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It is of opposite charge and it exerts an influence, only that is minus to plus so that is going to give me a minus term here.
它具有相反的电性，所以它也会有影响，只有这个负值相对正值来说会起到减少的作用,所以这里将会引入一个负值的量。
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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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it's going to be: 1 minus the probability they'll choose Right times 1, plus the probability that they choose Right times 4.
方程是,1减对手选右的概率再乘以1,加上对手选右的概率乘以4
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That's equal to seven minus six minus two 3 times 4.19, or 4.19 plus 2.3.
算出来是7-6-2*4。19的平方,或者4。19加2。
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Which is to say q1 plus q2 is equal to minus w1 plus w1 prime plus w2 plus w2 prime.
即q1加q2等于2,负的w1加w1一撇1,加w2加w2一撇。
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You got a plus f and a minus f, so you got to do what you got to do.
这里有一个正 f 和一个负 f,所以你知道该怎么办
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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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We know that this is going to be, by measurement, this is going to be plus and this is going to be minus.
我们知道通过测量，这儿将是正，这儿将是负。
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minus volatility to plus volatility it will return any value in here.
浮动值到+浮动值之间等概率地,去返回一个值。
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Anywhere from plus 0.5 to minus 0.5 angstroms.
从负0。5埃到正0。5埃的任何地方。
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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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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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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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And here d, of course, can take the values from plus three down to minus three.
而d层，当然，能够包括从+3到-3的所有值。
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It's going to be a dipole. This is plus; this is minus.
这是个偶极子，这里是正，这里是负。
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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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