“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。

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From the course by The Ohio State University

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。 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

审敛法

在第三个模块中，我们学习用于确定级数是否收敛的各种审敛法：特别地，我们将说明比值审敛法、根值审敛法和积分审敛法。

- Jim Fowler, PhDProfessor

Mathematics

Let's build bridges.

[NOISE] Normally, I'd build

a bridge across a river.

Let's suppose this blue area here, is a river.

And what would I need to build a bridge across this river?

Well I'll probably put pylons on either side of the two banks.

And then I build the bridge across those two pylons.

Here's a diagram.

All right I've got the earth,

I've got the river, here are the two banks of the river.

And then I can build a bridge that goes from one bank to another.

But it's attached to both banks.

But I want to think about a one-sided bridge.

Instead of being connected to both banks,

I just want to be connected to one of the banks.

Sure, so instead of building a bridge that's attached to both banks,

I want to build a bridge that's just attached to one bank, and

then I want to know how long can I make this bridge before it collapses.

Well, I mean look, if you'll [LAUGH] allow me to build the bridge

from some super strong metal like unobtanium.

Well, no problem.

I'll just build a really, really long bridge.

It's going to be no limit to how long that bridge could be.

But that's not really what I mean.

What I'm really going to ask you to do is to build the bridge out of these blocks.

So you're allowed just to stack blocks on top of each other.

And I want the thing to be stable, I don't want it to fall over.

And I want to know if you just start stacking blocks,

how long of an overhang can you get?

What's the maximum possible overhang that you can achieve if I give you n blocks?

You could try to get started just with some of these foam blocks.

Try to build a stack of these blocks and

see how far I can get these blocks to overhang before they fall over.

Why did my block tower just fall over?

Well, the issue has to do with center of mass.

My block tower's going to fall over if the center of mass isn't supported.

What I mean by that, the center of mass of the first block

isn't supported by the second block, then that first block will fall off.

If the center of mass of the just the first two blocks together,

which is maybe over here somewhere, isn't supported, if it isn't of the third block,

then those two blocks are going to fall, my tower is going to collapse.

So that's the issue.

I want to build a really tall tower of blocks

with a really long overhang and yet I want it to be stable.

Of course the easiest way to make it stable is just to make my block tower

perfectly vertical but then I don't have any overhang at all.

So these two forces are really working against each other, right?

My desire to have a really long overhang is really playing against

my desire to have my block tower not collapse.

Well, let me propose a specific configuration.

Here's the configuration I'm proposing.

So imagine that all these blocks are exactly the same.

They're all one block long.

And I've staggered them like this.

So this first block is offset by half a block width from the second block,

the second block is a quarter of a block width pushed in,

the next block is a sixth pushed in.

The next block is an eighth, this distance here is an eighth from a block.

The next block here, is a tenth of a block.

If I put another block under there, I'd offset it by a 12th of a block.

The next block would be by a 14th of a block,

then a 16th of a block and then so on.

I need to check that that configuration is stable.

The center of mass of the top block is smack in the middle of the top block.

And since I pushed the top block over half a block from the second block,

that puts the center of mass of the top block right above the edge

of the second block so it's stable.

What about the second block?

The center of mass of the second block is right in the middle of the second block.

That's not so relevant.

I mean yes the second block isn't tipping over, but what I really need to know is

what's the combined center of mass of the first block and the second block together.

So to figure that out I just need to remember that I pushed the second block

over a quarter of a block from the edge of the third block and

then I can compute the center of mass of the first and the second block together.

And I find out that in that case,

the center of mass is right there, which puts it right above the third block,

which means the first two blocks together are stable.

So the first block is stable, the first two blocks are stable,

but that's just the top two blocks.

I need to know this in general.

I need to compute the center of mass of top

n blocks relative to the right edge of the next block.

And to do this I'll start by averaging some center of masses.

All right I want to average the center of masses of the top and blocks.

I don't really need to know the y coordinate of the center of masses.

So I'm just going to add together the x coordinates of these center of masses,

where I put the origin at the right-hand edge of the block right under this stack.

All right, so I'm going to add together these x coordinates and then once I've

added together these x coordinates to average them I need to divide by n.

Okay so I just got to figure out where these blocks really are in space.

So first of all where's block number one.

Where's the top block relative to the next block under this

Colllection of the top N blocks.

Well that block's center of mass is right

here at one half plus a half is a fourth plus, plus 1 over 2n.

The center of mass is just a block by itself is at one half and

this records how far over I've pushed the top block

relative to the next block in the stack, after the top n blocks.

Okay, what about block number two.

Well that looks very similar, right?

Again, it's one-half because that's where the center of mass is just in

the block by itself, but the second block doesn't get pushed over half,

it gets pushed over fourth.

And then a sixth, and so on, until it's 1 over 2n.

So a fourth plus a sixth plus until I get to 1/2n,

that's how far over I pushed it relative to the next block.

And a half then moves me over to the middle of block number two.

Then block number three has a similar looking formula.

It's a half plus now a sixth plus an eighth until I get 1/2n.

And finally I get to the nth block which is right above the block that

I'm measuring from so its center of mass is at 1/2 + just how

much I just pushed over the nth block which is 1/2n.

So I gotta look at this and

see if there's anything I can say about this complicated sum.

Well, I got half, I got a half, I got a half, I got a half.

Every singe one of these n terms has a half so that gives me n over two.

I got n halves all together.

I've also got a one-half here and no extra one-halves.

So, I can had just this one half coming from right here.

I got a quarter here and quarter here and then no more quarters.

So, I got two quarters and I can add those.

How many sixths do I have?

Well I got a sixth and so the dot dot dot, I got a sixth here, I got a sixth here.

The next term doesn't have any sixths, so

I got three sixths all together and then I want to keep on going.

Right I can count how many eighths I have,

I could count how many tenths I have and so on.

And eventually I'll notice that I have got 1/2n and 1/2n and 1/2n and 1/2n and

I've got n/2nths and

that's it, that's all the terms in this sum.

So then I'm dividing this whole thing by n.

Well this is a half, this is also a half, this is a half, and this is a half.

Here I've got n halves.

So instead of writing a half + two fourths + three sixths and

everything I could just write n halves.

So now I've got n/2 + n/2 / n or

all together that's just n/n, that's just 1.

So relative to the right hand edge of the next block the center of mass of the top

n blocks is right on the left hand edge of the next block which means it's stable.

So it's stable, but

now what kind of overhang can I get with that configuration?

So in this case, I just used six blocks but

the total overhang is easy to calculate.

Right, I want to calculate the total overhang, it's the distance from the left

edge of the top block to the left edge of the bottom block.

And, it's just one-half, plus one-fourth, plus one-sixth,

plus one-eighth, plus one-tenth.

It's the total amount that I shifted all the blocks over by.

Now, if I built the same kind of configuration, but instead of six blocks,

I built it with some large number.

Call it big N number of blocks.

Then this total overhang is the sum little n from 1 to big N- 1, of 1/2n.

This is the total amount that I'd be shifting all the blocks over by.

That looks like the harmonic series, and indeed, the harmonic series diverges.

Well that means the sum of 1/2n, n goes from one to infinity,

also diverges but that means that by choosing

N big enough I can make this overhang as large as I desire.

Well here it is in the real world.

I've built one of these harmonic towers.

And you can see that I've got a pretty significant amount of overhang here.

I'll rotate it a little bit.

I've got this river here.

Anyway, I could make the overhang as long as I'd like,

as long as I'm willing to use more slabs.

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