k**t
发帖数: 12 | 1 1 suppose f is a continous function defined on [0, n], and f(0)=f(n)
(x,y) is a good pair if (1) x-y=Integer; (2) f(x)=f(y).
Prove there is at least n good pairs.
2 Define M_k=sup f^(k) ( f^(k): f's k-th derivative), f is a smooth function
on R.
Prove M_1^2<=2M_0M_2
到底有没有数学版啊 | I***e
发帖数: 1136 | 2
Assume that f'(0)=0, f'(a)=A>0 and f'(x) is positive between 0 and a. Then,
Sup_{0<=x<=a} | f(x) | > 1/2 Integral_{x=0 to a} f'(x) dx.
A^2 = f'(a)^2 = Integral_{x=0 to a} f'(x) f''(x) dx < Integral f'(x) dx *
sup(f''(x))
So A^2 < 2* Sup_[0,a] |f(x)| * Sup_[0,a] | f''(x) |
This easily generates to the desired result because of the following:
1. f has to be bounded. If f' has a zero point then the problem is solved.
2. if f' doesn't have a zero point, then this problem can be proven by
limiting s
【在 k**t 的大作中提到】
: 1 suppose f is a continous function defined on [0, n], and f(0)=f(n)
: (x,y) is a good pair if (1) x-y=Integer; (2) f(x)=f(y).
: Prove there is at least n good pairs.
: 2 Define M_k=sup f^(k) ( f^(k): f's k-th derivative), f is a smooth function
: on R.
: Prove M_1^2<=2M_0M_2
: 到底有没有数学版啊
| k**t
发帖数: 12 | 3 给一个几何证明
图中直线的斜率是－M_2,曲线是f'，曲线不能到三角形里面去，否则和M_2的定义矛盾，
所以曲线的积分面积大于三角形的面积。
假设f(0)＝0，f'(0)=M_1,
把刚才的意思翻译出来就得到不等式了。
一般情况下，要分情况讨论，idea就是这样
Then, | I***e
发帖数: 1136 | 4
Then,
Oops... you are right...
But the proof can be modified by considering f such that f'(0)=f'(b)=0 and
f'(a)=A is the local maximum of f'(x) between 0 and b, 0
that
f(a)>= min( integral_[0,a] f'(x), Integral_[a,b] f'(x) )
So there is no more 1/2 in the inequality for f(a). Thus we can still make the
conclusion...
Icare |
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