Summarize Timeline Top Qs Fact Check
f
0
,
f
1
,
f
2
,
…
{\displaystyle f_{0},f_{1},f_{2},\dots }
解析関数 の列で、任意の
i
>
0
{\displaystyle i>0}
f
i
−
1
−
f
i
=
k
i
z
f
i
+
1
{\displaystyle f_{i-1}-f_{i}=k_{i}\,z\,f_{i+1}}
を満たすものとする。ここで各
k
i
{\displaystyle k_{i}}
このとき
f
i
−
1
f
i
=
1
+
k
i
z
f
i
+
1
f
i
,
and so
f
i
f
i
−
1
=
1
1
+
k
i
z
f
i
+
1
f
i
{\displaystyle {\frac {f_{i-1}}{f_{i}}}=1+k_{i}z{\frac {f_{i+1}}{f_{i}}},{\text{ and so }}{\frac {f_{i}}{f_{i-1}}}={\frac {1}{1+k_{i}z{\frac {f_{i+1}}{f_{i}}}}}}
g
i
=
f
i
/
f
i
−
1
{\displaystyle g_{i}=f_{i}/f_{i-1}}
g
i
=
1
1
+
k
i
z
g
i
+
1
{\displaystyle g_{i}={\frac {1}{1+k_{i}zg_{i+1}}}}
より、
g
1
=
f
1
f
0
=
1
1
+
k
1
z
g
2
=
1
1
+
k
1
z
1
+
k
2
z
g
3
=
1
1
+
k
1
z
1
+
k
2
z
1
+
k
3
z
g
4
=
⋯
.
{\displaystyle g_{1}={\frac {f_{1}}{f_{0}}}={\cfrac {1}{1+k_{1}zg_{2}}}={\cfrac {1}{1+{\cfrac {k_{1}z}{1+k_{2}zg_{3}}}}}={\cfrac {1}{1+{\cfrac {k_{1}z}{1+{\cfrac {k_{2}z}{1+k_{3}zg_{4}}}}}}}=\cdots .\ }
これを無限に続けると、連分数展開
f
1
f
0
=
1
1
+
k
1
z
1
+
k
2
z
1
+
k
3
z
1
+
⋱
{\displaystyle {\frac {f_{1}}{f_{0}}}={\cfrac {1}{1+{\cfrac {k_{1}z}{1+{\cfrac {k_{2}z}{1+{\cfrac {k_{3}z}{1+{}\ddots }}}}}}}}}
が得られる。
ガウスの連分数では、関数列
f
i
{\displaystyle f_{i}}
0
F
1
{\displaystyle {}_{0}F_{1}}
1
F
1
{\displaystyle {}_{1}F_{1}}
2
F
1
{\displaystyle {}_{2}F_{1}}
f
i
−
1
−
f
i
=
k
i
z
f
i
+
1
{\displaystyle f_{i-1}-f_{i}=k_{i}zf_{i+1}}
恒等式 から生じる。これらの恒等式は、級数展開をして係数を比較したり、何度か微分して消えることを確かめるといった方法で示すことができる。
最も簡単なものは、関数
0
F
1
(
a
;
z
)
=
1
+
1
a
1
!
z
+
1
a
(
a
+
1
)
2
!
z
2
+
1
a
(
a
+
1
)
(
a
+
2
)
3
!
z
3
+
⋯
{\displaystyle \,_{0}F_{1}(a;z)=1+{\frac {1}{a\,1!}}z+{\frac {1}{a(a+1)\,2!}}z^{2}+{\frac {1}{a(a+1)(a+2)\,3!}}z^{3}+\cdots }
を使った恒等式
0
F
1
(
a
−
1
;
z
)
−
0
F
1
(
a
;
z
)
=
z
a
(
a
−
1
)
0
F
1
(
a
+
1
;
z
)
{\displaystyle \,_{0}F_{1}(a-1;z)-\,_{0}F_{1}(a;z)={\frac {z}{a(a-1)}}\,_{0}F_{1}(a+1;z)}
から始めるもので、
f
i
=
0
F
1
(
a
+
i
;
z
)
,
k
i
=
1
(
a
+
i
)
(
a
+
i
−
1
)
{\displaystyle f_{i}={}_{0}F_{1}(a+i;z),\,k_{i}={\tfrac {1}{(a+i)(a+i-1)}}}
と置けば次の展開が得られる。
0
F
1
(
a
+
1
;
z
)
0
F
1
(
a
;
z
)
=
1
1
+
1
a
(
a
+
1
)
z
1
+
1
(
a
+
1
)
(
a
+
2
)
z
1
+
1
(
a
+
2
)
(
a
+
3
)
z
1
+
⋱
{\displaystyle {\frac {\,_{0}F_{1}(a+1;z)}{\,_{0}F_{1}(a;z)}}={\cfrac {1}{1+{\cfrac {{\frac {1}{a(a+1)}}z}{1+{\cfrac {{\frac {1}{(a+1)(a+2)}}z}{1+{\cfrac {{\frac {1}{(a+2)(a+3)}}z}{1+{}\ddots }}}}}}}}}
または
0
F
1
(
a
+
1
;
z
)
a
0
F
1
(
a
;
z
)
=
1
a
+
z
(
a
+
1
)
+
z
(
a
+
2
)
+
z
(
a
+
3
)
+
⋱
{\displaystyle {\frac {\,_{0}F_{1}(a+1;z)}{a\,_{0}F_{1}(a;z)}}={\cfrac {1}{a+{\cfrac {z}{(a+1)+{\cfrac {z}{(a+2)+{\cfrac {z}{(a+3)+{}\ddots }}}}}}}}}
これらの展開は、2つの収束級数の比による有理型関数 に収束する（ただし当然、a は0や負の整数であってはいけない）。
次は、関数
1
F
1
(
a
;
b
;
z
)
=
1
+
a
b
1
!
z
+
a
(
a
+
1
)
b
(
b
+
1
)
2
!
z
2
+
a
(
a
+
1
)
(
a
+
2
)
b
(
b
+
1
)
(
b
+
2
)
3
!
z
3
+
⋯
{\displaystyle {}_{1}F_{1}(a;b;z)=1+{\frac {a}{b\,1!}}z+{\frac {a(a+1)}{b(b+1)\,2!}}z^{2}+{\frac {a(a+1)(a+2)}{b(b+1)(b+2)\,3!}}z^{3}+\cdots }
から生じる2通りの恒等式
1
F
1
(
a
;
b
−
1
;
z
)
−
1
F
1
(
a
+
1
;
b
;
z
)
=
(
a
−
b
+
1
)
z
b
(
b
−
1
)
1
F
1
(
a
+
1
;
b
+
1
;
z
)
{\displaystyle \,_{1}F_{1}(a;b-1;z)-\,_{1}F_{1}(a+1;b;z)={\frac {(a-b+1)z}{b(b-1)}}\,_{1}F_{1}(a+1;b+1;z)}
1
F
1
(
a
;
b
−
1
;
z
)
−
1
F
1
(
a
;
b
;
z
)
=
a
z
b
(
b
−
1
)
1
F
1
(
a
+
1
;
b
+
1
;
z
)
{\displaystyle \,_{1}F_{1}(a;b-1;z)-\,_{1}F_{1}(a;b;z)={\frac {az}{b(b-1)}}\,_{1}F_{1}(a+1;b+1;z)}
を交互に用いて、
f
0
(
z
)
=
1
F
1
(
a
;
b
;
z
)
,
{\displaystyle f_{0}(z)=\,_{1}F_{1}(a;b;z),}
f
1
(
z
)
=
1
F
1
(
a
+
1
;
b
+
1
;
z
)
,
{\displaystyle f_{1}(z)=\,_{1}F_{1}(a+1;b+1;z),}
f
2
(
z
)
=
1
F
1
(
a
+
1
;
b
+
2
;
z
)
,
{\displaystyle f_{2}(z)=\,_{1}F_{1}(a+1;b+2;z),}
f
3
(
z
)
=
1
F
1
(
a
+
2
;
b
+
3
;
z
)
,
{\displaystyle f_{3}(z)=\,_{1}F_{1}(a+2;b+3;z),}
f
4
(
z
)
=
1
F
1
(
a
+
2
;
b
+
4
;
z
)
{\displaystyle f_{4}(z)=\,_{1}F_{1}(a+2;b+4;z)}
…と置く。これより恒等式群
f
i
−
1
−
f
i
=
k
i
z
f
i
+
1
{\displaystyle f_{i-1}-f_{i}=k_{i}zf_{i+1}}
k
1
=
a
−
b
b
(
b
+
1
)
,
k
2
=
a
+
1
(
b
+
1
)
(
b
+
2
)
,
k
3
=
a
−
b
−
1
(
b
+
2
)
(
b
+
3
)
,
k
4
=
a
+
2
(
b
+
3
)
(
b
+
4
)
{\displaystyle k_{1}={\tfrac {a-b}{b(b+1)}},k_{2}={\tfrac {a+1}{(b+1)(b+2)}},k_{3}={\tfrac {a-b-1}{(b+2)(b+3)}},k_{4}={\tfrac {a+2}{(b+3)(b+4)}}}
1
F
1
(
a
+
1
;
b
+
1
;
z
)
1
F
1
(
a
;
b
;
z
)
=
1
1
+
a
−
b
b
(
b
+
1
)
z
1
+
a
+
1
(
b
+
1
)
(
b
+
2
)
z
1
+
a
−
b
−
1
(
b
+
2
)
(
b
+
3
)
z
1
+
a
+
2
(
b
+
3
)
(
b
+
4
)
z
1
+
⋱
{\displaystyle {\frac {{}_{1}F_{1}(a+1;b+1;z)}{{}_{1}F_{1}(a;b;z)}}={\cfrac {1}{1+{\cfrac {{\frac {a-b}{b(b+1)}}z}{1+{\cfrac {{\frac {a+1}{(b+1)(b+2)}}z}{1+{\cfrac {{\frac {a-b-1}{(b+2)(b+3)}}z}{1+{\cfrac {{\frac {a+2}{(b+3)(b+4)}}z}{1+{}\ddots }}}}}}}}}}}
または
1
F
1
(
a
+
1
;
b
+
1
;
z
)
b
1
F
1
(
a
;
b
;
z
)
=
1
b
+
(
a
−
b
)
z
(
b
+
1
)
+
(
a
+
1
)
z
(
b
+
2
)
+
(
a
−
b
−
1
)
z
(
b
+
3
)
+
(
a
+
2
)
z
(
b
+
4
)
+
⋱
{\displaystyle {\frac {{}_{1}F_{1}(a+1;b+1;z)}{b{}_{1}F_{1}(a;b;z)}}={\cfrac {1}{b+{\cfrac {(a-b)z}{(b+1)+{\cfrac {(a+1)z}{(b+2)+{\cfrac {(a-b-1)z}{(b+3)+{\cfrac {(a+2)z}{(b+4)+{}\ddots }}}}}}}}}}}
同様に
1
F
1
(
a
;
b
+
1
;
z
)
1
F
1
(
a
;
b
;
z
)
=
1
1
+
a
b
(
b
+
1
)
z
1
+
a
−
b
−
1
(
b
+
1
)
(
b
+
2
)
z
1
+
a
+
1
(
b
+
2
)
(
b
+
3
)
z
1
+
a
−
b
−
2
(
b
+
3
)
(
b
+
4
)
z
1
+
⋱
{\displaystyle {\frac {{}_{1}F_{1}(a;b+1;z)}{{}_{1}F_{1}(a;b;z)}}={\cfrac {1}{1+{\cfrac {{\frac {a}{b(b+1)}}z}{1+{\cfrac {{\frac {a-b-1}{(b+1)(b+2)}}z}{1+{\cfrac {{\frac {a+1}{(b+2)(b+3)}}z}{1+{\cfrac {{\frac {a-b-2}{(b+3)(b+4)}}z}{1+{}\ddots }}}}}}}}}}}
または
1
F
1
(
a
;
b
+
1
;
z
)
b
1
F
1
(
a
;
b
;
z
)
=
1
b
+
a
z
(
b
+
1
)
+
(
a
−
b
−
1
)
z
(
b
+
2
)
+
(
a
+
1
)
z
(
b
+
3
)
+
(
a
−
b
−
2
)
z
(
b
+
4
)
+
⋱
{\displaystyle {\frac {{}_{1}F_{1}(a;b+1;z)}{b{}_{1}F_{1}(a;b;z)}}={\cfrac {1}{b+{\cfrac {az}{(b+1)+{\cfrac {(a-b-1)z}{(b+2)+{\cfrac {(a+1)z}{(b+3)+{\cfrac {(a-b-2)z}{(b+4)+{}\ddots }}}}}}}}}}}
1
F
1
(
0
;
b
;
z
)
=
1
{\displaystyle {}_{1}F_{1}(0;b;z)=1}
a を0と置き、b + 1 を b で置き換えると次の単純化された特別な連分数が得られる。
1
F
1
(
1
;
b
;
z
)
=
1
1
+
−
z
b
+
z
(
b
+
1
)
+
−
b
z
(
b
+
2
)
+
2
z
(
b
+
3
)
+
−
(
b
+
1
)
z
(
b
+
4
)
+
⋱
{\displaystyle {}_{1}F_{1}(1;b;z)={\cfrac {1}{1+{\cfrac {-z}{b+{\cfrac {z}{(b+1)+{\cfrac {-bz}{(b+2)+{\cfrac {2z}{(b+3)+{\cfrac {-(b+1)z}{(b+4)+{}\ddots }}}}}}}}}}}}}
最後は
2
F
1
(
a
,
b
;
c
;
z
)
=
1
+
a
b
c
1
!
z
+
a
(
a
+
1
)
b
(
b
+
1
)
c
(
c
+
1
)
2
!
z
2
+
a
(
a
+
1
)
(
a
+
2
)
b
(
b
+
1
)
(
b
+
2
)
c
(
c
+
1
)
(
c
+
2
)
3
!
z
3
+
⋯
{\displaystyle {}_{2}F_{1}(a,b;c;z)=1+{\frac {ab}{c\,1!}}z+{\frac {a(a+1)b(b+1)}{c(c+1)\,2!}}z^{2}+{\frac {a(a+1)(a+2)b(b+1)(b+2)}{c(c+1)(c+2)\,3!}}z^{3}+\cdots }
から、再び2つの恒等式を交互に用いる。
2
F
1
(
a
,
b
;
c
−
1
;
z
)
−
2
F
1
(
a
+
1
,
b
;
c
;
z
)
=
(
a
−
c
+
1
)
b
z
c
(
c
−
1
)
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
z
)
{\displaystyle \,_{2}F_{1}(a,b;c-1;z)-\,_{2}F_{1}(a+1,b;c;z)={\frac {(a-c+1)bz}{c(c-1)}}\,_{2}F_{1}(a+1,b+1;c+1;z)}
2
F
1
(
a
,
b
;
c
−
1
;
z
)
−
2
F
1
(
a
,
b
+
1
;
c
;
z
)
=
(
b
−
c
+
1
)
a
z
c
(
c
−
1
)
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
z
)
{\displaystyle \,_{2}F_{1}(a,b;c-1;z)-\,_{2}F_{1}(a,b+1;c;z)={\frac {(b-c+1)az}{c(c-1)}}\,_{2}F_{1}(a+1,b+1;c+1;z)}
これら2式は a と b の入れ替えに対し本質的に不変である。
f
0
(
z
)
=
2
F
1
(
a
,
b
;
c
;
z
)
,
{\displaystyle f_{0}(z)=\,_{2}F_{1}(a,b;c;z),}
f
1
(
z
)
=
2
F
1
(
a
+
1
,
b
;
c
+
1
;
z
)
,
{\displaystyle f_{1}(z)=\,_{2}F_{1}(a+1,b;c+1;z),}
f
2
(
z
)
=
2
F
1
(
a
+
1
,
b
+
1
;
c
+
2
;
z
)
,
{\displaystyle f_{2}(z)=\,_{2}F_{1}(a+1,b+1;c+2;z),}
f
3
(
z
)
=
2
F
1
(
a
+
2
,
b
+
1
;
c
+
3
;
z
)
,
{\displaystyle f_{3}(z)=\,_{2}F_{1}(a+2,b+1;c+3;z),}
f
4
(
z
)
=
2
F
1
(
a
+
2
,
b
+
2
;
c
+
4
;
z
)
,
{\displaystyle f_{4}(z)=\,_{2}F_{1}(a+2,b+2;c+4;z),}
…とおく。これより恒等式群
f
i
−
1
−
f
i
=
k
i
z
f
i
+
1
{\displaystyle f_{i-1}-f_{i}=k_{i}zf_{i+1}}
k
1
=
(
a
−
c
)
b
c
(
c
+
1
)
,
k
2
=
(
b
−
c
−
1
)
(
a
+
1
)
(
c
+
1
)
(
c
+
2
)
,
k
3
=
(
a
−
c
−
1
)
(
b
+
1
)
(
c
+
2
)
(
c
+
3
)
,
k
4
=
(
b
−
c
−
2
)
(
a
+
2
)
(
c
+
3
)
(
c
+
4
)
{\displaystyle k_{1}={\tfrac {(a-c)b}{c(c+1)}},k_{2}={\tfrac {(b-c-1)(a+1)}{(c+1)(c+2)}},k_{3}={\tfrac {(a-c-1)(b+1)}{(c+2)(c+3)}},k_{4}={\tfrac {(b-c-2)(a+2)}{(c+3)(c+4)}}}
2
F
1
(
a
+
1
,
b
;
c
+
1
;
z
)
2
F
1
(
a
,
b
;
c
;
z
)
=
1
1
+
(
a
−
c
)
b
c
(
c
+
1
)
z
1
+
(
b
−
c
−
1
)
(
a
+
1
)
(
c
+
1
)
(
c
+
2
)
z
1
+
(
a
−
c
−
1
)
(
b
+
1
)
(
c
+
2
)
(
c
+
3
)
z
1
+
(
b
−
c
−
2
)
(
a
+
2
)
(
c
+
3
)
(
c
+
4
)
z
1
+
⋱
{\displaystyle {\frac {{}_{2}F_{1}(a+1,b;c+1;z)}{{}_{2}F_{1}(a,b;c;z)}}={\cfrac {1}{1+{\cfrac {{\frac {(a-c)b}{c(c+1)}}z}{1+{\cfrac {{\frac {(b-c-1)(a+1)}{(c+1)(c+2)}}z}{1+{\cfrac {{\frac {(a-c-1)(b+1)}{(c+2)(c+3)}}z}{1+{\cfrac {{\frac {(b-c-2)(a+2)}{(c+3)(c+4)}}z}{1+{}\ddots }}}}}}}}}}}
または
2
F
1
(
a
+
1
,
b
;
c
+
1
;
z
)
c
2
F
1
(
a
,
b
;
c
;
z
)
=
1
c
+
(
a
−
c
)
b
z
(
c
+
1
)
+
(
b
−
c
−
1
)
(
a
+
1
)
z
(
c
+
2
)
+
(
a
−
c
−
1
)
(
b
+
1
)
z
(
c
+
3
)
+
(
b
−
c
−
2
)
(
a
+
2
)
z
(
c
+
4
)
+
⋱
{\displaystyle {\frac {{}_{2}F_{1}(a+1,b;c+1;z)}{c{}_{2}F_{1}(a,b;c;z)}}={\cfrac {1}{c+{\cfrac {(a-c)bz}{(c+1)+{\cfrac {(b-c-1)(a+1)z}{(c+2)+{\cfrac {(a-c-1)(b+1)z}{(c+3)+{\cfrac {(b-c-2)(a+2)z}{(c+4)+{}\ddots }}}}}}}}}}}
2
F
1
(
0
,
b
;
c
;
z
)
=
1
{\displaystyle {}_{2}F_{1}(0,b;c;z)=1}
a を 0 と置き、c + 1 を c で置き換えると次の単純化された特別な連分数が得られる。
2
F
1
(
1
,
b
;
c
;
z
)
=
1
1
+
−
b
z
c
+
(
b
−
c
)
z
(
c
+
1
)
+
−
c
(
b
+
1
)
z
(
c
+
2
)
+
2
(
b
−
c
−
1
)
z
(
c
+
3
)
+
−
(
c
+
1
)
(
b
+
2
)
z
(
c
+
4
)
+
⋱
{\displaystyle {}_{2}F_{1}(1,b;c;z)={\cfrac {1}{1+{\cfrac {-bz}{c+{\cfrac {(b-c)z}{(c+1)+{\cfrac {-c(b+1)z}{(c+2)+{\cfrac {2(b-c-1)z}{(c+3)+{\cfrac {-(c+1)(b+2)z}{(c+4)+{}\ddots }}}}}}}}}}}}}
