Appellの超幾何級数F4[a,bc,d;x,y]:=∑0≤n,m(a,b)n+m(c)n(d)mn!m!xnymはJacobi多項式をPn(a,b)(x)=(a+1)nn!2F1[−n,a+b+n+1a+1;1−x2]として, 以下の展開公式を持つ.
F4[a,bc+1,d+1;t(1−x)(1−y)4,t(1+x)(1+y)4]=∑0≤nn!(a,b)n(c+1,d+1,c+d+n+1)ntn⋅2F1[a+n,b+nc+d+2n+2;t]Pn(c,d)(x)Pn(c,d)(y).
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F4[a,bc+1,d+1;t(1−x)(1−y)4,t(1+x)(1+y)4]=∑0≤n,m(a,b)n+m(c+1)n(d+1)mn!m!(t(1−x)(1−y)4)n(t(1+x)(1+y)4)m=∑0≤n(a,b)ntn∑k=0n1k!(c+1)k(n−k)!(d+1)n−k((1−x)(1−y)4)k((1+x)(1+y)4)n−k左辺は∑0≤nn!(a,b)n(c+1,d+1,c+d+n+1)ntn⋅2F1[a+n,b+nc+d+2n+2;t]Pn(c,d)(x)Pn(c,d)(y)=∑0≤nn!(a,b)n(c+1,d+1,c+d+n+1)ntn⋅∑0≤k(a+n,b+n)kk!(c+d+2n+2)ktkPn(c,d)(x)Pn(c,d)(y)=∑0≤n,kn!(a,b)n+k(c+d+2n+1)k!(c+1,d+1)n(c+d+n+1)n+k+1tn+kPn(c,d)(x)Pn(c,d)(y)=∑0≤n(a,b)ntn∑k=0nk!(c+d+2k+1)(n−k)!(c+1,d+1)k(c+d+k+1)n+1Pk(c,d)(x)Pk(c,d)(y)よって,∑k=0n1k!(c+1)k(n−k)!(d+1)n−k((1−x)(1−y)4)k((1+x)(1+y)4)n−k=∑k=0nk!(c+d+2k+1)(n−k)!(c+1,d+1)k(c+d+k+1)n+1Pk(c,d)(x)Pk(c,d)(y)を示せば良い. 右辺は∑k=0nk!(c+d+2k+1)(n−k)!(c+1,d+1)k(c+d+k+1)n+1Pk(c,d)(x)Pk(c,d)(y)=∑k=0n(−1)k(c+d+2k+1)k!(n−k)!(c+d+k+1)n+12F1[−k,c+d+k+1c+1;1−x2]2F1[−k,c+d+k+1d+1;1+y2]と表されるので, 1−x2,1+y2を改めてx,yとすることによって示すべき等式は∑k=0n1k!(c+1)k(n−k)!(d+1)n−k(x(1−y))k(y(1−x))n−k=∑k=0n(−1)k(c+d+2k+1)k!(n−k)!(c+d+k+1)n+12F1[−k,c+d+k+1c+1;x]2F1[−k,c+d+k+1d+1;y]両辺のxrysの係数を考えると(−1)r+s+n∑k=0n1k!(c+1)k(n−k)!(d+1)n−k(n−kr−k)(ks−n+k)=∑k=0n(−1)k(c+d+2k+1)k!(n−k)!(c+d+k+1)n+1(−k,c+d+k+1)rr!(c+1)r(−k,c+d+k+1)ss!(d+1)sを示せば良いことが分かる. 左辺はVandermondeの和公式より,(−1)r+s+n∑k=0n1k!(c+1)k(n−k)!(d+1)n−k(n−kr−k)(ks−n+k)=(−1)r+s+n(n−r)!(n−s)!∑k=0n1(r−k)!(c+1)k(s−n+k)!(d+1)n−k=(−1)r+s+n(n−r)!(n−s)!∑k=0r+s−n1k!(c+1)r−k(r+s−n−k)!(d+1)n−r+k=(−1)r+s+n(n−r)!(n−s)!(c+1)r(d+1)n−r(r+s−n)!2F1[n−r−s,−r−cd+1+n−r;1]=(−1)r+s+n(n−r)!(n−s)!(c+1)r(d+1)n−r(r+s−n)!(c+d+n+1)r+s−n(d+1+n−r)r+s−n=(−1)r+s+n(c+d+n+1)r+s−n(n−r)!(n−s)!(c+1)r(d+1)s(r+s−n)!.右辺は∑k=0n(−1)k(c+d+2k+1)k!(n−k)!(c+d+k+1)n+1(−k,c+d+k+1)rr!(c+1)r(−k,c+d+k+1)ss!(d+1)s=(−1)r+sr!s!(c+1)r(d+1)s∑k=0n(−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=(−1)r+s+nr!s!(c+1)r(d+1)s∑k=0n(−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=(−1)r+s+n(c+d+2n+1)r!s!(c+1)r(d+1)sn!(c+d+1)n+r(c+d+1)n+s(n−r)!(n−s)!(c+d+1)n(c+d+1)2n+1⋅∑k=0nc+d+2n−2k+1c+d+2n+1(r−n,s−n,−n−c−d,−1−2n−c−d)kk!(−n,−n−r−c−d,−n−s−c−d)k.ここで, Dougallの5F4和公式より,∑k=0nc+d+2n−2k+1c+d+2n+1(r−n,s−n,−n−c−d,−1−2n−c−d)kk!(−n,−n−r−c−d,−n−s−c−d)k=(−c−d−2n,−s)n−r(−n,−n−s−c−d)n−r=r!s!(1+n+r+c+d)n−rn!(r+s−n)!(1+r+s+c+d)n−r=r!s!(1+c+d)2n(1+c+d)r+sn!(r+s−n)!(1+c+d)n+r(1+c+d)n+sだから,(−1)r+s+n(c+d+2n+1)r!s!(c+1)r(d+1)sn!(c+d+1)n+r(c+d+1)n+s(n−r)!(n−s)!(c+d+1)n(c+d+1)2n+1⋅∑k=0nc+d+2n−2k+1c+d+2n+1(r−n,s−n,−n−c−d,−1−2n−c−d)kk!(−n,−n−r−c−d,−n−s−c−d)k=(−1)r+s+n(c+d+2n+1)r!s!(c+1)r(d+1)sn!(c+d+1)n+r(c+d+1)n+s(n−r)!(n−s)!(c+d+1)n(c+d+1)2n+1⋅r!s!(1+c+d)2n(1+c+d)r+sn!(r+s−n)!(1+c+d)n+r(1+c+d)n+s=(−1)r+s+n(c+d+n+1)r+s−n(n−r)!(n−s)!(c+1)r(d+1)s(r+s−n)!となって左辺に一致する.
証明からも分かるように, tの係数のPochhammer記号をいくつにしても成立する. (a,b)nだけ付けているのは左辺に現れる超幾何級数が扱いやすい2F1になるようにそのようになっているものと思われる.
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