“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

Cosine.

[SOUND]

[MUSIC]

I'd like to approximate cosine of x when x is near 0.

I'm really asking for

more than just say an approximation to cosine of 0.12 or something.

Right I want more than just an approximation that's good at

a single point.

This is really what I want to do here, this is going to be my goal.

It's going to be to find a polynomial, and

hopefully not a very high degree polynomial.

I'll call that polynomial p(x), so that the following happens.

The distance between p(x) and cosine of x

is less than a 100th whenever say x is between -1 and 1.

How can I get started?

So to do this we're going to use the Taylor series for cosine.

Taylor series for cosine of x is the sum n goes from 0 to infinity

of minus 1 to the n over 2n factorial times x to the 2n.

The question really boils down to trying to figure out how many terms I have to

take from the Taylor series.

What I am saying is that cosine of x is approximately the sum

n goes from 0 to some number big N of its Taylor series minus 1 to the n,

over 2n factorial times x to the 2n.

And the issue here is I need to know exactly how big that n needs to be,

and that depends on x.

I mean if x is really big I'm going to want to take more terms from my Taylor

series to get a good approximation.

But my goal here isn't to get an approximation for

cosine that's good everywhere, I just want an approximation for

cosine that's good on this interval, the interval from minus 1 to 1.

And then I've quantified exactly how good of an approximation I want.

I want that approximation to be within 1 100th.

I'll use Taylor's theorem to get some idea about the size of the remainder.

Okay, so the function that I'm studying here is cosine, so let's call that f.

And what does Taylor's theorem say?

Well, it says that the difference between cosine and

the big Nth partial sum of its Taylor series around 0.

So that's what I'm writing out here,

the nth derivative of f(0), divided by n factorial times x to the n.

Well this is the remainder term, big R sub big N(x).

And what Taylor's theorem tells me is it tells me something about big R sub N(x).

Tells me that big R sub N of X is equal to f,

the big N + 1th derivative of f, evaluated as point z,

divided by big N + 1 factorial, times x to the big N + 1.

And Z is just some number between 0,

it's the point that I'm taking the Taylor series around and x.

I'm assuming that x is between minus 1 and 1, and that assumption

is telling me something about the size of at least this term here, right?

So if x is in the closed interval between -1 and 1,

then what do I know about x to the big N+1 power?

Well then I know that the absolute value of x to the big n plus 1th power

is no bigger than 1.

And consequently the absolute value of big R sub n of x is no bigger than

the absolute value of the n plus first derivative evaluated as mystery point z.

Divided by n plus 1 factorial.

And I don't have to include this term because

an absolute value of this term is no bigger than 1.

I also know something about the derivatives of cosine.

Right, what I know is that the n plus first derivative of cosine at Z.

Well, what happens if I differentiate cosine a bunch of times?

I don't know how many times I'm differentiating it.

But as I differentiate cosine, all I'm going to get is maybe a plus or

minus sine of z or plus or maybe a plus or

minus cosine of z depending what big n plus 1 is actually equal to.

But look at these functions no matter what z is neither of these are larger than 1 in

absolute value.

So this is telling me that the absolute value of the big n plus 1th derivative of

f at the point z is no bigger than 1.

We'll use that fact to give a nicer bound on the remainder.

So putting together this fact and this fact, this is what I know.

I know that the remainder is no bigger in absolute value

than 1 over big N plus 1, factorial.

Now we're back to that question from the beginning.

How big does N have to be to guarantee that the remainder is bounded by 1 100th?

Well let me write that down.

All right I'm looking for a big N so

that 1 over big N + 1 factorial is less than 100th.

Well I know some factorials right?

I know that 4 factorial is 24 so 4 isn't a good choice for

big N + 1 but if this were 5 factorial all right,

that would be 1 over 5 factorial, and that's 1 over 120.

And 120th is less than 100th.

So if big N plus 1 is 5, that's surely good enough.

So N = 4 is good enough.

But let's think about the Taylor Series for cosine.

The Taylor Series for cosine is, cosine of x is equal to 1,

minus x squared over 2 plus x to the 4th over

24 minus x to the 6th over 6th factorial, which is 720 and it keeps on going.

The thing to notice here is that there is no x to the 5th term.

Why is that relevant?

Well it means that if I'm going to sum through the x to the 4th term,

if I'm going to set big N = 4, I might as well set big N = 5,

because that will improve the error bound, but it doesn't actually affect

the Taylor series at all because there is no x to the 5th term.

So let's just make this N = 5,

and that's going to improve my error estimate somewhat, for free.

Let's summarize what we've found.

Okay so what we've got is that cosine x is 1- x squared over

2 + x to the 4th over 24 + a remainder term where I'm imagining

that I'm summing through the nonexistent x to the 5th term.

So I can get away with using this remainder term as if I'm

summing through the x to the 5th term because there is no x to the 5th term..

The next term is actually x to the 6th.

All right.

And then I know something about how big this remainder term is.

It's estimated right up here so what do I know about R sub 5?

Well, R sub 5 of x is no bigger than 1 over 5 plus 1 factorial.

That's one over 6 factorial, that's 1 over 720.

And that was our goal.

Well, that was our goal.

I wanted to write down a polynomial, here it is.

1 minus x squared over 2 plus x to the fourth over 24.

And this polynomial is approximately cosine of x,

with an error no worse than 1 over 720.

And originally, I just wanted to get within a 100th, and

I've done way better than that.

I mean, this polynomial is an awfully good approximation of cosine as long as x is

between -1 and 1.

This is one of the reasons to love Taylor series.

We've been doing approximations but

thus far the approximations have really been at a single point.

Now, we're getting approximations that are good on an entire interval.

That's an extremely important idea and

if you want to read some more about this the phrase to

search for is uniform convergence.

[NOISE]

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