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The claim is, I can write any vector you give me as a real scale i plus a real scale j.
也就是说,我可以把任何你给我的矢量,写成 i 和 j 的实系数线性组合的形式
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Just swapping them, right? I temporarily hold on to what's in the i'th element so I can move the i plus first one in, and then replace that with the i'th element.
交换他们，对么？,临时的保存下第i个元素,然后把第i+1个元素移进来，把i+1的位置替换为第i个元素。
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I'm going to just put that in, and it's the cosine of the number times i plus sine of the number times j times R.
我要在式子加入这个量,这个式子就等于这个值的余弦乘以 i,加上这个值的正弦乘以 j 再乘以常数 R
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And using the text plus these archive notes,I think you will be able to piece together what you need.
使用这个课本，再加上存档笔记，我想,你们已经能组合出你们需要的了。
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So now, I'm gonna plug in 8 here and now finally, 24 that's 16 so that's plus 8, so that's 24.
这儿用8代入，就是16加8，等于。
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dU=dq+dw Well from the first law, du is equal to dq plus dw, and I wrote down everything I knew at the beginning here.
第一定律“,前面我们已经,看到了。
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When I started this company back almost nine, ten years ago, I started to recruit talents, recruit engineers, and I wrote in the job description that requires five-plus years of related experience.
大约是9到10年前，公司成立初期，我们开始招聘人才，招聘工程师，职位介绍由我亲自执笔,要求具有五年以上的相关工作经验。
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What I want is survival plus something else.
我希望拥有的是带有其他条件的存活。
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, okay. So 1 plus 1/4 of 0 is 1, so if Player II chooses 0, player I's best response is to choose 1.
是1,因为1+0/4=1,参与人II选0时I的最佳对策是1
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It's looking at the i'th and the i plus first element and it's saying, gee, if the i'th element is bigger than the i'th plus first element, what's the next set of three things doing?
他在观察第i个元素,和第i+1个元素,如果第i个元素大于第i+1个元素，接下来的要做哪三件事情？
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You can now collect the whole expression as something something i plus something something j.
你们现在可以推导出完整的表达式,用 i 和 j 相关的式子来表示
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And so the value of with i will be the value of i plus whatever I can get using the remaining items and decrementing the weight by the weight of i.
所以装上i之后的总价值,就是i的价值加上剩下物品中,符合条件者的所有价值,再在剩余重量中减去i的重量。
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Three plus five. I can take something to a power, double star, just take three to the fifth power.
+5，我也可以,求一个数的次方，例如求3的5次方。
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The power of linearity is F=k1+k2 if I come across f of x, y, z equals k1 plus k2, if it is a linear equation, I don't have to go and solve it all over again.
线性的威力是,一个方程，如果它是个线性方程，那么我就不用再去解他了。
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So what I'm saying is that du through path 1 plus du going through path 3.
于是1,因为能量。
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J++ Now here's the semicolon, J less than N, where N is this, J plus, plus; so what am I doing?
现在这里是一个分号，J小于N，N是这个，那么我在做什么？
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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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Being and void, being and void, so now I've got my plus two little projectile coming in, and plus two zooms right through.
有和无，有和无，现在带两个正电荷的发射体进来了，两个正电荷正好穿过。
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I want to then do, I need to find the square root b squared plus h squared, right?
我接下来要求b的平方和h的平方,的和的平方根对不对？
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It takes the number whose Fibonacci memo I want plus a memo.
它将我想要的斐波那契数,的值加上了参数。
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That means starting at the first one, I'm going to do something to it. And what am I'm going to do? I'm going to take that character, convert it back into an integer, and add it into some digits. And I've done a little short hand here, which is I should have said some digits is equal to some digits plus this.
这意味着从第一个字符开始，我要对他们进行一些操作,我要去做什么呢？我要取得这个字符，然后把它转换为整数,然后加到某些数上面去,我在这里用了一些缩写,我本来应该写一个数字等于这个数字。
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Therefore, the vector A that you gave me, I have managed to write as i times Ax plus j times Ay.
这样,你给我的矢量 A,我已把它写成 i ? Ax + j ? Ay的形式
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Initially, its location as a function of time is equal to i times x plus j times y.
初始时刻,它的位移作为时间的函数等于,i ? x + j ? y
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It says, in either case in general, t of b-- and this is where I'm going to abuse notation a little bit but I can basically bound it by t, 12 steps plus t of b over 2.
我可以用一个,比12+t的数代表,这里有点不准确的地方,具体的步数依赖于奇数偶数，但是你们可以看到在两个case中。
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There I have it Cp is equal to Cv plus R, right?
所以Cp=Cv+R，对吗？
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So I can concatenate techs plus ivys and assign that result to univs, and then when I print it you'll notice I just get a list of five strings.
和ivys数组用加号串联运算,并把结果赋值给univs数组，接下来我显示下univs的结果,你注意到我得到了一个。
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So similarly, I would find that ?2 2 equals 1 plus B S1 and this is the best response of Player II, as it depends on Player I's choice of effort S1.
同理可得?2等于1+B*S1,?2是参与人II的最佳对策,因为它与参与人I的策略S1有关
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The derivative of this guy will be 2t times i plus what?
这个式子的导数为 i ? 2t,加上什么呢
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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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And the way I remember that is I put a little cross here, and then I make that the plus end.
我还记得要,在箭头尾部打个叉，然后让加法完结了。
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