“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)

45 ratings

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

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!.

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]

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