“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。

Loading...

From the course by The Ohio State University

微积分二: 数列与级数 (中文版)

46 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。 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

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

Evidence for e to the x.

[SOUND]

[MUSIC]

I've already mentioned this remarkable result.

Let's define a function, f (x), which is equal to the value of

this power series, the sum n goes from zero to infinity.

Of x^n / n!

Now, earlier we said that the radius of convergence of this power series is

infinite.

So, this power series converges regardless of what value I choose for x.

So, this is a function whose domain is all the real numbers.

Now the big result here is that this function f(x)

ends up being equal to the more familiar function just e^x.

Let me now try to convince you that this is true.

Well first of all, f(0) is what?

What happens when I plug in 0 for x?

Then all of these terms vanish except for the n = 0 term.

because by convention 0 to the 0 is 1 so that's 1 / 0!, that's just one.

So the value of this power series when x=0 is 1.

And note that that's also e^0 power or it's also equal to 1.

Not only do these functions agree at a single value, but

they also satisfy the same differential equation.

Well let's see, so for the e^x,

what's the derivative of e^x?

Well, that's itself, right?

So e^x is a function which is its own derivative.

Now let's take a look at the derivative of the power series,

the sum n goes from 0 to infinity of x^n / n!, right?

I'm really just asking,

what's the derivative of this function that I'm calling f?

Well I'm differentiating a power series and I can do that term by term.

So the derivative of this power series is the sum and goes from 1 to infinity.

I can throw away the n = 0 term because a constant term in the derivative of

a constant is 0.

So it's the sum n goes from 1 to infinity of the derivative of x^n over n!.

So now I just have to differentiate each of the terms separately and

add all those up.

So this is the sum, n goes from 1 to infinity.

What's the derivative of x^n?

That's n · x^n- 1.

And then constant is just n!.

But I can simplify this.

This is n · x^n-1 over n!.

That is the same as the sum n goes from 1 to infinity of n / n!

That's just n- 1!

In the denominator.

And the numerator is x^n-1.

But if you think about what this series is, when I plug in n = 1, that's x^0 / 0!.

When I plug in n = 2, that's x^1 /1!.

When I plug in n- 3, that's just x^2 / 2!.

When I plug in n = 4, that's x^3 / 3!

This is actually the same as just the sum,

n goes from 0 to infinity of x^n / n!

So what I've shown is that f

is a function which is also it's own derivative just like e^x.

because if I differentiate f, if I differentiate this power series,

what I get back is just f again.

Two functions, both alike in dignity.

These two star-crossed functions agree at a single point, and

they're changing in the same way.

And consequently, they must be the same function.

And therefore, f(x)=e^x, right?

What I'm saying is that e^x is the sum and

goes from 0 to infinity of x^n / n!

[SOUND]

Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.