Table of Content
\(a, b\) 的角色可以互換（想當然爾）
\[\vartheta F(a_1, \cdots, a_m; b_1, \cdots, b_n; z) = \sum_{k \geq 0} k{a_1^{\bar k}, \cdots a_m^{\bar k} \over b_1^{\bar k}, \cdots b_n^{\bar k}} {z^k \over k!} \tag{2}\]Define operator \(D = {d \over dz},\ \vartheta = zD\)
底下比較有用：
\[(\vartheta + a_1)F(a_1, \cdots, a_m; b_1, \cdots, b_n; z) \\ = a_1F(a_1+1, \cdots, a_m; b_1, \cdots, b_n; z) \tag{3}\] \[(\vartheta + b_1 - 1)F(a_1, \cdots, a_m; b_1, \cdots, b_n; z) \\ = (b_1 - 1)F(a_1, \cdots, a_m; b_1-1, \cdots, b_n; z) \tag{4}\]綜合 \((1), (3), (4)\)，
\[D(\vartheta + b_1 - 1)\cdots(\vartheta + b_n - 1)F = (\vartheta + a_1)\cdots(\vartheta + a_m)F \tag{5}\]然後兩邊都乘上 \(z\)
\[{\vartheta}(\vartheta + b_1 - 1)\cdots(\vartheta + b_n - 1)F = z(\vartheta + a_1)\cdots(\vartheta + a_m)F \tag{6}\]第一個 \(\vartheta\) 可以想成 \((\vartheta + 1 - 1)\)，對應 term ratio 中分母的 \((k+1)\)；其餘依樣對應（互相記憶的好方法！）
這個微分方程可以將解表示為 hypergeometric function。
若
\[\vartheta G = z(\vartheta+1)^2G\]則
\[G = F(1,1;;z) \tag*{$\blacksquare$}\]或是用來驗證兩 identity 是否相同（是否符合同一微分方程）。
“Every new identity for hypergeometrics has consequences for binomial coefficients.” [CMath] p.222
但要注意退化
\(\vartheta\) 和 \(D\) 互相轉換
[待續] p.237