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Multiply.

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[MUSIC]

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

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[UNKNOWN]

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

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

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

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

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We can also multiply power series.

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Well, suppose, I want to multiply

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

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

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

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

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So I can write out the first few terms
here.

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That's a sub 0

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

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

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

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

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

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Well, I can try to think about how these

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terms might combine to give me various
powers of x.

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

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

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

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

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And then what's

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the next coefficient in the product?

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Well, I could think about how can I get
just

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

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

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

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

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

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1 times b sub 0.

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

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What about the x squared term, right?

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

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

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

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And a sub 2x squared times b sub 0 will

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

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

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

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are all different ways that I might get an
x squared

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when I multiply together these two power
series and

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then it would keep on going.

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The trouble is that the coefficients get
kind of complicated.

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

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Where's the pattern?

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

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

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

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

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Here look

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

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

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

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

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

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

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

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[LAUGH]

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

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

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this coefficient in front of x to the n.

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

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So imagining that I've got two functions,

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[INAUDIBLE]

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

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

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

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

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Now these power series might have
different radius of convergence.

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

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radii of convergence.

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

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

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

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

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the value of f of x times g of x,

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right?

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

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

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

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

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

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Now f and g have these power series and
the

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

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

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

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The radius of convergence of this power
series is

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

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

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[LAUGH]

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

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

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

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And

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[INAUDIBLE]

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

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These coefficients have indices that add
up to n.

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So certainly when I think about
multiplying a, a

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piece of this power series and a piece of
this

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power series, these are the terms that I
would

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expect to get in front of x to the n.

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

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

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

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

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

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

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

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

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

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

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2 factorial which is x squared over 2,
plug in

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

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

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

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Alright, lets e to the x squared.

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

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

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

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

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

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But I can also think about this by
multiplying

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the original power series by itself,
right?

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What do I get?

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

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Well I can think about the constant term,
I

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get the constant term by multiplying 1
times 1.

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

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

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

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

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

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

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

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And what's the next term, right?

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

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

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1 times x cubed

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

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

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

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

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And then I could keep on going.

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[SOUND]