By putting it in a function bug if I find a bug and I change my program I can just run the function again.
把我输入的值放到一个函数里,如果以后我在程序中发现了一个,并对程序进行更改的话,我可以直接。
Identify the best responses of each player as a function of the others and find out where they intersect.
把每个人的最佳对策看成别人策略的函数,然后找出函数的交点
Now remember, we went through before how it's a state function but to calculate it, you'd need to find a reversible path, along which you can figure this out.
请记住,我们不需要知道它是怎样的一个态函数,只计算就可以了,你需要找到一条可逆路径,沿着这条路径就能算得这个结果。
If I gave you the location of a particle as a function of time, you can find the velocity by taking derivatives.
如果我给出物体的位移是时间的函数,你可以通过求导来得到速度
Change the independent variable, find the change in the function, take the ratio and that's the derivative.
改变自变量,算出函数的变量,计算比值,这就是求导
So we'd really like to be able to find some sort of equation of state, or some sort of rather function of state that's going to relate the heat going in or out of the system with that function of state, because this isn't going to do it.
所以我们真的想要去,找到一些态方程或者态函数,通过这个态函数可以表示热量,在系统与外界的交换,因为这个不能表示它。
Now conversely, how do we find Firm 2's best response as a function of q1?
反过来如何由q1求公司2最佳对策的方程
The basic idea in solving these equations and integrating is you find one answer, so then when you take enough derivatives, the function does what it's supposed to do.
解决这类方程以及积分的基本思想就是,你求出一个解,然后进行多次求导,求导的结果就满足条件
Everyone knows from calculus that if you're trying to find a function about which you know only the derivative, you can always add a constant to one person's answer without changing anything.
学过微积分的人都知道,如果你想根据已知的导数,求出其原函数,你总是可以给某人的答案,随便加一个常数,且不影响结果
Find player I's payoff is a function of what player II's effort is.
参与人I的收益是参与人II付出的函数
Mathematically complete problem is that you can find the function x of t by saying that the second derivative of the function is equal to -k over m times the function.
这个数学问题就是,你可以求出函数 x,只要令函数的二阶导数,等于 -k 除以 m 再乘以函数
We have Player I's payoff as a function of the two efforts and now I want to find out what is Player I's best efforts given a particular choice of S2? Yeah.
参与人I的收益是两人付出的函数,那么对于某个给定的S2,怎么计算参与人I的最佳对策呢
What I did was I differentiated this fairly simple function with respect to q1 and since I want to find a maximum, what I'm going to do is I'm going to set this thing equal to 0.
只不过是对q1求导了,既然我们要求出最大值,只需要令导数等于0就可以了
I would say, okay, this guy wants me to find a function which reduces to the number a when I take two derivatives, and I know somewhere here, this result, which says that when I take a derivative, I lose a power of t.
出题者想要我找出一个函数,它在经过两次求导后得到数字a,我知道这里的某个地方,这个结论告诉我,我每求一次导,t就降一次幂
Find the best response function by differentiating this and maximizing the function for every q2, so we could just... Good, so what we're going to do is, all right good, we're going to... We're trying to maximize.
求导得出q2取最大值的情况下,参与人1最佳对策是q2的一个函数,说得很好,接下来,我们要求出最大值
Then, to find the meaning of b, we take one derivative of this, dx/dt, that's velocity as a function of time, and if you took the derivative of this guy, you will find as at+b. That's the velocity of the object.
接下来,为了弄清b的含义,我们取它的一阶导数,dx/dt,得到速度作为时间的函数,如果你对它求导的话,你会得到at+b,这就是物体的速度
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