“微积分二：数列与级数”将介绍数列、无穷级数、收敛判别法和泰勒级数。本课程不仅仅满足于得到答案，而且要做到知其然，并知其所以然。

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From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

45 ratings

The Ohio State University

45 ratings

“微积分二：数列与级数”将介绍数列、无穷级数、收敛判别法和泰勒级数。本课程不仅仅满足于得到答案，而且要做到知其然，并知其所以然。

From the lesson

幂级数

在第五个模块中，我们学习幂级数。截至目前为止，我们一次讲解了一种级数；对于幂级数，我们将讲解整个系列取决于参数 x 的级数。它们类似于多项式，因此易于处理。而且，我们关注的许多函数，如 e^x，也可表示为幂级数，因此幂级数将轻松的多项式环境带入棘手的函数域，如 e^x。

- Jim Fowler, PhDProfessor

Mathematics

Radius and center. [SOUND] Thus far, we've been considering power series that are centered zero. By which I mean, we've been looking at power series that look like this. The sum n goes from zero say to infinity of c sub n times just X to the n. But we can center our power series around some other point, around some other point, a. So instead of writing this, let's say I want to write down a power series that's centered around some other point, let's say a.

Well, then I write my power series like this. The sum N goes from zero to infinity of c sub n again. But now times, not just X to the N, but X minus A to the N-th power. In this case, the interval of convergence is centered around A as well. Well, let's find an interval on which this series converges. And the goal is that it shows that interval is really centered around a. So, it legitimately makes sense to think of this power series is centered around a. And to find the interval which this power series convergences, I'm going to use the ratio test. So let's do that now. So how does ratio test work here? Well, it asks me to look at a limit as n approaches infinity of the n plus first term, c sub n plus 1 times x minus a to the n plus first power divided by the nth term, which is c sub n times x minus a to the n. And I've got absolute value bars here because I'm really applying the ratio test to the sum of the absolute values. I'm using it to analyze absolute convergence. Okay, what's this limit? Well, I can simplify this limit a bit. I could write this as the limit and goes to infinity. Let me pull of the C sub n plus 1 and the C sub n. And it will be the fraction Cc sub n plus 1 over c sub n. And then here I've got n plus 1 copies of x minus a, and I've got n copies of x minus a, most of those will cancel. All that I'm left with is just one copy of x minus a in the numerator. So I'll write times absolute value of x minus a. And the ratio test tells me that if this is less than one, then the original series converges absolutely. And if this is bigger than one, then the original series diverges.

Well, how can I really analyze this? And I don't really know anything about c sub n plus 1 and c sub n. But let's just pretend, right? Let just pretend that the limit just of the c sub n plus 1 over c sub n ends up being 1 over r. And in this part here, the absolute value of x minus a, that's just a constant anyway. So I'll write times absolute value x minus a. I'm wondering when is that less than 1? I mean, I'm just playing make believe here. R is some positive number, I'll multiply both sides of this inequality by R since it is positive, it doesn't affect the inequality at all. Now I got the absolute value of x minus a is less than r after multiplying through by R. Well, what does this mean? This means the distance between x and a is less than r. Well, that really means that x is between a minus r and a plus r, and that at least when x is between a minus r and a plus r, I know that the series converges, absolutely. I don't know what happens at the end points. But the big deal here is that, look, this thing legitimately looks like an interval centered at the point a with the radius r. In what's to come, I'll usually be talking about power series that are centered around zero, but it's important to release that we can talk about power series that are centered around some other point. And there are certainly cases where that will come in handy. [NOISE] [SOUND]

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