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Today is going to be one of the
more difficult lectures of the

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term.
So, put on your thinking caps,

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as they would say in elementary
school.

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The topic is going to be what's
called a convolution.

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The convolution is something
very peculiar that you do to two

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functions to get a third
function.

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It has its own special symbol.
f of t asterisk is the

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universal symbol that's used for
that.

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So, this is a new function of
t, which bears very little

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resemblance to the ones,
f of t, that you started with.

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I'm going to give you the
formula for it,

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but first, there are two ways
of motivating it,

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and both are important.
There is a formal motivation,

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which is why it's tucked into
the section on Laplace

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transform.
And, the formal motivation is

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the following.
Suppose we start with the

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Laplace transform of those two
functions.

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Now, the most natural question
to ask is, since Laplace

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transforms are really a pain to
calculate is from old Laplace

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transforms, is it easy to get
new ones?

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And, the first thing,
of course, summing functions is

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easy.
That gives you the sum of the

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transforms.
But, a more natural question

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would be, suppose I want to
multiply F of t and G

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of t.
Is there, hopefully,

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some neat formula?
If I multiply the product of

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the, take the product of these
two, is there some neat formula

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for the Laplace transform of
that product?

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That would simply life greatly.
And, the answer is,

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there is no such formula.
And there never will be.

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Well, we will not give up
entirely.

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Suppose we ask the other
question.

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Suppose instead I multiply the
Laplace transforms.

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Could that be related to
something I cook up out of F of

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t and G of t?
Could it be the transform of

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something I cook up out of F of
t or G of t?

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And, that's what the
convolution is for.

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The answer is that F of s times
G of s turns out

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to be the Laplace transform of
the convolution.

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The convolution,
and that's one way of defining

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it, is the function of t you
should put it there in order

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that its Laplace transform turn
out to be the product of F of s

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times G of s.
Now, I'll give you,

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in a moment,
the formula for it.

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But, I'll give you one and a
quarter minutes,

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well, two minutes of motivation
as to why there should be such

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formula.
Now, I won't calculate this out

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to the end because I don't have
time.

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But, here's the reason why
there should be such formula.

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And, you might suspect,
and therefore it would be worth

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looking for.
It's because,

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remember, I told you where the
Laplace transform came from,

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that the Laplace transform was
the continuous analog of a power

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series.
So, when you ask a general

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question like that,
the place to look for is if you

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know an analogous idea,
say, does it work.

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something like that work there?
So, here I have a power series

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summation, (a)n x to the n.

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Remember, you can write this in
computer notation as a of n

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to make it look like f
of n,

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f of t.
And, the analog is turned into

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t when you turn a power series
into the Laplace transform,

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and x gets turned into e to the
negative s,

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and one formula just turns into
the other.

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Okay, so, there's a formula for
F of x.

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This is the analog of the
Laplace transform.

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And, similarly,
G of x here is

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summation (b)n x to the n.

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Now, again, the naīve question
would be, well,

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suppose I multiply the two
corresponding coefficients

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together, and add up that power
series, summation (a)n (b)n

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times x to the n.

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Is that somehow,
that sum related to F and G?

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And, of course,
everybody knows the answer to

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that is no.
It has no relation whatever.

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But, suppose instead I multiply
these two guys.

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In that case,
I'll get a new power series.

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I don't know what its
coefficients are,

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but let's write them down.
Let's just call them (c)n's.

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So, what I'm asking is,
this corresponds to the product

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of the two Laplace transforms.
And, what I want to know is,

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is there a formula which says
that (c)n is equal to something

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that can be calculated out of
the (a)i and the (b)j.

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Now, the answer to that is,
yes, there is.

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And, the formula for (c)n is
called the convolution.

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Now, you could figure out this
formula yourself.

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You figure it out.
Anyone who's smart enough to be

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interested in the question in
the first place is smart enough

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to figure out what that formula
is.

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And, it will give you great
pleasure to see that it's just

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like the formula for the
convolution of going to give you

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now.
So, what is that formula for

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the convolution?
Okay, hang on.

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Now, you are not going to like
it.

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But, you didn't like the
formula for the Laplace

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transform, either.
You felt wiser,

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grown-up getting it.
But it's a mouthful to swallow.

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It's something you get used to
slowly.

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And, you will get used to the
convolution equally slowly.

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So, what is the convolution of
f of t and g of t?

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It's a function calculated

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according to the corresponding
formula.

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It's a function of t.
It is the integral from zero to

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t of f of u, --
u is a dummy variable because

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it's going to be integrated out
when I do the integration,

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g of (t minus u) dt.

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That's it.

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I didn't make it up.
I'm just varying the bad news.

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Well, what do you do when you
see a formula?

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Well, the first thing to do,
of course, is try calculating

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just to get some feeling for
what kind of a thing,

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you know.
Let's try some examples.

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Let's see, let's calculate,
what would be a modest

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beginning?
Let's calculate the convolution

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of t with itself.
Or, better yet,

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let's calculate the convolution
just so that you could tell the

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difference, t with t squared,
t squared with t,

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to make it a little easier.
By the way, the convolution is

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symmetric.
f star g is the same thing as g

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star f.
Let's put that down explicitly.

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I forgot to last period.
So, tell all the guys who came

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to the one o'clock lecture that
you know something that they

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don't.
Now, that's a theory.

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It's commutative.
This operation is commutative,

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in other words.
Now, that has to be a theorem

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because the formula is not
symmetric.

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The formula does not treat f
and g equally.

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And therefore,
this is not obvious.

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It's at least not obvious if
you look at it that way,

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but it is obvious if you look
at it that way.

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Why?
In other words,

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f star g is the guy
whose Laplace transform is F of

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s times G of s.
Well, what would g star f?

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That would be the guy whose

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Laplace transform is G
times F.

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But capital F times capital G
is the same as capital G times

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capital F.
So, it's because the Laplace

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transforms are commutative.
Ordinary multiplication is

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commutative.
It follows that this has to be

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commutative, too.
So, I'll write that down,

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since F times G is equal to GF.
And, you have to understand

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that here, I mean that these are
the Laplace transforms of those

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guys.
But, it's not obvious from the

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formula.
Okay, let's calculate the

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Laplace transform of,
sorry, the convolution of t

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star,
let's do it by the formula.

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All right, by the formula,
I calculate integral zero to t.

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Now, I take the first function,
but I change its variable to

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the dummy variable,
u.

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So, that's u squared.
I take the second function and

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replace its variable by u minus
t.

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So, this is times t minus u,
sorry.

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Okay, do you see that to
calculate this is what I have to

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write down?
That's what the formula

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becomes.
Anything wrong?

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Oh, sorry, the du,
the integration's with expect

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to u, of course.
Thanks very much.

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Okay, let's do it.
So, it is, integral of u

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squared t is,
remember, it's integrated with

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respect to u.
So, it's u cubed over three

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times t.
The rest of it is the integral

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of u cubed,
which is u to the forth over

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four.
All this is to be evaluated

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between zero and t at the upper
limit.

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So, I put u equal t,
I get t to the forth over three

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minus t to the forth
over four.

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Of course, at the lower limit,
u is zero.

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So, both of these are terms of
zero.

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There's nothing there.
And, the answer is,

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therefore, t to the forth
divided by,

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a third minus a quarter is a
twelfth.

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So, that's doing it from the
formula.

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But, of course,
there is an easier way to do

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it.
We can cheat and use the

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Laplace transform instead.
If I Laplace transform it,

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the Laplace transform of t
squared is what?

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It's two factorial divided by s
cubed.

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The Laplace transform of t

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is one divided by s
squared.

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And so, because this is the
convolution of these,

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it should correspond to the
product of the Laplace

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transforms, which is two over s
to the 5th power.

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Well, is that the same as this?
What's the Laplace transform

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of, in other words,
what's the inverse Laplace

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transform of two over s to the
fifth?

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Well, the inverse Laplace
transform of four factorial over

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s to the fifth is how much?

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That's t to the forth,
right?

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Now, how does this differ?
Well, to turn that into that,

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I should divide by four times
three.

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So, this should be one twelfth
t to the forth,

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one over four times three
because this is 24,

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and that's two,
so, divide by 12 to determine

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what constant,
yeah.

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So, it works,
at least in that case.

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But now, notice that this is
not an ordinary product.

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The convolution of t squared
and t is not something

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like t cubed.
It's something like t to the

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forth, and there's a funny
constant in there,

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too, very unpredictable.
Let's look at the convolution.

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Let's take another example of
the convolution.

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Let's do something really
humble just assure you that

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this, even at the simplest
example, this is not trivial.

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Let's take the convolution of f
of t with one.

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Can you take,
yeah, one is a function just

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like any function.
But, you get something out of

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the convolution,
yes, yes.

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Let's just write down the
formula.

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00:13:05,000 --> 00:13:11,000
Now, I can't use the Laplace
transform here because you won't

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know what to do with it.
You don't have that formula

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yet.
It's a secret one that only I

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know.
So, let's do it.

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Let's calculate it out the way
it was supposed to.

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So, it's the integral from zero
to t of f of u,

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and now, what do I do with that
one?

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I'm supposed to take,
one is the function g of t,

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and wherever I see a
t, I'm supposed to plug in t

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minus u.
Well, I don't see any t there.

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But that's something for
rejoicing.

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There's nothing to do to make
the substitution.

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It's just one.
So, the answer is,

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00:13:52,000 --> 00:13:58,000
it's this curious thing.
The convolution of a function

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with one, you integrate it from
zero to t.

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Well, as they said in Alice in
Wonderland, things are getting

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curiouser and curiouser.
I mean, what is going on with

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this crazy function,
and where are we supposed to

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start with it?
Well, I'm going to prove this

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for you, mostly because the
proof is easy.

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In other words,
I'm going to prove that that's

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true.
And, as I give the proof,

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you'll see where the
convolution is coming from.

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That's number one.
And, number two,

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the real reason I'm giving you
the proof: because it's a

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00:14:38,000 --> 00:14:44,000
marvelous exercise in changing
the variables in a double

235
00:14:42,000 --> 00:14:48,000
integral.
Now, that's something you all

236
00:14:44,000 --> 00:14:50,000
know how to do,
even the ones who are taking

237
00:14:47,000 --> 00:14:53,000
18.02 concurrently,
and I didn't advise you to do

238
00:14:50,000 --> 00:14:56,000
that.
But, I've arranged the course

239
00:14:52,000 --> 00:14:58,000
so it's possible to do.
But, I knew that by the time we

240
00:14:56,000 --> 00:15:02,000
got to this, you would already
know how to change variables at

241
00:15:00,000 --> 00:15:06,000
a double integral.
So, and in fact,

242
00:15:04,000 --> 00:15:10,000
you will have the advantage of
remembering how to do it because

243
00:15:10,000 --> 00:15:16,000
you just had it about a week or
two ago, whereas all the other

244
00:15:15,000 --> 00:15:21,000
guys, it's something dim in
their distance.

245
00:15:19,000 --> 00:15:25,000
So, I'm reviewing how to change
variables at a double integral.

246
00:15:25,000 --> 00:15:31,000
I'm showing you it's good for
something.

247
00:15:29,000 --> 00:15:35,000
So, what we are out to try to
prove is this formula.

248
00:15:33,000 --> 00:15:39,000
Let's put that down in,
so you understand.

249
00:15:39,000 --> 00:15:45,000
Okay, let's do it.
Now, we'll use the desert

250
00:15:41,000 --> 00:15:47,000
island method.
So, you have as much time as

251
00:15:44,000 --> 00:15:50,000
you want.
You're on a desert island.

252
00:15:46,000 --> 00:15:52,000
In fact, I'm going to even go
it the opposite way.

253
00:15:49,000 --> 00:15:55,000
I'm going to start with--
you've got a lot of time on your

254
00:15:53,000 --> 00:15:59,000
hands and say,
gee, I wonder if I take the

255
00:15:56,000 --> 00:16:02,000
product of the Laplace
transforms, I wonder if there's

256
00:15:59,000 --> 00:16:05,000
some crazy function I could put
in there, which would make

257
00:16:03,000 --> 00:16:09,000
things work.
You've never heard of the

258
00:16:06,000 --> 00:16:12,000
convolution.
You're going to discover it all

259
00:16:09,000 --> 00:16:15,000
by yourself.
Okay, so how do you begin?

260
00:16:11,000 --> 00:16:17,000
So, we'll start with the left
hand side.

261
00:16:14,000 --> 00:16:20,000
We're looking for some nice way
of calculating that as the

262
00:16:17,000 --> 00:16:23,000
Laplace transform of a single
function.

263
00:16:19,000 --> 00:16:25,000
So, the way to begin is by
writing out the definitions.

264
00:16:23,000 --> 00:16:29,000
We couldn't use anything else
since we don't have anything

265
00:16:26,000 --> 00:16:32,000
else to use.
Now, looking ahead,

266
00:16:28,000 --> 00:16:34,000
I'm going to not use t.
I'm going to use two neutral

267
00:16:32,000 --> 00:16:38,000
variables when I calculate.
After all, the t is just a

268
00:16:35,000 --> 00:16:41,000
dummy variable anyway.
You will see in a minute the

269
00:16:40,000 --> 00:16:46,000
wisdom of doing this.
So, it's this times the

270
00:16:44,000 --> 00:16:50,000
integral, which gives the
Laplace transform of g.

271
00:16:48,000 --> 00:16:54,000
So, that's e to the negative s
v, let's say,

272
00:16:52,000 --> 00:16:58,000
times g of v,
dv.

273
00:16:53,000 --> 00:16:59,000
Okay, everybody can get that
far.

274
00:16:56,000 --> 00:17:02,000
But now we have to start
looking.

275
00:17:00,000 --> 00:17:06,000
Well, this is a single
integral, an 18.01 integral

276
00:17:03,000 --> 00:17:09,000
involving u, and this is an
18.01 integral involving v.

277
00:17:07,000 --> 00:17:13,000
But when you take the product
of two integrals like that,

278
00:17:11,000 --> 00:17:17,000
remember when you evaluate a
double integral,

279
00:17:14,000 --> 00:17:20,000
there's an easy case where it's
much easier than any other case.

280
00:17:19,000 --> 00:17:25,000
If you could write the inside,
if you are integrating over a

281
00:17:23,000 --> 00:17:29,000
rectangle, for example,
and you can write the integral

282
00:17:27,000 --> 00:17:33,000
as a product of a function just
of u, and a product of a

283
00:17:31,000 --> 00:17:37,000
function just as v,
then the integral is very easy

284
00:17:34,000 --> 00:17:40,000
to evaluate.
You can forget all the rules.

285
00:17:38,000 --> 00:17:44,000
You just take all the u part
out, all the v part out,

286
00:17:42,000 --> 00:17:48,000
and integrate them separately,
a to b, c to d.

287
00:17:44,000 --> 00:17:50,000
That's the easy case of
evaluating a double integral.

288
00:17:48,000 --> 00:17:54,000
It's what everybody tries to
do, even when it's not

289
00:17:51,000 --> 00:17:57,000
appropriate.
Now, here it is appropriate,

290
00:17:53,000 --> 00:17:59,000
except I'm going to use it
backwards.

291
00:17:56,000 --> 00:18:02,000
This is the result of having
done that.

292
00:17:58,000 --> 00:18:04,000
If this is the result of having
done it, what was the step just

293
00:18:02,000 --> 00:18:08,000
before it?
Well, I must have been trying

294
00:18:06,000 --> 00:18:12,000
to evaluate a double integral as
u runs from zero to infinity and

295
00:18:10,000 --> 00:18:16,000
v runs from zero to infinity,
of what?

296
00:18:13,000 --> 00:18:19,000
Well, of the product of these
two functions.

297
00:18:16,000 --> 00:18:22,000
Now, what is that?
e to the minus s u times e to

298
00:18:20,000 --> 00:18:26,000
the minus s v.

299
00:18:22,000 --> 00:18:28,000
Well, I must surely want to
combine those.

300
00:18:25,000 --> 00:18:31,000
e to the minus s u times e to
the minus s v.

301
00:18:30,000 --> 00:18:36,000
And, what's left?
Well, what gets dragged along?

302
00:18:33,000 --> 00:18:39,000
du dv.
This is the same as that

303
00:18:35,000 --> 00:18:41,000
because of that law I just gave
you this is the product of a

304
00:18:39,000 --> 00:18:45,000
function just of u,
and a function just of v.

305
00:18:42,000 --> 00:18:48,000
And therefore,
it's okay to separate the two

306
00:18:45,000 --> 00:18:51,000
integrals out that way because
I'm integrating sort of a

307
00:18:49,000 --> 00:18:55,000
rectangle that goes to infinity
that way and infinity that way.

308
00:18:54,000 --> 00:19:00,000
But, what I'm integrating is
over the plane,

309
00:18:57,000 --> 00:19:03,000
in other words,
this region of the plane as u,

310
00:19:00,000 --> 00:19:06,000
v goes from zero to infinity,
zero to infinity.

311
00:19:05,000 --> 00:19:11,000
Now, let's take a look.
What are we looking for?

312
00:19:10,000 --> 00:19:16,000
Well, we're looking for,
we would be very happy if u

313
00:19:15,000 --> 00:19:21,000
plus v were t.
Let's make it t.

314
00:19:20,000 --> 00:19:26,000
In other words,
I'm introducing a new variable,

315
00:19:25,000 --> 00:19:31,000
t, u plus v,
and it's suggested by the form

316
00:19:30,000 --> 00:19:36,000
in which I'm looking for the
answer.

317
00:19:35,000 --> 00:19:41,000
Now, of course you then have
to, we need another variable.

318
00:19:39,000 --> 00:19:45,000
We could keep either u or v.
Let's keep u.

319
00:19:43,000 --> 00:19:49,000
That means v,
we just gave a musical chairs.

320
00:19:46,000 --> 00:19:52,000
v got dropped out.
Well, we can't have three

321
00:19:50,000 --> 00:19:56,000
variables.
We only have room for two.

322
00:19:53,000 --> 00:19:59,000
But, we will remember it.
Rest in peace,

323
00:19:57,000 --> 00:20:03,000
v was equal to t minus u
in case we ever need him

324
00:20:02,000 --> 00:20:08,000
again.
Okay, let's now put in the

325
00:20:05,000 --> 00:20:11,000
limits.
Let's put in the integral,

326
00:20:08,000 --> 00:20:14,000
the rest of the change of
variable.

327
00:20:10,000 --> 00:20:16,000
So, I'm now changing it to
these new variables,

328
00:20:14,000 --> 00:20:20,000
t and u, so it's e to the
negative s t.

329
00:20:18,000 --> 00:20:24,000
Well, f of u I don't
have to do anything to.

330
00:20:22,000 --> 00:20:28,000
But, g of v,
I'm not allowed to keep v,

331
00:20:25,000 --> 00:20:31,000
so v has to be changed to t
minus u.

332
00:20:30,000 --> 00:20:36,000
Amazing things are happening.
Now, I want to change this to

333
00:20:34,000 --> 00:20:40,000
an integral du dt.
Now, for that,

334
00:20:37,000 --> 00:20:43,000
you have to be a little
careful.

335
00:20:40,000 --> 00:20:46,000
We have two things to do to
figure out this;

336
00:20:43,000 --> 00:20:49,000
what goes with that?
And, we have to put in the

337
00:20:47,000 --> 00:20:53,000
limits, also.
Now, those are the two

338
00:20:50,000 --> 00:20:56,000
nontrivial operations,
when you change variables in a

339
00:20:55,000 --> 00:21:01,000
double integral.
So, let's be really careful.

340
00:21:00,000 --> 00:21:06,000
Let's do the easier of the two,
first.

341
00:21:03,000 --> 00:21:09,000
I want to change from du dv to
du dt.

342
00:21:07,000 --> 00:21:13,000
And now, to do that,
you have to put in the Jacobian

343
00:21:12,000 --> 00:21:18,000
matrix, the Jacobian
determinant.

344
00:21:15,000 --> 00:21:21,000
Ah-ha!
How many of you forgot that?

345
00:21:19,000 --> 00:21:25,000
I won't even bother asking.
Oh, come on,

346
00:21:23,000 --> 00:21:29,000
you only lose two points.
It doesn't matter if you put it

347
00:21:28,000 --> 00:21:34,000
in the Jacobian.
As you see, you're going to

348
00:21:34,000 --> 00:21:40,000
forget something.
You will lose less credit for

349
00:21:39,000 --> 00:21:45,000
forgetting than anything else.
So, it's the Jacobian of u and

350
00:21:45,000 --> 00:21:51,000
v with respect to u and t.
So, to calculate that,

351
00:21:49,000 --> 00:21:55,000
you write u equals u,
v equals t minus u,

352
00:21:54,000 --> 00:22:00,000
and then the Jacobian is
the partial of the matrix,

353
00:22:00,000 --> 00:22:06,000
the determinant of partial
derivatives.

354
00:22:05,000 --> 00:22:11,000
So, it's the determinant whose
entries are the partial of u

355
00:22:09,000 --> 00:22:15,000
with respect to u,
the partial of u with respect

356
00:22:13,000 --> 00:22:19,000
to t, but these are independent
variables.

357
00:22:16,000 --> 00:22:22,000
So, that's zero.
The partial of v with respect

358
00:22:20,000 --> 00:22:26,000
to u is negative one.
The partial of v with respect

359
00:22:24,000 --> 00:22:30,000
to t is one.
So, the Jacobian is one.

360
00:22:27,000 --> 00:22:33,000
So, if you forgot it,
no harm.

361
00:22:31,000 --> 00:22:37,000
So, the Jacobian is one.
Now, more serious,

362
00:22:34,000 --> 00:22:40,000
and in some ways,
I think, for most of you,

363
00:22:37,000 --> 00:22:43,000
the most difficult part of the
operation, is putting in the new

364
00:22:41,000 --> 00:22:47,000
limits.
Now, for that,

365
00:22:43,000 --> 00:22:49,000
you look at the region over
which you're integrating.

366
00:22:46,000 --> 00:22:52,000
I think I'd better do that
carefully.

367
00:22:49,000 --> 00:22:55,000
I need a bigger picture.
That's really what I'm trying

368
00:22:53,000 --> 00:22:59,000
to say.
So, here's the (u,

369
00:22:54,000 --> 00:23:00,000
v) coordinates.
And, I want to change these to

370
00:22:58,000 --> 00:23:04,000
(u, t) coordinates.
The integration is over the

371
00:23:01,000 --> 00:23:07,000
first quadrant.
So, what you do is,

372
00:23:05,000 --> 00:23:11,000
when you do the integral,
the first step is u is varying,

373
00:23:10,000 --> 00:23:16,000
and t is held fixed.
So, in the first integration,

374
00:23:15,000 --> 00:23:21,000
u varies.
t is held fixed.

375
00:23:17,000 --> 00:23:23,000
Now, what is holding t fixed in
this picture mean?

376
00:23:22,000 --> 00:23:28,000
Well, t is equal to u plus v.

377
00:23:26,000 --> 00:23:32,000
So, u plus v is fixed,
is a constant,

378
00:23:29,000 --> 00:23:35,000
in other words.
Now, where are the curves along

379
00:23:34,000 --> 00:23:40,000
which u plus v is a
constant?

380
00:23:38,000 --> 00:23:44,000
Well, they are these lines.
These are the lines along which

381
00:23:43,000 --> 00:23:49,000
u plus v equals a constant,
or t is a constant.

382
00:23:47,000 --> 00:23:53,000
The reason I'm holding t a
constant is because the first

383
00:23:52,000 --> 00:23:58,000
integration only allows u to
change.

384
00:23:55,000 --> 00:24:01,000
t is held fixed.
Okay, you let u increase.

385
00:23:59,000 --> 00:24:05,000
As u increases,
and t is held fixed,

386
00:24:02,000 --> 00:24:08,000
I'm traversing these lines in
this direction.

387
00:24:08,000 --> 00:24:14,000
That's the direction on which u
is increasing.

388
00:24:11,000 --> 00:24:17,000
I integrate from the point,
from the u value where they

389
00:24:15,000 --> 00:24:21,000
leave the region.
And, to enter the region,

390
00:24:18,000 --> 00:24:24,000
what's the u value where they
enter the region?

391
00:24:21,000 --> 00:24:27,000
u is equal to zero.
Everybody would know that.

392
00:24:24,000 --> 00:24:30,000
Not so many people would be
able to figure out what to put

393
00:24:28,000 --> 00:24:34,000
for where it leaves the region.
What's the value of u when it

394
00:24:34,000 --> 00:24:40,000
leaves the region?
Well, this is the curve,

395
00:24:38,000 --> 00:24:44,000
v equals zero.
But, v equals zero is,

396
00:24:42,000 --> 00:24:48,000
in another language,
u equals t.

397
00:24:46,000 --> 00:24:52,000
t minus u equals zero,
or u equals t.

398
00:24:51,000 --> 00:24:57,000
In other words,
they enter the region where u

399
00:24:55,000 --> 00:25:01,000
equals zero,
and they leave where u is t,

400
00:25:00,000 --> 00:25:06,000
has the value of t.
And, how about the other guys?

401
00:25:05,000 --> 00:25:11,000
For which t's do I want to do
this?

402
00:25:07,000 --> 00:25:13,000
Well, I want to do it for all
these t values.

403
00:25:10,000 --> 00:25:16,000
Well, now, the t value here,
that's the starting one.

404
00:25:14,000 --> 00:25:20,000
Here, t is zero,
and here t is not zero.

405
00:25:16,000 --> 00:25:22,000
And, if I go out and cover the
whole first quadrant,

406
00:25:20,000 --> 00:25:26,000
I'll be letting t increase to
infinity.

407
00:25:23,000 --> 00:25:29,000
The sum of u and v,
I will be letting increase to

408
00:25:26,000 --> 00:25:32,000
infinity.
So, it's zero to infinity.

409
00:25:30,000 --> 00:25:36,000
So, all this is an exercise in
taking this double integral in

410
00:25:35,000 --> 00:25:41,000
(u, v) coordinates,
and changing it to this double

411
00:25:40,000 --> 00:25:46,000
integral, an equivalent double
integral over the same region,

412
00:25:46,000 --> 00:25:52,000
but now in (u,
t) coordinates.

413
00:25:48,000 --> 00:25:54,000
And now, that's the answer.
Somewhere here is the answer

414
00:25:54,000 --> 00:26:00,000
because, look,
since the first integration is

415
00:25:58,000 --> 00:26:04,000
with respect to u,
this guy can migrate outside

416
00:26:02,000 --> 00:26:08,000
because it doesn't involve u.
That only involves t,

417
00:26:08,000 --> 00:26:14,000
and t is only caught by the
second integration.

418
00:26:11,000 --> 00:26:17,000
So, I can put this outside.
And, what do I end up with?

419
00:26:15,000 --> 00:26:21,000
The integral from zero to
infinity of e to the negative s

420
00:26:18,000 --> 00:26:24,000
t times,

421
00:26:22,000 --> 00:26:28,000
what's left?
A funny expression,

422
00:26:24,000 --> 00:26:30,000
but you're on your desert
island and found it.

423
00:26:27,000 --> 00:26:33,000
This funny expression,
integral from zero to t,

424
00:26:30,000 --> 00:26:36,000
f of u, g of t minus u vu,

425
00:26:34,000 --> 00:26:40,000
in short,
the convolution,

426
00:26:37,000 --> 00:26:43,000
exactly the convolution.
So, all you have to do is get

427
00:26:42,000 --> 00:26:48,000
the idea that there might be a
formula, sit down,

428
00:26:45,000 --> 00:26:51,000
change variables and double
integral it, ego,

429
00:26:48,000 --> 00:26:54,000
you've got your formula.
Well, I would like to spend

430
00:26:52,000 --> 00:26:58,000
much of the rest of the
period--- in other words,

431
00:26:56,000 --> 00:27:02,000
that's how it relates to the
Laplace transform.

432
00:26:59,000 --> 00:27:05,000
That's how it comes out of the
Laplace transform.

433
00:27:04,000 --> 00:27:10,000
Here's how to use it,
calculate it either with the

434
00:27:07,000 --> 00:27:13,000
Laplace transform or directly
from the integral.

435
00:27:10,000 --> 00:27:16,000
And, of course,
you will solve problems,

436
00:27:13,000 --> 00:27:19,000
Laplace transform problems,
differential equations using

437
00:27:17,000 --> 00:27:23,000
the convolution.
But, I have to tell you that

438
00:27:20,000 --> 00:27:26,000
most people, convolution is very
important.

439
00:27:23,000 --> 00:27:29,000
And, most people who use it
don't use it in connection with

440
00:27:27,000 --> 00:27:33,000
the Laplace transform.
They use it for its own sake.

441
00:27:30,000 --> 00:27:36,000
The first place I learned that
outside of MIT people used a

442
00:27:34,000 --> 00:27:40,000
convolution was actually from my
daughter.

443
00:27:39,000 --> 00:27:45,000
She's an environmental
engineer, an environmental

444
00:27:41,000 --> 00:27:47,000
consultant.
She does risk assessment,

445
00:27:44,000 --> 00:27:50,000
and stuff like that.
But anyway, she had this paper

446
00:27:47,000 --> 00:27:53,000
on acid rain she was trying to
read for a client,

447
00:27:50,000 --> 00:27:56,000
and she said something about
calculating acid rain falls on

448
00:27:53,000 --> 00:27:59,000
soil.
And then, from there,

449
00:27:55,000 --> 00:28:01,000
the stuff leeches into a river.
But, things happen to it on the

450
00:27:58,000 --> 00:28:04,000
way.
Soil combines in various ways,

451
00:28:01,000 --> 00:28:07,000
reduces the acidity,
and things happen.

452
00:28:03,000 --> 00:28:09,000
Chemical reactions take place,
blah, blah, blah,

453
00:28:06,000 --> 00:28:12,000
blah.
Anyways, she said,

454
00:28:08,000 --> 00:28:14,000
well, then they calculated in
the end how much the river gets

455
00:28:11,000 --> 00:28:17,000
polluted.
But, she said it's convolution.

456
00:28:13,000 --> 00:28:19,000
She said, what's the
convolution?

457
00:28:15,000 --> 00:28:21,000
So, I told her she was too
young to learn about the

458
00:28:18,000 --> 00:28:24,000
convolution.
And she knows that I thought

459
00:28:20,000 --> 00:28:26,000
I'd better look it up first.
I mean, I, of course,

460
00:28:23,000 --> 00:28:29,000
knew the convolution was,
but I was a little puzzled at

461
00:28:26,000 --> 00:28:32,000
that application of it.
So, I read the paper.

462
00:28:29,000 --> 00:28:35,000
It was interesting.
And, in thinking about it,

463
00:28:33,000 --> 00:28:39,000
other people have come to me,
some guy with a problem about,

464
00:28:38,000 --> 00:28:44,000
they drilled ice cores in the
North Pole, and from the

465
00:28:41,000 --> 00:28:47,000
radioactive carbon and so on,
deducing various things about

466
00:28:46,000 --> 00:28:52,000
the climate 60 billion years
ago, and it was all convolution.

467
00:28:50,000 --> 00:28:56,000
He asked me if I could explain
that to him.

468
00:28:53,000 --> 00:28:59,000
So, let me give you sort of
all-purpose thing,

469
00:28:56,000 --> 00:29:02,000
a simple all-purpose model,
which can be adapted,

470
00:28:59,000 --> 00:29:05,000
which is very good way of
thinking of the convolution,

471
00:29:03,000 --> 00:29:09,000
in my opinion.
It's a problem of radioactive

472
00:29:08,000 --> 00:29:14,000
dumping.
It's in the notes,

473
00:29:11,000 --> 00:29:17,000
by the way.
So, I'm just,

474
00:29:13,000 --> 00:29:19,000
if you want to take a chance,
and just listen to what I'm

475
00:29:18,000 --> 00:29:24,000
saying rather that just
scribbling everything down,

476
00:29:23,000 --> 00:29:29,000
maybe you'll be able to figure
it out for the notes,

477
00:29:28,000 --> 00:29:34,000
also.
So, the problem is we have some

478
00:29:33,000 --> 00:29:39,000
factory, or a nuclear plant,
or some thing like that,

479
00:29:38,000 --> 00:29:44,000
is producing radioactive waste,
not always at the same rate.

480
00:29:44,000 --> 00:29:50,000
And then, it carts it,
dumps it on a pile somewhere.

481
00:29:49,000 --> 00:29:55,000
So, radioactive waste is
dumped, and there's a dumping

482
00:29:54,000 --> 00:30:00,000
function.
I'll call that f of t,

483
00:29:58,000 --> 00:30:04,000
the dump rate.
That's the dumping rate.

484
00:30:03,000 --> 00:30:09,000
Let's say t is in years.
You like to have units,

485
00:30:07,000 --> 00:30:13,000
and quantity,
kilograms, I don't know,

486
00:30:10,000 --> 00:30:16,000
whatever you want.
Now, what does the dumping rate

487
00:30:15,000 --> 00:30:21,000
mean?
The dumping rate means that if

488
00:30:18,000 --> 00:30:24,000
I have two times that are close
together, for example,

489
00:30:23,000 --> 00:30:29,000
two successive days,
midnight on two successive

490
00:30:27,000 --> 00:30:33,000
days, then there's a time
interval between them.

491
00:30:33,000 --> 00:30:39,000
I'll call that delta t.
To say the dumping rate is f of

492
00:30:38,000 --> 00:30:44,000
t means that the amount
dumped in this time interval,

493
00:30:43,000 --> 00:30:49,000
in the time interval from t1 to
t1 plus one is

494
00:30:49,000 --> 00:30:55,000
approximately,
not exactly,

495
00:30:52,000 --> 00:30:58,000
because the dumping rate isn't
even constant within this time

496
00:30:58,000 --> 00:31:04,000
interval.
But it's approximately the

497
00:31:02,000 --> 00:31:08,000
dumping rate times the time over
which the dumping is taking

498
00:31:09,000 --> 00:31:15,000
place.
That's what I mean by the dump

499
00:31:13,000 --> 00:31:19,000
rate.
And, it gets more and more

500
00:31:16,000 --> 00:31:22,000
accurate, the smaller the time
interval you take.

501
00:31:21,000 --> 00:31:27,000
Okay, now here's my problem.
The problem is,

502
00:31:26,000 --> 00:31:32,000
you start dumping at time t
equals zero.

503
00:31:33,000 --> 00:31:39,000
At time t equal t,
how much radioactive waste is

504
00:31:39,000 --> 00:31:45,000
in the pile?

505
00:31:55,000 --> 00:32:01,000
Now, what makes that problem
slightly complicated is

506
00:31:58,000 --> 00:32:04,000
radioactive waste decays.
If I put some at a certain day,

507
00:32:02,000 --> 00:32:08,000
and then go back several months
later and nothing's happened in

508
00:32:07,000 --> 00:32:13,000
between, I don't have the same
amount that I dumps because a

509
00:32:11,000 --> 00:32:17,000
fraction of it decayed.
I have less.

510
00:32:14,000 --> 00:32:20,000
And, our answer to the problem
must take account of,

511
00:32:18,000 --> 00:32:24,000
for each piece of waste,
how long it has been in the

512
00:32:22,000 --> 00:32:28,000
pile because that takes account
of how long it had to decay,

513
00:32:27,000 --> 00:32:33,000
and what it ends up as.
So, the calculation,

514
00:32:32,000 --> 00:32:38,000
the essential part of the
calculation will be that if you

515
00:32:37,000 --> 00:32:43,000
have an initial amount of this
substance, and it decays for a

516
00:32:43,000 --> 00:32:49,000
time, t, this is the amount left
at time t.

517
00:32:47,000 --> 00:32:53,000
This is the law of radioactive
decay.

518
00:32:51,000 --> 00:32:57,000
You knew that coming into
18.03, although,

519
00:32:55,000 --> 00:33:01,000
it's, of course,
a simple differential equation

520
00:33:00,000 --> 00:33:06,000
which produces it,
but I'll assume you simply know

521
00:33:05,000 --> 00:33:11,000
the answer.
k depends on the material,

522
00:33:10,000 --> 00:33:16,000
so I'm going to assume that the
nuclear plant dumps the same

523
00:33:14,000 --> 00:33:20,000
radioactive substance each time.
It's only one substance I'm

524
00:33:19,000 --> 00:33:25,000
calculating, and k is it.
So, assume the k is fixed.

525
00:33:23,000 --> 00:33:29,000
I don't have to change from one
k from one material to a k for

526
00:33:27,000 --> 00:33:33,000
another because it's mixing up
the stuff, just one material.

527
00:33:33,000 --> 00:33:39,000
Okay, and now let's calculate
it.

528
00:33:35,000 --> 00:33:41,000
Here's the idea.
I'll take the t-axis,

529
00:33:38,000 --> 00:33:44,000
but now I'm going to change its
name to the u-axis.

530
00:33:43,000 --> 00:33:49,000
You will see why in just a
second.

531
00:33:45,000 --> 00:33:51,000
It starts at zero.
I'm interested in what's

532
00:33:49,000 --> 00:33:55,000
happening at the time,
t.

533
00:33:51,000 --> 00:33:57,000
How much is left at time t?
So, I'm going to divide up the

534
00:33:56,000 --> 00:34:02,000
interval from zero to t on this
time axis into,

535
00:34:00,000 --> 00:34:06,000
well, here's u0,
the starting point,

536
00:34:03,000 --> 00:34:09,000
u1, u2, let's make this u1.
Oh, curses!

537
00:34:08,000 --> 00:34:14,000
u1, u2, u3, and so on.
Let's call this (u)n.

538
00:34:13,000 --> 00:34:19,000
So they're u(n + 1),
not that it matters.

539
00:34:18,000 --> 00:34:24,000
It doesn't matter.
Okay, now, the amount,

540
00:34:23,000 --> 00:34:29,000
so, what I'm going to do is
look at the amount,

541
00:34:28,000 --> 00:34:34,000
take the time interval from ui
to ui plus one.

542
00:34:36,000 --> 00:34:42,000
This is a time interval,

543
00:34:40,000 --> 00:34:46,000
delta u.
Divide it up into equal time

544
00:34:43,000 --> 00:34:49,000
intervals.
So, the amount dumped in the

545
00:34:46,000 --> 00:34:52,000
time interval from u(i) to u(i
plus one)

546
00:34:51,000 --> 00:34:57,000
is equal to approximately f of
u(i),

547
00:34:55,000 --> 00:35:01,000
the dumping function there,
times delta u.

548
00:35:00,000 --> 00:35:06,000
We calculated that before.
That's what the meaning of the

549
00:35:06,000 --> 00:35:12,000
dumping rate is.
By time t, how much has it

550
00:35:11,000 --> 00:35:17,000
decayed to?
It has decayed.

551
00:35:14,000 --> 00:35:20,000
How much is left,
in other words?

552
00:35:18,000 --> 00:35:24,000
Well, this is the starting
amount.

553
00:35:21,000 --> 00:35:27,000
So, the answer is going to be
it's f of (u)i times delta u

554
00:35:28,000 --> 00:35:34,000
times this factor,
which tells how much it decays,

555
00:35:34,000 --> 00:35:40,000
so, time.
So, this is the starting amount

556
00:35:39,000 --> 00:35:45,000
at time (u)i.
That's when it was first

557
00:35:41,000 --> 00:35:47,000
dumped, and this is the amount
that was dumped,

558
00:35:45,000 --> 00:35:51,000
times, multiply that by e to
the minus k times,

559
00:35:49,000 --> 00:35:55,000
now, what should I put up in
there?

560
00:35:51,000 --> 00:35:57,000
I have to put the length of
time that it had to decay.

561
00:35:55,000 --> 00:36:01,000
What is the length of time that
it had to decay?

562
00:36:00,000 --> 00:36:06,000
It was dumped at u(i).
I'm looking at time,

563
00:36:08,000 --> 00:36:14,000
t, it decayed for time length t
minus u i,

564
00:36:19,000 --> 00:36:25,000
the length of time it had all
the pile.

565
00:36:32,000 --> 00:36:38,000
So, the stuff that was dumped
in this time interval,

566
00:36:36,000 --> 00:36:42,000
at time t when I come to look
at it, this is how much of it is

567
00:36:41,000 --> 00:36:47,000
left.
And now, all I have to do is

568
00:36:44,000 --> 00:36:50,000
add up that quantity for this
time, the stuff that was dumped

569
00:36:49,000 --> 00:36:55,000
in this time interval plus the
stuff dumped in,

570
00:36:54,000 --> 00:37:00,000
and so on, all the way up to
the stuff that was dumped

571
00:36:58,000 --> 00:37:04,000
yesterday.
And, the answer will be the

572
00:37:01,000 --> 00:37:07,000
total amount left at time,
t, that is not yet decayed will

573
00:37:06,000 --> 00:37:12,000
be approximately,
you add up the amount coming

574
00:37:10,000 --> 00:37:16,000
from the first time interval
plus the amount coming,

575
00:37:15,000 --> 00:37:21,000
and so on.
So, it will be f of u(i),

576
00:37:19,000 --> 00:37:25,000
I'll save the delta u for the
end, times e to the minus k

577
00:37:23,000 --> 00:37:29,000
times t minus u(i) times
delta u.

578
00:37:27,000 --> 00:37:33,000
So, these two parts represent

579
00:37:29,000 --> 00:37:35,000
the amount dumped,
and this is the decay factor.

580
00:37:33,000 --> 00:37:39,000
And, I had those up as I runs
from, well, where did I start?

581
00:37:37,000 --> 00:37:43,000
From one to n,
let's say.

582
00:37:39,000 --> 00:37:45,000
And now, let delta t go to
zero, in other words,

583
00:37:42,000 --> 00:37:48,000
make this delta u go to zero,
make this more accurate by

584
00:37:46,000 --> 00:37:52,000
taking finer and finer
subdivisions.

585
00:37:48,000 --> 00:37:54,000
In other words,
instead of looking every month

586
00:37:51,000 --> 00:37:57,000
to see how much was dumped,
let's look every week,

587
00:37:55,000 --> 00:38:01,000
every day, and so on,
to make this calculation more

588
00:37:58,000 --> 00:38:04,000
accurate.
And, the answer is,

589
00:38:00,000 --> 00:38:06,000
this approach is the exact
amount, which will be the

590
00:38:04,000 --> 00:38:10,000
integral.
This sum is a Riemann sum.

591
00:38:08,000 --> 00:38:14,000
It approaches the integral from
zero to, well,

592
00:38:12,000 --> 00:38:18,000
I'm adding it up from u1 equals
zero to un equals t,

593
00:38:18,000 --> 00:38:24,000
the final value.
So, it will be the integral

594
00:38:22,000 --> 00:38:28,000
from the starting point to the
ending point of f of u e to the

595
00:38:28,000 --> 00:38:34,000
minus k times t minus u to u.

596
00:38:34,000 --> 00:38:40,000
That's the answer to the
problem.

597
00:38:36,000 --> 00:38:42,000
It's given by this rather funny
looking integral.

598
00:38:39,000 --> 00:38:45,000
But, from this point of view,
it's entirely natural.

599
00:38:42,000 --> 00:38:48,000
It's a combination of the
dumping function.

600
00:38:44,000 --> 00:38:50,000
This doesn't care what the
material was.

601
00:38:47,000 --> 00:38:53,000
It only wants to know how much
was put on everyday.

602
00:38:50,000 --> 00:38:56,000
And, this part,
which doesn't care how much was

603
00:38:53,000 --> 00:38:59,000
put on each day,
it just is an intrinsic

604
00:38:55,000 --> 00:39:01,000
constant of the material
involving its decay rate.

605
00:39:00,000 --> 00:39:06,000
And, this total thing
represents the total amount.

606
00:39:04,000 --> 00:39:10,000
And that is,
what is it?

607
00:39:06,000 --> 00:39:12,000
It's the convolution of f of t
with what function?

608
00:39:11,000 --> 00:39:17,000
e to the minus k t.
It's the convolution of the

609
00:39:16,000 --> 00:39:22,000
dumping function and the decay
function.

610
00:39:19,000 --> 00:39:25,000
And, the convolution is exactly
the operation that you have to

611
00:39:25,000 --> 00:39:31,000
have to do that.
Okay, so, I think this is the

612
00:39:28,000 --> 00:39:34,000
most intuitive physical approach
to the meaning of the

613
00:39:33,000 --> 00:39:39,000
convolution.
In this particular,

614
00:39:37,000 --> 00:39:43,000
you can say,
well, that's very special.

615
00:39:39,000 --> 00:39:45,000
Okay, so it tells you what the
meaning of the convolution with

616
00:39:43,000 --> 00:39:49,000
an exponential is.
But, what about the convolution

617
00:39:46,000 --> 00:39:52,000
with all the other functions
we're going to have to use in

618
00:39:50,000 --> 00:39:56,000
this course.
They can all be interpreted

619
00:39:52,000 --> 00:39:58,000
just by being a little flexible
in your approach.

620
00:39:55,000 --> 00:40:01,000
I'll give you two examples of
this, well, three.

621
00:39:58,000 --> 00:40:04,000
First of all,
I'll use it for,

622
00:40:00,000 --> 00:40:06,000
in the problem set I ask you
about a bank account.

623
00:40:05,000 --> 00:40:11,000
That's not something any of you
are interested in.

624
00:40:08,000 --> 00:40:14,000
Okay, so, suppose instead I
dumped garbage --

625
00:40:16,000 --> 00:40:22,000
-- undecaying.
So, something that doesn't

626
00:40:19,000 --> 00:40:25,000
decay at all,
what's the answer going to be?

627
00:40:22,000 --> 00:40:28,000
Well, the calculation will be
exactly the same.

628
00:40:26,000 --> 00:40:32,000
It will be the convolution of
the dumping function.

629
00:40:30,000 --> 00:40:36,000
The only difference is that now
the garbage isn't going to

630
00:40:34,000 --> 00:40:40,000
decay.
So, no matter how long it's

631
00:40:37,000 --> 00:40:43,000
left, the same amount is going
to be left at the end.

632
00:40:40,000 --> 00:40:46,000
In other words,
I don't want to exponential

633
00:40:42,000 --> 00:40:48,000
decay function.
I want to function,

634
00:40:44,000 --> 00:40:50,000
one, the constant function,
one, because once I stick it on

635
00:40:48,000 --> 00:40:54,000
the pile, nothing happens to it.
It just stays there.

636
00:40:51,000 --> 00:40:57,000
So, it's going to be the
convolution of this one because

637
00:40:54,000 --> 00:41:00,000
this is constant.
It's undecaying --

638
00:41:04,000 --> 00:41:10,000
-- by the identical reasoning.
And so, what's the answer going

639
00:41:07,000 --> 00:41:13,000
to be?
It's going to be the integral

640
00:41:09,000 --> 00:41:15,000
from zero to t of f of u du.

641
00:41:12,000 --> 00:41:18,000
Now, that's an 18.01 problem.

642
00:41:14,000 --> 00:41:20,000
If I dump with a dumping rate,
f of u,

643
00:41:17,000 --> 00:41:23,000
and I dump from time zero to
time t, how much is on the pile?

644
00:41:20,000 --> 00:41:26,000
They don't give it.
They always give velocity

645
00:41:23,000 --> 00:41:29,000
problems, and problems of how to
slice up bread loaves,

646
00:41:26,000 --> 00:41:32,000
and stuff like that.
But, this is a real life

647
00:41:28,000 --> 00:41:34,000
problem.
If that's the dumping rate,

648
00:41:32,000 --> 00:41:38,000
and you dump for t days from
zero to time t,

649
00:41:35,000 --> 00:41:41,000
how much do you have left at
the end?

650
00:41:37,000 --> 00:41:43,000
Answer: the integral of f of u
du from zero to t.

651
00:41:42,000 --> 00:41:48,000
I'll give you another example.

652
00:41:46,000 --> 00:41:52,000
Suppose I wanted a dumping
function, suppose I wanted a

653
00:41:50,000 --> 00:41:56,000
function, wanted to interpret
something which grows like t,

654
00:41:54,000 --> 00:42:00,000
for instance.
All I want is a physical

655
00:41:57,000 --> 00:42:03,000
interpretation.
Well, I have to think,

656
00:42:01,000 --> 00:42:07,000
I'm making a pile of something,
a metaphorical pile,

657
00:42:04,000 --> 00:42:10,000
we don't actually have to make
a physical pile.

658
00:42:07,000 --> 00:42:13,000
And, the thing should be
growing like t.

659
00:42:09,000 --> 00:42:15,000
Well, what grows like t?
Not bacteria,

660
00:42:11,000 --> 00:42:17,000
they grow exponentially.
Before the lecture,

661
00:42:14,000 --> 00:42:20,000
I was trying to think of
something.

662
00:42:16,000 --> 00:42:22,000
So, I came up with chickens on
a chicken farm.

663
00:42:19,000 --> 00:42:25,000
Little baby chickens grow
linearly.

664
00:42:21,000 --> 00:42:27,000
All little animals,
anyway, I've observed that

665
00:42:23,000 --> 00:42:29,000
babies grow linearly,
at least for a while,

666
00:42:26,000 --> 00:42:32,000
thank God.
After a while,

667
00:42:27,000 --> 00:42:33,000
they taper off.
But, at the beginning,

668
00:42:32,000 --> 00:42:38,000
they eat every four hours or
whatever.

669
00:42:35,000 --> 00:42:41,000
And they eat the same amount,
pretty much.

670
00:42:39,000 --> 00:42:45,000
And, that adds up.
So, let's suppose this

671
00:42:43,000 --> 00:42:49,000
represents the linear growth of
chickens, of baby chicks.

672
00:42:48,000 --> 00:42:54,000
That makes them sound cuter,
less offensive.

673
00:42:52,000 --> 00:42:58,000
Okay, so, a farmer,
chicken farmer,

674
00:42:56,000 --> 00:43:02,000
whatever they call them,
is starting a new brood.

675
00:43:02,000 --> 00:43:08,000
So anyway, the hens lay at a
certain rate,

676
00:43:05,000 --> 00:43:11,000
and each of those are
incubated.

677
00:43:08,000 --> 00:43:14,000
And after a while,
little baby chicks come out.

678
00:43:12,000 --> 00:43:18,000
So, this will be the production
rate for new chickens.

679
00:43:23,000 --> 00:43:29,000
Okay, and it will be the
convolution which will tell you

680
00:43:26,000 --> 00:43:32,000
at time, t, the number of
kilograms.

681
00:43:29,000 --> 00:43:35,000
We'd better do this in
kilograms, I'm afraid.

682
00:43:32,000 --> 00:43:38,000
Now, that's not as heartless as
it seems.

683
00:43:35,000 --> 00:43:41,000
The number of kilograms of
chickens times t.

684
00:43:38,000 --> 00:43:44,000
[LAUGHTER] It really isn't
heartless because,

685
00:43:41,000 --> 00:43:47,000
after all, why would the farmer
want to know that?

686
00:43:44,000 --> 00:43:50,000
Well, because a certain number
of pounds of chicken eat a

687
00:43:48,000 --> 00:43:54,000
certain number of pounds of
chicken feed,

688
00:43:51,000 --> 00:43:57,000
and that's how much he has to
dump, must have to give them

689
00:43:55,000 --> 00:44:01,000
every day.
That's how he calculates his

690
00:43:57,000 --> 00:44:03,000
expenses.
So, he will have to know the

691
00:44:01,000 --> 00:44:07,000
convolution is,
or better yet,

692
00:44:03,000 --> 00:44:09,000
he will hire you,
who knows what the convolution

693
00:44:07,000 --> 00:44:13,000
is.
And you'll be able to tell him.

694
00:44:09,000 --> 00:44:15,000
Okay, why don't we stop there
and go to recitation tomorrow.

695
00:44:13,000 --> 00:44:19,000
I'll be doing important things.