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Getting a grade like a B-plus or a C-minus adds or subtracts a few tenths of a point.
VOA: special.2009.03.05
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so it's important to note that it's not in b, now we're talking about b plus, because we've already taken an electron out here.
其中有一个非常重要的地方需要注意，不是硼，而是正一价硼离子，因为我们已经拿走了一个电子了。
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And what we say when we talk about the delta energy is I E 2 that this is going to be equal to i e 2, or the second ionization energy, or we could say the negative of the binding energy of a 2 s electron in b plus.
那么我们说,Δ，E，应该等于，或者说第二电离能，也就是正一价硼中，2，s，电子的,束缚能的负值。
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I want to then do, I need to find the square root b squared plus h squared, right?
我接下来要求b的平方和h的平方,的和的平方根对不对？
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pV=RT p plus a over v bar squared times v bar minus b equals r t. All right if you take a equal to zero, these are the two parameters, a and b. If you take those two equal to zero you have p v is equal to r t.
我们就回到,也就是理想气体,状态方程,下面我们来看看,这个方程。
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So the profit is 2 S1 plus S2 plus B S1 S2 minus S1 squared.
利润是2-S1平方
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So I take 2 plus 3 b steps to go through this loop OK. So if b is 300, it takes 902 steps.
好，如果b是300的话就是902步,如果b是3000那就有9002步。
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sb So just to say that it's 1 s squared plus 1 s b, all of that together squared.
这就是说它是1sa加上，这整个的平方。
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+3b I've said the product was is 2 plus 3 b.
我们已经说了答案是。
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And again, if we write out what all the terms are, we again have 1 s a squared plus 1 s b squared, but now what we're doing is we're actually subtracting the interference term.
同样，如果我们把所有的项都写出来，同样我们有1s平方加上1sb平方，先现在我们做的是,我们要减去干涉项。
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It says, in either case in general, t of b-- and this is where I'm going to abuse notation a little bit but I can basically bound it by t, 12 steps plus t of b over 2.
我可以用一个,比12+t的数代表,这里有点不准确的地方,具体的步数依赖于奇数偶数，但是你们可以看到在两个case中。
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sa So we have 1 s a, and we're drawing this as having a positive amplitude, but since we have destructive interference we're going to draw 1 s b as having the opposite sign, so we have a plus and a minus in terms of signs.
我们有，我们把这画成一个正的振幅，但因为我们是相消干涉,我们把1sb画成相反的符号，所以我们有一正一负两个符号。
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This is 3 plus 3 plus t of b minus 2. Right?
+3+t对不对？
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We can just rewrite that, if I divide through by 2 and rearrange, it's going to tell me that ?1, or if you like, ?1 is equal to 1 plus B S2.
我们这么写,如果我除以2然后整理,可以得出?1等于1+B*S2
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If I reduce that it would be 3 plus t of b minus 3, so in general 3*k+t this is 3 k plus t of b minus k. OK.
把b减去一个,在外面加个3就可以了,因此也就是。
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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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I'm going to have 2 still, and then this S1 is going to become a 1, and this S1 here is going to become a plus B S2, everyone happy with that?
还保留,S1求导后是1,这里的S1会变成B*S2,大家都会吧
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So similarly, I would find that ?2 2 equals 1 plus B S1 and this is the best response of Player II, as it depends on Player I's choice of effort S1.
同理可得?2等于1+B*S1,?2是参与人II的最佳对策,因为它与参与人I的策略S1有关
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I've got one test, I've got a subtraction, I've got a multiplication, that's three steps, plus whatever number of steps it takes to solve a problem of size b minus 1.
我进行了一次比较，一次减法，一次乘法，一共是三个步骤,再加上t的步骤数。
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And if I pull it out one more level, 12k it's 12 plus 12 plus t of b over 8, 12 k because I'll have 12 of those to add up, plus t of b over 2 to the k.
总结一下也就是说，在k步以后，总步数应该是，那这种情况什么时候才能停止呢？,才能到达最基本的情况呢？
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So what we're going to converge in on to, is the S1*, let's call it in this case, is equal to 1 plus B S2* and that S2* is equal to 1 plus B S1*.
交点的横坐标是S1,暂且这么叫吧,S1*=1+BS2*,而S2*=1+BS1
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