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There are some relative, the notion that the energy gap between n equals one and n equals two is greater than that for n equals two to n equals three. That is correctly represented.
还有很多与之相关的内容，比如说这个观点，第一能量级和第二能量级,之间的能量差要大于第二和第三能量级间的,能量差，而这已经被正确地表示出来了。
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n Alright. So if n is less than or equal to 1, return n. Well that's not right, right?
好吧，如果n小于等于1，它会返回，这里不对，是吧？
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And note that as Z increases, as the proton number increases the radius decreases for a given n number.
并注意到当Z不断增加，对于一个给定的n，即当质子数增加的时候,半径的n值就减小了。
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J++ Now here's the semicolon, J less than N, where N is this, J plus, plus; so what am I doing?
现在这里是一个分号，J小于N，N是这个，那么我在做什么？
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We'll get to discussing that, but what I want to point out here again is the fact that instead of just being dependent on n, the energy level is dependent on both n and l.
我们将要讨论它，但是我想指出的是事实上能级不只是与n有关,而是与n和l都有关系，而且n不再是。
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So, the optimal thing to do if you live in a world like this n is to get n as large possible and you can reduce the standard deviation of the portfolio very much and there's no cost in terms of expected return.
如果现实中也这样简单的话,那么你就尽量增大,这样就能让投资组合的标准差,就会大大降低,从预期收益率的角度来看,这样做的成本是零。
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RT So it's delta u plus RT, we can say T1 is the n temperature we've used here, delta n of the gas.
所以它应该是ΔU加,我们可以在这里用T1，气体的Δ
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It depends on n and k but obviously as n gets big, that one is going to be better.
但是很明显，当n取很大的值的时候,前者要比后者好。
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If I look at something that goes as n squared, if this is the edge of the nucleus here and if this is r1, 4 it says when n goes to two the radius goes to four.
如果在原子核外侧,我们发现某一半径和n的平方成正比，也就是说当n为2时半径等于。
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So, it would be a good question to ask n why are we limited clearly there's this relationship between l and n, and we can't get any higher than n equals one.
那我们要问了,为什么我们被限制了，很明显，l和,有关系，为什么不能取到，n-1更大的值。
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to the n, every value in the 1 bit vector we looked at last time is either 0 or 1. So it's a binary n number of n bits, 2 to the n.
从2到n，我们上次看到的,位向量的每个值不是0就是,所以它是n，比特的二进制数,从2到。
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If they type in bogus characters, it's going to yell at them and make them retry, and eventually I'm going to get handed back an int, which I'm storing in n. Well, if I actually want to judge this number based on its magnitude, well, I can say now, "If n is greater than or equal to one."
如果他们键入了不合法的字符，它将对他们叫喊,然后叫他们重试，最终我将,得到一个int数，它是存储在n中的，好的，如果我的确想要根据它的量级来判定这个数字，好的，现在我可以指明，“是否n大于等于1“
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One over two squared minus one over n squared 3 4 5 where n takes values three, four, five, six.
除以，2，的平方再减去1除以n的平方,将n赋值为。
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It at least does corroborate the claim that merge sort N*log N as we argue intuitively is in fact, N log N in running time.
但这至少证实了归并排序,的时间复杂度为。
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*t t of n minus 1 is 3 plus 2 t of n minus 2.
加上。
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If you have a sample with n observations, it's the summation I = 1 to n of xi/n--that's the average.
如果你有n个观测值,对Xi从i=1到n求和再除以n
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