“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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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

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

Limits.

[MUSIC]

Taylor series can provide some useful intuition for thinking about limits.

For example, you might remember thinking about this limit.

The limit as x approaches 0 of sine x over x.

And one way to think about this is via Taylor series, right.

And what the Taylor series tell us?

Well, Taylor series tells us that sine x, well,

that's equal to x- x cubed over 6 + x to the 5th over 5 factorial -,

and the series will keep on going.

But at least I know that sine x is approximately x +,

I'll just write higher order terms.

So if you believe this, you might try to evaluate this limit by making use of this,

admittedly at this point, very vague fact.

So what's the limit then as x approaches zero

of x + some higher order terms divided by x?

Well, that's the sort of limit that I could really approach, right?

I mean it's not a transcendental function anymore,

It's just looks like a polynomial as far as I'm concerned or I'm imagining here.

So I could think about how do I calculate the limit

as x approaches zero of x + higher order terms over x?

Well, what I would probably want to do is multiply the numerator and

the denominator by (1 over x).

And I've got these higher order terms.

The denominator's got x.

So this looks like the limit as x approaches zero of the numerator

now is 1 + these higher order terms.

And the denominator is just 1.

And if I'm taking the limit then as x approaches 0,

I'd be very tempted to say that this limit is just 1.

And that is not even close to a proof.

Well, what's wrong with this argument?

Well, basically this a circular argument.

I mean, how so?

So I'm trying to calculate the limit of sine x over x by

thinking about a Taylor series for sine.

And to find a Taylor series for sine, I have to find this Taylor series,

I need to be able to differentiate sine.

So what do you have to do to be able to differentiate sine?

Well, if you're trying to differentiate sin,

you in particular need to be able to differentiate sine at 0.

So let me just write down the limit of the difference quotient that calculates

the derivative of sine at 0.

That's the limit as h approaches 0 of sine (0+h)- sine 0, divided by h.

All right, this limit is calculating the derivative of the sine at 0,

because it's the ratio of how the output changes when you go from 0 to 0 + h.

That's how the input changes when you go from 0 to h.

Okay, but now how do I calculate this limit?

Well, what is this limit?

This is the limit as h goes to 0 of what?

Sine h- sine 0, sine 0 is 0 so the whole numerator is just sine h, divided by h.

[LAUGH] So that's exactly the limit that we've got here, just with x replaced by h.

So what happened here, right?

I'm imagining that I'm trying to calculate this limit

by thinking about Taylor series.

But to calculate the Taylor series for sine and to be able to differentiate sine.

But to be able to differentiate sine,

I really need to be able to calculate this limit.

So if you're thinking that you're really proving anything by this method,

you're really not.

This is just a circular argument.

Admittedly, it's a circular argument in a really big circle, so

maybe it looks like a straight line.

A reasonable argument.

But there really is something essentially circular going on here.

But the fact that the argument's circular shouldn't stop us from

making use of that kind of thinking where it's appropriate.

Let's find the limit as x approaches 0 of

(cosine x)-1, that'll be the numerator,

divided by (sine x) times (log (1-x)).

That's a very complicated looking limit question.

Let me try to approach that more complicated limit question

by using Taylor series.

So let's think about cosine.

So the Taylor series for cosine looks like this.

Cosine x is 1-x squared over 2 + those higher order terms,

so I'm going to write big O (x to the 4th).

And to be a little more pedantic, I'll say as x approaches 0.

And what is this mean?

Well, I don't want to talk about this too precisely yet.

But morally or intuitively at least, just think about it like this,

cosine of x is 1- x squared over 2 +, you can just think of these as being

the higher order terms in the Taylor series expansion of cosine around 0.

Now the next term is an x to the 4th term, but

there's also an x to the 6th term, and so on.

So what about (cosine x)-1?

So that means that (cosine of x)-1 is negative x squared over 2,

+ more terms (degree at least 4).

We can write sine of x in the same way.

Sine of x, now sine also has a Taylor series expansion, it's x.

And then those higher order terms, right?

So I'm just going to write, + higher order terms starting with an (x cubed term).

Let's do the log term.

Well, log (1-x), in the Taylor series expansion,

it starts like this.- x- x squared

over 2,- x cubed over 3- x to the 4th over 4, and it keeps on going.

And I could write that as just- x + terms (degree at least 2).

Let's put the sine and the log term together.

So let's see.

I'm going to multiply together things that involve these big O's.

So how does that work?

Well, I’m trying to calculate (sine of x) times (log (1-x)).

I want to write a Taylor series for that, but I don’t want to actually go through

the bother of calculating out what the Taylor series is going to be.

So I'm just going to try multiply together these Taylor series.

So (sine x) is x + terms (degree at least 3)), and

(log of (1-x)), not everywhere but at least around 0,

is given by (negative x + terms (degree at least 2)).

So what happens when I multiply it together, these two Taylor series?

Well, we got an (x) times a negative (x), so that gives me negative x squared.

But what other terms do I get?

Well, here I've got terms (degree at least 3) times x,

these will give me terms (degree at least 4).

Things (degree at least 3), times things (degree at least 2),

will give me things (degree at least 5).

And I've got x times things (degree at least 2).

So this x times possibly some x squared terms,

that will give me some things of (degree at least 3).

So all told, what I know is that this Taylor series when I multiply them

together, is going to start out -x squared.

And there's going to be some more terms (degree at least 3).

I'm not bothering to calculate those, right?

They're all sort of hidden in this big O notation.

Now, we're in a position to consider the original limit.

So thinking about a Taylor series for the numerator and

denominator, I might be tempted to write this is the limit as x approaches 0.

I've got a Taylor series for (cosine x)-1.

It's -x squared over 2 + terms (degree at least 4).

And I just found a Taylor series for the denominator by multiplying together

the Taylor series for sine and the Taylor series for (log (1-x)).

And that ended up being -x squared + some terms (degree at least 3).

Now, I could multiply this by just a disguised version of 1.

So (1 over x squared divided by 1 over x squared).

And this will give me what the limit as x approaches 0 of this numerator is now

minus a half plus terms (degree at least 2).

And the denominator is now just -1 + terms (degree at least one).

But what is this limit right?

I'm imagining that x is approaching zero, so

these terms and these terms are both approaching zero.

So the numerator has limit minus one-half and the denominator has limit minus 1.

And that this means that this limit, the limit of ratio is just one-half.

So maybe at this point you're not impressed.

You're thinking, I could have done that with L'Hospital's rule,

and you'd be right.

You could have use L'Hospital's rule.

So why are we thinking about Taylor series?

Well the point here isn't just that Taylor series are useful computationally.

My claim is that Taylor series really providing some deeper

insight into what's going on.

So we're presented with this fact, that this limit is equal to one-half,

and you're probably thinking who cares?

And you'd be right to think that.

Who cares if this limit is equal to one-half?

If a machine told me this was equal to one-half, I wouldn't care either.

What matters isn't that the machines are telling us that this is equal to one-half

or that we've got some horribly long algebraic calculation that

proves to us that this limit is equal to one-half.

What really matters here is that this limit is equal to one-half for

a reason that human beings can understand.

We can really comprehend why this limit should be equal to one-half.

We can think about Taylor series.

And Taylor series are telling us that this numerator looks like negative x squared

over 2 + higher order terms.

We think about Taylor series for sine, and

sine looks like (x + x cubed over 6 and so on), (x + higher order terms).

The Taylor series for log of 1-x starts off negative x -, and

then there's more terms.

And I can multiply together Taylor series, and when I multiply together these two

series, I'm getting a Taylor series for the whole denominator.

And the denominator then looks like negative x squared + higher order terms.

And now, I could bring to bear all of the intuition that I have about

limits of rational functions, limits of polynomials over polynomials.

And as x approaches zero, it makes perfect sense then, that if this is a polynomial,

it's really a power series, but if I'm thinking of this as a polynomial

that starts off negative x squared over 2 + higher order terms.

And the denominator looks like a polynomial, the power series, but

it's like a polynomial, negative x squared + higher order terms.

Well then what happens, right?

The limit is exactly the ratio of these leading terms.

It's negative one-half over negative 1.

So the fact that this limit equals one-half, you can understand a reason for

it by thinking about Taylor series.

Yeah, wow, and that's always the point, right?

The point isn't getting answers, it's getting insight, right?

The point of mathematics, it isn't truth, it's proof.

All right, it's not so much whether something is true but why is it true.

[SOUND]

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