“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

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

Taylor's theorem.

[NOISE] Well thus far we have been looking for

Taylor series with actually concerning ourselves with convergence.

So what I'd like to do is start with some function,

write down It's Taylor Series around zero.

Maybe I'll assume this function is infinitely differentiable so

I can actually write down it's Taylor Series around zero.

And I mean it's probably too much to hope that these are equal, but at least I could

replace this infinity with some big number N and maybe I can't hope for equality.

But I could at least hope that my function is being approximated

by this partial sum of its Taylor series.

Taylor's theorem is a result that's going to let us pull that off.

Well, here's a statement of Taylor's theorem or I mean, there's actually a lot

of results that are called Taylor's theorem that are all sort of like this.

This is another weak result, and I am just doing it around zero.

But anyhow, okay, Taylor's Theorem.

Here's a statement of Taylor's Theorem.

Suppose that I've got some function from the real numbers to the real numbers, and

just to make this a little bit easier, I'm going to say it's a smooth function.

So, it's infinitely differentiable.

And I've picked some number, big N, and then I'm writing out f(x) =

the big nth partial sum of it's Taylor series around the point 0.

And I'm saying that f(x) is equal to this plus this error term,

or what's usually called the remainder term, so big R for remainder.

And of course this is necessarily true.

I mean, f(x) is equal to anything plus some error.

So the big deal here isn't that I can write f(x) like this.

It's that I know something about the error term, and here's the upshot.

Then the error term or the remainder term is given by this formula,

the (N+1) derivative of f evaluated at some point z, (N+1)!

times x to the N + 1.

And this point z is just some pont between x and 0.

We'll eventually give a proof of Taylor's theorem.

But in the meantime, let's just see how we can choose Taylor's theorem.

So we've already considered the Taylor Series for sine x around the origin.

It's the sum, n goes from 0 to infinity of -1 to the n over

2n + 1 factorial times x to the 2n + 1.

And what I want to show now, right, what the big goal is, is to show that

sine x is actually equal to its Taylor series regardless of what x is.

To get there, we're going to use Taylor's theorem to control that error term, or

that remainder term.

Just to make it a little bit easier to talk about this.

So f of x equals sine x and it is right out,

then what Taylor's Theorem is telling us?

Taylor's Theorem is telling us that sine x- the big Nth partial sum,

so the sum little n goes from 0 to big N of the Taylor series which

is the nth derivative of f at 0 divided by n factorial times x to the n.

So just writing up a Taylor series in general, this is this remainder term,

big R sub N of x.

And what Taylor's Theorem tell us is that I've got a formula for

this remainder term.

I can control the remainder in terms of the next,

the big N plus 1th derivative of f.

All right, it's telling me that big R sub N

of x is equal to f (big N + 1th) derivative

of f at some point (z) / (big N + 1

factorial) times x to the big N plus 1 power.

For some point z which is between 0 and x.

But I know something about how big the big N plus 1 to derivative of f is.

Well, I don't know anything about big N but

I know that if I were to differentiate sine any number of times, right?

It doesn't matter how many times I differentiate sine.

I'm either getting plus or minus sine or maybe I'm getting plus or minus cosine.

But regardless of how many times they differentiate sin I'm getting one or

the other of this things maybe with a plus or minus sign.

And what that means is that regardless of what big N plus 1th is,

the absolute value of the big N + 1th derivative of sin

at some point x can be no bigger than 1 in absolute value.

So I can use this knowledge to say something about the remainder.

Because this is true about the big N + 1th derivative of f being no bigger than 1,

that tells me that at least an absolute value

the remainder is no bigger than, well how big could this be when absolute value's 1?

So it's no bigger than one / (N +1) factorial times x

to the N + 1th power in absolute value.

So I'd like that to be small or

even better, I'd like to say something about the limit.

So fix some value of x, I don't need to consider varying values of x,

I just want to consider the fixed value of x.

And I want to show that this is small in the limit.

I want to show that the limit as big N goes to infinity of the absolute

value of x to the big N + 1th power / (big N + 1) factorial.

I'd like to show that the limit of this is 0.

Why would thinking about the limit be a good idea?

Well if I know the limit of this is 0 then I know that another limit vanishes.

Then I know the limit as N approaches infinity

of something that's smaller than this, something closer to 0.

Well, here's something.

Then I know that the limit of the remainder term

is also equal to 0 because this is bigger than this.

But why would I care about knowing the limit of the remainder term is zero?

Well remember what this is measuring, right.

Big R is measuring the distance between the function and

the big Nth partial sum of its Taylor Series around zero.

So, if I know that in the limit as big N approaches infinity, this goes to zero.

That means in the limit this goes to this.

But what's the limit of this as big N goes to infinity?

Let me write that down, right?

What I'm saying is that sine of x must be equal to the sum.

N goes from zero and I'd write the big N here but

I'm taking the limit as big N goes to infinity.

And that's exactly what this series notation means.

All right, so what I'm saying is that if the remainder goes to zero in

the limit the sine is actually equal to it's Taylor Series around zero.

That means that sine of x is actually equal To a Taylor series.

And I know another way of writing on the Taylor series, right.

Then, I can say that sine of x is actually equal to the sum and

goes from 0 to infinity and instead of writing down this where I haven't actually

described what these coefficients are, I could write down (-1) to the n / (2n +

1) factorial times x to the 2n+1 Because we've already seen that if we

differentiate side a bunch of times, this is exactly what we get here.

So that is a great goal.

Right, so I've got to figure out some way to show that this limit is 0.

Well, how do I deal with that limit?

Well, there's a trick.

Claim.

The sum, big N goes from 0 to infinity of the absolute

value of x to the big N + 1 / (N + 1)!

converges.

If I can show that this series converges then I know that the limit

have this is equal to 0.

But how do I know that that series converges?

Well let's use the ratio test.

So I should look at the limit as big N goes to infinity.

It seems like it's going to be worse, but it's going to be better.

Of the N+1th term here, where N is replaced by N+1,

divided by this term here.

So let me write that out.

It's going to be a limit of x to the big N + 2 / (big N + 2) factorial.

And that's divided by x to the big N +1 / (big N +1)

factorial taking absolute value of this.

Now, how do I evaluate this limit?

This maybe seems worse, but there's a lot of fortuitous cancellation.

So, this limit ends up being the limit as big N approaches infinity.

Well, I've got an x to the big N + 2 / x to the big N + 1, so

this is just x, just one copy of x survives.

Here, I've got an N + 1 factorial in the denominator of the denominator, so

I'm just going to move that up to the numerator.

So I got N plus 1 factorial in the numerator.

And then in the denominator of the numerator, I've got N plus 2 factorial,

I'll just put that in the denominator.

All right so now I am going to calculate this limit, and what is this limit?

I've got an N+1 factorial and a numerator N+2 factorial in the denominator, so

this is the limit, as N approaches infinity of x / N+2.

And what happens now, x is fixed, big N is approaching infinity,

so this limit is equal to 0.

Zero is less than one, so by the ratio test,

this series converges, and therefore by the nth term test, if you like,

because this series converges, that means that this limit is equal to zero.

Let's put it all together.

So really, the argument starts when I fix some value for x,

I just pick a single value of x.

And I'm going to consider just that value of x from now on.

Now, what happens?

I fixed that value of x, and for fixed value of x, the limit is bigger and

approaches infinity of that fixed value of x, just a constant as far as big

N is concerned, divided by some number that's going to be enormous.

This is 0, but this is

also calculated in the limit of the ratio between subsequent terms in this series.

That means by the ratio test since this limit is 0 and

0 is less than 1, which is the cutoff for the ratio test.

That means that this series converges.

Now because that series converges, that means the limit of its nth term must be 0.

But because that limit is 0, I know something now about the remainder.

Right, the remainder is bounded above by this term.

So if this limit is 0, that tells me something about the remainder.

It's telling me that the limit of the remainder

as big N goes to infinity is equal to zero.

But if the remainder is going to zero in the limit, right.

That tells me that the difference between sine and

its big Nth partial sum is going to zero as big N goes to infinity.

But that's just to say that sine of x is actually equal to its Taylor series.

Because the error between sine of x and

the big nth partial sum over here is going to zero.

So in the limit, right?

When big N is going to infinity,

sine of x must actually be equal to its Taylor series.

We've already found its Taylor series, here it is.

And this true for all real X,

because at the beginning of this argument I just picked one value of X.

But I didn't use anything about that value of x, so

this must be true regardless of what x is.

That is truly a triumph.

This is a huge success.

This very mysterious function sine turns out to be given everywhere by this

relatively nice-looking power series.

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