“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

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 Series.

[NOISE] Suppose I know that f is represented by a power series.

For example, suppose that I that the function f is given by a power series.

The sum n goes from 0 to infinity of some coefficients an,

times x to the n at least say when x is between -R and R.

In other words, f(x) = a sub 0 + a sub

1 times x + a sub 2 times x squared and so on.

In this case, I can figure out what the coefficient,

the a sub ns must be equal to, in terms of the function f, Well here's what I mean.

Take a look at say, f(0).

Well if f is really given by this power series

on this interval, then f(0) = a0 + a1 times 0 +

a2 times 0 squared Plus a sub 3 times 0 cube and so on.

But all of these terms have a 0 in them.

A sub 1 times 0 is 0, a sub 2 times 0 squared is 0, a sub 3 times 0 cube is 0.

All these comes to 0 except for a sub 0.

And that means that I can conclude, That a's of 0 is just equal to f of 0.

So, if f has this nice power series representation,

the a's of 0 coefficient must actually be equal to the function's value at 0.

But wait, there's more The next coefficient, a sub 1,

that coefficient can be calculated in terms of the derivative of f.

What I can use is this power series again.

You can differentiate this power series, term by term.

And I've got a power series for the derivative of f,

at least valid on this interval.

So let me write that down.

So what I've got is that the derivative of f at x It's given by this power series,

the sum n goes from 1 to infinity of a's of n times the derivative

of x to the n which is n times x to the n minus 1 and

this is at least the valid when the absolute value of x is less than R.

Now, what am I going to do with this?

Well let me write out the first few terms of this power series,

just to give you an idea of what this looks like.

And of course it looks exactly like the derivative up here.

But, maybe I'll just plug in n=1, and

I get a sub 1 times 1 Times x to the 0th power, which is just 1.

So this is, the first term is just a sub 1 times 1.

The n = 2 term, well that's a sub 2 times 2 times x to the first power.

The n equals three term.

Well that's ace of three times three times x squared.

And then it keeps on going.

And you can see this is just the derivative of this.

Right? The derivative of ace of

zero is just the derivative of a constant which goes away

The derivative of a sub 1 times x is right here, a sub 1 times 1.

The derivative of a sub 2 times x squared is right here,

a sub 2 times the derivative of x squared is 2x.

It keeps on going.

Okay, now what happens when I evaluate this power series at x equals 0?

Well, every single term has an x in it, except for

the constant term, which is a sub 1.

So the derivative of f at 0 is just equal to a sub 1.

Which is telling me that if I've got a power series representation for

my function f, I know what the a sub 1 coefficient has to be.

It has to be the derivative of my function at zero.

And I can just keep on doing this.

The coefficient in front of x squared, a sub 2,

that coefficient can be calculated in terms of the second derivative of f.

Let me differentiate this again in order to get a power [INAUDIBLE]

representation for the second derivative of f.

So f double prime is the term by a term derivative of this power series, which

is the sum n goes, not from 1, but from 2 to infinity of the derivative of this.

Which is ace of n times n times the derivative of x to the n minus 1.

To the n minus 1 times X to the n minus 2.

And this is valid at least when the absolute value of X is less than R.

Now we'll look at the first few terms.

So when I n equals 2, I get what?

A sub 2 times 2 minus 1 times X to the 2 minus 2,

X to the 0, that's just times 1.

All right, so a's of 2 times 2 times 1 times 1 plus what happens

when I plug in n equals 3, a's of 3 times 3 times 3 minus 1,

so times 2 times x to the 3 minus 2, so just times x.

What about when n equals 4?

Well, that's a's of 4 times 4 times 4 minus 1,

times 3 Timex x to the four minus two times x squared.

And just going to keep on going.

And what happens, again, is that all of these terms have an x in there.

So when I plug x equals zero, this term dies, this term dies,

all the other terms die except for that constant term which is a of two times two.

So what is this telling me?

This is telling me that the 2nd derivative of f at 0, is just this constant term.

It's 2, times a sub 2.

And then, if I divide both sides by 2, what

I'm finding is that a sub 2, is the second derivative of f at 0, divided by 2.

Let's try to figure out A sub 3, I've got the original function, its derivative,

its second derivative, now I am going to calculate the third derivative.

So F triple prime at X is the sum N

goes not from 2 but 3 to infinity of the derivative of this.

Which is a sub n times n, times n minus 1,

times the derivative of x to the n minus 2.

So, times n minus 2, times x to the n minus 3rd power.

And this is valid at least Value of x is less than R.

Now what's the constant term?

Well the constant term is just the n equals 3 term.

because that's x to the 3 minus 3,

and that's exactly what the third derivative at 0 is equal to.

So let me write that down.

The third derivative at 0, is just the constant term in this power series, which

is the n equals 3 term, Which is a sub 3 times 3 times 3 minus 1 times 3 minus 2.

In other words, what this formula's telling me is that a sub 3 is

equal to the third derivative of f at zero divided by 6.

But it'll be better to think of that six.

As being three factorial.

Instead of writing down 6 let me write down three factorial.

Then I can write down the general fact.

I've got ace of n, the nth co-efficient in the power series for my function F.

Right. I'm assuming that my function F has this

nice power series representation.

And that a sub n is given by computing the nth derivative

of f at the point zero and dividing that by n factorial.

So yeah, in general, the coefficient on x to the n that a sub n,

that could be calculated in terms of the nth derivative of f.

But why is this important, right?

Why do we care about this?

Well, the whole point is that now we're in a position to work backwards.

So I mean that I'll assume that my functions and

function that I'm interested in has a power series representation and

it's given by say this power series.

On the inner roll from minus R to R.

Now if I assume that that's the case, then by differentiating,

I can figure out what these coefficients have to be, all right?

And that's exactly what I'm claiming, that the nth coefficient

is given by calculating the nth derivative of f at zero.

And dividing by n factorial.

Let me say this a bit differently.

If this function f can be written as a power series on the interval from minus R

to R, then [LAUGH] f(x) is equal to this power series,

the sum n goes from 0 to infinity.

Of the nth derivative of f at the point 0 divided by n factorial.

That's what the coefficient has to be.

Times x to the n.

This has a name.

Or rather, it has two names.

One of those names is Maclaurin Series.

You'll often see this thing called the Maclaurin Series for f.

The other name that you'll see, you'll see this thing called the Taylor Series for

f, centered around 0.

Regardless of what you want to call this, I should warn you about something.

Suppose that you just start with a function that

you don't know very much about.

Maybe all you know about the function f is that you can differentiate it as many

times as you like.

Well, then you could write down a coherence series for f.

You could write down this.

The Taylor series for f center at 0.

All you need to be able to do is differentiate f as many times as you like

at the point 0 in order to write down this power series.

[LAUGH] The issue though, is that nothing as we've said so far actually tells

you that this power series is related To the original function f in any way.

This whole story has been predicated on an assumption,

it's been predicated on the assumption that f is equal to a power series and

if f is equal to a power series then this must be the Series that f is equal to.

But, if you don't know that, if you're just starting with some random function,

then there's no guarantee whatsoever that if you write this thing down,

that it's going to have any relationship at all to the original function.

And relating f and this, it's [INAUDIBLE] series, or its Taylor series [INAUDIBLE]

at zero, it's going to be a huge part of what we do the rest of this week.

[SOUND]

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