Hypergeometric Functions II

#Concrete Mathematics, Hypergeometric

2022/09/01

---

Table of Content

• Hypergeometric Transformation
  • Reflection Law
  • Differentiation
    • Example $\rm I$
  • More on Differential Operators
• Partial Hypergeometric Sums

---

Hypergeometric Transformation

Reflection Law

$${1 \over (1-z)^a}F(a, b; c; {-z \over 1-z}) = F(a, c-b; c; z)$$

$a, b$ 的角色可以互換（想當然爾）

Differentiation

$${d \over dz}F(a_1, \cdots, a_m; b_1, \cdots, b_n; z) \\ = {a_1 \cdots a_m \over b_1 \cdots b_n}F(a_1+1, \cdots, a_m+1; b_1+1, \cdots, b_n+1; z) \tag{1}$$

Define operator $D = {d \over dz},\ \vartheta = zD$

$$\vartheta F(a_1, \cdots, a_m; b_1, \cdots, b_n; z) = \sum_{k \geq 0} k{a_1^{\bar k}, \cdots a_m^{\bar k} \over b_1^{\bar k}, \cdots b_n^{\bar k}} {z^k \over k!} \tag{2}$$

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。

Example $\rm I$

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

More on Differential Operators

$$\vartheta^n = \sum_k\begin{Bmatrix}n \\ k\end{Bmatrix}z^kD^k$$ $$z^nD^n = \sum_k\begin{bmatrix}n \\ k\end{bmatrix}(-1)^{n-k}\vartheta^k$$

$\vartheta$ 和 $D$ 互相轉換

---

Partial Hypergeometric Sums

[待續] p.237