Holdemanの公式のa類似

かなり前の記事でLegendre多項式の変数付きの拡張として, $a$ -Legendre多項式を導入した, それは $[a]_{n} := \frac{(a)_{n}}{n!}$ として,
$\begin{aligned} ρ_{n}^{(a)} (x) & = (- 1)^{n} [a]_{n} \sum_{k = 0}^{n} \frac{(- n, a + n)_{k}}{k! (a)_{k}} x^{k} \end{aligned}$
によって定義される. 今回は, Holdemanによる公式
$\begin{aligned} _{2} F_{1} [\begin{array}{c} b, c \\ 1 \end{array}; \frac{1 + x}{2}] & = \frac{Γ (2 - b - c)}{Γ (2 - b) Γ (2 - c)} \sum_{0 \leq n} \frac{(2 n + 1) (b, c)_{n}}{(2 - b, 2 - c)_{n}} P_{n} (x) \end{aligned}$
の $a$ 類似となる以下の公式を示したいと思う.

a

-Holdemanの公式

$\begin{aligned} _{2} F_{1} [\begin{array}{c} b, c \\ a \end{array}; x] & = \frac{Γ (a) Γ (1 + a - b - c)}{Γ (1 + a - b) Γ (1 + a - c)} \sum_{0 \leq n} \frac{(2 n + a) (b, c)_{n}}{(1 + a - b, 1 + a - c)_{n}} ρ_{n}^{(a)} (x) \end{aligned}$

実際に $a = 1$ のとき, $ρ^{(1)} (x) = P_{n} (2 x - 1)$ より
$\begin{aligned} _{2} F_{1} [\begin{array}{c} b, c \\ 1 \end{array}; x] & = \frac{Γ (2 - b - c)}{Γ (2 - b) Γ (2 - c)} \sum_{0 \leq n} \frac{(2 n + 1) (b, c)_{n}}{(2 - b, 2 - c)_{n}} P_{n} (2 x - 1) \end{aligned}$
となって, Holdemanの公式が得られる.

項別積分とSaalschützの和公式によって,
$\begin{aligned} \int_{0}^{1} x^{c - 1} ρ_{n}^{(a)} (x) d x & = (- 1)^{n} [a]_{n} \sum_{k = 0}^{n} \frac{(- n, a + n)_{k}}{k! (a)_{k} (k + c)} \\ = \frac{(- 1)^{n} [a]_{n}}{c}_{3} F_{2} [\begin{array}{c} - n, a + n, c \\ a, c + 1 \end{array}; 1] \\ = \frac{(- 1)^{n} [a]_{n}}{c} \frac{(1, a - c)_{n}}{(a, c + 1)_{n}} \\ = \frac{(- 1)^{n} (a - c)_{n}}{(c)_{n + 1}} \end{aligned}$
である. よって, 項別積分とGaussの超幾何定理より,
$\begin{aligned} \int_{0}^{1} x^{a - 1}_{2} F_{1} [\begin{array}{c} b, c \\ a \end{array}; x] ρ_{n}^{(a)} (x) d x & = \int_{0}^{1} \sum_{0 \leq k} \frac{(b, c)_{k}}{k! (a)_{k}} x^{k + a - 1} ρ_{n}^{(a)} (x) d x \\ = \sum_{0 \leq k} \frac{(b, c)_{k}}{k! (a)_{k}} \frac{(- 1)^{n} (- k)_{n}}{(a + k)_{n + 1}} \\ = \sum_{0 \leq k} \frac{(b, c)_{k}}{(k - n)! (a)_{n + k + 1}} \\ = \frac{(b, c)_{n}}{(a)_{2 n + 1}}_{2} F_{1} [\begin{array}{c} b + n, c + n \\ a + 2 n + 1 \end{array}; 1] \\ = \frac{(b, c)_{n}}{(a)_{2 n + 1}} \frac{Γ (a + 2 n + 1) Γ (1 + a - b - c)}{Γ (1 + a - b + n) Γ (1 + a - c + n)} \\ = \frac{Γ (a) Γ (1 + a - b - c)}{Γ (1 + a - b) Γ (1 + a - c)} \frac{(b, c)_{n}}{(1 + a - b, 1 + a - c)_{n}} \end{aligned}$
である. よって, $ρ_{n}^{(a)} (x)$ の直交性,
$\begin{aligned} \int_{0}^{1} ρ_{n}^{(a)} (x) ρ_{m}^{(a)} (x) x^{a - 1} d x & = \frac{1}{2 n + a} δ_{n, m} \end{aligned}$
によって, 定理が得られる.

さて, 証明の過程におけるGaussの超幾何定理を用いたところで, より一般にSaalschützの和公式を用いることを考えると, 以下を得る.

$\begin{aligned} _{3} F_{2} [\begin{array}{c} b, c, - N \\ a, b + c - N - a \end{array}; x] & = \frac{(1 + a - b, 1 + a - c)_{N}}{(a)_{N + 1} (1 + a - b - c)_{N}} \sum_{0 \leq n} \frac{(- 1)^{n} (2 n + a) (b, c, - N)_{n}}{(1 + a - b, 1 + a - c, 1 + a + N)_{n}} ρ_{n}^{(a)} (x) \end{aligned}$

先ほどと同様の方針で, Saalschützの和公式を用いると
$\begin{aligned} \int_{0}^{1} x^{a - 1}_{3} F_{2} [\begin{array}{c} b, c, - N \\ a, b + c - N - a \end{array}; x] ρ_{n}^{(a)} (x) d x \\ = \sum_{0 \leq k} \frac{(b, c, - N)_{k}}{(k - n)! (a)_{n + k + 1} (b + c - N - a)_{k}} \\ = \frac{(b, c, - N)_{n}}{(a)_{2 n + 1} (b + c - N - a)_{n}}_{2} F_{1} [\begin{array}{c} b + n, c + n, n - N \\ a + 2 n + 1, b + c - N - a + n \end{array}; 1] \\ = \frac{(b, c, - N)_{n}}{(a)_{2 n + 1} (b + c - N - a)_{n}} \frac{(1 + a - b + n, 1 + a - c + n)_{N - n}}{(a + 2 n + 1, 1 + a - b - c)_{N - n}} \\ = \frac{(1 + a - b, 1 + a - c)_{N}}{(a)_{N + 1} (1 + a - b - c)_{N}} \frac{(- 1)^{n} (b, c, - N)_{n}}{(1 + a - b, 1 + a - c, 1 + a + N)_{n}} \end{aligned}$
となることから分かる.