11/07/30 22:49:10.99
>>461
(略証)
g(x) = 1 - (1-x)^n とおくと (左辺) = 1 -g(x^m) + {g(x)}^m.
g(x) の逆函数を f(z) と書くと、 f(0)=0, f(1)=1 かつ
f(z) = 1 - (1-z)^(1/n) = (1/n)z + (1/2n)(1-1/n)z^2 + (1/3n)(1-1/n)(1-1/2n)z^3 + ……
a_k = {(k-1)/k}･{1 -1/(k-1)n}･a_{k-1} > 0.
∴ f(z) は下記の【命題268】の条件をみたす。
∴ f(z^m) ≧ {f(z)}^m,
∴ z^m ≧ g({f(z)}^m),
∴ {g(x)}^m ≧ g(x^m),
[初代スレ.563(7), 973]
[第２章.21, 346-347, 353]