作者Dirichlet ( )
看板trans_math
标题Re: 连续的问题
时间Thu Jul 28 14:30:22 2005
※ 引述《JimCroce (我要下五子棋 N )》之铭言:
: Let f be a continuous function and "n" a positive integer.
: Show that if 0<= f(x) <=1 for all x (- [0,1] ,
: then there exists at least one point c in [0,1] for which f(c) = c^n
: 抱歉 可以问一下好心人 是要把定义弄熟做这些show的题目才会好上手嘛
Define g(x) = f(x) - x^n for all x in [0,1],
take a,b in [0,1] s.t. f(a) = 0 and f(b) = 1
case1. a=0 or b=1 (easy)
case2. a≠0 and b≠1 => g(a) < 0 and g(b) > 0, I.V.T. => there is one point c
in [a,b] s.t. g(c) = f(c) - c^n = 0 i.e. f(c) = c^n
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