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It is indicated by experimention that the outcome of this algorithm is very efficient,especially in those problem who fulfil triangle inequation.

经过实验表明,法求解近似度很高,其是在满足三角不等题中,误差更低。

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可汗院:统计(视版)

But we can operate on all of them at the same time, this entire inequality.

但我们能够同时对所有这些进行操作,即整个不。

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高数之美

But of course, for those of you who want some more passive viewing, don't forget that Henry and I just put out a video on Bell's inequalities over on MinutePhysics.

但当然, 对于那些想要一些被动观看的朋友们,别忘了亨利和我刚刚在分钟物理上发布了一段关于贝尔不的视。

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耶鲁大公开课:博弈论

So at the beginning of the game, when people are out here, if we ask the question, is P1D plus P2D-1, is it bigger than 1?

因此,在游戏期, 当人们还在场时,如果我们提出疑问:P1D 加上 P2D-1 否大于 1? 在场的情况下,这个不否成立呢?

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可汗院:统计(视版)

Have this inequality expressed in terms of mu.

将这个不用μ表示。

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耶鲁大公开课:博弈论

There's going to be a critical point when, for the first time, the inequality switches from being incorrect to correct, from being false to being true.

当不首次从错误变为正确,从假变为真时, 将出现一个关键点。

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高数之美

Well, he and I just made a video on a certain quantum mechanical topic - " Bell's inequalities" .

好吧,他和我刚刚做了一个关于某个量子力主题的视——“贝尔不”。

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耶鲁大公开课:博弈论

We're saying this inequality, which I'm claiming is going to be important, but none of you know why yet, I'm claiming is going to be important.

我们正在讨论这个不, 我断言它将非常重要,尽管你们现在还不知原因, 但我坚持认为它至关重要。

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可汗院:统计(视版)

And then we still have our inequalities-- is going to be less than 17.17 minus the mean, which is less than 2.13.

然后我们仍然有这些不——它将小于17.17减去平均值,这又小于2.13。

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耶鲁大公开课:博弈论

So once you cross this inequality, you can't go back.

因此,一旦你跨越了这个不,就无法回头。

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可汗院:统计(视版)

So if I do that-- so let me just do that right over-- so if I multiply this entire-- this is really two equations or two inequalities I should say.

所以,如果我这么做——让我就在这里直接进行——如果我将整个子——这实际上两个方程或两个不,我应该说。即这个量大于那个量,而那个量又大于另一个量。

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耶鲁大公开课:博弈论

This inequality is not met, not met, not met, not met, not met, not met, not met, not met, not met, and then it's met, met, met, met, met, met.

这个不未被满足,未被满足,未被满足,未被满足,未被满足,未被满足,未被满足,未被满足,未被满足,然后它被满足了,满足了,满足了,满足了,满足了,满足了。

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可汗院:统计(视版)

So what we want to do is multiply this entire inequality by this value right over here.

因此,我们希望将整个不乘以这个值。

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耶鲁大公开课:博弈论

It's because if you look at those inequalities, it's monotone.

这因为如果你审视那些不,它们单调的。

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可汗院:统计(视版)

If you do that, if you multiply the entire thing times negative 1, this quantity right here, this negative 2.13 will become a positive 2.13.

如果你这样做, 将整个不乘以负1,这里的这个量, 即负2.13, 将会变成正2.13。

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耶鲁大公开课:博弈论

I've just added something to both sides of an inequality.

我刚刚在不的两边都加上了一些东西。

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可汗院:统计(视版)

So what I'm going to do is multiply this entire inequality by negative 1.

因此,我打算将这个不的两边同时乘以负1。

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耶鲁大公开课:博弈论

So, in here, the inequality is met: the sum is bigger than 1; and out here it's less than 1.

因此,在这里,不得到满足:总和大于 1;而在这里它小于 1。

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可汗院:统计(视版)

Just be clear, I'm just multiplying all three sides of this inequality by this number right over here.

只为了清楚起见,我只将这个不的三边都乘以了这里的这个数。

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可汗院:统计(视版)

But since we are multiplying an inequality by a negative number you have to swap the inequality sign.

但由于我们在不两边乘以一个负数,你必须调换不的符号。

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3G, 401(K), a,
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