“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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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

Radius of convergence.

[MUSIC]

Let's think about a not too mysterious function.

Let's think about the function f of x equals 1 over 1 plus x squared.

I can write down a power series for this function around 0.

So, I'll begin with a power series for 1 over 1 minus x.

That's the sum n goes from 0 to infinity of x to the n.

Now, I'll replace x by negative x.

And then, I'll find that 1 over 1 plus x.

Is the sum n goes from 0 to infinity of -x to the n, which I could write

as the sum n goes from 0 to infinity of -1 to the n times x to the n.

And now, to get from 1 over 1+x to 1 over 1+x squared,

I'll just replace x by x squared.

So 1 over 1 plus x squared is the sum and

goes from 0 to infinity of negative 1 to the n times x squared to the n.

Which I could also write as the sum and

goes from 0 to infinity of negative 1 to the n times x to the 2n power.

What's the radius of convergence of that power series?

I was being super sloppy when I wrote this out.

I wrote 1/1-x equals this power series, but

that's not true unless I include this statement.

I should have written this down, if the absolute value of x is less than one and

if that's true, then these are equal.

The radius of convergence of this series for 1/1+x squared is also 1.

And we can see that in the graph.

Here I've graphed the function 1 over 1+x squared, and

take a look at its Taylor series expansion around a 0.0.

Here's just the first term which is the constant term.

Here's through the quadratic term, here's through the x to the 4th term,

through the x to the 6th term, and so on.

And you can see that it's doing an increasingly good job of approximating

the function between minus one and one.

But over here it's actually doing an increasingly bad job.

1/1+x squared looks like a really nice function.

So if it's such a nice function why isn't the Taylor series centered around 0

doing a better job of approximating the function far away?

Well here I've graphed y equals sine of x.

And honestly, I mean qualitatively at least,

this isn't so different looking from the graph of 1/1+x squared.

The graph of 1/1+x squared had kind of a big bump in the middle,

and this thing's just got lots of bumps.

And yet the Taylor series is just totally different experience.

Let's start writing down the Taylor series around the origin.

So here's the point around which I'm expanding.

Here's just the constant term in red.

Here's to the linear term.

Here's to the cubic term.

And I'm going to keep on going.

And as I add more and more terms in my Taylor series expansion I'm getting

increasingly good approximations to sine of x everywhere.

The sine of x is equal to its Taylor series centered around 0 for

all values of x.

Whereas 1/1+x squared, its Taylor series around

0 is only equal to 1/1+x squared if x is less than 1.

And we know other examples like that.

Like this, this is the graph y=1/1-x.

And I can start right around the Taylor series around this point.

Here's the constant term.

The linear term.

Take more and more terms and yeah.

I mean, the radius of convergence here is 1.

But maybe that isn't so surprising.

I mean, 1/1-x has a bad point at 1, right?

I don't want to plug in x=1 because then I'll be dividing by 0.

This function is not defined at 1.

So if I start a Taylor series around 0 and

I imagine how big do I really expect that radius of convergence to be?

Well, you know an interval of radius 1 centered around 0 bumps into the problem

point.

So maybe I don't really expect the radius of convergence to be more than 1 at 0.

So maybe 1/1-x has a problem when x equals 1.

But 1/1+x squared doesn't have any problem at 1.

It doesn't have any problem anywhere.

This function is defined for all x.

So if 1/1+x squared looks just as nice as sine of x, why is the Taylor series for

this function centered around 0 so much worse than the Taylor series for

the sine of x centered around 0?

Taylor series for sine of x converges to sine of x everywhere and

the Taylor series for 1 / 1 + x squared is only good in an interval of radius 1.

Well let's see what happens if we write down a Taylor series expansion for

1 / 1 + x squared but not centered around 0 but centered around some other point.

Let's write in the Taylor series expansion center out x = 1 and see what happens.

So here we go, there's the constant term, there's the linear terms.

And we keep on going, we add more and more terms to the Taylor series expansion

centered around 1 and I mean we're doing an increasingly good job but

not over the whole real line again.

But at least the radius of convergence is bigger now.

Well how much bigger?

The Taylor series for 1 / 1 + x squared centered at 0 has radius 1.

The Taylor series centered at 1 turns out to have radius the square root of 2.

And we had this idea that maybe places where the function is undefined,

these bad points where the function doesn't exist,

maybe those are somehow to blame for the radius of convergence.

And that's what happened in the case of 1 / 1- x.

The radius of convergence around x equals 0 was 1 because this function has it's

bad point at x equals 1.

Function is not defined if I plug in x equals 1.

In the end, bumping into a bad point here too.

Well let's diagram the situation this way.

Here's the real line and around 0,

I've got this, well it's an interval, but I've drawn it as a circle.

It's a circle of radius 1, so that's representing the radius of convergence

of my Taylor series around the point 0 for the function 1 / 1 + x squared.

Now around the point 1, all right.

I've got a different radius and

convergence that it's bigger it's a square root of two.

So here is a circle of radius the square root of 2.

And I'm placing it so that its center is at 1.

And the idea here is to try to get a sense of whether or

not maybe I'm bumping into a bad point.

Is those some point on the real line which is distance 1 from 0 and

distance the square root of 2, from 1.

But there isn't such a point on the real line.

Okay but when I draw the circles, the circles are touching at these two points.

So where is that point?

Well those points are in the complex plane.

That point there is i and that point there is -i.

Those are square roots of minus 1.

Those are imaginary numbers.

Well what happens if I evaluate 1/1+x squared when x equals i or x equals -i?

So i squared is -1.

So what's f of i, well f is 1/1 plus its input squared,

so that would be 1/1 plus what's i squared, its -1.

1/1 + -1, that is not define, that divide by 0.

So there is bad point.

There is a point where the function is undefined.

It's just that the bad point for this function isn't a real point.

It's an imaginary input.

Right, if I evaluate this function i or -i, it's undefined.

And yet that bad point in the complex plane is messing up my radius and

convergence even along the real line.

We're beginning to get a glimpse of the important role that complex numbers play

even in the theory of just real value, the Taylor Series.

Additional evidence comes for example by this equation, e to ix = cos x + i sin x.

And again, i is a square root of- 1 and

you can interpret this through this statement about power series.

So here I've written down the Taylor series for e to the x but

with x replaced by ix.

Here, I've just written down cosine of x and here I've written down sine x but

I multiplied by i.

And you can expand this out using the fact that i squares to -1 to get this equality.

And you can do kind of fun things like substitute in, say pi for x and

conclude that e to the i pi is cosine pi plus i sine pi.

So cosine pi is minus 1, sine pi is 0 so e to the i pi is negative 1.

Those series aren't just a jumping off point for calculus and for

miracle approximations.

Taylor's theories area also our first step into the theory of complex analysis and

that theory is more natural that it might seem at at first.

I mean if the complex numbers are affecting the radius of convergence of my

real power series then they must be really there.

I mean they are not, as imaginary as people might think.

So, Taylor series aren't just a jumping off point for our calculus and for

numerical approximations.

Taylor series are also our first step into the theory of complex analysis and

that theory is more natural than it might seem at first.

I mean, if the complex numbers are affecting the radius of convergence of my

real power series, then they must be really there.

I mean they're not as imaginary

as people might think.

[SOUND]

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