There is a Homburg and there is a Borsalino, Here's Bohr mixing it up with royalty.
一个是汉堡帽,一个是博尔萨利诺帽,这是波尔和皇室人员。
At my maximum, I'll put a hat over it to indicate this is the argmax; at my maximum I'm going to set this thing equal to 0.
给每个最大值都标注上一个帽来,在最大值处导数方程等于0
Here's one climbing out of the pool: First her face appeared, long and cadaverous, with a bandage-like bathing cap coming down almost to her eyes, and sharp teeth protruding from her mouth.
有一个女人从游泳池中出来了,首先是她的脸,修长,而惨白,如绷带般的泳帽从她的头部,耷拉到她的眼睛,锋利的牙齿从她的口中突出。
Why not just put our own orgasmitron hat on with a little-- not rat lever, but now human lever-- with the electrodes stimulating our own brains so that we get this intense burst of pleasure?
我们为何不带上电子高潮帽-,把老鼠杠杆变成人类杠杆-,用电极刺激我们的大脑,获得强烈的快感?
And h with the carrot or the hat here, well, that carrot or hat tell us it must be an operator, and this is called the Hamiltonian operator.
而且带帽的H,好,这个帽告诉我们它肯定是一个算符,这个被称为哈密顿算符。
Pretend the hat isn't there a second.
暂时先不要考虑这里的帽
To make this a first order condition, I'm going to say "at the best response," put a hat over the 1.
为了达到一阶条件,我说在最佳对策下,在S1上写个帽
I know that Player 2's best response for every possible choice of Player 1, which if we had done it would be q2 hat is going to a -c over 2b--q1 over 2, right?
参与人1不同策略下参与人2的最佳对策,即q2帽等于/2b - q1/2
So the best response for Player 1, as a function of what Player 2 chooses, q2, is just equal to the q1 hat in this expression and if I solve that out carefully, I will no doubt make a mistake, but let's try it.
这个就是参与人1的最佳对策,它是参与人2策略q2的一个函数,它和之前的q1帽那个表达式是相等的,虽然我是很仔细地计算的,还是有可能算错的,我来验证一下
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