“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

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

Fibonacci.

[SOUND] Long time ago,

we met the Fibonacci numbers.

So that's a sequence that goes, I will start with 0 and then 1.

And then each subsequent term is the sum of the previous 2, so 0 plus 1 is 1.

1 plus 1 is 2, 1 plus 2 is 3, 2 plus 3 is 5,

3 plus 5 is 8, 5 plus 8 is 13, and then it keeps on going.

I'd like a formula for the Fibonacci numbers.

So in this case, a0 is 0,

a1 is 1, a2 is 1.

A3, 0, 1, 2, 3, that's 2.

A4 is 3, and a5, that's 0, 1,

2, 3, 4, 5, a5 is 5.

And what we're looking for is a formula for the nth term in terms of n.

I want to know that an is equal to something but

I want the something to involve ns.

And maybe you're thinking don't we already have a formula?

You're right, there's this formula an is an-1 + an-2.

This is the formula that we use to define the Fibonacci Sequence.

Each term is the sum of previous 2.

The trouble is that this isn't a very useful formula right to calculate

the thousands term using this formula.

I'd have to calculate the 999th term and the 998th term.

And we're looking for some kind of formula that just involves the usual kinds of

maths function that I'm used to maybe powers multiplying, adding,

subtracting and by that sort of things.

And the number n, I don't want a formula that depends on my calculating all

the previous terms to calculate the next term.

Is that even possible?

How am I going to go about finding such a formula?

Here's the trick.

The trick is to assemble the Fibonacci Sequence into a power series.

What do I mean?

We should think about this function, function f(x) and

it will be the sum n goes from 0 to infinity of an x to the n, but

here these ans are the Fibonacci numbers.

So a0 is 0, but I'm not going to write that, but

a1 the first Fibonacci number is 1.

So it's x a sub 2 next Fibonacci number is also 1 so

+ x squared a3, the third Fibonacci number is 2, so 2x cubed.

The next Fibonacci number is 3, so 3x to the 4th,

the next Fibonacci number is 5, so 5x to the 5th then it'll keep on going.

Now, you can set up an equation that f satisfies.

Well here's how it goes, we're going to multiply f(x) by x.

So x times f(x) is what?

Well I multiply this by x, the x gives me an x squared,

the x squared gives me an x cubed, the 2x cubed gives me a 2x to the 4th.

The 3x to the 4th gives me a 3x to the 5th, and we keep on going.

And then I'm going to multiply this by x squared,

so what's x squared times f(x).

Well then x times x squared gives me an x cubed,

x squared times x squared gives me an x to the 4th.

2x cube times x squared gives me a 2x to the 5th and I'm going to keep on going.

Now I'll do some subtraction, so

we got f(x) minus x times f(x) minus x squared times f(x) and

that's equal to will be x survives, x squared minus x squared that cancels.

2x cubed minus x cubed minus x cubed that cancels.

3x to the 4th minus 2x to the 4th that cancels.

5x to the 5th minus 3x to the 5th minus 2x to the 5th that cancels.

Well everything cancels, and that's not a coincidence.

The defining feature of this sequence of numbers the coefficients an,

it's a Fibonacci Sequence.

So each number is the sum of the previous 2, 5 is 3 plus 2.

The next number over here is what?

It's 8, and 8 is 5 plus 3 because each

term in a Fibonacci Sequence is the sum of the previous 2.

I've rigged it so that f(x) minus x f(x) minus x squared f(x) is just equal to x.

What does that even mean?

So just totally ignoring issues of conversions,

just thinking about this entirely formally.

What I've got is that f(x)-x times f(x)-x2

times f(x) is just x, where f is the function.

Given my power series use coefficients are the Fibonacci numbers.

I'll factor out f(x),

got f(x) times (1-x-x squared) = x.

Now the divide both sides by this.

So f(x) is x over 1-x-x squared.

Here's the plan.

I'm going to find another power series for that rational function and

then I'm going to equate the two power series.

The new power series that I found and the old power series,

the coefficients of which are just a Fibonacci numbers.

It turns out that when those two power series are equal

their coefficients are equal.

And that way, I'll get a formula for the Fibonacci numbers.

Now I'm going to rewrite this function in, and

I think will initially will seem like a much more complicated way.

First, I'm going to define this other number that's actually traditionally

called phi, and it's the golden ratio.

Phi is 1 + the square root of 5 over 2.

So that's this number phi, and

I'm going to use phi to rewrite this function f.

So it turns out that after quite a lengthy calculation

you convince yourself that this 1 over the square

root of 5 divided by 1-x times phi + -1 over

square root of 5 divided by 1-x times not phi but 1-phi.

How is that better?

It looks like I just made a total mess of things.

Well the trick now is that I can write these each as power series.

Here we go.

This is 1 over the square root of 5,

times 1 over 1-x times phi + negative

1 over square root of 5 times 1 over 1-x times 1-phi.

But I have a power series for 1 over 1 minus something.

So this is 1 over the squared of 5 times the sum,

n goes from 0 to infinity of x times phi

of the nth power + -1 over the square

root of 5 times the sum n goes from 0 to

infinity of x times 1-phi to the nth power.

Now write it as a single series.

So this is 1 over the square root of 5 times

the sum n goes to 0 infinity about this is phi to the n times x to the n.

Plus the -1 over the square root of 5 times the sum and

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

And how I'm going to write this just a single power series,

I'm going to collect together with the coefficients here in front of x to the n.

So this is the sum n goes from 0 to infinity

of 1 over the square root of 5 times phi to

the n-1 over the square root of 5 times

1-phi to the n, all of this times x to the n.

I can simplify that a bit.

Yeah, I could write this as the sum n goes from 0

to infinity of phi to the n-1-phi to the n over

the square root of 5 times x to the nth power.

And here's the punchline,

well admittedly we haven't been concerned about conversions at all.

But at least remember where this came from.

We derived this through a series of perhaps unjustified operations.

But we started with just this power series where these coefficients were

the Fibonacci numbers.

Now I've got a new power series.

And the upshot or what I'm hoping here if these two power series are equal

then there coefficients must be equal.

And that means that this is a formula for this,

this is a formula for the nth Fibonacci number.

For real, that is a formula for the Fibonacci numbers.

Well let's try it, here's the 100th Fibonacci number.

It's this pretty big number.

So then the idea here, is that instead of trying to calculate a100 by hand,

I'm just going to plug in n = 100, and computer this coefficient.

Let's estimate that without using a calculator.

Well instead of writing 1.6, that's about what phi is, let's write 16/10.

Instead of writing 16/10, let's approximate phi by 16/10 written

this way to the 4th for 16 so 2 to the 4th over 10.

But if I write it that way it makes it a little bit easier to do the estimation.

Because 2 to the 4th over 10 to the 100th power well that's 2 to

the 400th over 10 to the 100th.

But now 2 to 400th, I can write that as 2 to the 10th to the 40th.

And that's a smart move because 2 to the 10th I happen to know is 1,024.

So 2 to the 10th is practically 10 to the 3rd.

It's about 1,000, but 10 to the 3rd to the 40th, well that's 10 to the 120th.

So all together, the 100th Fibonacci number is about 10

to the 120th power divided by 10 to the 100th power.

All together then,

I can guess that the 100th Fibonacci number is about 10 to the 20th.

Is that even close to being accurate?

Well the actual retail value, if you remember, here it is.

Here is the 100th Fibonacci number.

And although it is kind of a pain to count the digits,

that 100th Fibonacci number is approximately 3.54x times 10 to the 20th.

Rock on rock star.

We have estimated the 100th Fibonacci number armed with nothing, but our wits.

Now this technique of analyzing a sequence by building the associative power series,

that technique has a name.

And this sort of trick where I analyzed a sequence of numbers.

In this case the Fibonacci sequence, by building a power series and playing around

with it like this and in the most formal way, this is called generating functions.

One reason why this is a powerful technique is that it lets you carry

a problem from one area of mathematics into a whole nother are of mathematics.

We started with a very common feeling problemm, the discrete problem,

a problem about sequences.

And by applying this method of generating functions,

we transported that problem into the realm of calculus.

And I think this shows mathematics at its deepest levels is not a collection of

disconnected subjects.

Mathematics is a single unified whole.

And all of these different areas of mathematics are connected together.

[SOUND]

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