c*w
发帖数: 4736 | 1 Here's a math problem from my friend. It spent me more than an hour
to work it out.
At first, I want to ask you a question: selecting randomly n+1 points
on a
n-dimesion sphere, what is the probability that they are on one
semi-sphere?
Please start with n=2, then show me your genius in general statistics.
hehe.
NOTE and HINT: if n points on a n-dimension sphere, then the
probability that
they are on the same semi-sphere is 1!!! | c*w
发帖数: 4736 | 2 I'm sorry! What I meant by n-dimension sphere is:
in a (zhi jiao) coordinate x1, x2, ...., xn,
the function of a n-dimensional sphere is:
x1*x1 + x2*x2 + .... + xn*xn = R*R
Just now a Da Xia told me the standard defination of a m-dimensional
sphere is like: 1-dimension sphere is a circle. Then I know that
I've made a stupid mistake. heehee....
~{!>~} ~{TZ~} cyw (~{E#~}) ~{5D4sWwVPLa5=~}: ~{!?~}
points
statistics. | C******a
发帖数: 115 | 3 在|z|=1上归纳定义多项式P_n(z)和Q_n(z)，n=0,1,2,....
P_0(z)=Q_0(z)=1, P_{m+1}(z)=P_m(z)+z^{2^m}*Q_m(z),
Q_{m+1}(z)=P_m(z)-z^{2^m}*Q_m(z).
则P_n和Q_n的次数是2^n-1，且所有系数都是1或-1。
用平行四边形法则可以证明，对所有的|z|=1，
|P_{m+1}(z)|^2+|Q_{m+1}(z)|^2=2*(|P_m(z)|^2+|Q_m(z)|^2)。
所以|P_m(z)|^2+|Q_m(z)|^2=2^{m+1}，进而
|P_m(z)|, |Q_m(z)|<=2^{(m+1)/2}。
现在将Q_m(z)写成\sum_{i=0}^{2^m-1} a_{m,i}*z^i的形式。
这里a_{m,i}都是1或者-1。如果取上述和式的一部分求和，
例如i从k*2^n到(k+1)*2^n-1，其中0<=n
暂时称这样的和式为n级的标准块吧。
这个和式必然是z^{k*2^n}*P_n(z), z^{k*2^n}*Q_n(z),
-z^{k*2^n | s******r
发帖数: 9 | 4 Find two sequences An and Bn of integers
such that
limAn(n->Wu Qiong Da)
=limBn(n->Wu Qiong Da)
=Wu Qiong Da
and
neither An=O(Bn) nor Bn=O(An) holds. (Big O)
Thanks! | w*********l
发帖数: 7 | 5 A=X^2
B=X^3 when X odd;
B=X when X even.
【在 s******r 的大作中提到】
: Find two sequences An and Bn of integers
: such that
: limAn(n->Wu Qiong Da)
: =limBn(n->Wu Qiong Da)
: =Wu Qiong Da
: and
: neither An=O(Bn) nor Bn=O(An) holds. (Big O)
: Thanks!
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