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sort of takes the cyclicality out of earnings and you can see over history how that multiple has varied.
一定程度上在收益上避免了周期性的,而且你可以看到随着历史的进程,这个倍数发生了多大的变化。
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So multiplying a vector by a number means stretch it by that factor.
所以用数字乘以矢量,就相当于将它伸长了相应的倍数
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If the product is just the sum or multiple of a sum of the inputs, there wouldn't be much of a point working in a team at all.
如果总收益只是投入的和或者倍数,那这样合作也没什么意思了
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I don't happen to like this, but he thinks that is, give a combination of a number multiplication in a string, this will in fact give us back a new string with that many replicas, if you like, of the string concatenated together.
我不喜欢这么做但是他认为,也就是说，返回一个字符串的,某个特定数字的倍数，这实际上会返回一个新字符串,一个原来的字符串的很多复制品，连接到一块的一个字符串。
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And so we might need to work out various, more complicated, formulas here, where we multiply the pain times its duration and take into account its intensity, get the sheer quantity of pain that way.
我们需要编制出各种，更复杂的公式，以便增加痛苦的倍数，延续时间,还要考虑痛苦的强度，获得痛苦的纯量。
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There's nothing you can throw at me that I cannot handle with some multiple of i and some multiple of j.
你们找不出任何一个我无法,用 i 和 j 的倍数来表示的矢量
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When I give you multiple of i and another multiple of i, there's some has got as its coefficient the sum of the two coefficients.
如果我给你两个不同 i 矢量的倍数,就可以得到某个矢量,且它的系数是这两个系数的和
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If the Ax and Ay, some are positive and some are negative, this is the way by which we have learned we should combine multiples of i.
如果 Ax 和 Ay 有正有负,这就要用到我们所学过的方法,将所有 i 的倍数加起来
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This part, being parallel to j, has to be some multiple of j.
而平行于 j 的那部分,也必然是 j 的倍数
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We know that because you can stretch i by whatever factor you like.
因为你可以让 i 伸长任意的倍数
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This part, being parallel to i, has to be a multiple of i.
平行于 i 的这部分必然是 i 的倍数
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Now, I've not given you any other example besides the displacement vector, but at the moment, we'll define a vector to be any object which looks like some multiple of i plus some multiple of j.
除了位移矢量之外,我就不再举其他的例子了,但是现在我们要定义一个矢量,可以表示为 i 的倍数加上 j 的倍数
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