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Remember, we don't do a one-to-one correlation, because p x and p y are some linear combination of the m plus 1 and m minus 1 orbital.
记住，我们不需要把它们一一对应，因为ｐｘ和ｐｙ轨道是,ｍ等于正负１轨道的线性组合。
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STUDENT: PROFESSOR: Oh, I'm sorry. So it's n minus l minus 1.
学生：,教授：不好意思，这是ｎ减去ｌ减去１．
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So I'm first going to look for something that's not in the list, I'm going to see, is minus 1 in this list, so it's going to be at the far end, and if I do that in the basic case, bam.
如果我试试第一种最基本的方法,噢，一下就完成了对不对？,因为这种方法查了下第一个元素，然后发现目标数比较下，因为目标数小于第一个元素。
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Right? In other words I've got the stoichiometric coefficients in there and the values, and I'm subtracting the reactants from products -1652kJ/mol wind up with minus 1652 kilojoules per mole.
对吧？换句话说这里我用了化学,计量系数和生成热的值，从生成物中减去反应物,最后得到。
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And I'm going to get minus 2, which came from here: minus 2 and that is in fact negative, which is what I wanted to know.
最后只剩下-2了,2是一个负数,这正是我想要的
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That's true, but especially because of my parenthesization 32 over here, because I'm saying, "Do f minus 32, but then multiply it by the division on the left."
是那样的，但是特别地，因为我这里加上了括号,因为我指明，“f减去，然后它乘以左边的那个除法的结果“
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1 Here I have m is one, zero, minus one.
再有，m为1，0，或。
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uniform So here, I'm going to look at the thing random dot uniform, for example, between minus volatility and plus volatility.
所以在这里，我会看到random。,比方说，在-浮动性值+浮动值之间。
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And I'll just put a minus indicating I'm done.
我会放一个-在这里,表示我没有物品可选了。
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So basically what I'm going to do here is I'm going to set my price to equal his price, minus a little bit, I'll just undercut him a little bit and by just undercutting him a little bit, I'm going to get the whole of the market and I'll make as much money as I can on those sales.
所以基本上我要做的是,设定我的价格等于他的价格,减去一点点,我的价格将仅仅比他的低一点点,并且通过比他的价格低一点点,我将占领整个市场,那些销售带给我尽可能多的钱
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The end test-- and I've got the wrong glasses on but it's up here, where I'm looking to see, is last minus first less than or equal to 2?
最后一次比较，我看错了,在上面，last减去first是不是小于2呢？,好的，既然我得到了一个？
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So we can have, if we have the final quantum number m equal plus 1 or minus 1, we're dealing with a p x or a p y orbital.
所以如果我们有,磁量子数ｍ等于正负１,我们讨论的就是ｐｘ或者ｐｙ轨道。
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q2 So I'm going to write this as q2 over q1 over minus one plus q2 over q1.
因此这等于q2除以q1，除以负1加上q2除以,我这么做的目的是。
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So finally, if we talk about our last example of when n is going to be equal 2, l -1 and then minus 1 for m.
最后我们要讲的,这个例子，是n等于2时，等于1，we，can，have，2，，1，for，l,而m等于。
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1 Similarly, if m is equal to either plus 1 or minus 1, py we would in turn call it the p y orbital, or the p x orbital.
类似的，如果m等于+1或，我们可以叫它，或者px轨道。
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So I'm done when b minus k equals 1, or k equals b minus 1.
因此b-k=1或k-b=1的时候，我就可以停下来了。
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I've got to test to see if I'm in the base case, and if I'm not, then I need to move a tower of size n minus 1, I need to move a tower of size 1, and I need to move a second-sorry about that a second tower of size n minus 1.
首先我看看我是不是在最基本的情况，如果不是的话，我得先做一个N-1个，圆盘的移动，移动一个圆盘,然后再做一次N-1个圆盘的移动。
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And I can solve them, and if I solve them out, I'm going to get something like, let me just be careful, I'm going to get something like: 1 minus B S1* is equal to 1, or S1* equals S2* is equal to 1 over .
我们都会解这个方程,如果解完方程,我们会得到,我得仔细点别算错,我们会得到,1-BS1*=1,或S1*=S2*=1/
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That's just the those-- let me use my pointer that's just these two lines here. I checked the value, and in one case, I'm changing the last to be mid minus 1, which is the case I'm in here and I just call again. All right?
来看看ppt，这里有两条线,我检查了值，在一种情况下,我把last指向了中点减一,就是这种情况,我们再来调用一下怎么样？
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On the next step though, this, we get substituted by that. Right, on the next step, I'm back in the even case, it's going to take six more steps, plus t of b minus 1. Oops, sorry about that, over 2.
这一步就是偶数了，这一步会让我们得到，6+t这样的结果，因为b-1现在是偶数了,别忽略这里的细节。
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If I want to move this stack here, I'm going to take a stack of size n minus 1 move it to the spare spot, now I can move the base disk over, and then I'm going to move that stack of size n minus 1 to there. That's literally what I did, OK?
如果我想移动这些圆盘，我先把从n-1个圆盘1,移动到多余的柱子上去，这样我就能把最底下的圆盘放到这儿了，然后再把从n-1个圆盘放到这儿来,这就是确切的我怎么做的对不对？
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