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.. I started out with the equally-weighted-- I was talking about stocks-- about n stocks that all have the same variance and are all independent of each other.
开始的时候我讲了等权重的-,我开始时讲了股票-,几支拥有相同方差的股票,彼此间相互独立。
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And an electron is something where, i n fact, we might be able to, if we calculate it and see how that works out, actually observe some of its wave-like properties.
如果我们对电子做计算,并且知道如何算出来的,那么我们是可以观测到,电子的一些波动性质的。
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It turns out there's a new function strlen S-T-R-L-E-N that you probably have not used yet called strlen, S-T-R-L-E-N, programmers early on and still like to be succinct but communicates sufficiently what they mean.
结果这里有个新的函数,你可能没有使用过，叫做，早期的和现在的程序员喜欢简洁地,但是能充分的传达他们的意思。
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n*pi/L It turns out it is n pi over l.
结果是。
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And then one of the things that I suggested was that if we could figure out some way to order it, and in particular, if we could order it in n log n time, and we still haven't done that, but if we could do that, then we said the complexity changed a little bit.
这就涉及到了排序,如果可以想出一种来将其进行排序,甚至可以在n，log，n的时间内完成,虽然目前我们没做这件事,但是一旦开始做这件事,那么复杂性就是发生一些变化。
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So if you're trying to make a more complicated organic molecule carbon-carbon bonds are one of the most difficult things to make in organic chemistry, and it turns out that c n minus is a very reactive molecule, so it's a good way, even though we'll go over some drawbacks in a second, it is a good way to make carbon-carbon bonds.
如果你要合成一个更复杂的有机分子,碳碳键是有机化学中,最难制造的键之一，而实际上氰离子是一种具有很高活性的分子，用它是一个好办法，尽管我们一会儿将看到它的一些缺点，但它的确是一个制造碳碳键的好方法。
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So, it turns out that n is not the only quantum number needed to describe a wave function, however. There's two more you can see come out of it.
事实上，n不是描述一个波函数需要的,唯一的量子数，你们可以看到，还需要，两个量子数。
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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 instead you'd have to maybe if you start with wavelength, go over there, and then figure out velocity and do something more like kinetic energy equals 1/2 n b squared to get there.
这时你要先从波长开始，到这，然后算出速度,然后像动能等于1/2nb平方得到这。
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psi So we're going to for psi, and before that, we're going to figure out that instead of n just that one quantum number n, we're going to have a few other quantum numbers that fall out of solving the Schrodinger equation for what psi is.
我们要讲到，但在这之前，我们已经知道了,主量子数，现在我们需要知道，其他一些,解psi的薛定谔方程,所需要的量子数。
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So there are two electron configurations in the n equals one shell, if we follow according to the selection rules that we spelled out last day.
如果根据上次课,我们阐明的原子光谱选择定则，我们就会知道在n等于1的那一层,有两种电子图像构型。
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So what we'll do is this problem here, which is let's calculate out what the wavelength of radiation n would be emitted from a hydrogen atom if we start at the n equals 3 level and we go down to the n equals 2 level.
我们来做这个问题,让我们来计算一下,从n等于3到,等于2能级氢原子辐射的波长是多少。
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And we can generalize to figure out, based on any principle quantum number n, how many orbitals we have of the same energy, n and what we can say is that for any shell n, there are n squared degenerate orbitals.
我们可以总结出来，在，主量子数为n的情况下，同一个能量上，有多少个轨道，我们可以说，对任何壳层，有n平方个简并轨道。
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I was just finding very tunnel vision-like, the smallest elements at that moment in time which means I don't know anything about the other elements other than they are not the smallest and so no matter what with Selection Sort I had to repeat this again and again and again and if you do out the math it's roughly N squared steps in the worst case as well.
我只有一个狭窄的视野，只知道某时刻的最小元素,就意味着我并不知道其他元素的任何情况,只知道它们不是最小的,所以不管怎样，在选择排序中,我就得一遍一遍地重复选择过程,在最坏情况下,大概需要N的平方次比较。
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