Feldheimの双線形公式

右辺は
\begin{align} &F_4\left[\begin{matrix}a,b\\c+1,d+1\end{matrix};\frac{t(1-x)(1-y)}4,\frac{t(1+x)(1+y)}4\right]\\ &=\sum_{0\leq n,m}\frac{(a,b)_{n+m}}{(c+1)_n(d+1)_mn!m!}\left(\frac{t(1-x)(1-y)}4\right)^n\left(\frac{t(1+x)(1+y)}4\right)^m\\ &=\sum_{0\leq n}(a,b)_nt^n\sum_{k=0}^n\frac 1{k!(c+1)_k(n-k)!(d+1)_{n-k}}\left(\frac{(1-x)(1-y)}4\right)^k\left(\frac{(1+x)(1+y)}4\right)^{n-k} \end{align}
左辺は
\begin{align} &\sum_{0\leq n}\frac{n!(a,b)_n}{(c+1,d+1,c+d+n+1)_{n}}t^n\\ &\qquad\cdot \F21{a+n,b+n}{c+d+2n+2}t P^{(c,d)}_n(x)P^{(c,d)}_n(y)\\ &=\sum_{0\leq n}\frac{n!(a,b)_n}{(c+1,d+1,c+d+n+1)_{n}}t^n\\ &\qquad\cdot\sum_{0\leq k}\frac{(a+n,b+n)_k}{k!(c+d+2n+2)_k}t^k P^{(c,d)}_n(x)P^{(c,d)}_n(y)\\ &=\sum_{0\leq n,k}\frac{n!(a,b)_{n+k}(c+d+2n+1)}{k!(c+1,d+1)_n(c+d+n+1)_{n+k+1}}t^{n+k}P^{(c,d)}_n(x)P^{(c,d)}_n(y)\\ &=\sum_{0\leq n}(a,b)_nt^n\sum_{k=0}^n\frac{k!(c+d+2k+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{align}
よって,
\begin{align} &\sum_{k=0}^n\frac 1{k!(c+1)_k(n-k)!(d+1)_{n-k}}\left(\frac{(1-x)(1-y)}4\right)^k\left(\frac{(1+x)(1+y)}4\right)^{n-k}\\ &=\sum_{k=0}^n\frac{k!(c+d+2k+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{align}
を示せば良い. 右辺は
\begin{align} &\sum_{k=0}^n\frac{k!(c+d+2k+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+2k+1)}{k!(n-k)!(c+d+k+1)_{n+1}}\F21{-k,c+d+k+1}{c+1}{\frac{1-x}2}\F21{-k,c+d+k+1}{d+1}{\frac{1+y}2} \end{align}
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{align} &\sum_{k=0}^n\frac 1{k!(c+1)_k(n-k)!(d+1)_{n-k}}\left(x(1-y)\right)^k\left(y(1-x)\right)^{n-k}\\ &=\sum_{k=0}^n\frac{(-1)^k(c+d+2k+1)}{k!(n-k)!(c+d+k+1)_{n+1}}\F21{-k,c+d+k+1}{c+1}{x}\F21{-k,c+d+k+1}{d+1}{y} \end{align}
両辺の$x^ry^s$の係数を考えると
\begin{align} &(-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+2k+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{align}
を示せば良いことが分かる. 左辺はVandermondeの和公式より,
\begin{align} &(-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)!}\F21{n-r-s,-r-c}{d+1+n-r}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{align}
右辺は
\begin{align} &\sum_{k=0}^n\frac{(-1)^k(c+d+2k+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)^kk!(c+d+2k+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+2n-2k+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)_{2n-k+1}}\\ &=\frac{(-1)^{r+s+n}(c+d+2n+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)_{2n+1}}\\ &\cdot\sum_{k=0}^n\frac{c+d+2n-2k+1}{c+d+2n+1}\frac{(r-n,s-n,-n-c-d,-1-2n-c-d)_k}{k!(-n,-n-r-c-d,-n-s-c-d)_k}. \end{align}
ここで, Dougallの${}_5F_4$和公式より,
\begin{align} &\sum_{k=0}^n\frac{c+d+2n-2k+1}{c+d+2n+1}\frac{(r-n,s-n,-n-c-d,-1-2n-c-d)_k}{k!(-n,-n-r-c-d,-n-s-c-d)_k}\\ &=\frac{(-c-d-2n,-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)_{2n}(1+c+d)_{r+s}}{n!(r+s-n)!(1+c+d)_{n+r}(1+c+d)_{n+s}} \end{align}
だから,
\begin{align} &\frac{(-1)^{r+s+n}(c+d+2n+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)_{2n+1}}\\ &\cdot\sum_{k=0}^n\frac{c+d+2n-2k+1}{c+d+2n+1}\frac{(r-n,s-n,-n-c-d,-1-2n-c-d)_k}{k!(-n,-n-r-c-d,-n-s-c-d)_k}\\ &=\frac{(-1)^{r+s+n}(c+d+2n+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)_{2n+1}}\\ &\cdot\frac{r!s!(1+c+d)_{2n}(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{align}
となって左辺に一致する.