“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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“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

幂级数

在第五个模块中，我们学习幂级数。截至目前为止，我们一次讲解了一种级数；对于幂级数，我们将讲解整个系列取决于参数 x 的级数。它们类似于多项式，因此易于处理。而且，我们关注的许多函数，如 e^x，也可表示为幂级数，因此幂级数将轻松的多项式环境带入棘手的函数域，如 e^x。

- Jim Fowler, PhDProfessor

Mathematics

Integrate.

[SOUND] We've already seen how we

can differentiate term by term.

Well can we integrate term by term?

Well yes, and here's a theorem.

Suppose I got some function f and f(x) is this power series,

the sum n goes from 0 to infinity of an x to the n.

And big R is the radius of convergence of this power series.

Well then, I can integrate this function term by term.

Meaning at the integral, f(x) dx, x goes from 0 to some perimeter t.

E is the sum and goes from 0 to infinity of this,

a sub n times t to the n + 1 over n + 1.

And this is true for any value of t between (-R,R).

If you want to be a little more specific, turns out the radius of convergence of

this series is exactly the same as the radius of convergence of this series,

if you are wondering where this term comes from,

that is exactly what you will get just by integrating the term of AnX to the n,

the inner role of AnX to the n Dx X goes from 0 to t.

Well that just is a sub n times t to the n plus 1 over n plus 1.

Armed with this result we can do some pretty great stuff.

For example we know that the sum n goes from 0 to infinity of x to the n.

Well that's 1 over 1 minus x as long as the absolute value of x is less than one.

In other words right in this particular instance the radius of convergence,

big R is 1.

Well let's integrate that.

So the integral from x equals 0 to t Of 1 over 1- x dx.

Why is in that in a row.

Well I can evaluate them by making in used of substitutions or let say u = 1- x.

And that case du is just -dx.

I don't see -dx there, but

I can all refresh create the -dx with some carefully placed minus signs there.

All right, now I've got a negative dx there.

That looks great.

So this is negative the integral.

X goes from zero to t of, and I've just got to du over one minus x, but that's u.

So I want to know how do I differentiate one over u

well this is negative log the absolute value of u

then evaluated at x equals t and x equals 0 and I take the difference.

So that's negative log over what's u absolute value,

1 minus x and I'm evaluating this at x equals 0 and

t So this is negative log of the absolute value of 1 minus t.

And then it would be subtracting this with x equals 0.

And I'm subtracting a negative so

I'm going to add the log of the absolute value of 1 minus 0.

Well what's log of 1, log of 1 is just 0.

So I don't need to include this term.

So all told if I integrate 1 over 1 minus x from 0 to t

what I'm getting is negative the natural log the absolute value of 1 minus t.

Now we can integrate the power series term by term.

Alright, we just showed negative log the absolute value of 1 minus t is

the integral x goes from 0 to t of 1 over 1 minus x d x.

And I'm going to replace this 1 over 1 minus x with a power series.

So this is, equal to the integral.

X goes from 0 to t instead of 1 over 1- x.

The sum n goes from 0 to infinity of x to the n.

Because 1 over 1- x is equal to this,

while as long as the absolute value of x is less than 1, dx.

Now in integrating a power series, and by the theorem I can do that term by term.

So this is the sum n goes from 0 to infinity of the integral

of x going from 0 to t of x to the n dx.

So I just have to be able to do this integration problem.

So this is the sum n goes to from 0 to infinity of what's this?

Well I'm integrating x to the n d x from x equals 0 to t.

So that's t to the n plus 1 over n plus 1.

So there it is right.

I found a series for this function for

negative natural log of the absolute value of 1 minus t.

Here's a power series for that function.

And it's valid for which values of t.

Well, as long as t is less than one in absolute value.

Before we get too excited, it's worth mentioning a fine point here.

Well one case when you'd like to use this formula is when t equals -1 because,

in that case, what do you get?

Then we get negative the natural log of the absolute value of

one minus negative, is equal to what?

Well, is equal to this series.

The sum n goes from 0 to infinity, none negative 1,

to the n plus 1 over n plus 1.

And I can simplify this a bit, right?

What is this?

Negative the natural log of one minus negative one.

Well, this is just two.

So this is negative the natural log of two is given by the series.

Or is it?

Right, the issue here is that I'm going to plug in t equals negative one but

strictly speaking that's not one of the values of t that I am allowed to plug in.

If you want to dig deeper,

there are some other theorems that you might want to look into.

For example, it turns out that this is true.

All right. It turns out that negative the natural log

of two is given by this alternating series.

But to justify that, you do a little bit more.

One way to do it is by invoking Abel's Theorem.

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