Re: 不動點
※ 引述《aacvbn (救我的普物阿~~(?o?))》之銘言:
: Let f:[0,1]-->[0,1] is a continuous function ,
: then there exists a point c (- [0,1]
: such that f(c)=1-c^(2)
: 謝謝
Consider g(x) = f(x) - (1 - x^2).
g(0) = f(0) - 1≦0,
g(1) = f(1) ≧ 0.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.112.218.142
※ 編輯: plover 來自: 140.112.218.142 (11/08 00:17)
推
11/08 00:24, , 1F
11/08 00:24, 1F