WEBVTT
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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
00:10:10.000 --> 00:10:16.000
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.
00:12:19.000 --> 00:12:25.000
But now, notice that this is
not an ordinary product.
00:12:22.000 --> 00:12:28.000
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.
00:12:41.000 --> 00:12:47.000
Let's do something really
humble just assure you that
00:12:45.000 --> 00:12:51.000
this, even at the simplest
example, this is not trivial.
00:12:50.000 --> 00:12:56.000
Let's take the convolution of f
of t with one.
00:12:54.000 --> 00:13:00.000
Can you take,
yeah, one is a function just
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like any function.
But, you get something out of
00:13:01.000 --> 00:13:07.000
the convolution,
yes, yes.
00:13:03.000 --> 00:13:09.000
Let's just write down the
formula.
00:13:05.000 --> 00:13:11.000
Now, I can't use the Laplace
transform here because you won't
00:13:09.000 --> 00:13:15.000
know what to do with it.
You don't have that formula
00:13:13.000 --> 00:13:19.000
yet.
It's a secret one that only I
00:13:15.000 --> 00:13:21.000
know.
So, let's do it.
00:13:16.000 --> 00:13:22.000
Let's calculate it out the way
it was supposed to.
00:13:19.000 --> 00:13:25.000
So, it's the integral from zero
to t of f of u,
00:13:23.000 --> 00:13:29.000
and now, what do I do with that
one?
00:13:25.000 --> 00:13:31.000
I'm supposed to take,
one is the function g of t,
00:13:29.000 --> 00:13:35.000
and wherever I see a
t, I'm supposed to plug in t
00:13:32.000 --> 00:13:38.000
minus u.
Well, I don't see any t there.
00:13:39.000 --> 00:13:45.000
But that's something for
rejoicing.
00:13:43.000 --> 00:13:49.000
There's nothing to do to make
the substitution.
00:13:48.000 --> 00:13:54.000
It's just one.
So, the answer is,
00:13:52.000 --> 00:13:58.000
it's this curious thing.
The convolution of a function
00:13:58.000 --> 00:14:04.000
with one, you integrate it from
zero to t.
00:14:04.000 --> 00:14:10.000
Well, as they said in Alice in
Wonderland, things are getting
00:14:08.000 --> 00:14:14.000
curiouser and curiouser.
I mean, what is going on with
00:14:12.000 --> 00:14:18.000
this crazy function,
and where are we supposed to
00:14:16.000 --> 00:14:22.000
start with it?
Well, I'm going to prove this
00:14:19.000 --> 00:14:25.000
for you, mostly because the
proof is easy.
00:14:23.000 --> 00:14:29.000
In other words,
I'm going to prove that that's
00:14:26.000 --> 00:14:32.000
true.
And, as I give the proof,
00:14:29.000 --> 00:14:35.000
you'll see where the
convolution is coming from.
00:14:32.000 --> 00:14:38.000
That's number one.
And, number two,
00:14:34.000 --> 00:14:40.000
the real reason I'm giving you
the proof: because it's a
00:14:38.000 --> 00:14:44.000
marvelous exercise in changing
the variables in a double
00:14:42.000 --> 00:14:48.000
integral.
Now, that's something you all
00:14:44.000 --> 00:14:50.000
know how to do,
even the ones who are taking
00:14:47.000 --> 00:14:53.000
18.02 concurrently,
and I didn't advise you to do
00:14:50.000 --> 00:14:56.000
that.
But, I've arranged the course
00:14:52.000 --> 00:14:58.000
so it's possible to do.
But, I knew that by the time we
00:14:56.000 --> 00:15:02.000
got to this, you would already
know how to change variables at
00:15:00.000 --> 00:15:06.000
a double integral.
So, and in fact,
00:15:04.000 --> 00:15:10.000
you will have the advantage of
remembering how to do it because
00:15:10.000 --> 00:15:16.000
you just had it about a week or
two ago, whereas all the other
00:15:15.000 --> 00:15:21.000
guys, it's something dim in
their distance.
00:15:19.000 --> 00:15:25.000
So, I'm reviewing how to change
variables at a double integral.
00:15:25.000 --> 00:15:31.000
I'm showing you it's good for
something.
00:15:29.000 --> 00:15:35.000
So, what we are out to try to
prove is this formula.
00:15:33.000 --> 00:15:39.000
Let's put that down in,
so you understand.
00:15:39.000 --> 00:15:45.000
Okay, let's do it.
Now, we'll use the desert
00:15:41.000 --> 00:15:47.000
island method.
So, you have as much time as
00:15:44.000 --> 00:15:50.000
you want.
You're on a desert island.
00:15:46.000 --> 00:15:52.000
In fact, I'm going to even go
it the opposite way.
00:15:49.000 --> 00:15:55.000
I'm going to start with--
you've got a lot of time on your
00:15:53.000 --> 00:15:59.000
hands and say,
gee, I wonder if I take the
00:15:56.000 --> 00:16:02.000
product of the Laplace
transforms, I wonder if there's
00:15:59.000 --> 00:16:05.000
some crazy function I could put
in there, which would make
00:16:03.000 --> 00:16:09.000
things work.
You've never heard of the
00:16:06.000 --> 00:16:12.000
convolution.
You're going to discover it all
00:16:09.000 --> 00:16:15.000
by yourself.
Okay, so how do you begin?
00:16:11.000 --> 00:16:17.000
So, we'll start with the left
hand side.
00:16:14.000 --> 00:16:20.000
We're looking for some nice way
of calculating that as the
00:16:17.000 --> 00:16:23.000
Laplace transform of a single
function.
00:16:19.000 --> 00:16:25.000
So, the way to begin is by
writing out the definitions.
00:16:23.000 --> 00:16:29.000
We couldn't use anything else
since we don't have anything
00:16:26.000 --> 00:16:32.000
else to use.
Now, looking ahead,
00:16:28.000 --> 00:16:34.000
I'm going to not use t.
I'm going to use two neutral
00:16:32.000 --> 00:16:38.000
variables when I calculate.
After all, the t is just a
00:16:35.000 --> 00:16:41.000
dummy variable anyway.
You will see in a minute the
00:16:40.000 --> 00:16:46.000
wisdom of doing this.
So, it's this times the
00:16:44.000 --> 00:16:50.000
integral, which gives the
Laplace transform of g.
00:16:48.000 --> 00:16:54.000
So, that's e to the negative s
v, let's say,
00:16:52.000 --> 00:16:58.000
times g of v,
dv.
00:16:53.000 --> 00:16:59.000
Okay, everybody can get that
far.
00:16:56.000 --> 00:17:02.000
But now we have to start
looking.
00:17:00.000 --> 00:17:06.000
Well, this is a single
integral, an 18.01 integral
00:17:03.000 --> 00:17:09.000
involving u, and this is an
18.01 integral involving v.
00:17:07.000 --> 00:17:13.000
But when you take the product
of two integrals like that,
00:17:11.000 --> 00:17:17.000
remember when you evaluate a
double integral,
00:17:14.000 --> 00:17:20.000
there's an easy case where it's
much easier than any other case.
00:17:19.000 --> 00:17:25.000
If you could write the inside,
if you are integrating over a
00:17:23.000 --> 00:17:29.000
rectangle, for example,
and you can write the integral
00:17:27.000 --> 00:17:33.000
as a product of a function just
of u, and a product of a
00:17:31.000 --> 00:17:37.000
function just as v,
then the integral is very easy
00:17:34.000 --> 00:17:40.000
to evaluate.
You can forget all the rules.
00:17:38.000 --> 00:17:44.000
You just take all the u part
out, all the v part out,
00:17:42.000 --> 00:17:48.000
and integrate them separately,
a to b, c to d.
00:17:44.000 --> 00:17:50.000
That's the easy case of
evaluating a double integral.
00:17:48.000 --> 00:17:54.000
It's what everybody tries to
do, even when it's not
00:17:51.000 --> 00:17:57.000
appropriate.
Now, here it is appropriate,
00:17:53.000 --> 00:17:59.000
except I'm going to use it
backwards.
00:17:56.000 --> 00:18:02.000
This is the result of having
done that.
00:17:58.000 --> 00:18:04.000
If this is the result of having
done it, what was the step just
00:18:02.000 --> 00:18:08.000
before it?
Well, I must have been trying
00:18:06.000 --> 00:18:12.000
to evaluate a double integral as
u runs from zero to infinity and
00:18:10.000 --> 00:18:16.000
v runs from zero to infinity,
of what?
00:18:13.000 --> 00:18:19.000
Well, of the product of these
two functions.
00:18:16.000 --> 00:18:22.000
Now, what is that?
e to the minus s u times e to
00:18:20.000 --> 00:18:26.000
the minus s v.
00:18:22.000 --> 00:18:28.000
Well, I must surely want to
combine those.
00:18:25.000 --> 00:18:31.000
e to the minus s u times e to
the minus s v.
00:18:30.000 --> 00:18:36.000
And, what's left?
Well, what gets dragged along?
00:18:33.000 --> 00:18:39.000
du dv.
This is the same as that
00:18:35.000 --> 00:18:41.000
because of that law I just gave
you this is the product of a
00:18:39.000 --> 00:18:45.000
function just of u,
and a function just of v.
00:18:42.000 --> 00:18:48.000
And therefore,
it's okay to separate the two
00:18:45.000 --> 00:18:51.000
integrals out that way because
I'm integrating sort of a
00:18:49.000 --> 00:18:55.000
rectangle that goes to infinity
that way and infinity that way.
00:18:54.000 --> 00:19:00.000
But, what I'm integrating is
over the plane,
00:18:57.000 --> 00:19:03.000
in other words,
this region of the plane as u,
00:19:00.000 --> 00:19:06.000
v goes from zero to infinity,
zero to infinity.
00:19:05.000 --> 00:19:11.000
Now, let's take a look.
What are we looking for?
00:19:10.000 --> 00:19:16.000
Well, we're looking for,
we would be very happy if u
00:19:15.000 --> 00:19:21.000
plus v were t.
Let's make it t.
00:19:20.000 --> 00:19:26.000
In other words,
I'm introducing a new variable,
00:19:25.000 --> 00:19:31.000
t, u plus v,
and it's suggested by the form
00:19:30.000 --> 00:19:36.000
in which I'm looking for the
answer.
00:19:35.000 --> 00:19:41.000
Now, of course you then have
to, we need another variable.
00:19:39.000 --> 00:19:45.000
We could keep either u or v.
Let's keep u.
00:19:43.000 --> 00:19:49.000
That means v,
we just gave a musical chairs.
00:19:46.000 --> 00:19:52.000
v got dropped out.
Well, we can't have three
00:19:50.000 --> 00:19:56.000
variables.
We only have room for two.
00:19:53.000 --> 00:19:59.000
But, we will remember it.
Rest in peace,
00:19:57.000 --> 00:20:03.000
v was equal to t minus u
in case we ever need him
00:20:02.000 --> 00:20:08.000
again.
Okay, let's now put in the
00:20:05.000 --> 00:20:11.000
limits.
Let's put in the integral,
00:20:08.000 --> 00:20:14.000
the rest of the change of
variable.
00:20:10.000 --> 00:20:16.000
So, I'm now changing it to
these new variables,
00:20:14.000 --> 00:20:20.000
t and u, so it's e to the
negative s t.
00:20:18.000 --> 00:20:24.000
Well, f of u I don't
have to do anything to.
00:20:22.000 --> 00:20:28.000
But, g of v,
I'm not allowed to keep v,
00:20:25.000 --> 00:20:31.000
so v has to be changed to t
minus u.
00:20:30.000 --> 00:20:36.000
Amazing things are happening.
Now, I want to change this to
00:20:34.000 --> 00:20:40.000
an integral du dt.
Now, for that,
00:20:37.000 --> 00:20:43.000
you have to be a little
careful.
00:20:40.000 --> 00:20:46.000
We have two things to do to
figure out this;
00:20:43.000 --> 00:20:49.000
what goes with that?
And, we have to put in the
00:20:47.000 --> 00:20:53.000
limits, also.
Now, those are the two
00:20:50.000 --> 00:20:56.000
nontrivial operations,
when you change variables in a
00:20:55.000 --> 00:21:01.000
double integral.
So, let's be really careful.
00:21:00.000 --> 00:21:06.000
Let's do the easier of the two,
first.
00:21:03.000 --> 00:21:09.000
I want to change from du dv to
du dt.
00:21:07.000 --> 00:21:13.000
And now, to do that,
you have to put in the Jacobian
00:21:12.000 --> 00:21:18.000
matrix, the Jacobian
determinant.
00:21:15.000 --> 00:21:21.000
Ah-ha!
How many of you forgot that?
00:21:19.000 --> 00:21:25.000
I won't even bother asking.
Oh, come on,
00:21:23.000 --> 00:21:29.000
you only lose two points.
It doesn't matter if you put it
00:21:28.000 --> 00:21:34.000
in the Jacobian.
As you see, you're going to
00:21:34.000 --> 00:21:40.000
forget something.
You will lose less credit for
00:21:39.000 --> 00:21:45.000
forgetting than anything else.
So, it's the Jacobian of u and
00:21:45.000 --> 00:21:51.000
v with respect to u and t.
So, to calculate that,
00:21:49.000 --> 00:21:55.000
you write u equals u,
v equals t minus u,
00:21:54.000 --> 00:22:00.000
and then the Jacobian is
the partial of the matrix,
00:22:00.000 --> 00:22:06.000
the determinant of partial
derivatives.
00:22:05.000 --> 00:22:11.000
So, it's the determinant whose
entries are the partial of u
00:22:09.000 --> 00:22:15.000
with respect to u,
the partial of u with respect
00:22:13.000 --> 00:22:19.000
to t, but these are independent
variables.
00:22:16.000 --> 00:22:22.000
So, that's zero.
The partial of v with respect
00:22:20.000 --> 00:22:26.000
to u is negative one.
The partial of v with respect
00:22:24.000 --> 00:22:30.000
to t is one.
So, the Jacobian is one.
00:22:27.000 --> 00:22:33.000
So, if you forgot it,
no harm.
00:22:31.000 --> 00:22:37.000
So, the Jacobian is one.
Now, more serious,
00:22:34.000 --> 00:22:40.000
and in some ways,
I think, for most of you,
00:22:37.000 --> 00:22:43.000
the most difficult part of the
operation, is putting in the new
00:22:41.000 --> 00:22:47.000
limits.
Now, for that,
00:22:43.000 --> 00:22:49.000
you look at the region over
which you're integrating.
00:22:46.000 --> 00:22:52.000
I think I'd better do that
carefully.
00:22:49.000 --> 00:22:55.000
I need a bigger picture.
That's really what I'm trying
00:22:53.000 --> 00:22:59.000
to say.
So, here's the (u,
00:22:54.000 --> 00:23:00.000
v) coordinates.
And, I want to change these to
00:22:58.000 --> 00:23:04.000
(u, t) coordinates.
The integration is over the
00:23:01.000 --> 00:23:07.000
first quadrant.
So, what you do is,
00:23:05.000 --> 00:23:11.000
when you do the integral,
the first step is u is varying,
00:23:10.000 --> 00:23:16.000
and t is held fixed.
So, in the first integration,
00:23:15.000 --> 00:23:21.000
u varies.
t is held fixed.
00:23:17.000 --> 00:23:23.000
Now, what is holding t fixed in
this picture mean?
00:23:22.000 --> 00:23:28.000
Well, t is equal to u plus v.
00:23:26.000 --> 00:23:32.000
So, u plus v is fixed,
is a constant,
00:23:29.000 --> 00:23:35.000
in other words.
Now, where are the curves along
00:23:34.000 --> 00:23:40.000
which u plus v is a
constant?
00:23:38.000 --> 00:23:44.000
Well, they are these lines.
These are the lines along which
00:23:43.000 --> 00:23:49.000
u plus v equals a constant,
or t is a constant.
00:23:47.000 --> 00:23:53.000
The reason I'm holding t a
constant is because the first
00:23:52.000 --> 00:23:58.000
integration only allows u to
change.
00:23:55.000 --> 00:24:01.000
t is held fixed.
Okay, you let u increase.
00:23:59.000 --> 00:24:05.000
As u increases,
and t is held fixed,
00:24:02.000 --> 00:24:08.000
I'm traversing these lines in
this direction.
00:24:08.000 --> 00:24:14.000
That's the direction on which u
is increasing.
00:24:11.000 --> 00:24:17.000
I integrate from the point,
from the u value where they
00:24:15.000 --> 00:24:21.000
leave the region.
And, to enter the region,
00:24:18.000 --> 00:24:24.000
what's the u value where they
enter the region?
00:24:21.000 --> 00:24:27.000
u is equal to zero.
Everybody would know that.
00:24:24.000 --> 00:24:30.000
Not so many people would be
able to figure out what to put
00:24:28.000 --> 00:24:34.000
for where it leaves the region.
What's the value of u when it
00:24:34.000 --> 00:24:40.000
leaves the region?
Well, this is the curve,
00:24:38.000 --> 00:24:44.000
v equals zero.
But, v equals zero is,
00:24:42.000 --> 00:24:48.000
in another language,
u equals t.
00:24:46.000 --> 00:24:52.000
t minus u equals zero,
or u equals t.
00:24:51.000 --> 00:24:57.000
In other words,
they enter the region where u
00:24:55.000 --> 00:25:01.000
equals zero,
and they leave where u is t,
00:25:00.000 --> 00:25:06.000
has the value of t.
And, how about the other guys?
00:25:05.000 --> 00:25:11.000
For which t's do I want to do
this?
00:25:07.000 --> 00:25:13.000
Well, I want to do it for all
these t values.
00:25:10.000 --> 00:25:16.000
Well, now, the t value here,
that's the starting one.
00:25:14.000 --> 00:25:20.000
Here, t is zero,
and here t is not zero.
00:25:16.000 --> 00:25:22.000
And, if I go out and cover the
whole first quadrant,
00:25:20.000 --> 00:25:26.000
I'll be letting t increase to
infinity.
00:25:23.000 --> 00:25:29.000
The sum of u and v,
I will be letting increase to
00:25:26.000 --> 00:25:32.000
infinity.
So, it's zero to infinity.
00:25:30.000 --> 00:25:36.000
So, all this is an exercise in
taking this double integral in
00:25:35.000 --> 00:25:41.000
(u, v) coordinates,
and changing it to this double
00:25:40.000 --> 00:25:46.000
integral, an equivalent double
integral over the same region,
00:25:46.000 --> 00:25:52.000
but now in (u,
t) coordinates.
00:25:48.000 --> 00:25:54.000
And now, that's the answer.
Somewhere here is the answer
00:25:54.000 --> 00:26:00.000
because, look,
since the first integration is
00:25:58.000 --> 00:26:04.000
with respect to u,
this guy can migrate outside
00:26:02.000 --> 00:26:08.000
because it doesn't involve u.
That only involves t,
00:26:08.000 --> 00:26:14.000
and t is only caught by the
second integration.
00:26:11.000 --> 00:26:17.000
So, I can put this outside.
And, what do I end up with?
00:26:15.000 --> 00:26:21.000
The integral from zero to
infinity of e to the negative s
00:26:18.000 --> 00:26:24.000
t times,
00:26:22.000 --> 00:26:28.000
what's left?
A funny expression,
00:26:24.000 --> 00:26:30.000
but you're on your desert
island and found it.
00:26:27.000 --> 00:26:33.000
This funny expression,
integral from zero to t,
00:26:30.000 --> 00:26:36.000
f of u, g of t minus u vu,
00:26:34.000 --> 00:26:40.000
in short,
the convolution,
00:26:37.000 --> 00:26:43.000
exactly the convolution.
So, all you have to do is get
00:26:42.000 --> 00:26:48.000
the idea that there might be a
formula, sit down,
00:26:45.000 --> 00:26:51.000
change variables and double
integral it, ego,
00:26:48.000 --> 00:26:54.000
you've got your formula.
Well, I would like to spend
00:26:52.000 --> 00:26:58.000
much of the rest of the
period--- in other words,
00:26:56.000 --> 00:27:02.000
that's how it relates to the
Laplace transform.
00:26:59.000 --> 00:27:05.000
That's how it comes out of the
Laplace transform.
00:27:04.000 --> 00:27:10.000
Here's how to use it,
calculate it either with the
00:27:07.000 --> 00:27:13.000
Laplace transform or directly
from the integral.
00:27:10.000 --> 00:27:16.000
And, of course,
you will solve problems,
00:27:13.000 --> 00:27:19.000
Laplace transform problems,
differential equations using
00:27:17.000 --> 00:27:23.000
the convolution.
But, I have to tell you that
00:27:20.000 --> 00:27:26.000
most people, convolution is very
important.
00:27:23.000 --> 00:27:29.000
And, most people who use it
don't use it in connection with
00:27:27.000 --> 00:27:33.000
the Laplace transform.
They use it for its own sake.
00:27:30.000 --> 00:27:36.000
The first place I learned that
outside of MIT people used a
00:27:34.000 --> 00:27:40.000
convolution was actually from my
daughter.
00:27:39.000 --> 00:27:45.000
She's an environmental
engineer, an environmental
00:27:41.000 --> 00:27:47.000
consultant.
She does risk assessment,
00:27:44.000 --> 00:27:50.000
and stuff like that.
But anyway, she had this paper
00:27:47.000 --> 00:27:53.000
on acid rain she was trying to
read for a client,
00:27:50.000 --> 00:27:56.000
and she said something about
calculating acid rain falls on
00:27:53.000 --> 00:27:59.000
soil.
And then, from there,
00:27:55.000 --> 00:28:01.000
the stuff leeches into a river.
But, things happen to it on the
00:27:58.000 --> 00:28:04.000
way.
Soil combines in various ways,
00:28:01.000 --> 00:28:07.000
reduces the acidity,
and things happen.
00:28:03.000 --> 00:28:09.000
Chemical reactions take place,
blah, blah, blah,
00:28:06.000 --> 00:28:12.000
blah.
Anyways, she said,
00:28:08.000 --> 00:28:14.000
well, then they calculated in
the end how much the river gets
00:28:11.000 --> 00:28:17.000
polluted.
But, she said it's convolution.
00:28:13.000 --> 00:28:19.000
She said, what's the
convolution?
00:28:15.000 --> 00:28:21.000
So, I told her she was too
young to learn about the
00:28:18.000 --> 00:28:24.000
convolution.
And she knows that I thought
00:28:20.000 --> 00:28:26.000
I'd better look it up first.
I mean, I, of course,
00:28:23.000 --> 00:28:29.000
knew the convolution was,
but I was a little puzzled at
00:28:26.000 --> 00:28:32.000
that application of it.
So, I read the paper.
00:28:29.000 --> 00:28:35.000
It was interesting.
And, in thinking about it,
00:28:33.000 --> 00:28:39.000
other people have come to me,
some guy with a problem about,
00:28:38.000 --> 00:28:44.000
they drilled ice cores in the
North Pole, and from the
00:28:41.000 --> 00:28:47.000
radioactive carbon and so on,
deducing various things about
00:28:46.000 --> 00:28:52.000
the climate 60 billion years
ago, and it was all convolution.
00:28:50.000 --> 00:28:56.000
He asked me if I could explain
that to him.
00:28:53.000 --> 00:28:59.000
So, let me give you sort of
all-purpose thing,
00:28:56.000 --> 00:29:02.000
a simple all-purpose model,
which can be adapted,
00:28:59.000 --> 00:29:05.000
which is very good way of
thinking of the convolution,
00:29:03.000 --> 00:29:09.000
in my opinion.
It's a problem of radioactive
00:29:08.000 --> 00:29:14.000
dumping.
It's in the notes,
00:29:11.000 --> 00:29:17.000
by the way.
So, I'm just,
00:29:13.000 --> 00:29:19.000
if you want to take a chance,
and just listen to what I'm
00:29:18.000 --> 00:29:24.000
saying rather that just
scribbling everything down,
00:29:23.000 --> 00:29:29.000
maybe you'll be able to figure
it out for the notes,
00:29:28.000 --> 00:29:34.000
also.
So, the problem is we have some
00:29:33.000 --> 00:29:39.000
factory, or a nuclear plant,
or some thing like that,
00:29:38.000 --> 00:29:44.000
is producing radioactive waste,
not always at the same rate.
00:29:44.000 --> 00:29:50.000
And then, it carts it,
dumps it on a pile somewhere.
00:29:49.000 --> 00:29:55.000
So, radioactive waste is
dumped, and there's a dumping
00:29:54.000 --> 00:30:00.000
function.
I'll call that f of t,
00:29:58.000 --> 00:30:04.000
the dump rate.
That's the dumping rate.
00:30:03.000 --> 00:30:09.000
Let's say t is in years.
You like to have units,
00:30:07.000 --> 00:30:13.000
and quantity,
kilograms, I don't know,
00:30:10.000 --> 00:30:16.000
whatever you want.
Now, what does the dumping rate
00:30:15.000 --> 00:30:21.000
mean?
The dumping rate means that if
00:30:18.000 --> 00:30:24.000
I have two times that are close
together, for example,
00:30:23.000 --> 00:30:29.000
two successive days,
midnight on two successive
00:30:27.000 --> 00:30:33.000
days, then there's a time
interval between them.
00:30:33.000 --> 00:30:39.000
I'll call that delta t.
To say the dumping rate is f of
00:30:38.000 --> 00:30:44.000
t means that the amount
dumped in this time interval,
00:30:43.000 --> 00:30:49.000
in the time interval from t1 to
t1 plus one is
00:30:49.000 --> 00:30:55.000
approximately,
not exactly,
00:30:52.000 --> 00:30:58.000
because the dumping rate isn't
even constant within this time
00:30:58.000 --> 00:31:04.000
interval.
But it's approximately the
00:31:02.000 --> 00:31:08.000
dumping rate times the time over
which the dumping is taking
00:31:09.000 --> 00:31:15.000
place.
That's what I mean by the dump
00:31:13.000 --> 00:31:19.000
rate.
And, it gets more and more
00:31:16.000 --> 00:31:22.000
accurate, the smaller the time
interval you take.
00:31:21.000 --> 00:31:27.000
Okay, now here's my problem.
The problem is,
00:31:26.000 --> 00:31:32.000
you start dumping at time t
equals zero.
00:31:33.000 --> 00:31:39.000
At time t equal t,
how much radioactive waste is
00:31:39.000 --> 00:31:45.000
in the pile?
00:31:55.000 --> 00:32:01.000
Now, what makes that problem
slightly complicated is
00:31:58.000 --> 00:32:04.000
radioactive waste decays.
If I put some at a certain day,
00:32:02.000 --> 00:32:08.000
and then go back several months
later and nothing's happened in
00:32:07.000 --> 00:32:13.000
between, I don't have the same
amount that I dumps because a
00:32:11.000 --> 00:32:17.000
fraction of it decayed.
I have less.
00:32:14.000 --> 00:32:20.000
And, our answer to the problem
must take account of,
00:32:18.000 --> 00:32:24.000
for each piece of waste,
how long it has been in the
00:32:22.000 --> 00:32:28.000
pile because that takes account
of how long it had to decay,
00:32:27.000 --> 00:32:33.000
and what it ends up as.
So, the calculation,
00:32:32.000 --> 00:32:38.000
the essential part of the
calculation will be that if you
00:32:37.000 --> 00:32:43.000
have an initial amount of this
substance, and it decays for a
00:32:43.000 --> 00:32:49.000
time, t, this is the amount left
at time t.
00:32:47.000 --> 00:32:53.000
This is the law of radioactive
decay.
00:32:51.000 --> 00:32:57.000
You knew that coming into
18.03, although,
00:32:55.000 --> 00:33:01.000
it's, of course,
a simple differential equation
00:33:00.000 --> 00:33:06.000
which produces it,
but I'll assume you simply know
00:33:05.000 --> 00:33:11.000
the answer.
k depends on the material,
00:33:10.000 --> 00:33:16.000
so I'm going to assume that the
nuclear plant dumps the same
00:33:14.000 --> 00:33:20.000
radioactive substance each time.
It's only one substance I'm
00:33:19.000 --> 00:33:25.000
calculating, and k is it.
So, assume the k is fixed.
00:33:23.000 --> 00:33:29.000
I don't have to change from one
k from one material to a k for
00:33:27.000 --> 00:33:33.000
another because it's mixing up
the stuff, just one material.
00:33:33.000 --> 00:33:39.000
Okay, and now let's calculate
it.
00:33:35.000 --> 00:33:41.000
Here's the idea.
I'll take the t-axis,
00:33:38.000 --> 00:33:44.000
but now I'm going to change its
name to the u-axis.
00:33:43.000 --> 00:33:49.000
You will see why in just a
second.
00:33:45.000 --> 00:33:51.000
It starts at zero.
I'm interested in what's
00:33:49.000 --> 00:33:55.000
happening at the time,
t.
00:33:51.000 --> 00:33:57.000
How much is left at time t?
So, I'm going to divide up the
00:33:56.000 --> 00:34:02.000
interval from zero to t on this
time axis into,
00:34:00.000 --> 00:34:06.000
well, here's u0,
the starting point,
00:34:03.000 --> 00:34:09.000
u1, u2, let's make this u1.
Oh, curses!
00:34:08.000 --> 00:34:14.000
u1, u2, u3, and so on.
Let's call this (u)n.
00:34:13.000 --> 00:34:19.000
So they're u(n + 1),
not that it matters.
00:34:18.000 --> 00:34:24.000
It doesn't matter.
Okay, now, the amount,
00:34:23.000 --> 00:34:29.000
so, what I'm going to do is
look at the amount,
00:34:28.000 --> 00:34:34.000
take the time interval from ui
to ui plus one.
00:34:36.000 --> 00:34:42.000
This is a time interval,
00:34:40.000 --> 00:34:46.000
delta u.
Divide it up into equal time
00:34:43.000 --> 00:34:49.000
intervals.
So, the amount dumped in the
00:34:46.000 --> 00:34:52.000
time interval from u(i) to u(i
plus one)
00:34:51.000 --> 00:34:57.000
is equal to approximately f of
u(i),
00:34:55.000 --> 00:35:01.000
the dumping function there,
times delta u.
00:35:00.000 --> 00:35:06.000
We calculated that before.
That's what the meaning of the
00:35:06.000 --> 00:35:12.000
dumping rate is.
By time t, how much has it
00:35:11.000 --> 00:35:17.000
decayed to?
It has decayed.
00:35:14.000 --> 00:35:20.000
How much is left,
in other words?
00:35:18.000 --> 00:35:24.000
Well, this is the starting
amount.
00:35:21.000 --> 00:35:27.000
So, the answer is going to be
it's f of (u)i times delta u
00:35:28.000 --> 00:35:34.000
times this factor,
which tells how much it decays,
00:35:34.000 --> 00:35:40.000
so, time.
So, this is the starting amount
00:35:39.000 --> 00:35:45.000
at time (u)i.
That's when it was first
00:35:41.000 --> 00:35:47.000
dumped, and this is the amount
that was dumped,
00:35:45.000 --> 00:35:51.000
times, multiply that by e to
the minus k times,
00:35:49.000 --> 00:35:55.000
now, what should I put up in
there?
00:35:51.000 --> 00:35:57.000
I have to put the length of
time that it had to decay.
00:35:55.000 --> 00:36:01.000
What is the length of time that
it had to decay?
00:36:00.000 --> 00:36:06.000
It was dumped at u(i).
I'm looking at time,
00:36:08.000 --> 00:36:14.000
t, it decayed for time length t
minus u i,
00:36:19.000 --> 00:36:25.000
the length of time it had all
the pile.
00:36:32.000 --> 00:36:38.000
So, the stuff that was dumped
in this time interval,
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
00:36:41.000 --> 00:36:47.000
left.
And now, all I have to do is
00:36:44.000 --> 00:36:50.000
add up that quantity for this
time, the stuff that was dumped
00:36:49.000 --> 00:36:55.000
in this time interval plus the
stuff dumped in,
00:36:54.000 --> 00:37:00.000
and so on, all the way up to
the stuff that was dumped
00:36:58.000 --> 00:37:04.000
yesterday.
And, the answer will be the
00:37:01.000 --> 00:37:07.000
total amount left at time,
t, that is not yet decayed will
00:37:06.000 --> 00:37:12.000
be approximately,
you add up the amount coming
00:37:10.000 --> 00:37:16.000
from the first time interval
plus the amount coming,
00:37:15.000 --> 00:37:21.000
and so on.
So, it will be f of u(i),
00:37:19.000 --> 00:37:25.000
I'll save the delta u for the
end, times e to the minus k
00:37:23.000 --> 00:37:29.000
times t minus u(i) times
delta u.
00:37:27.000 --> 00:37:33.000
So, these two parts represent
00:37:29.000 --> 00:37:35.000
the amount dumped,
and this is the decay factor.
00:37:33.000 --> 00:37:39.000
And, I had those up as I runs
from, well, where did I start?
00:37:37.000 --> 00:37:43.000
From one to n,
let's say.
00:37:39.000 --> 00:37:45.000
And now, let delta t go to
zero, in other words,
00:37:42.000 --> 00:37:48.000
make this delta u go to zero,
make this more accurate by
00:37:46.000 --> 00:37:52.000
taking finer and finer
subdivisions.
00:37:48.000 --> 00:37:54.000
In other words,
instead of looking every month
00:37:51.000 --> 00:37:57.000
to see how much was dumped,
let's look every week,
00:37:55.000 --> 00:38:01.000
every day, and so on,
to make this calculation more
00:37:58.000 --> 00:38:04.000
accurate.
And, the answer is,
00:38:00.000 --> 00:38:06.000
this approach is the exact
amount, which will be the
00:38:04.000 --> 00:38:10.000
integral.
This sum is a Riemann sum.
00:38:08.000 --> 00:38:14.000
It approaches the integral from
zero to, well,
00:38:12.000 --> 00:38:18.000
I'm adding it up from u1 equals
zero to un equals t,
00:38:18.000 --> 00:38:24.000
the final value.
So, it will be the integral
00:38:22.000 --> 00:38:28.000
from the starting point to the
ending point of f of u e to the
00:38:28.000 --> 00:38:34.000
minus k times t minus u to u.
00:38:34.000 --> 00:38:40.000
That's the answer to the
problem.
00:38:36.000 --> 00:38:42.000
It's given by this rather funny
looking integral.
00:38:39.000 --> 00:38:45.000
But, from this point of view,
it's entirely natural.
00:38:42.000 --> 00:38:48.000
It's a combination of the
dumping function.
00:38:44.000 --> 00:38:50.000
This doesn't care what the
material was.
00:38:47.000 --> 00:38:53.000
It only wants to know how much
was put on everyday.
00:38:50.000 --> 00:38:56.000
And, this part,
which doesn't care how much was
00:38:53.000 --> 00:38:59.000
put on each day,
it just is an intrinsic
00:38:55.000 --> 00:39:01.000
constant of the material
involving its decay rate.
00:39:00.000 --> 00:39:06.000
And, this total thing
represents the total amount.
00:39:04.000 --> 00:39:10.000
And that is,
what is it?
00:39:06.000 --> 00:39:12.000
It's the convolution of f of t
with what function?
00:39:11.000 --> 00:39:17.000
e to the minus k t.
It's the convolution of the
00:39:16.000 --> 00:39:22.000
dumping function and the decay
function.
00:39:19.000 --> 00:39:25.000
And, the convolution is exactly
the operation that you have to
00:39:25.000 --> 00:39:31.000
have to do that.
Okay, so, I think this is the
00:39:28.000 --> 00:39:34.000
most intuitive physical approach
to the meaning of the
00:39:33.000 --> 00:39:39.000
convolution.
In this particular,
00:39:37.000 --> 00:39:43.000
you can say,
well, that's very special.
00:39:39.000 --> 00:39:45.000
Okay, so it tells you what the
meaning of the convolution with
00:39:43.000 --> 00:39:49.000
an exponential is.
But, what about the convolution
00:39:46.000 --> 00:39:52.000
with all the other functions
we're going to have to use in
00:39:50.000 --> 00:39:56.000
this course.
They can all be interpreted
00:39:52.000 --> 00:39:58.000
just by being a little flexible
in your approach.
00:39:55.000 --> 00:40:01.000
I'll give you two examples of
this, well, three.
00:39:58.000 --> 00:40:04.000
First of all,
I'll use it for,
00:40:00.000 --> 00:40:06.000
in the problem set I ask you
about a bank account.
00:40:05.000 --> 00:40:11.000
That's not something any of you
are interested in.
00:40:08.000 --> 00:40:14.000
Okay, so, suppose instead I
dumped garbage --
00:40:16.000 --> 00:40:22.000
-- undecaying.
So, something that doesn't
00:40:19.000 --> 00:40:25.000
decay at all,
what's the answer going to be?
00:40:22.000 --> 00:40:28.000
Well, the calculation will be
exactly the same.
00:40:26.000 --> 00:40:32.000
It will be the convolution of
the dumping function.
00:40:30.000 --> 00:40:36.000
The only difference is that now
the garbage isn't going to
00:40:34.000 --> 00:40:40.000
decay.
So, no matter how long it's
00:40:37.000 --> 00:40:43.000
left, the same amount is going
to be left at the end.
00:40:40.000 --> 00:40:46.000
In other words,
I don't want to exponential
00:40:42.000 --> 00:40:48.000
decay function.
I want to function,
00:40:44.000 --> 00:40:50.000
one, the constant function,
one, because once I stick it on
00:40:48.000 --> 00:40:54.000
the pile, nothing happens to it.
It just stays there.
00:40:51.000 --> 00:40:57.000
So, it's going to be the
convolution of this one because
00:40:54.000 --> 00:41:00.000
this is constant.
It's undecaying --
00:41:04.000 --> 00:41:10.000
-- by the identical reasoning.
And so, what's the answer going
00:41:07.000 --> 00:41:13.000
to be?
It's going to be the integral
00:41:09.000 --> 00:41:15.000
from zero to t of f of u du.
00:41:12.000 --> 00:41:18.000
Now, that's an 18.01 problem.
00:41:14.000 --> 00:41:20.000
If I dump with a dumping rate,
f of u,
00:41:17.000 --> 00:41:23.000
and I dump from time zero to
time t, how much is on the pile?
00:41:20.000 --> 00:41:26.000
They don't give it.
They always give velocity
00:41:23.000 --> 00:41:29.000
problems, and problems of how to
slice up bread loaves,
00:41:26.000 --> 00:41:32.000
and stuff like that.
But, this is a real life
00:41:28.000 --> 00:41:34.000
problem.
If that's the dumping rate,
00:41:32.000 --> 00:41:38.000
and you dump for t days from
zero to time t,
00:41:35.000 --> 00:41:41.000
how much do you have left at
the end?
00:41:37.000 --> 00:41:43.000
Answer: the integral of f of u
du from zero to t.
00:41:42.000 --> 00:41:48.000
I'll give you another example.
00:41:46.000 --> 00:41:52.000
Suppose I wanted a dumping
function, suppose I wanted a
00:41:50.000 --> 00:41:56.000
function, wanted to interpret
something which grows like t,
00:41:54.000 --> 00:42:00.000
for instance.
All I want is a physical
00:41:57.000 --> 00:42:03.000
interpretation.
Well, I have to think,
00:42:01.000 --> 00:42:07.000
I'm making a pile of something,
a metaphorical pile,
00:42:04.000 --> 00:42:10.000
we don't actually have to make
a physical pile.
00:42:07.000 --> 00:42:13.000
And, the thing should be
growing like t.
00:42:09.000 --> 00:42:15.000
Well, what grows like t?
Not bacteria,
00:42:11.000 --> 00:42:17.000
they grow exponentially.
Before the lecture,
00:42:14.000 --> 00:42:20.000
I was trying to think of
something.
00:42:16.000 --> 00:42:22.000
So, I came up with chickens on
a chicken farm.
00:42:19.000 --> 00:42:25.000
Little baby chickens grow
linearly.
00:42:21.000 --> 00:42:27.000
All little animals,
anyway, I've observed that
00:42:23.000 --> 00:42:29.000
babies grow linearly,
at least for a while,
00:42:26.000 --> 00:42:32.000
thank God.
After a while,
00:42:27.000 --> 00:42:33.000
they taper off.
But, at the beginning,
00:42:32.000 --> 00:42:38.000
they eat every four hours or
whatever.
00:42:35.000 --> 00:42:41.000
And they eat the same amount,
pretty much.
00:42:39.000 --> 00:42:45.000
And, that adds up.
So, let's suppose this
00:42:43.000 --> 00:42:49.000
represents the linear growth of
chickens, of baby chicks.
00:42:48.000 --> 00:42:54.000
That makes them sound cuter,
less offensive.
00:42:52.000 --> 00:42:58.000
Okay, so, a farmer,
chicken farmer,
00:42:56.000 --> 00:43:02.000
whatever they call them,
is starting a new brood.
00:43:02.000 --> 00:43:08.000
So anyway, the hens lay at a
certain rate,
00:43:05.000 --> 00:43:11.000
and each of those are
incubated.
00:43:08.000 --> 00:43:14.000
And after a while,
little baby chicks come out.
00:43:12.000 --> 00:43:18.000
So, this will be the production
rate for new chickens.
00:43:23.000 --> 00:43:29.000
Okay, and it will be the
convolution which will tell you
00:43:26.000 --> 00:43:32.000
at time, t, the number of
kilograms.
00:43:29.000 --> 00:43:35.000
We'd better do this in
kilograms, I'm afraid.
00:43:32.000 --> 00:43:38.000
Now, that's not as heartless as
it seems.
00:43:35.000 --> 00:43:41.000
The number of kilograms of
chickens times t.
00:43:38.000 --> 00:43:44.000
[LAUGHTER] It really isn't
heartless because,
00:43:41.000 --> 00:43:47.000
after all, why would the farmer
want to know that?
00:43:44.000 --> 00:43:50.000
Well, because a certain number
of pounds of chicken eat a
00:43:48.000 --> 00:43:54.000
certain number of pounds of
chicken feed,
00:43:51.000 --> 00:43:57.000
and that's how much he has to
dump, must have to give them
00:43:55.000 --> 00:44:01.000
every day.
That's how he calculates his
00:43:57.000 --> 00:44:03.000
expenses.
So, he will have to know the
00:44:01.000 --> 00:44:07.000
convolution is,
or better yet,
00:44:03.000 --> 00:44:09.000
he will hire you,
who knows what the convolution
00:44:07.000 --> 00:44:13.000
is.
And you'll be able to tell him.
00:44:09.000 --> 00:44:15.000
Okay, why don't we stop there
and go to recitation tomorrow.
00:44:13.000 --> 00:44:19.000
I'll be doing important things.