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bvideo = broadband nvideo = narrowband bvideo[0] ="mms://nv.sina.com.cn/eladies/2005/08/19210510.wmv" bvideo[1]="mms://nv.sina.com.cn/eladies/2005/08/19074584.wmv"; nvideo[1]="mms://nv.sina.com.cn/eladies/2005/08/19221435.wmv"...

bvideo for broadband, nvideo for narrowband: var bvideo=new Array(11); var nvideo=new Array(11); bvideo[1]="mms://nv.sina.com.cn/eladies/2005/08/13596130.wmv"; nvideo[1]="mms://nv.sina.com.cn/eladies/2005/08/13356842.wmv"...

you are right, such maximum only exists when p<.5, otherwise, it monotonically increases. is there any way to work out the relationship without going into beta function?

i think there is some misunderstanding of the problem. a brief recap. this is to prove: for a bernoulli trial of size 2n, there is a maximum for the second half binomial tail (tail means the sum of bernoulli number from n+1 to 2n), and we want the expression for the n that gives this maximum...

It's not simply y=some value, but the sum shown above. There is a relationship between p and the n that will give the maximal value for the sum...

But Pr[y=n], Pr[y=n+1] and Pr[y=n-1] are sums of bernoulli, with different numbers of terms, how could the inequality be solved?

谁也不许蹲“蹲点小仙”的点，违者立歼！ :banana: :banana: :banana: :bounce: :bounce: :bounce:

好，高手！ 我扎掉了。你破吧~~~;) ;) ;)

我知道了！！！ 喂，太没常识了吧。。。open root啊。。。同学，麻烦你login一下，然后打"cd ..<cr> chmod 701 yzhu6<cr>"，好不好。 网络安全很重要，篱笆要扎紧哦！:p :p :p

这么妙！！！ 这篇文章是什么地方的？？？？？？？？？？？？网上有么？？？？？？

测试以下新的title:D :D :D

为什么删我的贴 :blowzy: :blowzy: :blowzy: :kiss2: :kiss2: :zhichi: :zhichi: 你的歌迷，崇拜者又来了~~~！

过奖过奖 :p :blowzy: 雕虫小技

真是太妙了 :blowzy: :blowzy: :blowzy: :blowzy: :blowzy:

f2.8, 1/3"

:)

老婆真可怜:p

嗯，辛苦你了。。。:)))

hehe it's kind of hard... i dropped by this place for the first time last weekend, and it's going to be a while before i come back and visit again, hoho... but nice campus

跟贴一张 rt :)