“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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微积分二: 数列与级数 (中文版)

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“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

Multiply.

[MUSIC]

We can multiply polynomials.

Yeah, I mean I have got a polynomials in a 1 + x squared say,

maybe multiply that times 3- x.

I can multiply polynomials and get a new polynomial, right?

1 x 3- x + 3x squared- x cubed is what I get

when I just multiply these two polynomials.

We can also multiply power series.

Let's suppose I want to multiply the sum angles from zero to infinity

of a sub n times x to the n by another power series.

Maybe the sum n goes from 0 to infinity of b sub n x to the n.

Well how do I even get started.

Well one way to at least get started on this is to just expand these out, right?

So I can write out the first few terms here.

That's a sub 0 + a sub 1 x + a sub 2 x squared and I will keep on going.

And then I want to multiply that by all item a first few terms in this blue

power surge with the b sub n, so b sub 0 + b sub 1

x + b sub 2 x squared and it would keep on going.

And what do I get when I multiply this?

Well, I can try to think about

how these terms might combine that give me various powers of x.

So I have to pick something here and multiply by something here.

So the only way that I got a constant term, a term about an x,

is that I multiply a sub 0 by b sub 0.

So I can start by writing that down, a sub 0 times b sub 0.

Then what's the next coefficient in the product?

Well I have to think about how can I get just a linear term,

a term with just an x it.

And there's two different ways I could do that.

I could multiply a sub 0 by b sub 1 x or I could multiply a sub 1 x by b sub 0.

So I should write down both of those terms.

So a sub 0 times b sub 1.

Plus a sub 1 times b sub 0 plus a sub 1 times b sub 0 and

also the only way that I can get a term with just a single x.

What about the x squared term, right?

[LAUGH] There's actually three different ways I can earn x squared term,

a sub 0 times b sub 2 x squared will give me an x squared.

A sub-one x times b sub-one x would give me an x squared.

And a sub 2 x squared times b sub 0 will also give me an x squared, so

let me write down all three of those.

So I got a sub 0 times b sub 2 plus a sub 1 times

b sub 1, plus a sub 2 times b sub 0, and

those are all different ways that I might get

an x squared when I multiply together this

two indices, and then we keep on going.

>> The trouble is that the coefficients get kind of complicated.

>> Of course not all hope is lost, there is a pattern in here.

What's the pattern?

Well here, I've got the constant term, the x to the 0 term if you like.

And these are both 0.

Here I've got the x to the first term and

these indices add up to 1, 0 + 1 and 1 + 0.

Here I've got the x squared term and these indices also all up to 2.

0 + 2 is 2, 1 + 1 is 2, 2 + 0 is 2.

And you might guess them.

Well what's the coefficients on x cubed?

It's going to be combinations of the a sub n and

the b sub n where the indices add up to 3.

Let's write down at least what our guess is then for the formula in general.

So this will be the sum, n goes from 0 to infinity of another [LAUGH] series.

The sum i goes from 0 to n of a sub i

times b sub n- i, and

it's this coefficient in front of x to the n.

>> I need to say a little bit more about what this even means.

>> So I'm imagining that I've got two functions.

Maybe I've got a function little f, which is given is the sum and

goes from 0 infinity of a sub n x to n.

And got a function little g which is the sum and

goes from 0 to infinity b sub n x to n.

Now these power series might have different radius of convergence.

So let's just have big R be the minimum of

their radii of convergence.

>> It isn't just that the series with the convolved coefficients converges.

>> I mean, I've got this product series, but I'm not just saying that this

series converges when the absolute value of x is less than R.

I' m actually saying that this series converges to

the value of f(x) time g(x), right.

I'm making a claim that f(x) times g(x) is equal to the value of this series.

>> Let me put it all together.

>> Well, here's a precise theorem.

So I got two functions, a function f and a function g.

F of x is the sum n goes from zero to infinity of a sub n x to the n,

and g of x is the sum n goes from zero to infinity of b sub n x to the n.

Now, f and

g have these powers series and the power series have some radius of conversions.

Now let's assume that both of those radii of convergence is greater than or

equal to big R.

Well, then I've got a new power series here.

The radius of convergence of this power series is at least big R.

And here's what I know.

f(x)g(x) is equal the sum angles from 0 to infinity and

this is weird convolved coefficient, the sum of little i from

zero to little n, of a sub i, b sub n, minus i, times x to the n.

And this quality that least holds when x is between -R and R.

In any sort of the reasonable looking, I mean, what was going on here?

These coefficients have indices that add up to n.

So certainly when I think about multiplying a piece of this power series

and a piece of this power series,

these are the terms that I expect to get in front of x to the n.

>> I should warn you that we're not going to prove this result, but

I hope it's plausible and I hope you'll play around with it and

try to get a sense of some of the consequences of this theorem.

I mean here's one example, kind of a cool thing that you can do with it.

>> For example, we've thought a little bit about e to the x, and e to the x has this

really nice power series representation of the sum and goes from zero to

infinity of x to the n over n factorial, and I can write out the first few term.

If I plug in n = 0 got 1, plug in n = 1, got x plug in n = 2,

I got x squared over 2 factorial and just over 2 plug in n = 3,

I've got x to the third over 3 factorial which is 6 +...

Now I can think about what happens when I multiply this power series by

itself, right?

What's e to the x squared?

Well I secretly know what the answer is right just because of how exponents work.

E to the x squared is e to the 2 x.

So this should be 1 plus I replace all these x by 2 x.

2x plus 2x quantity squared over 2 is 2x squared.

Quantity 2x cubed divided by 6 is 8x cubed over 6 plus dot dot.

But I can also think about this by multiplying

the original power series by itself.

All right, what do I get?

So one + x + x squared over two dot dot dot dot squared, right?

Well, I can think about the constant term,

I get the constant term by multiplying one times one.

How do I get the next term, the term in front of the x?

Well, that's one times x or x times one, so that's 2x.

How do I get the term in front of x squared?

Well, there's three different ways to get that.

1 times x squared over a half, so a half.

X times x, so 1 or x squared over 2 times 1, so I put a half here.

I'm sure enough that a half plus 1 plus a half is two.

What is the next term, right?

I'm looking for the coefficient in front of x cubed.

There's four other way to get x cubed, one times x cubed over six, so it's six.

X times x squared over two, so it's a half.

X squared over two times (X), that's a half, and

then (X) cubed over six times one, that's a sixth, and

in sure enough as six plus a half, plus a half, plus a sixth, that's eight-sixths.

And then I could keep on going.

[SOUND]

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