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Jacobi多項式の母関数に関するBaileyの公式

区間 $(0, 1)$ における重み関数 $t^{a - 1} (1 - t)^{b - 1}$ に対するJacobi多項式を
$\begin{array}{r} ρ_{n}^{(a, b)} (x) := (- 1)^{n} \frac{(a)_{n}}{n!} \sum_{k = 0}^{n} \frac{(- n, a + b + n - 1)_{k}}{k! (a)_{k}} x^{k} \end{array}$
とする.

Watson(1922)

$\begin{aligned} _{} F_{4} [\begin{array}{c} - n, a + n \\ b, 1 + a - b \end{array}; x y, (1 - x) (1 - y)] & = \frac{(- 1)^{n} (b)_{n}}{(1 + a - b)_{n}}_{2} F_{1} [\begin{array}{c} - n, a + n \\ b \end{array}; x]_{2} F_{1} [\begin{array}{c} - n, a + n \\ b \end{array}; y] \end{aligned}$

前の記事で示したBaileyの公式
$\begin{aligned} _{} F_{4} [\begin{array}{c} a, b \\ c, 1 + a + b - c \end{array}; x (1 - y), y (1 - x)] & =_{2} F_{1} [\begin{array}{c} a, b \\ c \end{array}; x]_{2} F_{1} [\begin{array}{c} a, b \\ 1 + a + b - c \end{array}; y] \end{aligned}$
において, $a \mapsto - n, b \mapsto a + n, c \mapsto b, y \mapsto 1 - y$ として,
$\begin{aligned} _{2} F_{1} [\begin{array}{c} - n, a + n \\ 1 + a - b \end{array}; 1 - y] & = \frac{(- 1)^{n} (b)_{n}}{(1 + a - b)_{n}}_{2} F_{1} [\begin{array}{c} - n, a + n \\ b \end{array}; y] \end{aligned}$
を用いればよい.

Bailey(1938)

$s = \frac{4 t}{(1 + t)^{2}}$ として,
$\begin{aligned} \sum_{0 \leq n} \frac{n! (a + b - 1)_{n}}{(a, b)_{n}} ρ_{n}^{(a, b)} (x) ρ_{n}^{(a, b)} (y) t^{n} & = \frac{1}{(1 + t)^{a + b - 1}}_{} F_{4} [\begin{array}{c} \frac{a + b - 1}{2}, \frac{a + b}{2} \\ a, b \end{array}; s x y, s (1 - x) (1 - y)] \end{aligned}$
が成り立つ.

右辺は
$\begin{aligned} \frac{1}{(1 + t)^{a + b - 1}}_{} F_{4} [\begin{array}{c} \frac{a + b - 1}{2}, \frac{a + b}{2} \\ a, b \end{array}; s x y, s (1 - x) (1 - y)] & = \sum_{0 \leq m, n} \frac{{(\frac{a + b - 1}{2}, \frac{a + b}{2})}_{m + n}}{m! (a)_{m} n! (b)_{n}} \frac{(4 t)^{m + n}}{(1 + t)^{2 m + 2 n + a + b - 1}} (x y)^{m} ((1 - x) (1 - y))^{n} \\ = \sum_{0 \leq m, n, k} \frac{(a + b - 1)_{2 m + 2 n}}{m! (a)_{m} n! (b)_{n}} \frac{(2 m + 2 n + a + b - 1)_{k}}{k!} (- 1)^{k} t^{m + n + k} (x y)^{m} ((1 - x) (1 - y))^{n} \\ = \sum_{0 \leq m, n, m + n \leq k} \frac{(a + b - 1)_{m + n + k}}{m! (a)_{m} n! (b)_{n} (k - m - n)!} (- 1)^{k - m - n} t^{k} (x y)^{m} ((1 - x) (1 - y))^{n} \end{aligned}$
より, 両辺の $t^{N}$ の関数を考えると,
$\begin{aligned} \frac{N! (a + b - 1)_{N}}{(a, b)_{N}} ρ_{N}^{(a, b)} (x) ρ_{N}^{(a, b)} (y) & = \sum_{0 \leq m, n} \frac{(a + b - 1)_{m + n + N}}{m! (a)_{m} n! (b)_{n} (N - m - n)!} (- 1)^{N - m - n} (x y)^{m} ((1 - x) (1 - y))^{n} \end{aligned}$
を示せばよい. これは
$\begin{aligned} \frac{(a)_{N}}{(b)_{N}}_{2} F_{1} [\begin{array}{c} - N, a + b + N - 1 \\ a \end{array}; x]_{2} F_{1} [\begin{array}{c} - N, a + b + N - 1 \\ a \end{array}; y] & = (- 1)^{N}_{} F_{4} [\begin{array}{c} - N, a + b + N - 1 \\ a, b \end{array}; x y, (1 - x) (1 - y)] \end{aligned}$
と同値であるので, 前の命題より従う.

元々のBaileyによる結果は区間 $(- 1, 1)$ におけるJacobi多項式 $P_{n}^{(a, b)}$ で書かれているが, 個人的に区間を $(0, 1)$ にした方が扱いやすいのでそのようにした.

$y = 1 - x$ とすると
$\begin{aligned} _{} F_{4} [\begin{array}{c} a, b \\ c_{1}, c_{2} \end{array}; x, x] & =_{4} F_{3} [\begin{array}{c} a, b, \frac{c_{1} + c_{2} - 1}{2}, \frac{c_{1} + c_{2}}{2} \\ c_{1}, c_{2}, c_{1} + c_{2} - 1 \end{array}; 4 x] \end{aligned}$
なので, 以下の系を得る.

$\begin{aligned} \sum_{0 \leq n} \frac{n! (a + b - 1)_{n}}{(a, b)_{n}} ρ_{n}^{(a, b)} (x) ρ_{n}^{(b, a)} (x) (- t)^{n} & =_{4} F_{3} [\begin{array}{c} \frac{a + b - 1}{2}, \frac{a + b - 1}{2}, \frac{a + b}{2}, \frac{a + b}{2} \\ a, b, a + b - 1 \end{array}; 4 x (1 - x) \frac{4 t}{(1 + t)^{2}}] \end{aligned}$

さらに $a = b$ としてGegenbauer多項式, Legendre多項式で書き表すと以下を得る

$\begin{aligned} \sum_{0 \leq n} \frac{n!}{(2 a)_{n}} C_{n}^{(a)} (2 x - 1)^{2} (- t)^{n} & =_{2} F_{1} [\begin{array}{c} a, a \\ 2 a \end{array}; 4 x (1 - x) \frac{4 t}{(1 + t)^{2}}] \\ \sum_{0 \leq n} P_{n} (2 x - 1)^{2} (- t)^{n} & =_{2} F_{1} [\begin{array}{c} \frac{1}{2}, \frac{1}{2} \\ 1 \end{array}; 4 x (1 - x) \frac{4 t}{(1 + t)^{2}}] \end{aligned}$