Whippleの4F3変換公式

Eulerの変換公式
$$\begin{array}{r}{}_2{F}_1[\begin{array}{c}a,b\\ c\end{array};x]=(1-x{)}^{c-a-b}{}_2{F}_1[\begin{array}{c}c-a,c-b\\ c\end{array};x]\end{array}$$
より$c-a-b=f-d-e$のとき,
$$\begin{array}{r}(1-x{)}^{c-a-b}{}_2{F}_1[\begin{array}{c}c-a,c-b\\ c\end{array};x]{}_2{F}_1[\begin{array}{c}d,e\\ f\end{array};x]=(1-x{)}^{f-d-e}{}_2{F}_1[\begin{array}{c}a,b\\ c\end{array};x]{}_2{F}_1[\begin{array}{c}f-d,f-e\\ f\end{array};x]\end{array}$$
であるから,
$$\begin{array}{r}{}_2{F}_1[\begin{array}{c}c-a,c-b\\ c\end{array};x]{}_2{F}_1[\begin{array}{c}d,e\\ f\end{array};x]={}_2{F}_1[\begin{array}{c}a,b\\ c\end{array};x]{}_2{F}_1[\begin{array}{c}f-d,f-e\\ f\end{array};x]\end{array}$$
である. 両辺の$x^n$の係数を比較すると
$$\begin{array}{r}\sum _{k=0}^n\frac{(c-a,c-b{)}_k}{k!(c{)}_k}\frac{(d,e{)}_{n-k}}{(n-k)!(f{)}_{n-k}}=\sum _{k=0}^n\frac{(a,b{)}_k}{k!(c{)}_k}\frac{(f-d,f-e{)}_{n-k}}{(n-k)!(f{)}_{n-k}}\end{array}$$
を得る. これを書き換えると,
$$\begin{array}{r}{}_4{F}_3[\begin{array}{c}c-a,c-b,1-n-f,-n\\ c,1-n-d,1-n-e\end{array};1]=\frac{(f-d,f-e{)}_n}{(d,e{)}_n}{}_4{F}_3[\begin{array}{c}a,b,1-n-f,-n\\ c,1-n-f+d,1-n-f+e\end{array};1]\end{array}$$
である. $a\mapsto c-a,b\mapsto c-b,d\mapsto 1-n-d,e\mapsto 1-n-e,f\mapsto 1-n-f$とすると, 条件は$1+a+b+f=c+d+e+n$となり,
$$\begin{array}{rl}{}_4{F}_3[\begin{array}{c}a,b,f,-n\\ c,d,e\end{array};1]& =\frac{(d-f,e-f{)}_n}{(1-n-d,1-n-e{)}_n}{}_4{F}_3[\begin{array}{c}c-a,c-b,f,-n\\ c,1-n+f-d,1-n+f-e\end{array};1]\\ & =\frac{(d-f,e-f{)}_n}{(d,e{)}_n}{}_4{F}_3[\begin{array}{c}c-a,c-b,f,-n\\ c,1-n+f-d,1-n+f-e\end{array};1]\end{array}$$
が得られる. 変数を入れ替えて定理を得る.