We can't actually go ahead and derive this equation of the wave function squared, because no one ever derived it, it's just an interpretation, but it's an interpretation that works essentially perfectly.
从这个方程中,导出,波函数的平方,没有人可以这样做,这仅仅是一种解释,但这种解释,能解释的很好,自从它第一次被提出来之后。
Occasionally, you'll find you need to cancel out units, because, of course, you're always doing unit analysis as you solve your problems, and sometimes you'll need to convert joules to kilogram meters square per second squared.
偶然地,你会发现需要消除单位,因为在解题时,经常要做单位分析,所以有时候需要把,焦耳换做,一千克乘以米的平方除以秒的平方。
We have seen log, linear, quadratic, and exponential.
平方级的和指数级复杂度的方法,再说一遍,可能会有些常量。
It weights big deviations a lot because the square of a big number is really big.
使偏离的权重更大,一个数的平方是一个更大的数
A joule is a kilogram meter squared per second squared.
一个焦耳等于,一千克乘以米的平方除以秒的平方。
So let's actually just simplify this to the other version of the Rydberg constant, since we can use that here.
除以n初始的平方,我们把它简化成,另一种形式的Rydberg常数。
Got a simple square procedure,.
其实是一个很简单的求一个数的平方的过程。
I'm given an integer that's a perfect square, and I want to write a little piece of code that's going to find the square root of it. All right so I'm cheating a little, I know it's a perfect square, somebody's given it to me, we'll come back in a second to generalizing it, so what would the steps be that I'd use to walk through it?
完美平方数的整数,我想写一段代码来求这个数的平方根,好,我这儿有点儿作弊了,我知道这是一个完美的平方数了,他们给我的,我们后面会讲怎么产生这个数的,那么我想解决这个问题,需要什么步骤呢?
And the word degenerate simply means same energy, are of equal energy when they're degenerate.
简并“一词指的是,能量相同,你有n平方个等能轨道,是简并的。
And obviously, when I get to something whose square is equal to x, I've got the answer I want, and I kick it out.
大于x的时候,我已经经过了,想要的答案了,很明显,当我得到一个数的平方等于x的时候。
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