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So, The Norton Anthology, this book, this heavy book, I order it as a way to, well, reduce your expenses.
叫《一首诗的心路历程》，就在这本诺顿诗集里，我自己订购了一本，这样可以减少你们的花费。
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I didn't order this.
这菜不是我点的。
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All right. In order to deal with this, let me show you an example, and I hope that comes up, great.
好，为了说明这些内容，让我们来看个例子，我希望这能出来。
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And so in theory after just 2 or 3 steps you were in order of magnitude farther along in this counting process than I would be because I'd still be on person literally 3 or 4.
因此从理论上说,经过这样短短的两三个步骤,你们就有了分级次序，比我数数快多了,因为我才数到第三四个人。
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And if you remember this phrase, I don't know why you would, but this will help you learn the lanthanides in order.
如果你能记得这句话的话，我不管你们是怎么做到的，但是这个确实会帮助你按顺序记住镧系元素。
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And what I'm imagining is how you're going to respond to these words in this order.
我会想象，当你看到这一连串文字时,你会作何反应。
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So if we differentiate this object, I'm gonna find a first order condition in a second.
想要求它的导数,先让我想想一阶条件
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Now I think in order to make this viewpoint seem plausible, we probably should study it for a moment a little bit more philosophically.
现在为了让这个观点听起来更合理一些，我们可能,应该更哲学地学习一下。
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I really didn't know and I needed to have an answer to this question. So I created Star Festival with this team from MIT in order to try to answer this question.
我不知道，我需要一个答案,所以我和我来自麻省的团队,创造了，“七夕“项目,以回答这个问题。
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The only thing I need to say in order to distinguish this particular part of the chemistry of a DNA or an RNA molecule is to say 'it's DNA or its RNA', and 'what the base is'.
想要区别DNA和RNA,这两种化学结构不同的核酸的话,我只要说明它是DNA还是RNA,它的碱基是什么就够了
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You can tell him, "I didn't order this."
你可以告诉他，“I didn't order this”。
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why did I work so hard in order to get into this place?
我为什么要拼命入读这所学校？
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One question is this: Is it really true that in order to think about the perfectly straight, I must have somehow, somewhere at some point come up against, had direct knowledge of, the perfectly straight?
一个问题是,难道真的说,为了想象一条完美的直线,我必须,在某处直接碰到,并直接了解一条完美的直线吗
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OK. What order of growth? What's complexity of this? I've got to get rid of this candy.
那么增长率怎么样呢,复杂度呢？我想发个糖果了。
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In a moment, I'm gonna drag and drop in order to induce this right to do something.
一会，我要拖拽这些东西,来进行一些操作。
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So I would choose something that have a wave lines off You wouldn't measure the dimension of a human hair with a yardstick, so I need something on this order.
所以我会选些东西,你将不会用一个码尺去量,人的头发有多大，所以我需要一些这个级数上的东西。
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What is this? This is a wonderful piece of abstraction. It is saying, you don't need to know squat about what's inside the body of this function. You don't have to worry about the parameter names, because they're going to be preserved, you don't need to worry about how I'm doing it, this tells you how you can use this, in order to use it correctly. Of course, I can then close it off, and off we go.
但是这能告诉你一些信息,这是什么？,这是一段完美的抽象,它的意思是你不需要知道,这个函数的内部构造，你不必担心参数是否重名，因为它们已经被预存了,你不需要担心我如何使用它,它告诉了你你该如何操作，从而正确的使用它，当然我也可以把它关掉。
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Really there's no work to be done if I am handed all in sorted order so, you know, There's no work to be done if I'm handed all of the arrays in sorted order so, you know, if I demand that you give me this assumption that the cups are already sorted and then I'll sort them for you, I mean, this is kind of a cyclical argument.
如果杯子是有序排列的,那就没必要再对它进行排序了，同样如果给出的序列本身就是有序的,那也不必再做什么，如果给出这样的假设：,杯子已经有序，但仍需要对其进行排序，这像是个循环的论点。
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I differentiate a second time and check the sign, so the second order condition, I differentiate this expression again with respect to q1.
我们对它进行二次求导然后看符号,这个式子的二阶导数,就是一阶导数再对q1进行求导
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I may not want to do the full range of this, but I may decide I have to use a lot of gigabytes of space in order to do a trade off.
哈希成一个整数，我不想完整的去做这件事，但是我可以想像我需要用许多G的空间,去完成空间和时间之间的权衡。
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And if I'm walking down the list, this is probably order of the length of the list s because I'm looking at each element once.
这可能大概就是数组的长度，因为我会遍历数组中的每个元素一次,现在你可能会想，等等，数组已经排好序了。
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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
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And I have this, to write it out, this is order the length of the list squared, OK?
我得写下来,这是把列表的长度平方，对么？
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In the linear case, meaning in the unsorted case what's the complexity of this? k times n, right? Order n to do the search, and I've got to do it k times, so this would be k times n.
复杂度是多少？k的n次方，对吧？,在序列n中做搜索，要做k次，所以是k的n次方次,如果先排序后搜索。
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