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标题: Find an counterexample [打印本页]

作者: rongrong 时间: 2006-10-25 20:57
标题: Find an counterexample

f is a continuous function, suppose x1 and x2 are any element of (a, b) which is subset of R. If x1f(x2), so we can conclude that for all x is the element of (a,b), f '(x)<0. (false)

Can you guys find an counterexample to disprove this statement? Just need to find an example to show that f '(x) would be greater or equal to 0. Thank you.

作者: 野菜花 时间: 2006-10-25 21:21
标题: 回复：Find an counterexample

f(x)=-x³ is decreasing on (-1,1)

but f '(x)=-3x² is zero at x=0

(it is impossible to find an x-value such that f '(x)>0)

作者: rongrong 时间: 2006-10-27 18:58
标题: 回复：回复：Find an counterexample

Awesome, you are reallly professional. Thank you very much.

作者: constant 时间: 2006-10-27 21:22
标题: 能不能做一个不为常数的可微函数，在一个稠密集上导数为0？（不知道答案，瞎问[:P]）

能不能做一个不为常数的可微函数，在一个稠密集上导数为0？（不知道答案，瞎问

）

作者: yinyin 时间: 2006-10-28 07:16
标题: 回复：能不能做一个不为常数的可微函数，在一个稠密集上导数为0？---No.

No. This can be proved via a "proof by contradiction" where the Finite-Covering Theorem is used.

作者: yma16 时间: 2006-10-28 08:19
标题: 回复：能不能做一个不为常数的可微函数，在一个稠密集上导数为0？（不知道答案，瞎问[:P]）

f(x)=x^3 if x>0 and f(x)=0 if x<=0

If it does not work, please let me know. I don't remember these stuff.

作者: yinyin 时间: 2006-10-28 09:21
标题: 回复：回复：能不能做一个不为常数的可微函数，在一个稠密集上导数为0？（不知道答案，瞎问[:P]）

The dense set in Constant's question should be understood as "the set is dense in the whole domain of this function". Otherwise, the question is trivial as shown in your example.

作者: yma16 时间: 2006-10-28 16:41
标题: I got it-eom

I got it-eom

作者: constant 时间: 2006-10-28 18:23
标题: Can you write out your proof? It is most likely

not that simple, probably beyong calculus. I can construct a function that is continuous, non-constant, differentiable almost everywhere, and yet has 0 derivative on a dense set.

作者: yinyin 时间: 2006-10-28 20:44
标题: 回复：Can you write out your proof? It is most likely

In case you use "differentiable almost everywhere" to replace "differentiable everywhere", the question is trivial. For example, a function being zero except at the origin (x=0).

作者: constant 时间: 2006-10-28 20:51
标题: Read carefully. "continuous"

Read carefully. "continuous"

作者: yinyin 时间: 2006-10-28 21:06
标题: 回复：Can you write out your proof? It is most likely

You are right. The proof is "not simple" and "beyond calculus". The following are the steps (sorry for omitting the details) of the proof.

(1) Assume that function f is not a constant. Then there are two points a and b such that a0.www.ddhw.com

(2) Choose a dense set D at which f '=0.

(3) For each point d in D, choose an open interval I_d (may be very small) such that |[f(d)- f(x)]/( d-x)|< h/2(b-a) for any x in D.www.ddhw.com

(4) The union of these open intervals covers the closed interval [a, b]. By the Open Covering Theorem, we can choose finitely many intervals from these open intervals such that their union still covers [a, b]. www.ddhw.com

(5) Use (3) and (4) to show that |f(a)-f(b)|

作者: yinyin 时间: 2006-10-28 21:19
标题: 回复：Can you write out your proof? It is most likely

Sorry. I thought that the continuity is also almost everywhere.

作者: yinyin 时间: 2006-10-28 21:22
标题: 回复：Can you write out your proof? It is most likely

Yes. You may use Cantor's set to form such an Example.

作者: yinyin 时间: 2006-10-28 21:23
标题: 回复：回复：Can you write out your proof? It is most lik

Sorry. There is some problem in step 4. Let me consider it again.

作者: 野菜花 时间: 2006-10-28 22:27
标题: 康大帝，你不知道答案还要来为难我呀，我脑筋已经生锈了。不过对你构造的那个函

康大帝，你不知道答案还要来为难我呀，我脑筋已经生锈了不过对你构造的那个函数还是很有兴趣看的.www.ddhw.com

抱歉昨天没上网，刚看到。看来 yinyin 也是个数学高手啊!

作者: constant 时间: 2006-10-29 00:12
标题: 回复：康大帝，你不知道答案还要来为难我呀，我脑筋已经生锈了。不过对你构造的那个函

大致上是在[0,1]上作一个测度<1的集S，使得S与任何子区间的交有正测度。然后让f(x)在S上为0，其余为1。再作F(x)为 f 的积分。F是单调函数，因此几乎处处可微，并且导数与 f 几乎处处相等。

作者: 野菜花 时间: 2006-10-29 03:09
标题: 很妙，佩服！[:-Q][:-Q][@};-]

很妙，佩服！

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