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Feldheimの双線形公式

Appellの超幾何級数
$\begin{aligned} F_{4} [\begin{array}{c} a, b \\ c, d \end{array}; x, y] := \sum_{0 \leq n, m} \frac{(a, b)_{n + m}}{(c)_{n} (d)_{m} n! m!} x^{n} y^{m} \end{aligned}$
はJacobi多項式を
$\begin{aligned} P_{n}^{(a, b)} (x) & = \frac{(a + 1)_{n}}{n!}_{2} F_{1} [\begin{array}{c} - n, a + b + n + 1 \\ a + 1 \end{array}; \frac{1 - x}{2}] \end{aligned}$
として, 以下の展開公式を持つ.

Feldheim(1941)

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

原論文にはアクセスできなかったので, 代わりに思いついた証明を書くことにする.

右辺は
$\begin{aligned} F_{4} [\begin{array}{c} a, b \\ c + 1, d + 1 \end{array}; \frac{t (1 - x) (1 - y)}{4}, \frac{t (1 + x) (1 + y)}{4}] \\ = \sum_{0 \leq n, m} \frac{(a, b)_{n + m}}{(c + 1)_{n} (d + 1)_{m} n! m!} {(\frac{t (1 - x) (1 - y)}{4})}^{n} {(\frac{t (1 + x) (1 + y)}{4})}^{m} \\ = \sum_{0 \leq n} (a, b)_{n} t^{n} \sum_{k = 0}^{n} \frac{1}{k! (c + 1)_{k} (n - k)! (d + 1)_{n - k}} {(\frac{(1 - x) (1 - y)}{4})}^{k} {(\frac{(1 + x) (1 + y)}{4})}^{n - k} \end{aligned}$
左辺は
$\begin{aligned} \sum_{0 \leq n} \frac{n! (a, b)_{n}}{(c + 1, d + 1, c + d + n + 1)_{n}} t^{n} \\ \cdot_{2} F_{1} [\begin{array}{c} a + n, b + n \\ c + d + 2 n + 2 \end{array}; t] P_{n}^{(c, d)} (x) P_{n}^{(c, d)} (y) \\ = \sum_{0 \leq n} \frac{n! (a, b)_{n}}{(c + 1, d + 1, c + d + n + 1)_{n}} t^{n} \\ \cdot \sum_{0 \leq k} \frac{(a + n, b + n)_{k}}{k! (c + d + 2 n + 2)_{k}} t^{k} P_{n}^{(c, d)} (x) P_{n}^{(c, d)} (y) \\ = \sum_{0 \leq n, k} \frac{n! (a, b)_{n + k} (c + d + 2 n + 1)}{k! (c + 1, d + 1)_{n} (c + d + n + 1)_{n + k + 1}} t^{n + k} P_{n}^{(c, d)} (x) P_{n}^{(c, d)} (y) \\ = \sum_{0 \leq n} (a, b)_{n} t^{n} \sum_{k = 0}^{n} \frac{k! (c + d + 2 k + 1)}{(n - k)! (c + 1, d + 1)_{k} (c + d + k + 1)_{n + 1}} P_{k}^{(c, d)} (x) P_{k}^{(c, d)} (y) \end{aligned}$
よって,
$\begin{aligned} \sum_{k = 0}^{n} \frac{1}{k! (c + 1)_{k} (n - k)! (d + 1)_{n - k}} {(\frac{(1 - x) (1 - y)}{4})}^{k} {(\frac{(1 + x) (1 + y)}{4})}^{n - k} \\ = \sum_{k = 0}^{n} \frac{k! (c + d + 2 k + 1)}{(n - k)! (c + 1, d + 1)_{k} (c + d + k + 1)_{n + 1}} P_{k}^{(c, d)} (x) P_{k}^{(c, d)} (y) \end{aligned}$
を示せば良い. 右辺は
$\begin{aligned} \sum_{k = 0}^{n} \frac{k! (c + d + 2 k + 1)}{(n - k)! (c + 1, d + 1)_{k} (c + d + k + 1)_{n + 1}} P_{k}^{(c, d)} (x) P_{k}^{(c, d)} (y) \\ = \sum_{k = 0}^{n} \frac{(- 1)^{k} (c + d + 2 k + 1)}{k! (n - k)! (c + d + k + 1)_{n + 1}}_{2} F_{1} [\begin{array}{c} - k, c + d + k + 1 \\ c + 1 \end{array}; \frac{1 - x}{2}]_{2} F_{1} [\begin{array}{c} - k, c + d + k + 1 \\ d + 1 \end{array}; \frac{1 + y}{2}] \end{aligned}$
と表されるので, $\frac{1 - x}{2}, \frac{1 + y}{2}$ を改めて $x, y$ とすることによって示すべき等式は
$\begin{aligned} \sum_{k = 0}^{n} \frac{1}{k! (c + 1)_{k} (n - k)! (d + 1)_{n - k}} {(x (1 - y))}^{k} {(y (1 - x))}^{n - k} \\ = \sum_{k = 0}^{n} \frac{(- 1)^{k} (c + d + 2 k + 1)}{k! (n - k)! (c + d + k + 1)_{n + 1}}_{2} F_{1} [\begin{array}{c} - k, c + d + k + 1 \\ c + 1 \end{array}; x]_{2} F_{1} [\begin{array}{c} - k, c + d + k + 1 \\ d + 1 \end{array}; y] \end{aligned}$
両辺の $x^{r} y^{s}$ の係数を考えると
$\begin{aligned} (- 1)^{r + s + n} \sum_{k = 0}^{n} \frac{1}{k! (c + 1)_{k} (n - k)! (d + 1)_{n - k}} (\binom{n - k}{r - k}) (\binom{k}{s - n + k}) \\ = \sum_{k = 0}^{n} \frac{(- 1)^{k} (c + d + 2 k + 1)}{k! (n - k)! (c + d + k + 1)_{n + 1}} \frac{(- k, c + d + k + 1)_{r}}{r! (c + 1)_{r}} \frac{(- k, c + d + k + 1)_{s}}{s! (d + 1)_{s}} \end{aligned}$
を示せば良いことが分かる. 左辺はVandermondeの和公式より,
$\begin{aligned} (- 1)^{r + s + n} \sum_{k = 0}^{n} \frac{1}{k! (c + 1)_{k} (n - k)! (d + 1)_{n - k}} (\binom{n - k}{r - k}) (\binom{k}{s - n + k}) \\ = \frac{(- 1)^{r + s + n}}{(n - r)! (n - s)!} \sum_{k = 0}^{n} \frac{1}{(r - k)! (c + 1)_{k} (s - n + k)! (d + 1)_{n - k}} \\ = \frac{(- 1)^{r + s + n}}{(n - r)! (n - s)!} \sum_{k = 0}^{r + s - n} \frac{1}{k! (c + 1)_{r - k} (r + s - n - k)! (d + 1)_{n - r + k}} \\ = \frac{(- 1)^{r + s + n}}{(n - r)! (n - s)! (c + 1)_{r} (d + 1)_{n - r} (r + s - n)!}_{2} F_{1} [\begin{array}{c} n - r - s, - r - c \\ d + 1 + n - r \end{array}; 1] \\ = \frac{(- 1)^{r + s + n}}{(n - r)! (n - s)! (c + 1)_{r} (d + 1)_{n - r} (r + s - n)!} \frac{(c + d + n + 1)_{r + s - n}}{(d + 1 + n - r)_{r + s - n}} \\ = \frac{(- 1)^{r + s + n} (c + d + n + 1)_{r + s - n}}{(n - r)! (n - s)! (c + 1)_{r} (d + 1)_{s} (r + s - n)!} . \end{aligned}$
右辺は
$\begin{aligned} \sum_{k = 0}^{n} \frac{(- 1)^{k} (c + d + 2 k + 1)}{k! (n - k)! (c + d + k + 1)_{n + 1}} \frac{(- k, c + d + k + 1)_{r}}{r! (c + 1)_{r}} \frac{(- k, c + d + k + 1)_{s}}{s! (d + 1)_{s}} \\ = \frac{(- 1)^{r + s}}{r! s! (c + 1)_{r} (d + 1)_{s}} \sum_{k = 0}^{n} \frac{(- 1)^{k} k! (c + d + 2 k + 1) (c + d + 1)_{k + r} (c + d + 1)_{k + s}}{(k - r)! (k - s)! (n - k)! (c + d + 1)_{k} (c + d + 1)_{n + k + 1}} \\ = \frac{(- 1)^{r + s + n}}{r! s! (c + 1)_{r} (d + 1)_{s}} \sum_{k = 0}^{n} \frac{(- 1)^{k} (n - k)! (c + d + 2 n - 2 k + 1) (c + d + 1)_{n - k + r} (c + d + 1)_{n - k + s}}{(n - k - r)! (n - k - s)! k! (c + d + 1)_{n - k} (c + d + 1)_{2 n - k + 1}} \\ = \frac{(- 1)^{r + s + n} (c + d + 2 n + 1)}{r! s! (c + 1)_{r} (d + 1)_{s}} \frac{n! (c + d + 1)_{n + r} (c + d + 1)_{n + s}}{(n - r)! (n - s)! (c + d + 1)_{n} (c + d + 1)_{2 n + 1}} \\ \cdot \sum_{k = 0}^{n} \frac{c + d + 2 n - 2 k + 1}{c + d + 2 n + 1} \frac{(r - n, s - n, - n - c - d, - 1 - 2 n - c - d)_{k}}{k! (- n, - n - r - c - d, - n - s - c - d)_{k}} . \end{aligned}$
ここで, Dougallの $_{5} F_{4}$ 和公式より,
$\begin{aligned} \sum_{k = 0}^{n} \frac{c + d + 2 n - 2 k + 1}{c + d + 2 n + 1} \frac{(r - n, s - n, - n - c - d, - 1 - 2 n - c - d)_{k}}{k! (- n, - n - r - c - d, - n - s - c - d)_{k}} \\ = \frac{(- c - d - 2 n, - s)_{n - r}}{(- n, - n - s - c - d)_{n - r}} \\ = \frac{r! s! (1 + n + r + c + d)_{n - r}}{n! (r + s - n)! (1 + r + s + c + d)_{n - r}} \\ = \frac{r! s! (1 + c + d)_{2 n} (1 + c + d)_{r + s}}{n! (r + s - n)! (1 + c + d)_{n + r} (1 + c + d)_{n + s}} \end{aligned}$
だから,
$\begin{aligned} \frac{(- 1)^{r + s + n} (c + d + 2 n + 1)}{r! s! (c + 1)_{r} (d + 1)_{s}} \frac{n! (c + d + 1)_{n + r} (c + d + 1)_{n + s}}{(n - r)! (n - s)! (c + d + 1)_{n} (c + d + 1)_{2 n + 1}} \\ \cdot \sum_{k = 0}^{n} \frac{c + d + 2 n - 2 k + 1}{c + d + 2 n + 1} \frac{(r - n, s - n, - n - c - d, - 1 - 2 n - c - d)_{k}}{k! (- n, - n - r - c - d, - n - s - c - d)_{k}} \\ = \frac{(- 1)^{r + s + n} (c + d + 2 n + 1)}{r! s! (c + 1)_{r} (d + 1)_{s}} \frac{n! (c + d + 1)_{n + r} (c + d + 1)_{n + s}}{(n - r)! (n - s)! (c + d + 1)_{n} (c + d + 1)_{2 n + 1}} \\ \cdot \frac{r! s! (1 + c + d)_{2 n} (1 + c + d)_{r + s}}{n! (r + s - n)! (1 + c + d)_{n + r} (1 + c + d)_{n + s}} \\ = \frac{(- 1)^{r + s + n} (c + d + n + 1)_{r + s - n}}{(n - r)! (n - s)! (c + 1)_{r} (d + 1)_{s} (r + s - n)!} \end{aligned}$
となって左辺に一致する.