Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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From the course by The Ohio State University

Calculus Two: Sequences and Series

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Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Power Series

In this fifth module, we study power series. Up until now, we had been considering series one at a time; with power series, we are considering a whole family of series which depend on a parameter x. They are like polynomials, so they are easy to work with. And yet, lots of functions we care about, like e^x, can be represented as power series, so power series bring the relaxed atmosphere of polynomials to the trickier realm of functions like e^x.

- Jim Fowler, PhDProfessor

Mathematics

Multiply.

[MUSIC]

We can multiply polynomials. Yeah, I mean, I have

[UNKNOWN]

polynomials, you know, 1 plus x squared say maybe multiply that times 3 minus x.

Well, I can multiply polynomials and get a new polynomial, right?

1 times 3 minus x plus 3x squared minus x cubed.

It's what I get when I multiply these two polynomials.

We can also multiply power series.

Well, suppose, I want to multiply

the sum, n goes from 0 to infinity, of a sub n times x to the

n by another power series, maybe the sum, n goes from 0 to

infinity, of b sub n x to the n. Well, how am I going to get started?

Well, one way to at least get started on this is to just expand this out, alright?

So I can write out the first few terms here.

That's a sub 0

plus a sub 1x plus a sub 2x squared and that will keep on going.

And then I want to multiply that by, I'll write then the first

few terms of this blue power series with the b sub n's.

So b sub 0 plus b sub 1x plus b sub 2x squared and

it would keep on going. And what do I get when I multiply these?

Well, I can try to think about how these

terms might combine to give me various powers of x.

So I have to pick something here and multiply it by something here.

So the only way that I get a constant term, a term without

an x, is that I multiply a sub 0 by by b sub 0.

So I can start by writing that down, a sub 0 times b sub 0.

And then what's

the next coefficient in the product?

Well, I could think about how can I get just

a linear term, a term with just an x in it.

And there's two different ways I could do that.

I can multiply a sub 0 by b sub 1x or I could multiply

a sub 1x by b sub 0. So I should write down both of those

terms. So a sub 0 times b sub 1, plus a sub

1 times b sub 0.

And those are the only ways that I can get a term just a single x.

What about the x squared term, right?

Well, there is actually three different ways I get an x squared term.

A sub 0 times b sub 2x squared would give me an x squared.

A sub 1x times b sub 1x would give me an x squared.

And a sub 2x squared times b sub 0 will

also give me an x squared. So let me write down all three of those.

So I've got a sub 0 times b sub 2, plus a sub 1

times b sub 1, plus a sub 2 times b sub 0, and those

are all different ways that I might get an x squared

when I multiply together these two power series and

then it would keep on going.

The trouble is that the coefficients get kind of complicated.

Of course, not all hopes is lost, there is a pattern in here.

Where's the pattern?

Well here, we've got the constant term that x to

the 0 term if you like, and these are both 0.

Here I've got the x to the first term and these

indices add up to 1, 0 plus 1 and 1 plus 0.

Here look

at the x squared term and these indices also all add up to 2.

0 plus 2 is 2, 1 plus 1 is 2, 2 plus 0 is 2.

And you might guess them, well, what's the coefficients on x cubed?

It's going to be combinations of the a sub n's and

the b sub n's where the indices add up to 3.

We'll try it down at least what our guess is then for the formula in general.

So this will be the sum, n goes from 0 to infinity, of

[LAUGH]

another series, the sum i goes from 0 to n,

of a sub i, times b sub n minus i, and it's

this coefficient in front of x to the n.

I, I need to say a little bit more about what this even means.

So imagining that I've got two functions,

[INAUDIBLE]

I got a function little f, which is given as the sum,

n goes from 0 to infinity of a sub n x to the

n, and I've got a function little g, which is the sum, n

goes from 0 to infinity of b sub n x to the n.

Now these power series might have different radius of convergence.

So let's just have big R, be the minimum of their

radii of convergence.

It isn't just that the series with the convolved coefficients, converges.

I mean, I've got this product series but I'm not just saying that

this series converges when the absolute value of x is less than R.

Right, I'm actually saying that this series converges to

the value of f of x times g of x,

right?

I'm making it claim that f of x times g of x is equal to the value of this series.

Now, you put it all together. Well, here's a precise theorem.

So I got two functions, a function f and a function g.

F of x is the sum, n goes from 0 to infinity of a sub n x to the n

and g of x is the sum, n goes from 0 to infinity of b sub n x to the n.

Now f and g have these power series and the

power series have some radius of convergence, and I'll assume that both of

those are radii of convergence, is greater than or equal to big R.

Well then I've got a new power series here.

The radius of convergence of this power series is

at least big R and here's what I know.

F of x times g of x is equal to the sum n goes from 0 to infinity and

[LAUGH]

it's this weird convolved coefficient, the sum, little i

from 0 to little n, of a sub i b sub n minus i, times x to the n and

this equality that leads towards when x is between minus R and R.

And

[INAUDIBLE]

sort of reasonable looking I mean look what's going on here.

These coefficients have indices that add up to n.

So certainly when I think about multiplying a, a

piece of this power series and a piece of this

power series, these are the terms that I would

expect to get in front of x to the n.

I should warn you that were not going to prove this result,

but I hope its possible and I hope you will play around

with it, try to get a sense of some of the consequences of this theorem.

I mean here is one example, kind of cool from what you can do with it.

For example, we've thought a little bit about e to the x

and e to the x has this really nice power series representation.

It's the sum, n goes from 0 to infinity of x to the

n over n factorial, and I can write out the first few terms.

If I plugin n equals 0, got 1, plugin n equals 1, I've just got

x, plugin n equals 2, I've got x squared over

2 factorial which is x squared over 2, plug in

n equals 3, I've got x to the third over

3 factorial which is 6 and I plus dot dot dot.

Now I can think about what happens when I multiply this power series by itself.

Alright, lets e to the x squared.

Because I secretly know what the answer is, right, just because

of how exponents work, e to the x squared is e to the 2x.

So if it should be 1 plus, I am going to replace all these x's by 2x,

2x plus 2x quantity squared over 2 is 2x squared, quantity

2x cubed divided by 6 is 8x cubed over 6 and then plus dot dot.

But I can also think about this by multiplying

the original power series by itself, right?

What do I get?

So 1 plus x plus x squared over 2 plus dot, dot, dot, squared, right.

Well I can think about the constant term, I

get the constant term by multiplying 1 times 1.

How do I get the next term, the term in front of the x?

Well, that's 1 times x or x times 1, so that's 2x.

How do I get the term in front of x squared?

Well, there's three different ways to get that.

1 times x squared over a half, so a half.

X times x, so 1, or x squared over 2 times 1, so put a half here.

And sure enough, a half plus 1 plus a half is 2.

And what's the next term, right?

I'm looking for the coefficient in front of x cubed.

Well, there's four different ways to get x cubed.

1 times x cubed

over 6, so a 6. X times x squared over 2, so it's a half.

X squared over 2 times x, that's a half.

And then x cubed over 6 times 1, that's a sixth, and then sure

enough, a sixth plus a half, plus a half plus a sixth, that's 8 sixths.

And then I could keep on going.

[SOUND]

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