline{m}}}{n^m} = 1,从而 (n+mn)m!nm=1\displaystyle \binom{n+m}{n} \cdot \frac{m!}{n^m} = 1

m=zm = z,即可证明

这是为了引出阶乘加倍公式

x!(x12)!=(2x)!(12)!/22x(5)\displaystyle x! \left(x - \frac{1}{2}\right)! = (2x)! \left(-\frac{1}{2} \right)! \bigg/ 2^{2x} \quad \quad \textbf{(5)}

证明并不难,在 (4)\textbf{(4)} 中分别令 z=1/2,z=x,z=x1/2z = -1/2, \quad z = x, \quad z = x-1/2

(1/2)!=1/(n1/2n)n1/2\displaystyle (-1/2)! = 1 \bigg/ \binom{n-1/2}{n} \cdot n^{-1/2}

x!(x1/2)!=(n+xn)nxnx+1/2(n+x1/2n)\displaystyle x!(x-1/2)! = \binom{n+x}{n} \cdot n^{-x} \cdot n^{-x+1/2} \cdot \binom{n+x-1/2}{n}

从而有 (1/2)!x!(x1/2)!=limn+(n+xn)(n+x1/2n)n2x/(n1/2n)(1.0)\displaystyle \frac{(-1/2)!}{x!(x - 1/2)!} = \lim_{n \to +\infty} \binom{n+x}{n} \binom{n+x-1/2}{n} \cdot n^{-2x} \bigg/ \binom{n-1/2}{n} \quad \textbf{(1.0)}

这里我们要依据两个已知的恒等式rk(r12)k=(2r)2k/22k(1.1)\displaystyle r^{\underline{k}} \left(r- \frac{1}{2} \right)^{\underline{k}} = (2r)^{\underline{2k}} \bigg / 2^{2k} \quad (1.1)

(n12n)=(2nn)/22n(1.2)\displaystyle \binom{n-\frac{1}{2}}{n} = \binom{2n}{n} \bigg/ 2^{2n} \quad (1.2)

(1.1)(1.1) 中,令 r=n+x, k=nr = n+x, \ k = n,可以得到 (n+x)nn!(n+x1/2)nn!=(2n+2x)2n(2n)!(2n)!n!n!22n\displaystyle \frac{(n+x)^{\underline{n}}}{n!} \cdot \frac{(n+x-1/2)^{\underline{n}}}{n!} = \frac{(2n+2x)^{\underline{2n}}}{(2n)!} \cdot \frac{(2n)!}{n!n!} \cdot 2^{-2n}

同时 (1.2)(1.2) 给出 (n1/2n)=(2nn)22n\displaystyle \binom{n-1/2}{n} = \binom{2n}{n} \cdot 2^{-2n}

(1.0)\textbf{(1.0)} 化简成 limn(2x+2n2n)n2x\displaystyle \lim_{n \to \infty}\binom{2x+2n}{2n} \cdot n^{-2x}

同时在 (4)\textbf{(4)} 中令 z=2x, n2nz = 2x, \ n \leftarrow 2n,综上所得,可知

x!(x12)!=(12)!(2n+2x2n)n2x,(2x)!=(2n)2x(2n+2x2n)\displaystyle x!(x-\frac{1}{2})! = \frac{(-\frac{1}{2})!}{\binom{2n+2x}{2n}} \cdot n^{2x}, \quad \quad (2x)! = \frac{(2n)^{2x}}{\binom{2n+2x}{2n}}

相乘即可得到 (5)\textbf{(5)}

超几何级数与二项式还有一个推论

km(mkn)(m+n+1k)(12)k=((m+n)/2n)2nm[m+n是偶数],mn0(6)\displaystyle \sum_{k \leqslant m}\binom{m-k}{n} \binom{m+n+1}{k} \left(\frac{-1}{2} \right)^k = \binom{(m+n)/2}{n} 2^{n-m} [m+n \text{是偶数}], \quad m \geqslant n \geqslant 0 \quad \textbf{(6)}

首先不难将二项式写成超几何形式 limε0(mn)F(nm,nm1+αεm+ε| 12)\displaystyle \lim_{\varepsilon \to 0}\binom{m}{n}F\left(\begin{gathered} n-m, -n-m-1+\alpha \varepsilon \\ -m + \varepsilon \end{gathered} \middle| \ \frac{1}{2} \right)

α=2\alpha = 2,这样可以用高斯恒等式
limε0(mn)F(2(n/2m/2),2(n/2m/21/2+ε)m+ε| 12)=limε0(mn)F(n/2m/2,n/2m/21/2+εm+ε| 1)\displaystyle \lim_{\varepsilon \to 0}\binom{m}{n}F\left(\begin{gathered} 2(n/2-m/2), 2(-n/2-m/2-1/2+ \varepsilon) \\ -m +\varepsilon \end{gathered} \middle| \ \frac{1}{2} \right) = \lim_{\varepsilon \to 0}\binom{m}{n}F\left(\begin{gathered} n/2-m/2, -n/2-m/2-1/2+ \varepsilon \\ -m +\varepsilon \end{gathered} \middle| \ 1 \right)

1) m+n1)\ m+n 是奇数,m+n=2N1m+n = 2N-1,那么 F=F(Nm1/2,N+εm+ε| 1)\displaystyle F = F\left(\begin{gathered} N-m-1/2, -N + \varepsilon \\ -m +\varepsilon \end{gathered} \middle| \ 1 \right),只要证明 limε0F=0\displaystyle \lim_{\varepsilon \to 0} F = 0

\quad \quad 根据欧拉恒等式的推广,F(a,bc| 1)=Γ(cab)Γ(c)Γ(ca)Γ(cb),RcRa+Rb\displaystyle F\left(\begin{gathered} a, b \\ c \end{gathered} \middle| \ 1 \right) = \frac{\Gamma(c-a-b)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}, \quad \mathbb{R}c \geqslant \mathbb{R}a + \mathbb{R}b

\quad \quad 此时 Γ(cb)=Γ(Nm)\displaystyle \Gamma(c-b) = \Gamma(N-m),因为 2N2m=nm+102N-2m = n-m+1 \leqslant 0
\quad \quad 所以 NmN \leqslant m,因此分母 Γ(Nm)\Gamma(N-m) \to \infty,从而 F0F \to 0

2) m+n2)\ m+n 是偶数,nm=2Nn-m = -2N,那么令 n=m2Nn = m - 2N,可以得到
\quad \quad limε0F(N,Nm1/2+εm+ε| 1)\displaystyle \lim_{\varepsilon \to 0} F\left(\begin{gathered} -N, N-m-1/2+\varepsilon \\ -m+\varepsilon \end{gathered} \middle| \ 1 \right)

\quad \quad 而这个形式恰好符合恒等式 F(n,ac| 1)=(ca)ncn\displaystyle F\left(\begin{gathered} -n, a \\ c \end{gathered} \middle| \ 1 \right) = \frac{(c-a)^{\overline{n}}}{c^{\overline{n}}},代入
limε0F=(N1/2)NmN\quad \quad \displaystyle \lim_{\varepsilon \to 0}F = \displaystyle \frac{(N-1/2)^{\underline{N}}}{m^{\underline{N}}}

\quad \quad 此时只需要证明 (mm2N)(N1/2)!(1/2)!(mN)!m!=(mNm2N)22N\displaystyle \binom{m}{m-2N} \frac{(N-1/2)!}{(-1/2)!} \cdot \frac{(m-N)!}{m!} = \binom{m-N}{m-2N}2^{-2N}
\quad \quad(5)\textbf{(5)} 中令 x=Nx = N,即可证明

(mN)!(m2N)!(N)!N!(2N)!(N1/2)!(1/2)!=(mNm2N)22N\displaystyle \quad \quad \frac{(m-N)!}{(m-2N)!(N)!} \cdot \frac{N!}{(2N)!} \cdot \frac{(N-1/2)!}{(-1/2)!} = \binom{m-N}{m-2N} \cdot 2^{-2N}

至此,超几何函数只剩下差分,机械求和法两个部分没有讨论了

q-binomial

q-binomial 定义和常用结论

定义
[n]q=i=0n1qi=limxq1xn1x,[n]!q=i=1n[i]q\displaystyle [n]_q = \sum_{i = 0}^{n-1} q^i = \lim_{x \to q} \frac{1-x^n}{1-x}, \quad [n]!_q = \prod_{i = 1}^n [i]_q

[nm]q=[n]!q[m]!q[nm]!q\displaystyle {n \brack m}_q = \frac{[n]!_q}{[m]!_q [n-m]!_q}

性质1

[nm]1=(nm)\displaystyle {n \brack m}_1 = \binom{n}{m}

展开可得

[nm]q=limxqi=1n1xi1xi=1m1xi1xi=1nm1xi1x=limxqi=nm+1n(1xi)i=1m(1xi)\displaystyle {n \brack m}_q = \displaystyle \lim_{x \to q} \displaystyle \frac{\displaystyle \prod_{i = 1}^{n} \displaystyle\frac{1-x^i}{1-x}}{\displaystyle \prod_{i = 1}^{m} \displaystyle \frac{1-x^i}{1-x} \displaystyle \prod_{i = 1}^{n - m} \displaystyle \frac{1-x^i}{1-x} } = \displaystyle \lim_{x \to q} \displaystyle \frac{\displaystyle \prod_{i = n-m+1}^{n} \displaystyle (1-x^i)}{\displaystyle \prod_{i = 1}^{m} (1-x^i)}

性质2

[nm]q=[nnm]q(1)\displaystyle {n \brack m}_q = {n \brack {n-m}}_q \quad \textbf{(1)}

n1n \geqslant 1,那么有

[nm]q=[n1m1]q+qm[n1m]q(2)\displaystyle {n \brack m}_q = {n-1 \brack{m-1}}_q + q^m {n-1 \brack m}_q \quad \textbf{(2)}

证明,只需要展开即可验证
[n1m1]q+qm[n1m]q\displaystyle {n-1 \brack{m-1}}_q + q^m {n-1 \brack m}_q

=i=nm+1(1qi)i=1m1(1qi)+qmi=nmn1(1qi)i=1m(1qi)\quad \quad \displaystyle = \displaystyle \frac{\displaystyle \prod_{i = n-m+1}(1-q^i)}{\displaystyle \prod_{i = 1}^{m-1} (1-q^i)} + q^m \cdot \displaystyle \frac{\displaystyle \prod_{i = n-m}^{n-1} (1-q^i) }{\displaystyle \prod_{i = 1}^{m} (1-q^i)}

=(1qm)i=nm+1n1(1qi)+qm(1qnm)i=nm+1n1(1qi)i=1m(1qi)\quad \quad \displaystyle = \displaystyle \frac{(1-q^m)\displaystyle \prod_{i = n-m+1}^{n-1} (1-q^i) + q^m(1-q^{n-m}) \displaystyle \prod_{i = n-m+1}^{n-1}(1-q^i)}{\displaystyle \prod_{i = 1}^{m} (1-q^i)}

整理得到,i=nm+1n(1qi)i=1m(1qi)=[nm]q\displaystyle \frac{\displaystyle \prod_{i = n-m+1}^{n} (1-q^i)}{\displaystyle \prod_{i = 1}^{m} (1-q^i)} = {n \brack m}_q

于此同时,令 m=nmm = n-m 还可以得到

[nm]q=qnm[n1m1]q+[n1m]q\displaystyle {n \brack m}_q = q^{n-m} {n-1 \brack {m-1}}_q + {n-1 \brack{m}}_q

这个式子还有一个组合意义,建立一个纵向 [0,n][0, n],横向 [0,m][0, m]n×mn \times m 的网格
(0,0)(0, 0) 开始,每一次往上或者往右走一步,最后到达 (nm,m)(n-m, m)
在所有可能经过的路径中,路径右下方网格数cnt\text{cnt},那么 [nm]q\displaystyle {n \brack m}_q 就表示 qcnt\displaystyle \sum {q^{\text{cnt}}}

证明用递推,从 (0,0)(nm,m)(0, 0) \to (n-m, m) 的路径中,到达 (nm,m)(n-m, m)