“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。 注意：此课程的注册将在2018年3月30日结束。如果您在该日期之前注册，您将可以在2018年9月之前访问该课程。

Loading...

From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

46 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。 注意：此课程的注册将在2018年3月30日结束。如果您在该日期之前注册，您将可以在2018年9月之前访问该课程。

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

Transform.

[MUSIC]

We have some series that we already know about.

For example, I know that the sum n goes from 0 to infinity of x to the n,

that's a power series for a function we already know.

It's just the power series for 1 over 1- x and

we've got some operations that we can apply to power series.

Well we can differentiate a power series, we can integrate a power series term

by term, I can multiply together power series, and I can divide power series.

Now, we can put together the series that we know and

the operations that we can do in order to build new power series.

Well here's our goal.

I'd like to find a power series for the function 1 over 1- x, quantity squared.

We can do this in two different ways.

Let's notice something about this.

This is the derivative of 1 over 1- x.

Why is that?

Well, let's just work it out, right?

This is the derivative, I'm claiming, of 1- x to the -1 power.

Well, differentiating this using the power rule, that's -1 times

1- x to the -2 power times the derivative of the inside by the chain rule,

which is just multiplying this by another copy of -1.

So, this is -1 time 1- x to the -2 power times -1.

Well, what happens?

This -1's cancel, and what I'm left with is just 1- x to the -2 power.

Well, I could write that as 1 over 1- x squared.

That's exactly what I've got here.

So, the derivative of 1 over 1- x is this function that I'm interested in.

So, let's differentiate the power series term by term.

So this is the derivative of the sum n goes from 0 to infinity of x to the n and

then, like I said, differentiating term by term, this is the sum n goes from 1

to infinity of the derivative of the nth term, the derivative of x to the n.

Note that I changed from n = 0, n = 1.

Because then I differentiate the n = 0 term,

that's the derivative of the constant which is just 0.

So I'm going to start here with n = 1.

All right, so this is the sum n goes from 1 to infinity.

What's the derivative of x to the n?

By the power rule, that's n times x to the n- 1.

And now if I don't like having the n- 1 there,

I could re-index this series, right?

What happens when I plug in n = 1?

Well that's one times x to the zero.

And what happens when I plug in x equals two?

Well that's two times x to the first.

What happens when I plug in x equals three?

It's three times x squared and so on.

So I could re-write this series as just the sum,

n goes from zero to infinity of n + 1 times x to the nth power.

[LAUGH] And there we go.

I've written down a power series for the function one over 1- x quantity squared.

Of course, if I don't like thinking like this,

if I don't like differentiating, I could also approach this problem but

it's trying to multiply together two copies of the original power series.

Well what I mean is that the sum of x to n,

n goes from zero to infinity, is 1 over 1- x.

[LAUGH] That means that if I square this side and square this side,

I should get a formula then, a power series for one over 1- x quantity squared.

But how do I write that as a power series?

Well, let's at least get started.

Let's at least figure out what the first few terms are.

So I'm just going to write down the sum of x to the n, or the first few terms of it.

So that's (1 + x + x squared + x cube +...) And

I'm squaring that, so I'm multiplying it by itself, so

I'll just write down the same thing again, (1 + x + x squared + x cube +...).

Now I want to figure out what happens when I multiply these things together.

So, what do I get?

Well, 1 times 1 that's 1, and how many x terms do I get here?

Well, I got a 1 times an x or we got an x times a 1,

but there's not other way to multiply these things and get an x.

So there's plus 2x, how many x squareds do I get out?

Well, I can multiply this 1 times this x squared, multiply this x times this x, or

I can multiply this x sqaured times this 1.

So, I've got 3x squared.

There's no other way to get an x squared out.

How can I get an x cube?

Well, I can multiply 1 times x cube or x times x squared, or

this x squared times this x, or this x cubed times one.

So that gives me four ways of getting an x cubed term and

I'll write plus dot dot dot.

Now maybe you believe that that patterning continues.

Well it certainly looks like this is giving me the sum n

goes from 0 to infinity of n + 1 times x to the nth power.

I mean n = 0 is 1 times x to the 0, n = 1 is 2 times x to the 1st,

n equals two is 3 times x squared, n equals three is 4 times x cubed and so on.

To make that more rigorous, we probably have to talk about induction.

Or, we could just bring up a theorem on multiplying power series.

Well here's our theorem for multiplying power series, the product of this power

series and this power series is given by this power series.

It's a little bit complicated to see how the coefficient's affected, right?

Here the coefficients are a sub n.

Here the coefficients are b sub n.

And when I multiply these together, I get this convolved series,

the sum i goes from 0 to n of a sub i times b sub n minus i.

Now we can apply that theorem.

Well, in this case, our formula for

1 over 1- x is just for all the coefficients are 1.

So, if i multiply 1 over 1- x by 1 over 1- x,

then I get 1 over 1- x squared.

So that means all the a sub i's and b sub n- i's are all just 1.

So here is a power series for 1 over 1- x squared.

Let's simplify.

Well, the sum i goes from zero to n of just one,

well that's one plus one plus one.

But it's n plus one, once.

So I can rewrite this power series as just the sum and

goes from zero to infinity of n + 1 times x to the n,

and that's exactly what we got before By using derivatives.

[SOUND]

Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.