“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

泰勒级数

在最后一个模块中，我们介绍泰勒级数。与从幂级数开始并找到其代表的函数的更好描述不同，我们将从函数开始，并尝试为其寻找幂级数。无法保证一定会成功！但令人难以置信的是，许多我们最喜欢的函数都具有幂级数表达式。有时，梦想会成真。和许多梦想相似，多数不说为妙。我希望对泰勒级数的这一简介能激起你学习更多微积分的欲望。

- Jim Fowler, PhDProfessor

Mathematics

Value. [SOUND] Remember the mean value theorem? Well here's what the theorem says, suppose I got some function f that's defined in the closed interval a to b and it's values are real numbers. It's a continuous function, so it's continuous in the closed interval between a and b. And on the open interval between a and b it's differentiable. Now with these assumptions here's what happens. Then there's some pont c between a and b, it doesn't tell me how to find that point. But there is one, so that this happens the derivative of the function at the point c is equal to f(b)- f(a) divided by b-a. Let me rearrange this a bit. So we're going to change the names of some of these variables. So I'll write f prime not of c but of z is equal f not of b but of x, and I'm going to keep the name of a the same. So I'll call that f(a), okay and I'll again divide by x-a. Now I'm going to multiply both sides of this by the quantity x-a. And what do I'll get? Well I'll get f prime of z times the quantity x-a and that's equal to this numerator here, f(x)-f(a). So we'll put an x in there, and an a in there. And I'm going to add f(a) to both sides and I'm going to start with f(x) on this side. So I just going to write this f(x) = f(a) + this quantity here, f'(z) times (x-a). Why would I want to write it that way? Well I could write this in yet another way, I could write it like this. Is the same thing here but just written a little bit differently. F(x) = f(a) + a remainder term R sub 0 of x. Where r sub 0 of x is this, the derivative of f at some point z between x and a divided by 1 factorial which is just 1, times x- a to the 1st power. But when read it like this it looks exactly like an instance of Taylor's theorem. Well let's remember what Taylor's theorem says. Suppose that you got some function f, takes real inputs, produces real outputs, and I'm just going to assume it's smooth. I don't want to worry too much about the exact differentiability conditions that I need. And f(x) is the sum, little n goes from 0 to big N of the little nth derivative of f at the point a divided by n factorial times x-a to the nth power, plus this remainder term, big R sub big N of x. Then this happens, big R sub big N(x) is given by this, the big N plus first derivative of f at some point z between x and a. Divided by big N + 1 factorial times (x-a) to the big N + 1st power. And think about what happens here when you plug in big N = 0. And in that case, what you get is exactly what we had before. It's f(x) equals this is just the N = 0 term here, plus a remainder term. And that remainder term then when big N = 0 is exactly the derivative of that some point Z divided 1 factorial times x-a to the 1st power. It looks exactly like the mean value theorem. So Taylor's theorem is a whole lot like a souped version of the mean value theorem. The mean value therom tells you something about your first derivative and Taylor's therom is telling you information about your higher derivatives. And this isn't just a theoretical story, you can actually see this in the real world. Let's imagine the following scenario, well let's let f(t) be your position at time t seconds after the beginning of the experiment. And when the experiment begins with f(0), where are you? F(0) is 0 so you're at the origin and you're not moving. Your velocity at time 0 is 0. So the derivative of your position at time 0 is 0. And I prefer that we survive the experience, so I'd like to control the acceleration that we experience. So the second derivative of f at any time t, we know bigger than 250 meters per second square this is 25 gs or so. And I think humans don't do so well above 25 g, so this is the reasonable thing if you want to make this a healthy trip. Another question is how big can f(60 seconds) be? How far can you get starting at 0 not moving, how far away can you be after 60 seconds? I want to bound our acceleration so that we survive the trip. Now this problem doesn't have to be approached using Taylor series. But we can do it that way, so let's start down that path. So I'll write this down, f(t) is f(0) + f prime (0) times (t-0). Those are the first two terms in the Taylor series expansion for f, + R sub 1(t), this is the remainder term. And what do I know about the remainder term? Well R sub 1(t) = the second derivative, some point z divided by 2 factorial times (t-0) squared. I can bound that remainder term. Because I don't want to be injured during this trip, I don't want to be accelerating more than about 25 gs. I can now control something about R sub 1 and I know something of how big R sub 1 is. The absolute value of R sub 1 of t is less than or equal to how big in he second derivative b at any point 250 divided by 2 factorial times t squared. Now I can say something about how large f(t) is. So f(t) is less then or equal to f(0) plus f prime of 0 times t plus this quantity, 250 divided by 2 factorial is 125 time t squared. Now I know what f(0) is, I'm assuming that I'm starting at a point I'm calling the origin and I know something about the derivative of f(0). I'm assuming that I start my journey motionless. So this quantity here is equal to 0 + 0 times t plus just 125t squared. I gave us one minute of travel time. So f(60 seconds) is no bigger than 125 times 60 squared which is equal to 450,000, and this is meters, so f(60 seconds) is no bigger than 450 kilometers.

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