So, from going from the shell of n equals 2, let's say, to the shell of n equals 3.

比如说，从ｎ等于２到ｎ,等于３壳层如何变化。

麻省理工公开课 - 化学原理课程节选

So kind of that strange cursive r, and our n final is 2, R so 1 over 2 squared minus n initial, so 1 over 3 squared.

因为我们可以在这里用到它，这个有点奇怪的花体。

麻省理工公开课 - 化学原理课程节选

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。

麻省理工公开课 - 固态化学导论课程节选

We said each of the merge operations O was of order n. But n is different. Right?

注意这里发生了什么,我们说过每一次合并操作的复杂度都是？

麻省理工公开课 - 计算机科学及编程导论课程节选

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。

哈佛公开课 - 计算机科学课程节选

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

耶鲁公开课 - 金融市场课程节选

That energy will be absorbed by the hydrogen atom, n=1 the electron will rise from n equals one n=2 to n equals two.

这能量将会被氢原子吸收，这个电子会从,上升到。

麻省理工公开课 - 固态化学导论课程节选

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。

麻省理工公开课 - 固态化学导论课程节选

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？

麻省理工公开课 - 计算机科学及编程导论课程节选

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。

麻省理工公开课 - 固态化学导论课程节选

*t t of n minus 1 is 3 plus 2 t of n minus 2.

加上。

麻省理工公开课 - 计算机科学及编程导论课程节选

So it looks a lot less messy if we just draw our Lewis structure like this for h c n, where we have h bonded to c triple bonded to n, and then a lone pair on the nitrogen there.

这看起来整洁了不少,如果我们把氰化氢的路易斯结构画成这样的话，这样我们就有氢与碳之间的单键和碳与氮之间三键，然后还有一对孤对电子在氮这里。

麻省理工公开课 - 化学原理课程节选

So in other words, every time I merge the point that I kept emphasizing verbally there and that I'm only touching each number once, means that we have to account for the amount of time it takes to merge N which is going to be just N. Now, this is again one of these cyclical answers.

换言之，之前在做合并时,我不停地强调,对每个数字我只碰了一次，这就是说,我们要记录合并所花的时间量,也就是这里的,这又是一种循环性的答案。

哈佛公开课 - 计算机科学课程节选

N log N is not nearly as good as log N. As a sanity check, what algorithm have we seen that runs in log N time?

而N，log，N和log，N并不一样，我们之前探讨过的哪个算法其时间复杂度是log，N呢？

哈佛公开课 - 计算机科学课程节选

We're getting further away from the nucleus because we're jumping, for example, from the n equals 2 to the n equals 3 shell, or from the n equals 3 to the n equals 4 shell.

我们将会离原子核越来越远,因为我们在跃迁，比如从，n，等于，2，的壳层到n等于，3，的壳层，或者从，n，等于，3，的壳层到n等于，4，的壳层。

麻省理工公开课 - 化学原理课程节选

b The repulsive term goes as some constant lower case b divided by R to the n. N is not the quantum number.

这种斥力很想一个固定的小写字母,被R到n分开的话，N不是量子数。

麻省理工公开课 - 固态化学导论课程节选