作者bks (bks)
看板Math
标题[微积] Rolle's thm
时间Fri Mar 23 13:52:58 2012
题目是这样的
Suppose that f and g are differentiable functions and f(x)g'(x)-g(x)f'(x) has
no zeros on some intervals I. Assume that there are numbers a and b in I with
a<b for which f(a)=f(b)=0 and that f has no zeros in (a,b). Prove that if
g(a)≠0 and g(b)≠0, then g has exactly one zero in (a,b).
题目给的hint是：解 h=f/g, 再解 k=g/f
h=f/g 这部份我写这样
Suppose that g has no zeros in (a,b) and consider h(x)=f(x)/g(x).
Since g(x)≠0 in [a,b] => h(x) is conti. on [a,b] and diff. on (a,b) and
h(a)=h(b)=0. Therefore by Rolle's thm, ∃c in (a,b) s.t. h'(c)=0 -><-
=> g has at least one zero in (a,b).
可是第二部分就看不懂提示了， 令 k=g/f
因为 f(a)=f(b)=0, 所以 k 在 a,b 两点就没有定义了 要怎麽用 Rolle's thm ？
谢谢
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◆ From: 140.115.221.192
※ 编辑: bks 来自: 140.115.221.192 (03/23 14:22)
1F：→ empty24 :假设g在(a,b)上有两相异根c,d(c<d) 对h用rolle's 03/23 14:45
2F：→ empty24 :试试看 最後会弄出与"f'g-gf'在I上没有根"矛盾 03/23 14:46
3F：→ empty24 :打错了抱歉~~是对k 还有f'g-fg' 03/23 14:54
4F：→ bks :谢谢、我想通了 03/23 16:58