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PROFESSOR STRANG: OK, thank
you for coming today.

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The day before Thanksgiving.
 

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Day before my
birthday, actually.

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So it's a special day.
 

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Everybody gets an
A for showing up.

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Even you.
 

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So, let's see.
 

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Last time, I wrote down these
formulas for the Fourier

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integral transform.
 

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And I thought I'd just write
them again so you kind of

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photograph them and
remember them.

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00:00:52 --> 00:00:54
They're easy to remember.
 

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As always, you take the
function, you multiply by

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e^(-ikx), and you integrate.
 

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To get the amount - so k
is my frequency variable.

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It could well have been omega
or some other variable.

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I stayed with k because it
was k in the Fourier series.

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So that's the calculation which
as always, I mean, these are

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00:01:21 --> 00:01:23
integrals that we may be able
to do if the function is

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00:01:23 --> 00:01:26
especially nice, or we may not.
 

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But that's the formula.
 

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00:01:28 --> 00:01:33
And then to reconstruct the
function, we combine all they

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e^(ikx)'s in that amount
to get f(x) back.

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OK, nice formula.
 

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So I did one example last
time, and now could

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I just double it up?
 

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This is also in the textbook
and so this is now going

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to be an even function.
 

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00:01:55 --> 00:02:02
Last time the example I
did was zero, up to x=0.

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00:02:04 --> 00:02:06
This time I'll make
it symmetric, make

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00:02:06 --> 00:02:08
the function even.
 

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00:02:08 --> 00:02:10
And then I have two pieces.
 

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In the integral.
 

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And if you remember what it
was, you remember that this,

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I'll just remind
you what we did.

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We wrote that it's e^-(a+ik)x.
 

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00:02:22 --> 00:02:27
 
 

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That was clear.
 

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00:02:29 --> 00:02:33
And then when we integrated we
got that same function divided

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by -(a+ik). a And then
we put in the limits.

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And the answer was, let
me maybe write the

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answer down here.
 

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Was just at x equal infinity
the limit was zero because

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this thing is tailing off.
 

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At x=0 this is one.
 

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It comes in with a minus
because that's the lower limit.

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So it was 1/(a+ik)
for the first half.

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And we were not surprised to
see this 1/k because the first

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half all by itself has that
jump, from zero to one.

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00:03:15 --> 00:03:20
So we see that jump
reflected in slow decay.

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Alright, but now I'm
making it even.

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What are you going to guess for
the rate of decay of f hat

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of k for this function?
 

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This function no
longer has a jump.

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But it does have - I don't know
were we saying ramp, or corner?

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This is not a smooth point
here, because the derivative

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going up is plus a, so I'll
just put a circle right, the

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derivative is a e^(ax),
and at x=0 that would

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be plus a going up.
 

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And here the derivative
is minus a, e^(-ax).

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Put in x=0, and the derivative
coming down is minus x.

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So there's a jump
in the derivative.

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So what, just before we see it,
what will you expect for the

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rate of decay of the transform?
 

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1/k to what power, now?
 

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So it didn't have a
jump, a jump was 1/k.

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This has a jump in the
derivative, so we're expecting

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1/k squared. k squared, it'll
be one order smoother.

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OK, you can easily
see that happen.

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Because this part, well this
is just e^(a-ik)x, which I'm

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going to integrate to get
this thing over a-ik.

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00:04:59 --> 00:05:03
And I'm going to plug in the
limits minus infinity and zero.

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00:05:03 --> 00:05:07
And at minus infinity I'll get
nothing, this e^(ax), that

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minus infinity will be zero.
 

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Where the function starts.
 

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Way down at zero.
 

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So and at x=0, this is a one.
 

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So I just get 1/(a+ik).
 

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Over a minus, thank you.
 

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Right, over a minus.
 

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And somehow it can't be an
accident that this is the

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complex conjugate
of that somehow.

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That's not a surprise.
 

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OK, so let's put those together
into a single fraction

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00:05:46 --> 00:05:47
and see what we have.
 

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So the denominator of
that fraction will

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be this times this.
 

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And that's the most
basic multiplication

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of complex numbers.
 

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That one times its
conjugate gives me what?

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00:06:02 --> 00:06:05
It gives me an a squared.
 

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00:06:05 --> 00:06:11
And what else? a+k squared
because i times minus i is

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plus k squared, and
no imaginary part.

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00:06:18 --> 00:06:23
There's a plus i k a and a
minus i k a, all we're seeing

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here is the sum of squares.
 

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The usual z times z bar.
 

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And in the numerator,
let's see.

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00:06:32 --> 00:06:36
When I put it over this, so
this was putting it over

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this common denominator.
 

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00:06:37 --> 00:06:43
So I should have an a-ik
going up on top there.

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And an a+ik going
up on top here.

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00:06:48 --> 00:06:49
Right?
 

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00:06:49 --> 00:06:50
Those are my two fractions.
 

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00:06:50 --> 00:06:53
That over this, and
that over this.

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And now that numerator
simplifies, oh look it's great.

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I'm getting a real answer.
 

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And because the minus ik and
the plus ik cancel, and

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00:07:05 --> 00:07:08
it's just the two way.
 

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00:07:08 --> 00:07:12
And probably no surprise
that somehow that that's

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00:07:12 --> 00:07:14
the jump in slope.
 

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00:07:14 --> 00:07:17
That must have something
to do with that 2a.

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00:07:17 --> 00:07:27
So we got a real even, Fourier
f hat from my real even f.

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And it decays like k squared.
 

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OK, so that's another
good example.

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A very useful example.
 

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00:07:35 --> 00:07:36
Right.
 

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I could add other examples.
 

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One quite, before I use that
in application, let's do

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just a few more examples.
 

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Suppose f(x) is the
delta function.

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00:07:56 --> 00:08:00
What's f hat of k?
 

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00:08:00 --> 00:08:07
Can you just plug in
f(x)=delta(x) here?

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00:08:07 --> 00:08:11
Do that integration, and what
does f hat of k come out to be?

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One.
 

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00:08:12 --> 00:08:17
Because if it's a delta
function in there at x, at x=0

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the spike is at x=0, so I
plug in x=0, I get one.

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00:08:21 --> 00:08:26
So we're kind of, we'd be
surprised if it wasn't

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00:08:26 --> 00:08:28
a constant, right?
 

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00:08:28 --> 00:08:33
A delta function in physical
space goes in, has all

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00:08:33 --> 00:08:35
frequencies in equal amounts.
 

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00:08:35 --> 00:08:39
And it's a constant
in frequency space.

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00:08:39 --> 00:08:44
Then there's one more that
takes a little trick to do, but

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00:08:44 --> 00:08:50
it's a very neat one. f(x)
is e to the minus x

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00:08:50 --> 00:08:53
squared over two.
 

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00:08:53 --> 00:08:58
Do you recognize that as an
important function, e to the

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00:08:58 --> 00:09:02
minus x squared or it usually
has that e to the minus x

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00:09:02 --> 00:09:05
squared over two or sometimes
an e to the minus x

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00:09:05 --> 00:09:09
squared over two sigma
squared, a rescaling?

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00:09:09 --> 00:09:15
But this would be the
bell shaped curve.

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00:09:15 --> 00:09:16
The bell shaped curve.
 

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00:09:16 --> 00:09:24
It decays, very quickly the
variance, because that's a two

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and not a two sigma squared,
the standard deviation

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00:09:28 --> 00:09:29
is one here.
 

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00:09:29 --> 00:09:31
The variance is one.
 

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00:09:31 --> 00:09:36
So it's a bell shaped curve
that has about 2/3 of its area

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00:09:36 --> 00:09:37
between minus one and one.
 

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00:09:37 --> 00:09:43
This is the all important
function for probability.

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00:09:43 --> 00:09:47
The normal distribution, the
Gaussian, both of those words

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00:09:47 --> 00:09:51
are used, it's the most
important probability

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00:09:51 --> 00:09:52
distribution.
 

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00:09:52 --> 00:09:58
I need a one over square root
of 2pi to make the total

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00:09:58 --> 00:09:59
probability be one.
 

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00:09:59 --> 00:10:02
But let me just leave it there.
 

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00:10:02 --> 00:10:04
That's a very, very
important function.

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00:10:04 --> 00:10:09
It's also going to be important
in the heat equation.

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00:10:09 --> 00:10:13
In math finance, shows
up all over the place.

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00:10:13 --> 00:10:19
And its integral would not be
easy to do from zero to one.

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00:10:19 --> 00:10:23
The integral of that function,
from zero to one, we

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00:10:23 --> 00:10:24
have tables of it.
 

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00:10:24 --> 00:10:26
To the nth place.
 

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00:10:26 --> 00:10:34
But so there's no simple,
elementary function whose

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00:10:34 --> 00:10:35
derivative is this.
 

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00:10:35 --> 00:10:39
That x squared is what's
making the integral tricky.

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00:10:39 --> 00:10:44
So from zero to one, we just
have to give it a name.

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00:10:44 --> 00:10:46
So, error function.
 

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00:10:46 --> 00:10:49
This would be ERF, error
function, the integral of that

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00:10:49 --> 00:10:52
thing correctly normalized.
 

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00:10:52 --> 00:10:57
I'm just saying, important,
important function.

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00:10:57 --> 00:11:02
And it turns out that integrals
from minus infinity to infinity

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00:11:02 --> 00:11:06
can be done so beautifully
by some trickery.

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00:11:06 --> 00:11:09
We can find the
transform of this.

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00:11:09 --> 00:11:12
We can find the transform of
this, we can do this integral

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00:11:12 --> 00:11:15
from minus infinity to
infinity, where we could not

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00:11:15 --> 00:11:16
do it from zero to one.
 

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00:11:16 --> 00:11:22
So I'll just write down
the answer for this guy.

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00:11:22 --> 00:11:25
Only because it's
such a key example.

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00:11:25 --> 00:11:30
It's some constant that
involves 2pi times e to the

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00:11:30 --> 00:11:34
minus k squared over two.
 

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00:11:34 --> 00:11:36
Boy, that's pretty amazing.
 

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00:11:36 --> 00:11:44
Right, the Fourier integral
transform, f hat of k, has the

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00:11:44 --> 00:11:48
same form as the function.
 

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00:11:48 --> 00:11:53
And of course this function
is infinitely smooth.

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00:11:53 --> 00:11:58
So its transform to k
is infinitely fast.

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00:11:58 --> 00:12:04
Yeah, there's no problems like
one over k squared here, there

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00:12:04 --> 00:12:10
are no bumps in the and
bell shaped curve.

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00:12:10 --> 00:12:15
So I won't push that example
except I'll use it.

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00:12:15 --> 00:12:19
What else should I say just
to, like, emphasize that

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00:12:19 --> 00:12:23
this is such an important
distribution in probability?

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00:12:23 --> 00:12:25
Why is it important
in probability?

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00:12:25 --> 00:12:26
That's the question.
 

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00:12:26 --> 00:12:32
Why does everybody assume if
you can get away with it and

201
00:12:32 --> 00:12:40
don't have any natural
alternative, everybody assumes

202
00:12:40 --> 00:12:44
that noise, whatever, is coming
with a normal distribution.

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00:12:44 --> 00:12:52
So, in other words, with a
sigma squared in there.

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00:12:52 --> 00:12:56
So a normal distribution, that
has mean zero because it's

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absolutely centered at the
origin and it has variance one,

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00:13:01 --> 00:13:05
but I could change the variance
and that would just spread out

207
00:13:05 --> 00:13:08
or tighten the bell
shaped curve.

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00:13:08 --> 00:13:12
Why is the bell shaped
curve so important?

209
00:13:12 --> 00:13:16
That's certainly, we're not
going to launch into theory

210
00:13:16 --> 00:13:20
of probability but it's
the central limit theorem.

211
00:13:20 --> 00:13:22
So let me just use those words.
 

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00:13:22 --> 00:13:26
The central limit theorem that
says that if I start with other

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00:13:26 --> 00:13:30
probability distributions,
like I'm flipping a coin.

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00:13:30 --> 00:13:34
I flip a coin a million times.
 

215
00:13:34 --> 00:13:39
Then the mean, and let's say
zero for tails, one for heads.

216
00:13:39 --> 00:13:42
OK, so I flip, flip, flip.
 

217
00:13:42 --> 00:13:48
Well, the mean of that, the
expected mean is what,

218
00:13:48 --> 00:13:50
half a million, right?
 

219
00:13:50 --> 00:13:53
Half tails, half heads.
 

220
00:13:53 --> 00:13:56
So if I give zero for tails,
one for heads and flip a

221
00:13:56 --> 00:13:58
million times the mean would
be about half a million.

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00:13:58 --> 00:14:04
And then, so let me
center the mean.

223
00:14:04 --> 00:14:07
I could have centered it by
taking minus one and one.

224
00:14:07 --> 00:14:08
That would have been smarter.
 

225
00:14:08 --> 00:14:11
Minus one and one.
 

226
00:14:11 --> 00:14:13
Minus one for tails, one
for heads would have

227
00:14:13 --> 00:14:16
had a mean of zero.
 

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00:14:16 --> 00:14:20
And then it would be natural if
I have a million of these to

229
00:14:20 --> 00:14:25
divide by a thousand, I think.
 

230
00:14:25 --> 00:14:27
Of course, the answer
won't be zero, right?

231
00:14:27 --> 00:14:30
If I do a million flips it's
not going to come out exactly

232
00:14:30 --> 00:14:34
half a million and
half a million.

233
00:14:34 --> 00:14:35
I'm remembering.
 

234
00:14:35 --> 00:14:37
I used to have a long
discussion with a

235
00:14:37 --> 00:14:41
nice guy in college.
 

236
00:14:41 --> 00:14:45
He ran for Mayor of
Boston, actually.

237
00:14:45 --> 00:14:53
But he had the idea that after
a million flips, suppose there

238
00:14:53 --> 00:14:56
had been more heads than tails.
 

239
00:14:56 --> 00:15:03
Then the next flip, he figured,
was more likely to be tails.

240
00:15:03 --> 00:15:06
I couldn't convince them that
this was not mathematically the

241
00:15:06 --> 00:15:09
right thing to think about.
 

242
00:15:09 --> 00:15:12
And all I did was say
don't go to Las Vegas.

243
00:15:12 --> 00:15:16
I mean, if you're thinking
that way save your money.

244
00:15:16 --> 00:15:18
So, anyway.
 

245
00:15:18 --> 00:15:20
But this is much studied.
 

246
00:15:20 --> 00:15:24
The variation what that
curve looks like, that's

247
00:15:24 --> 00:15:26
quite interesting.
 

248
00:15:26 --> 00:15:30
But my point is, that as the
number gets bigger and bigger

249
00:15:30 --> 00:15:34
and we scale it properly,
the distribution will

250
00:15:34 --> 00:15:38
approach the norm.
 

251
00:15:38 --> 00:15:39
All sorts of distributions.
 

252
00:15:39 --> 00:15:44
If I just repeat and repeat
experiments and scale it, the

253
00:15:44 --> 00:15:47
central limit theorem says
you're always going to

254
00:15:47 --> 00:15:49
the normal distribution.
 

255
00:15:49 --> 00:15:51
So that's highly important.
 

256
00:15:51 --> 00:15:54
OK, and it comes up
different places.

257
00:15:54 --> 00:15:57
And it's quite a neat function.
 

258
00:15:57 --> 00:16:02
OK, so that's some examples.
 

259
00:16:02 --> 00:16:06
Now, let me use, like every
topics that I introduce,

260
00:16:06 --> 00:16:08
I want to find a use for.
 

261
00:16:08 --> 00:16:11
So now, can I start
on this one?

262
00:16:11 --> 00:16:14
Constant coefficient
differential equations.

263
00:16:14 --> 00:16:16
I'm going to write down a
differential equation, which

264
00:16:16 --> 00:16:19
will look pretty much like
the ones we started

265
00:16:19 --> 00:16:22
this course with.
 

266
00:16:22 --> 00:16:27
And I could, well, let
me write it down.

267
00:16:27 --> 00:16:31
Minus d second u/dx squared,
we're used to that.

268
00:16:31 --> 00:16:33
Now let me put in an a^2*u.
 

269
00:16:35 --> 00:16:39
Which is a lower lower order
term, we could deal with that.

270
00:16:39 --> 00:16:40
Equals sum f(x).
 

271
00:16:43 --> 00:16:46
And now, because I want
to do Fourier integrals,

272
00:16:46 --> 00:16:50
I'm thinking all x.
 

273
00:16:50 --> 00:16:51
We're on the whole line.
 

274
00:16:51 --> 00:16:55
Instead of the interval (0,1)
where I might use Fourier

275
00:16:55 --> 00:17:00
series and have sine series or
cosine series, depending on

276
00:17:00 --> 00:17:01
the boundary conditions.
 

277
00:17:01 --> 00:17:04
Here, the boundary condition
is just everything

278
00:17:04 --> 00:17:06
drops off at infinity.
 

279
00:17:06 --> 00:17:07
And minus infinity.
 

280
00:17:07 --> 00:17:13
So all these functions we
can do these integrals.

281
00:17:13 --> 00:17:15
OK, so there's a good question.
 

282
00:17:15 --> 00:17:19
What's the solution?
 

283
00:17:19 --> 00:17:28
We could tackle it, but I want
to suggest to use Fourier.

284
00:17:28 --> 00:17:31
So it's not the only way,
but it's one way to see it.

285
00:17:31 --> 00:17:34
So now if I use, what do
I mean by using Fourier?

286
00:17:34 --> 00:17:37
It means I'm going to take
the Fourier integral

287
00:17:37 --> 00:17:41
transform of every term.
 

288
00:17:41 --> 00:17:43
So when I take the Fourier
transform of the right-hand

289
00:17:43 --> 00:17:46
side, I'm going to get
f hat of k, whatever.

290
00:17:46 --> 00:17:47
This is known, of course.
 

291
00:17:47 --> 00:17:50
This guy is given.
 

292
00:17:50 --> 00:17:51
That's the source term.
 

293
00:17:51 --> 00:17:54
And u is the unknown.
 

294
00:17:54 --> 00:17:59
OK, so I'm going to take the
Fourier transform of every

295
00:17:59 --> 00:18:03
term, well this is, a is a
constant. a had to be a

296
00:18:03 --> 00:18:09
constant, or I couldn't do, you
know if a depended on x this

297
00:18:09 --> 00:18:15
would be some multiplication
and the transform

298
00:18:15 --> 00:18:17
would be a mess.
 

299
00:18:17 --> 00:18:20
Fourier applies when you've
got constant coefficients and

300
00:18:20 --> 00:18:22
nice boundary conditions.
 

301
00:18:22 --> 00:18:24
And here our boundary
conditions are nice, they

302
00:18:24 --> 00:18:27
just go to zero fast.
 

303
00:18:27 --> 00:18:30
OK, so the transform
of this is, that's a

304
00:18:30 --> 00:18:34
constant. u hat of k.
 

305
00:18:34 --> 00:18:37
And what's the
transform of that?

306
00:18:37 --> 00:18:41
So this is our chance to use
probably the most important

307
00:18:41 --> 00:18:49
rule for Fourier integrals.
 

308
00:18:49 --> 00:18:52
Maybe you'll tell
me what it is.

309
00:18:52 --> 00:18:53
You should think what it is.
 

310
00:18:53 --> 00:18:56
If I take a derivative,
that's the rule.

311
00:18:56 --> 00:19:00
If I take a derivative
of the function, what's

312
00:19:00 --> 00:19:03
happening in frequencies?
 

313
00:19:03 --> 00:19:05
I could make that happen here.
 

314
00:19:05 --> 00:19:09
If I took the derivative, yeah.
 

315
00:19:09 --> 00:19:13
So maybe if I take the
derivative here, so here it's

316
00:19:13 --> 00:19:15
just remembering the rule.
 

317
00:19:15 --> 00:19:20
Suppose I take the derivative
of this equation.

318
00:19:20 --> 00:19:23
I get this integral,
and what would f' be?

319
00:19:23 --> 00:19:25
What would f hat, sorry.
 

320
00:19:25 --> 00:19:30
If I take the x derivative
of this, if I take the x

321
00:19:30 --> 00:19:32
derivative of this equation,
what happens on the

322
00:19:32 --> 00:19:36
right-hand side when I
take the x derivative?

323
00:19:36 --> 00:19:38
Down comes i k.
 

324
00:19:38 --> 00:19:40
Down comes ik.
 

325
00:19:40 --> 00:19:42
So when ik is coming down,
I won't even finish

326
00:19:42 --> 00:19:44
that equation.
 

327
00:19:44 --> 00:19:49
And ik is coming down, when
I take the derivative.

328
00:19:49 --> 00:19:51
So the derivative, the
transform, is multiplied by

329
00:19:51 --> 00:19:56
ik, higher frequencies are
emphasized now because

330
00:19:56 --> 00:19:57
of that k factor.
 

331
00:19:57 --> 00:20:01
And now if I take two
derivatives, I bring ik twice.

332
00:20:01 --> 00:20:04
Because i squared k squared,
the i squared and the

333
00:20:04 --> 00:20:07
minus give me a plus.
 

334
00:20:07 --> 00:20:12
So that's just k
squared. u hat, of k.

335
00:20:12 --> 00:20:16
OK with that?
 

336
00:20:16 --> 00:20:22
And now, we get an immediate
formula for u hat

337
00:20:22 --> 00:20:25
of k, the solution.
 

338
00:20:25 --> 00:20:28
Well, it's the solution but
it's in frequency space.

339
00:20:28 --> 00:20:31
If we wanted to know it in
x space, as we do, we've

340
00:20:31 --> 00:20:33
got to transform back.
 

341
00:20:33 --> 00:20:34
But what do we get here?
 

342
00:20:34 --> 00:20:38
It's just f hat of k.
 

343
00:20:38 --> 00:20:42
Divided by, this is just
multiplied by a squared

344
00:20:42 --> 00:20:47
plus k squared.
 

345
00:20:47 --> 00:20:52
OK, so that's the answer.
 

346
00:20:52 --> 00:20:55
In frequency space.
 

347
00:20:55 --> 00:20:56
That was simple.
 

348
00:20:56 --> 00:21:01
And then if I wanted it
in x space, I take the

349
00:21:01 --> 00:21:02
reverse transform.
 

350
00:21:02 --> 00:21:08
Notice that this is, it here
are the same three steps that I

351
00:21:08 --> 00:21:12
emphasize all the time about
using eigenvectors

352
00:21:12 --> 00:21:13
and eigenvalues.
 

353
00:21:13 --> 00:21:16
Do you remember those
three steps for solving

354
00:21:16 --> 00:21:17
differential equations?
 

355
00:21:17 --> 00:21:20
Difference equations, linear
equations, whatever?

356
00:21:20 --> 00:21:26
The three steps were, find
that coefficients, expand

357
00:21:26 --> 00:21:29
everything in eigenfunctions.
 

358
00:21:29 --> 00:21:31
I won't write, I'll talk.
 

359
00:21:31 --> 00:21:35
The three steps were expand in
eigenfunctions, follow each

360
00:21:35 --> 00:21:38
eigenfunction function
separately, that was the

361
00:21:38 --> 00:21:41
trivial step with just a
division like this division.

362
00:21:41 --> 00:21:48
And then use those coefficients
of the eigenfunctions, combine

363
00:21:48 --> 00:21:51
them all back to
get the answer.

364
00:21:51 --> 00:21:51
Right?
 

365
00:21:51 --> 00:21:55
Step one, write it in the
right basis, step two

366
00:21:55 --> 00:21:57
easy in that basis.
 

367
00:21:57 --> 00:22:02
Step three go back to
your physical space.

368
00:22:02 --> 00:22:04
We're doing exactly
the same thing here.

369
00:22:04 --> 00:22:09
These e^(ikx)'s are the
eigenfunctions of this thing.

370
00:22:09 --> 00:22:11
They're the eigenfunctions.
 

371
00:22:11 --> 00:22:15
And here the eigenvalue of
this stuff is k squared

372
00:22:15 --> 00:22:16
plus a squared.
 

373
00:22:16 --> 00:22:19
And that's what we divided by.
 

374
00:22:19 --> 00:22:22
And then the final job of
going back from u hat

375
00:22:22 --> 00:22:28
to u, so write u there.
 

376
00:22:28 --> 00:22:37
Can I do, this f now, it's
really u that I'm wanting to

377
00:22:37 --> 00:22:39
bring back to physical space.
 

378
00:22:39 --> 00:22:47
So just for the sake of your --
I see -- it, let me put a u in.

379
00:22:47 --> 00:22:54
So that's the answer in a way.
 

380
00:22:54 --> 00:22:57
It's the answer, it's a
formula for the answer.

381
00:22:57 --> 00:23:02
It did depend on our being
able to do two integrals.

382
00:23:02 --> 00:23:07
By the integral from f to f hat
may not have been easy, and

383
00:23:07 --> 00:23:11
then the integral from u hat
back to u, this integral,

384
00:23:11 --> 00:23:12
might not have been easy.
 

385
00:23:12 --> 00:23:14
So it's a formula.
 

386
00:23:14 --> 00:23:18
OK, now I want to go with
it a little longer.

387
00:23:18 --> 00:23:23
Because I want to show you how
the delta function pays off.

388
00:23:23 --> 00:23:29
So let me do the example where
f(x) is the delta function.

389
00:23:29 --> 00:23:35
So now we're really close to
where this course began.

390
00:23:35 --> 00:23:38
Differential equation
with a delta function.

391
00:23:38 --> 00:23:41
The only new thing is, we're
not on an interval we're

392
00:23:41 --> 00:23:43
on the whole line.
 

393
00:23:43 --> 00:23:47
So I take transforms, so
what's the transform now

394
00:23:47 --> 00:23:51
of this specific f(x) is?
 

395
00:23:51 --> 00:23:52
One.
 

396
00:23:52 --> 00:23:54
We just saw.
 

397
00:23:54 --> 00:23:57
OK, so now we get a one there.
 

398
00:23:57 --> 00:23:59
Now we just divide by here.
 

399
00:23:59 --> 00:24:04
And we've got a one here.
 

400
00:24:04 --> 00:24:08
So we were able to go, this was
an integral we could easily do,

401
00:24:08 --> 00:24:13
to get from delta to delta
hat, which was just one.

402
00:24:13 --> 00:24:19
And fantastically, this is an
integral to go back to u(x),

403
00:24:19 --> 00:24:27
to go back to u(x), that's
an integral we can do.

404
00:24:27 --> 00:24:29
Well, you may ask
how can we do it.

405
00:24:29 --> 00:24:33
How do I find the u(x)
that has this transform?

406
00:24:33 --> 00:24:41
Well, I either use complex
variables to do integrals like

407
00:24:41 --> 00:24:46
this, residue methods that are
in Chapter 5, or I

408
00:24:46 --> 00:24:48
look in a table.
 

409
00:24:48 --> 00:24:50
Or I look at the
blackboard over there.

410
00:24:50 --> 00:24:52
I think that's the best way.
 

411
00:24:52 --> 00:24:54
Look at this blackboard.
 

412
00:24:54 --> 00:24:54
Right?
 

413
00:24:54 --> 00:24:58
Because this is the
answer we got.

414
00:24:58 --> 00:25:02
We got that same answer apart
from a constant factor 2a.

415
00:25:03 --> 00:25:05
So this is our function.
 

416
00:25:05 --> 00:25:08
This is our solution,
u(x) is this.

417
00:25:08 --> 00:25:10
What am I going to call that?
 

418
00:25:10 --> 00:25:12
Two-sided pulse?
 

419
00:25:12 --> 00:25:15
I'll call that the
two-sided pulse?

420
00:25:15 --> 00:25:20
Maybe I should give it a name
but I'll just write out those

421
00:25:20 --> 00:25:21
words two-sided pulse.
 

422
00:25:21 --> 00:25:24
That's it divided by 2a.
 

423
00:25:25 --> 00:25:28
So we've got the answer.
 

424
00:25:28 --> 00:25:30
Let me just make a
little more space here.

425
00:25:30 --> 00:25:36
This was one over a squared
plus k squared, and now having

426
00:25:36 --> 00:25:40
seen that already, I just
say yep, that must be it.

427
00:25:40 --> 00:25:47
It's the two-sided pulse,
and I have to divide by 2a.

428
00:25:49 --> 00:25:53
Do you see that that's
the correct answer?

429
00:25:53 --> 00:25:57
We can substitute that in the
equation and see that it works.

430
00:25:57 --> 00:25:59
I mean, so we have
solved the problem.

431
00:25:59 --> 00:26:04
We have solved the problem when
the right side was delta.

432
00:26:04 --> 00:26:05
Let's put it into the equation.
 

433
00:26:05 --> 00:26:08
So this is just because we
did this, it's nice to do it

434
00:26:08 --> 00:26:10
again after all this time.
 

435
00:26:10 --> 00:26:13
So I put it in the equation.
 

436
00:26:13 --> 00:26:15
This two-sided pulse over 2a.
 

437
00:26:15 --> 00:26:18
So what's my equation?
 

438
00:26:18 --> 00:26:22
Well, this is zero
most of the time.

439
00:26:22 --> 00:26:26
So I believe that if I
plug in this function, it

440
00:26:26 --> 00:26:29
will give me the zero.
 

441
00:26:29 --> 00:26:30
Do you want to just plug it in?
 

442
00:26:30 --> 00:26:36
I believe that if I plug in
a^(-x), or a^x, either one,

443
00:26:36 --> 00:26:38
can I just check that
you try u=e^(-ax).

444
00:26:38 --> 00:26:41
 
 

445
00:26:41 --> 00:26:44
Put it in and just
see that I get zero.

446
00:26:44 --> 00:26:48
Because yeah, two derivatives
bring down a squared

447
00:26:48 --> 00:26:49
with a minus.
 

448
00:26:49 --> 00:26:51
And there's a squared
with a plus.

449
00:26:51 --> 00:26:53
It works, right?
 

450
00:26:53 --> 00:26:58
Two derivatives of this
function bring down minus a

451
00:26:58 --> 00:27:01
twice, so that's a squared.
 

452
00:27:01 --> 00:27:04
So it's minus a squared,
plus a squared.

453
00:27:04 --> 00:27:05
Works.
 

454
00:27:05 --> 00:27:10
And then, of course, the
important point is x=0

455
00:27:10 --> 00:27:12
where the spike is.
 

456
00:27:12 --> 00:27:18
What happens at the
spike, going back to the

457
00:27:18 --> 00:27:20
beginning of the course?
 

458
00:27:20 --> 00:27:23
This term is going to be
unimportant compared

459
00:27:23 --> 00:27:25
to this term.
 

460
00:27:25 --> 00:27:28
What do I see?
 

461
00:27:28 --> 00:27:34
With -u'' equal a spike, what
was the solution to that?

462
00:27:34 --> 00:27:36
u had a corner, right?
 

463
00:27:36 --> 00:27:41
The slope of u, what
did the slope of u do?

464
00:27:41 --> 00:27:43
It dropped by one,
was that right?

465
00:27:43 --> 00:27:46
The slope of u dropped by one.
 

466
00:27:46 --> 00:27:51
We used to have corners going
up and down and the difference

467
00:27:51 --> 00:27:53
between the slopes was one.
 

468
00:27:53 --> 00:27:58
And here, the difference
between the slopes, ah, look.

469
00:27:58 --> 00:28:06
The slope has dropped by 2a,
and when we divided by the

470
00:28:06 --> 00:28:12
2a, it was just right.
 

471
00:28:12 --> 00:28:18
And now, when I divide by the
2a, this has a slope of 1/2.

472
00:28:18 --> 00:28:21
This has a slope of minus
1/2, the drop is one.

473
00:28:21 --> 00:28:23
And we're right.
 

474
00:28:23 --> 00:28:25
It solves the equation.
 

475
00:28:25 --> 00:28:27
Nobody doubted that.
 

476
00:28:27 --> 00:28:29
OK, so that's great.
 

477
00:28:29 --> 00:28:35
We have found the solution
to this equation, when

478
00:28:35 --> 00:28:39
the right side is delta.
 

479
00:28:39 --> 00:28:40
Good.
 

480
00:28:40 --> 00:28:43
Now, can I ask you do
you remember the name?

481
00:28:43 --> 00:28:47
There's a special name for
the solution when the right

482
00:28:47 --> 00:28:51
side is a delta function.
 

483
00:28:51 --> 00:28:53
Whose name is
associated with that?

484
00:28:53 --> 00:28:56
So that this particular
u, I'm going to give

485
00:28:56 --> 00:28:57
it another letter.
 

486
00:28:57 --> 00:29:04
It's the particular u, the
special u when the right side

487
00:29:04 --> 00:29:09
is delta, and whose name is
associated with that solution?

488
00:29:09 --> 00:29:11
Green.
 

489
00:29:11 --> 00:29:13
It's the Green's function.
 

490
00:29:13 --> 00:29:14
Green's function.
 

491
00:29:14 --> 00:29:15
The famous Green's function.
 

492
00:29:15 --> 00:29:19
Green's function is just like
an inverse to the problem.

493
00:29:19 --> 00:29:22
This is like having an identity
on the right-hand side.

494
00:29:22 --> 00:29:24
It's like there it is.
 

495
00:29:24 --> 00:29:27
So let me just use G
for Green's function.

496
00:29:27 --> 00:29:31
So that's the Fourier transform
of the Green's function, and

497
00:29:31 --> 00:29:33
this is the Green's function.
 

498
00:29:33 --> 00:29:35
This is the Green's function.
 

499
00:29:35 --> 00:29:42
Now I can give it its name,
Green's function, when

500
00:29:42 --> 00:29:43
I divide by the 2a.
 

501
00:29:45 --> 00:29:51
And now the slope is 1/2 going
up, and minus 1/2 coming down.

502
00:29:51 --> 00:29:53
And it's all right.
 

503
00:29:53 --> 00:29:56
OK, so we found the
Green's function.

504
00:29:56 --> 00:29:59
We found the fundamental
solution to the equation,

505
00:29:59 --> 00:30:03
and this is it.
 

506
00:30:03 --> 00:30:05
It was straight lines, right?
 

507
00:30:05 --> 00:30:12
It was straight lines in the
first weeks of the course.

508
00:30:12 --> 00:30:16
But now there's an exponential
drop-off caused by

509
00:30:16 --> 00:30:20
this additional term.
 

510
00:30:20 --> 00:30:25
OK, good.
 

511
00:30:25 --> 00:30:27
So that's straightforward.
 

512
00:30:27 --> 00:30:32
Depending on our being able to
recognize or do the transform

513
00:30:32 --> 00:30:36
back to the x space.
 

514
00:30:36 --> 00:30:41
Now comes the question, what
about the original f(x)?

515
00:30:41 --> 00:30:43
 
 

516
00:30:43 --> 00:30:45
How can the Green's
function be used?

517
00:30:45 --> 00:30:51
So you're seeing now what use
is this Green's function?

518
00:30:51 --> 00:30:54
With that right-hand side,
when the right-hand side

519
00:30:54 --> 00:30:56
is something different?
 

520
00:30:56 --> 00:30:58
When the right-hand side
is some different f(x)?

521
00:30:58 --> 00:31:08
So let me go back to
an f(x) on the right.

522
00:31:08 --> 00:31:14
And then there's an f hat
of k, after the transform.

523
00:31:14 --> 00:31:21
How can I use the Green's
function for a general source?

524
00:31:21 --> 00:31:24
The general source
term, a general load?

525
00:31:24 --> 00:31:30
This is a fundamental idea.
 

526
00:31:30 --> 00:31:33
I would say fundamental.
 

527
00:31:33 --> 00:31:37
How do you use the
Green's function?

528
00:31:37 --> 00:31:40
And remember, the Green's
function is like telling

529
00:31:40 --> 00:31:42
you the inverse matrix.
 

530
00:31:42 --> 00:31:44
So it can't be too hard.
 

531
00:31:44 --> 00:31:47
It's like solving a linear
system when you know

532
00:31:47 --> 00:31:52
the inverse matrix.
 

533
00:31:52 --> 00:31:56
So that's the analogy, but
let's just focus on the

534
00:31:56 --> 00:32:00
particular question.
 

535
00:32:00 --> 00:32:04
I think the intuition, you
should have an intuition for

536
00:32:04 --> 00:32:07
how the Green's function works.
 

537
00:32:07 --> 00:32:09
So the Green's function
was the solution when the

538
00:32:09 --> 00:32:12
source term was a delta.
 

539
00:32:12 --> 00:32:14
And here's the intuition.
 

540
00:32:14 --> 00:32:19
It's rough, but it works.
 

541
00:32:19 --> 00:32:26
Any source term, f(x),
is in some way a

542
00:32:26 --> 00:32:29
combination of delta.
 

543
00:32:29 --> 00:32:35
If f(x) is a combination of
deltas, then our answer u(x) is

544
00:32:35 --> 00:32:39
the same combination of the
Green's function, right?

545
00:32:39 --> 00:32:45
If the right-hand side is some
combination of this special

546
00:32:45 --> 00:32:48
delta, then the solution
will be the same.

547
00:32:48 --> 00:32:50
This is just linearity.
 

548
00:32:50 --> 00:32:55
Super position, whatever short
or long word you like to use.

549
00:32:55 --> 00:32:59
So if I can make sense of
that statement, that

550
00:32:59 --> 00:33:08
f(x) is a combination of
deltas, then I'm in.

551
00:33:08 --> 00:33:12
Now, what do I mean by a
combination of deltas?

552
00:33:12 --> 00:33:18
I mean, well, those deltas are
going to be shifted deltas.

553
00:33:18 --> 00:33:21
Obviously the single delta,
delta(x), is a spike

554
00:33:21 --> 00:33:23
at the origin.
 

555
00:33:23 --> 00:33:26
That's only one point.
 

556
00:33:26 --> 00:33:30
I want to combine delta
of x and its shifts.

557
00:33:30 --> 00:33:35
So I'm going to have to
expecting to be using G(x),

558
00:33:35 --> 00:33:37
and its shifts, right?
 

559
00:33:37 --> 00:33:40
OK so now I'll just
say this again.

560
00:33:40 --> 00:33:43
I'm thinking of f(x) as a
combination of delta and its

561
00:33:43 --> 00:33:48
shifts, and then the solution u
will be the same combination

562
00:33:48 --> 00:33:51
of G(x) and its shifts.
 

563
00:33:51 --> 00:33:55
So now you just have to
tell me what combination.

564
00:33:55 --> 00:34:01
What combination of
delta and its shifts?

565
00:34:01 --> 00:34:03
Maybe you'll allow me.
 

566
00:34:03 --> 00:34:09
Let me just do this
maybe on that board.

567
00:34:09 --> 00:34:15
I just can't help writing
down the discrete case.

568
00:34:15 --> 00:34:20
So, in the discrete case, the
delta vector corresponds to

569
00:34:20 --> 00:34:22
something like .
 

570
00:34:22 --> 00:34:23
Right?
 

571
00:34:23 --> 00:34:26
That was a typical delta
vector with a one in

572
00:34:26 --> 00:34:28
the zeroth position.
 

573
00:34:28 --> 00:34:33
Then its shifts would be <0,
1, 0, 0>, that'd be a shift.

574
00:34:33 --> 00:34:37
And another shift would
be .

575
00:34:37 --> 00:34:41
And another shift would
be .

576
00:34:41 --> 00:34:44
So now there is the delta
vector and its shifts.

577
00:34:44 --> 00:34:46
These four guys.
 

578
00:34:46 --> 00:34:50
OK, now suppose my f,
my right-hand side,

579
00:34:50 --> 00:34:54
is .
 

580
00:34:54 --> 00:34:59
I want to write that as a
combination of those deltas.

581
00:34:59 --> 00:35:03
This would be in the case when
if I know the solution for each

582
00:35:03 --> 00:35:07
of these guys, the Green's
function, the inverse matrix.

583
00:35:07 --> 00:35:11
Everybody sees that if I know
the solution to those four,

584
00:35:11 --> 00:35:13
I know the inverse matrix.
 

585
00:35:13 --> 00:35:14
Right?
 

586
00:35:14 --> 00:35:16
Because if I can solve with
those four right-hand sides,

587
00:35:16 --> 00:35:21
those four solutions are the
columns of the inverse matrix.

588
00:35:21 --> 00:35:22
Right?
 

589
00:35:22 --> 00:35:25
You remember that if I had a
matrix A and I was looking

590
00:35:25 --> 00:35:29
for its inverse, I solve
A A inverse equal I.

591
00:35:29 --> 00:35:35
And I is just these four guys.
 

592
00:35:35 --> 00:35:42
So a inverse is the solution
from these four guys.

593
00:35:42 --> 00:35:46
OK, now everybody's going to
tell me what's the solution

594
00:35:46 --> 00:35:51
for this right-hand
side ?

595
00:35:51 --> 00:35:57
Suppose this guy has solution,
so a inverse, the columns of

596
00:35:57 --> 00:36:02
a inverse are this
Green's function.

597
00:36:02 --> 00:36:05
This Green's function
with a shift.

598
00:36:05 --> 00:36:10
Maybe SG, Green's
function with a shift.

599
00:36:10 --> 00:36:14
Yeah, S squared and G, Green's
function with a double shift.

600
00:36:14 --> 00:36:17
S cubed G; I'm just
cooking up, I never used

601
00:36:17 --> 00:36:18
these letters before.
 

602
00:36:18 --> 00:36:20
But what's the answer?
 

603
00:36:20 --> 00:36:24
Then u is what?
 

604
00:36:24 --> 00:36:28
It's one times the
Green's function.

605
00:36:28 --> 00:36:34
And it's two times
the guy, right?

606
00:36:34 --> 00:36:39
This f is just, this f is just,
is one times the Green's

607
00:36:39 --> 00:36:42
function, two times
the shifted.

608
00:36:42 --> 00:36:45
Three times the double
shifted, and seven

609
00:36:45 --> 00:36:48
times the triple shift.
 

610
00:36:48 --> 00:36:49
Right?
 

611
00:36:49 --> 00:36:56
Just taking four minutes to do
something simple because over

612
00:36:56 --> 00:37:01
there, when I use the
continuous case it'll look a

613
00:37:01 --> 00:37:04
little strange, but
here it's so easy.

614
00:37:04 --> 00:37:08
That'll involve integrals,
this involves a sum.

615
00:37:08 --> 00:37:09
So what is it?
 

616
00:37:09 --> 00:37:13
I have G, twice the shift of G.
 

617
00:37:13 --> 00:37:17
Three times the double shift
of G, and seven times

618
00:37:17 --> 00:37:20
the triple shift of G.
 

619
00:37:20 --> 00:37:21
Right?
 

620
00:37:21 --> 00:37:29
By linearity, by superposition,
if this is my f, this is my u.

621
00:37:29 --> 00:37:31
Everybody's with
me here, right?

622
00:37:31 --> 00:37:34
That if I take a combination
of f's, the answer

623
00:37:34 --> 00:37:37
is the corresponding
combination of G's.

624
00:37:37 --> 00:37:42
OK, good.
 

625
00:37:42 --> 00:37:45
I didn't mean that.
 

626
00:37:45 --> 00:37:49
That wasn't so great.
 

627
00:37:49 --> 00:37:54
That shouldn't have been
called, I didn't mean to call

628
00:37:54 --> 00:37:56
those the Green's functions.
 

629
00:37:56 --> 00:37:59
I meant to call those
the deltas, right?

630
00:37:59 --> 00:38:01
And the Green's functions
were the answers.

631
00:38:01 --> 00:38:05
So this a shift to delta, this
is a doubly shift to delta,

632
00:38:05 --> 00:38:08
this is a triply shift to
delta, and now this is one of

633
00:38:08 --> 00:38:14
the delta, two of the
shifted delta, three of the

634
00:38:14 --> 00:38:17
doubly shifted delta.
 

635
00:38:17 --> 00:38:21
The f is a combination of
deltas and its shifts.

636
00:38:21 --> 00:38:25
The u is a combination of
Green's function, second shift.

637
00:38:25 --> 00:38:29
Apologies for making that
mistake, but maybe it's

638
00:38:29 --> 00:38:31
brought us back to the point.
 

639
00:38:31 --> 00:38:40
So the point is, express your
function f, your source as a

640
00:38:40 --> 00:38:43
combination of deltas,
just exactly our plan.

641
00:38:43 --> 00:38:48
Then u is the same combination
of the G's with the same shift.

642
00:38:48 --> 00:38:52
Alright, now back here.
 

643
00:38:52 --> 00:38:59
How do I express f(x),
in what way is f(x) a

644
00:38:59 --> 00:39:03
combination of deltas?
 

645
00:39:03 --> 00:39:08
I mean, slow down just
to see that point.

646
00:39:08 --> 00:39:18
How much of delta, of the spike
at x=3, how much of delta(x-3),

647
00:39:18 --> 00:39:24
so that's a spike
at three, right?

648
00:39:24 --> 00:39:27
How much of that do you
think I need in f(x)?

649
00:39:27 --> 00:39:31
 
 

650
00:39:31 --> 00:39:34
f of? f(3).
 

651
00:39:36 --> 00:39:40
Whatever f is at that
point, x=3, that's the

652
00:39:40 --> 00:39:42
amount I need there.
 

653
00:39:42 --> 00:39:43
So this would be f(3).
 

654
00:39:45 --> 00:39:51
So that that part would sort of
have the right pep, the right

655
00:39:51 --> 00:39:56
punch at the point three.
 

656
00:39:56 --> 00:40:01
Now, three could be
any point, that's any

657
00:40:01 --> 00:40:03
point along the line.
 

658
00:40:03 --> 00:40:07
Let me, I can't use x for other
points on the line because

659
00:40:07 --> 00:40:09
I've got an x in the formula.
 

660
00:40:09 --> 00:40:12
Let me use t.
 

661
00:40:12 --> 00:40:17
So this was t equal to three.
 

662
00:40:17 --> 00:40:20
Now I want to do
it at all points.

663
00:40:20 --> 00:40:24
I want to take f(2), and
f(pi), and f at everything

664
00:40:24 --> 00:40:25
and put them together.
 

665
00:40:25 --> 00:40:28
And of course putting them
together in the continuous

666
00:40:28 --> 00:40:32
case means not sum, as I
did there, but integrate.

667
00:40:32 --> 00:40:38
So I have to, now I'm going to
change three to a t, because

668
00:40:38 --> 00:40:54
this is the amount of f of, so
that's the amount, that's the

669
00:40:54 --> 00:40:59
delta functions which spike at
t, multiplied by f(t), and now

670
00:40:59 --> 00:41:01
how do I get f(x) out of this?
 

671
00:41:01 --> 00:41:05
I put them together. dt.
 

672
00:41:05 --> 00:41:08
I add up, this is the
combination I've

673
00:41:08 --> 00:41:10
been talking about.
 

674
00:41:10 --> 00:41:13
So this is like any f.
 

675
00:41:13 --> 00:41:20
This is like a crazy
delta function identity.

676
00:41:20 --> 00:41:23
Actually, it's not crazy,
it's the identity we've

677
00:41:23 --> 00:41:25
used our whole lives.
 

678
00:41:25 --> 00:41:28
Or at least our whole
18.085 lives, which is

679
00:41:28 --> 00:41:31
all that matters, right?
 

680
00:41:31 --> 00:41:35
OK, so I'm integrating
a delta function.

681
00:41:35 --> 00:41:39
And I want to see, do
I get that answer?

682
00:41:39 --> 00:41:41
And you're going to say
absolutely, clearly

683
00:41:41 --> 00:41:43
you get that answer.
 

684
00:41:43 --> 00:41:44
No.
 

685
00:41:44 --> 00:41:47
Everybody knows that if I
integrate something times the

686
00:41:47 --> 00:41:52
delta function that I plug in
at the point t=x, where that

687
00:41:52 --> 00:41:57
spike happens, I put t=x,
I get f(x), correct.

688
00:41:57 --> 00:42:00
So that's like any, that's
an identity or whatever

689
00:42:00 --> 00:42:02
you would like to call it.
 

690
00:42:02 --> 00:42:06
And now just tell me
the final answer.

691
00:42:06 --> 00:42:09
Let me put it on
the board above.

692
00:42:09 --> 00:42:14
Now, so this was, this
expressed my f(x) as a

693
00:42:14 --> 00:42:16
combination of delta.
 

694
00:42:16 --> 00:42:20
Just the way over here I
expressed  as

695
00:42:20 --> 00:42:22
a combination of delta.
 

696
00:42:22 --> 00:42:24
Now, what's that u?
 

697
00:42:24 --> 00:42:25
What's the function u?
 

698
00:42:25 --> 00:42:29
The solution that
comes from f(x).

699
00:42:30 --> 00:42:35
Just erase here so that
I can put all of them.

700
00:42:35 --> 00:42:39
The point of the whole
example now is for you

701
00:42:39 --> 00:42:41
to tell me what's u(x)?
 

702
00:42:41 --> 00:42:50
 
 

703
00:42:50 --> 00:42:51
Can you do it?
 

704
00:42:51 --> 00:42:53
You see what u(x) is
going to come out?

705
00:42:53 --> 00:42:56
Going to come out nicely.
 

706
00:42:56 --> 00:42:58
I've written the
right-hand side f as a

707
00:42:58 --> 00:43:01
combination of deltas.
 

708
00:43:01 --> 00:43:05
I know the answer, for delta.
 

709
00:43:05 --> 00:43:08
It's G.
 

710
00:43:08 --> 00:43:12
What's the answer when the
delta is shifted along?

711
00:43:12 --> 00:43:17
What's the Green's function
when the spike is moved

712
00:43:17 --> 00:43:20
along to a point, t?
 

713
00:43:20 --> 00:43:25
Just because this constant
coefficient translation and

714
00:43:25 --> 00:43:31
variant LTI problem is shifting
in variant, the answer, when

715
00:43:31 --> 00:43:35
the spike is moved along, is
just the answer G moved along.

716
00:43:35 --> 00:43:38
So here's my answer.
 

717
00:43:38 --> 00:43:47
All this, this is my input and
my output is the same integral,

718
00:43:47 --> 00:43:51
this from minus infinity to
infinity, of where it

719
00:43:51 --> 00:43:55
was f, now it's G.
 

720
00:43:55 --> 00:43:58
Well no, what do I want?
 

721
00:43:58 --> 00:44:01
Help me out here.
 

722
00:44:01 --> 00:44:06
No, f(t) is just the amount
of the delta, but now

723
00:44:06 --> 00:44:08
what's the solution?
 

724
00:44:08 --> 00:44:10
What do I write now?
 

725
00:44:10 --> 00:44:13
G of? x-t.
 

726
00:44:13 --> 00:44:24
 
 

727
00:44:24 --> 00:44:27
That's it.
 

728
00:44:27 --> 00:44:29
You see why that works?
 

729
00:44:29 --> 00:44:32
Because this was the
input, G's the output.

730
00:44:32 --> 00:44:36
The problem was shift in
variant, so if I shifted the

731
00:44:36 --> 00:44:38
input I shifted the output.
 

732
00:44:38 --> 00:44:44
It was a linear, so if I add up
a bunch of deltas, the solution

733
00:44:44 --> 00:44:47
is add up a bunch of G.
 

734
00:44:47 --> 00:44:53
That's the answer.
 

735
00:44:53 --> 00:44:56
Oh, I could just say
one thing more.

736
00:44:56 --> 00:44:59
But you've got it
if you see math.

737
00:44:59 --> 00:45:02
So that's the point, that
if you know the Green's

738
00:45:02 --> 00:45:06
function, well yeah.
 

739
00:45:06 --> 00:45:11
Maybe from a practical point
of view, what have we done?

740
00:45:11 --> 00:45:17
The original way we did it
involved our computing

741
00:45:17 --> 00:45:18
two integrals.
 

742
00:45:18 --> 00:45:22
We had to, if we were given an
f(x), we had to find it's

743
00:45:22 --> 00:45:27
transform f hat of k, that we
weren't sure we could do

744
00:45:27 --> 00:45:28
with pencil and paper.
 

745
00:45:28 --> 00:45:32
And then we got an answer with
an f hat of k up here and then

746
00:45:32 --> 00:45:34
we had to transform back.
 

747
00:45:34 --> 00:45:36
Step one and step
three, we had to do.

748
00:45:36 --> 00:45:40
Now this is better.
 

749
00:45:40 --> 00:45:48
This is like better, because
we were able to get an

750
00:45:48 --> 00:45:52
explicit answer, G,
when this was a delta.

751
00:45:52 --> 00:45:56
You could say I've got it
down to one integral.

752
00:45:56 --> 00:45:58
Well, for whatever
that's worth.

753
00:45:58 --> 00:46:01
I was going to say, probably
can't do that one either

754
00:46:01 --> 00:46:03
depending what f(x) is.
 

755
00:46:03 --> 00:46:10
But that's a nice way to see
the answer, you have to admit.

756
00:46:10 --> 00:46:14
And it's because the problem is
is a shift in variance and I

757
00:46:14 --> 00:46:16
can write the answer that way.
 

758
00:46:16 --> 00:46:22
OK, and now one more
thing about this.

759
00:46:22 --> 00:46:26
And I've written the word,
the key word, down here.

760
00:46:26 --> 00:46:30
Convolution.
 

761
00:46:30 --> 00:46:37
Do you see the nice way
to write that answer?

762
00:46:37 --> 00:46:41
It's the convolution
of f with G.

763
00:46:41 --> 00:46:44
We didn't do integral
convolutions, we just did the

764
00:46:44 --> 00:46:50
discrete sums, but I mentioned
that in the integral case, you

765
00:46:50 --> 00:46:56
have this same thing. t and x-t
adding up to x, just the way we

766
00:46:56 --> 00:47:00
had k and n-k adding up to n.
 

767
00:47:00 --> 00:47:03
In other words, this
is just notation.

768
00:47:03 --> 00:47:10
But I'm just going to write
the answer in a nice way.

769
00:47:10 --> 00:47:15
So after all that lecture, the
answer to the differential

770
00:47:15 --> 00:47:20
equations is in three
symbols f star G.

771
00:47:20 --> 00:47:23
Where this just
simply means this.

772
00:47:23 --> 00:47:29
This is the continuous
convolution, not cyclic,

773
00:47:29 --> 00:47:38
and there's the answer.
 

774
00:47:38 --> 00:47:41
I'll allow myself
one more thing.

775
00:47:41 --> 00:47:46
Here we had a convolution
for the right-hand side.

776
00:47:46 --> 00:47:48
We started with this,
this is a convolution.

777
00:47:48 --> 00:47:55
Now, what are the three
symbols that I write down?

778
00:47:55 --> 00:47:59
For the shorthand
for this equation?

779
00:47:59 --> 00:48:01
So this was to be
true for any f.

780
00:48:01 --> 00:48:07
And now can I write down, how
do I write? f(x) is equal to,

781
00:48:07 --> 00:48:14
what is this right-hand side?
 

782
00:48:14 --> 00:48:19
It's f convolved with?
 

783
00:48:19 --> 00:48:29
So any f is the same f
convolved with delta.

784
00:48:29 --> 00:48:33
In convolution, delta is one.
 

785
00:48:33 --> 00:48:39
Because when I go to the
other space and I get

786
00:48:39 --> 00:48:44
a multiplication, it
really is one, right?

787
00:48:44 --> 00:48:46
In the other space.
 

788
00:48:46 --> 00:48:52
So in the other space, so this
convolution in x space turns

789
00:48:52 --> 00:48:54
into multiplication
in frequency space.

790
00:48:54 --> 00:48:59
And it just tells me that
f hat is f hat times one.

791
00:48:59 --> 00:49:04
So that's the way to look
at it in physical space.

792
00:49:04 --> 00:49:11
And this is the way to
look at the solution.

793
00:49:11 --> 00:49:16
So, one more sod and
I'll come back to that.

794
00:49:16 --> 00:49:21
So this G, the Green's
function, this is what a

795
00:49:21 --> 00:49:25
CAT scan does, what an
X-ray telescope does.

796
00:49:25 --> 00:49:27
What all sorts of
physical things do.

797
00:49:27 --> 00:49:31
Provided we can assume this
translation in variance, which

798
00:49:31 --> 00:49:35
is never perfectly true because
the telescope is finite.

799
00:49:35 --> 00:49:40
But a telescope takes the star,
takes the light signal,

800
00:49:40 --> 00:49:44
convolves it with the
telescope's own little

801
00:49:44 --> 00:49:45
Green's function.
 

802
00:49:45 --> 00:49:48
It blurs it, it's the
point spread function.

803
00:49:48 --> 00:49:51
It's the blurring function, G.
 

804
00:49:51 --> 00:49:54
The Green's function on a
telescope is somehow, that's

805
00:49:54 --> 00:49:57
what's you're convolving with.
 

806
00:49:57 --> 00:50:03
And if you want to find that
star as a bright, single point.

807
00:50:03 --> 00:50:06
You've got to do deconvolution.
 

808
00:50:06 --> 00:50:09
You've got to do a division
to get the G out.

809
00:50:09 --> 00:50:13
May I just say those words
and then it's Thanksgiving?

810
00:50:13 --> 00:50:22
That a machine, a sensor which
is translation in variant, or

811
00:50:22 --> 00:50:27
you could say close enough to
pretend it is, because nothing

812
00:50:27 --> 00:50:29
is going to be perfect
translation in variant all

813
00:50:29 --> 00:50:31
the way out to infinity.
 

814
00:50:31 --> 00:50:34
But if it's near the star.
 

815
00:50:34 --> 00:50:40
So I look at this point star in
the telescope I see a blur.

816
00:50:40 --> 00:50:46
That's because the telescope
has convolved the correct

817
00:50:46 --> 00:50:51
thing that I should
have seen, with G.

818
00:50:51 --> 00:50:54
It's blurred it by its
point spread function.

819
00:50:54 --> 00:50:58
So what if the person, the
factory where the telescope

820
00:50:58 --> 00:51:01
was built can test this
whole thing on points.

821
00:51:01 --> 00:51:04
And it can find the
point spread function.

822
00:51:04 --> 00:51:07
And if I knew G, then
I could undo it.

823
00:51:07 --> 00:51:10
And get a clear picture.
 

824
00:51:10 --> 00:51:16
So it's that step that won the
Nobel Prize for the CAT scan,

825
00:51:16 --> 00:51:21
and I'm sure is winning Nobel
Prizes for astronomers.

826
00:51:21 --> 00:51:24
OK, have a great Thanksgiving,
I'll see you Monday.

827
00:51:24 --> 00:51:25
Good.
 

828
00:51:25 --> 00:51:26