Re: WRONG! Re: PROOF -- Re: EE challenge# CS - 计算机科学
f*r
1 楼
1/N,
In fact, z(0)+z(1/N)+ z(2/N) + .... z(N-1/N)=0 has used 1/N. Otherwise, it can
not be zero. Anyway, melo's idea is cool.
It seems that the problem's second part is more challenge. I'm not math guy,
hope some guy can give an answer. But if it involves too advanced knowledge,
then I will lose interest to it.
am
y(t).