In either case, they regard him as surreptitiously taking the side of the people against the nobles.
二者都把他看作是秘密地,站在人民的立场上反抗权贵的人。
And in either case if we first talk about constructive interference, what again we're going to see is that where these two orbitals come together, we're going to see increased wave function in that area, so we saw constructive interference.
在任何情况下,如果我们首先讨论相长干涉的话,我们同样会看到,当这两个轨道靠拢的时候,我们看到这个区域有波函数增加,所以我们看到的是相长干涉。
the abuse is, you know, it's not quite right, it depends upon whether it's all ready, but you can see in either case, after 12 steps, 2 runs through this and down to a problem size b over 2.
在12步以后,两轮过后,这个问题的范围b被缩小了一倍,这为什么很不错呢?,这意味着再来12步。
And of course, in either case, w delta u is q plus w, so it's q irreversible plus w irreversible, u being a state function it's the same in either case.
当然了,在这两个过程中,Δu都等于q加,因此现在是q不可逆加w不可逆,同时u在这两个过程中都是态函数。
Either you open a nice hole between yourself and the rest of your army, in which case somebody's going to get very badly killed and you're going to be on the run very soon, or you have to somehow communicate to the rest of the army, "Everybody come over this way," which will still leave that flank open to being flanked by the other side.
或者在你和自己的军队之间,形成一个缺口,这样有会造成惨重的伤亡,你很快会全军溃逃,或者你以某种方式和部队沟通,全都过来这边,这样仍会使一侧对敌人敞开,从而受到对方攻击
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。
We don't always want to go and solve the Schrodinger equation, and in fact, once we start talking about molecules, I can imagine none of you, as much as you love math or physics, want to be trying to solve this Schrodinger equation in that case either. So, what Lewis structures allow us to do is over 90% of the time be correct in terms of figuring out what the electron configuration is.
我们并不想每次都去解薛定谔方程,而且实际上,一旦我们开始讨论分子,我可以想象,你们中没有一个人,不管你有多么热爱数学或物理,会想去解这种情况下的薛定谔方程,总之,路易斯结构能让我们,有超过,90%,的概率判断出正确的,电子排布。
In either case, I take that value and multiply back by two, if it was even I get back the original number, if it was odd, I'm not going to get back the original number, so I can just check to see if they're the same.
在两种情况中,我都把结果值,再乘以2,如果x是偶数我就,得到了原来的值,如果x是奇数那么就不会得到,原来的值,因此我看看得到的值,和原来的值相等不相等就可以了。
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