“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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From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

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The Ohio State University

45 ratings

“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

Sine.

[MUSIC]

It's not too painful to differentiate sine even if we have got to do it a whole

bunch of times.

Here we go.

Here is my function I'll call it f.

So f of x is sine x.

And now let's start differentiating.

The derivative of f, derivative of sin is cosine.

The derivative of cosine is negative sin, so that's the second derivative of sin.

The derivative of negative sin is negative cosine so

that's the third derivative of sin.

The fourth derivative of f is the derivative of negative cosine

which is sine again because it's negative negative sine.

So, it's just sine.

The fifth derivative of sine is the derivative of sine again,

which is just cosine of x again.

Now, let's evaluate those derivatives at zero.

Well, f of 0 is 0.

F prime of 0 cosine of 0 is 1.

F double prime of 0, so what's the -sin of 0 that's 0.

The third derivative at 0 is -cosine of 0 that's -1.

The fourth derivative at zero is sine of zero, that's zero.

The fifth derivative at zero is cosine of zero, which is one.

We're seeing a pattern.

What's really going on is that the fourth derivative of sine is itself.

So the fifth derivative of sine is the same as the first derivative of sine.

It means the sixth derivative of sine is the same as the second derivative of sine.

So for just looking at these derivatives at 0, we're seeing 0, 1, 0, -1, 0, 1.

What's the sixth derivative at 0?

It's going to be 0 again.

What's the seventh derivative at zero.

It's gotta be minus one.

Because it's zero one zero- one, and

the fourth derivative is the same as the zero derivative.

Right, the fourth derivative is the function itself again.

So it's zero one zero minus one zero one zero minus one

The eighth derivative at 0, would be 0.

The ninth derivative at 0 would be 1, the tenth derivative at 0 would be 0.

The eleventh derivative at 0 would be -1, and

it just keep on going, 0, 1, 0, -1, 0, and [LAUGH] so on and so forth.

Finding that pattern lets us write down the Taylor series around 0.

Well here's what it ends up being, or at least,

this is one way of writing down what it ends up being.

So the Taylor series around 0, so centered around 0, for sine is the following.

It's the sum, n goes from 0 to infinity Of -1^n / (2n+1)!

times X ^ (2n+1).

But why does that work?

Well, think about this.

I made this table here.

This is a little table that shows N,

and below here is the Nth derivative of sin evaluated at 0.

All right, so the zero-th derivative is just a function sine at 0.

The first derivative is cosine at 0, which is 1, and so on.

These are the numbers that we computed before.

And I've got this pattern that we saw, 0, 1, 0, -1, 0, 1, 0, -1, and so on.

But what you should notice is that all the even derivatives are 0.

So this thing doesn't include any terms with X to an even power and

that's why I've written X to 2 N plus one.

This is kind of a sneaky trick just for according all of the odd powers, and

what about those terms with X to an odd power.

Well an N is an odd number I'm flip flopping in sign in S I G N and

that's exactly what this minus 1 to the N Accomplishes for me.

This is all pretty great, but it's a raising a very serious question.

Is this power series actually equal to sine of x on any interval?

That is the problem.

We're assuming that sine has a power series representation valid on an interval

around 0.

And if it can be represented as a power series then we've

figured out what the power series has to be.

But how do we know that sine has any power series representation.

That's the problem that remains.

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