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PROFESSOR STRANG: Well, hope
you had a good Thanksgiving.

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So this is partly review
today, even, Wednesday

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even more review.
 

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Wednesday evening.
 

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Or Wednesday at 4 I'll be
here for any questions.

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And then the exam is
Thursday at 7:30 in Walker.

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Top floor of Walker this
time, not the same 54-100.

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OK, and then, no
lectures after that.

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Holiday, whatever.
 

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Yes.
 

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Right, you get a chance
to do something.

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Catch up with all those other
courses that are being

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neglected in favor
of 18.085 Right.

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OK, so here's a bit of
review right away.

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We really had four cases.
 

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We started with Fourier series,
that was periodic functions.

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And then discrete Fourier
series, also periodic in a way.

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Because w^n was one.
 

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So that we have n numbers
and then we could repeat

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them if we wanted.
 

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So those are the
two that repeat.

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This is the f(x), this is
all x, so that would be

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the Fourier integral that
we did just last week.

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Fourier integral transform.
 

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And this was the - well,
these are all, this is

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the discrete all the way.
 

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So that's - oh, you can see
these pair off, right?

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The periodic function, the 2pi
periodic function has Fourier

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coefficients for all k, so
that's the pair that

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we started with.
 

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Section 4.1.
 

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This sort of pairs, I
don't know whether

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to say with itself.
 

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I mean, we start with n numbers
and we end with n numbers.

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We have n numbers
in physical space.

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And we have n numbers
in frequency space.

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Right, so we call those,
so those went to

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c_0 up to c_(N-1).
 

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And this, all x pair for
the function, paired

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off with itself.
 

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Or with this went to f -
well maybe I use small f.

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I guess I did in last week.
 

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So that, and I called
its Fourier transform

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f hat of k, all k.
 

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So that's the pairing kind of
inside n-dimensional space

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with the Fourier matrix.
 

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This is the pairing of the
formula for f, and its similar

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formula for f hat, and these
are the guys that connect

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with each other.
 

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OK, so that's what we know.
 

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What we haven't done is
anything into two dimensions.

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So I would like to
include that today.

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I think my real message about
2-D, and I'm not going to

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include it on the exam, but
you might wonder, OK, can I

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have a function of x and y?
 

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And will the whole setup work.
 

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And the answer is yes.
 

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So really, my message is not to
be afraid in any way of 2-D.

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It's just the same formulas
with x,y or two indices, k,l.

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Yeah.
 

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You'll see that.
 

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OK, now for the new part.
 

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What's a convolution equation?
 

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That's my word for an equation
where instead of doing a

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convolution and finding the
right-hand side, instead we're

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given the right-hand side.
 

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And the unknown is
in the convolution.

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So let me write examples
of convolution equation.

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Every one of these would
allow a convolution.

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So the convolution equation
would be something the integral

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of F(t)u, for the unknown,
at x-t is - oh no, sorry.

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F will be the right-hand side.
 

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F of, well, can I - yeah,
better if I put it on

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the right-hand side.
 

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Wouldn't want to call it
the right-hand side.

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So this would be some, shall
I call it often K for kernel

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is sometimes the word.
 

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So what I'm saying is
equations come this way.

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This is really K
convolved with u.

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Equals F.
 

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You see, the only novelty
is the unknown is here.

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So that's why the word d
convolution is up there.

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Because that's what
we have to do.

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We have to undo the
convolution, this unknown

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function is convolved with a
known, K is known, some known

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kernel that tells us the point
spread of the telescope or

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whatever we're doing.
 

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And gives us the output
that we're looking at.

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And then we have to
find the input.

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OK, can I write down the
similar equations for

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the other three here?
 

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And then we'll just think
how would we find u, how

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would we solve them?
 

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So the equation here might
be that some kernel

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circle, convolved with
the unknown u is some y.

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These are now vectors.
 

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This is known.
 

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This is known.
 

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And those, the end-components
of u, are unknown.

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OK, so that would be
the same problem here.

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What would be here?
 

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Same thing.
 

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Now the integral will go from
- the only difference is the

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integral will go from minus
infinity to infinity,

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K(t)u(x-t)dt=f(x).
 

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u(x-t) And finally
regular convolution.

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What am I going to call it?
 

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K would be a sequence, maybe
I should call it a, known

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convolved with u, unknown,
is some c, known.

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Yeah.
 

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So those would be
four equations.

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You might say, wait a minute
where is Professor Strang come

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up with these problems at
the last week of the course.

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But, these are exactly the
type of problems that

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we know and love.
 

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These come from constant
coefficient time invariant,

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shift invariant.
 

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Linear problems.
 

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LTI, linear time and variant.
 

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And my lecture Wednesday, just
before Thanksgiving, took a

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differential equation for u and
found, and put it in this form.

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I'll come back to that.
 

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So suddenly we're seeing, I
mean, we're actually seeing

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some new things but also it
includes all the old ones.

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These are all of the best
problems in the world.

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These linear constant
coefficient problems.

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Time in variant, of
any of these types.

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This one was an integral from
minus pi to pi, where this

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one went all the way.
 

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So this is not brand new stuff.
 

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But it sort of looks new.
 

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And now the question is, so my
immediate question is, before

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doing any example, how would
you solve such an equation.

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And I saw on old exams, some
of this sort for example,

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let me focus on this one.
 

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Let me, instead of K there,
I'm not used to using K

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for a vector, I'm used to,
well maybe I'll use c.

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For the vector there.
 

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So this is N equations,
N unknowns.

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Oops, capital N is
our usual here.

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For the number.
 

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N u, N unknown u's.
 

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It's a matrix equation
with a circulant matrix.

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So all these equations are sort
of the special best kind.

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Because they're convolutions.
 

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And now tell me the main point.
 

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How do we solve
equations like this?

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How do we do a deconvolution,
so the unknown is convolved

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with c here, it's convolved
with K, it's convolved with

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a, how do we deconvolve
it to get u by itself?

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So what's the
central idea here?

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Central idea: go into
frequency space.

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Use the convolution rule.
 

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In frequency space, where these
transform, we're looking

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at multiplication.
 

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And multiplication,
we can undo.

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We can de-multiply.
 

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De-multiply is just a big
word for divide, right?

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So that's the point.
 

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Get into that space.
 

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That's what we've been
doing all the time.

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I better get one example, the
example from the problem

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Wednesday, just up here.
 

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Just so you see it.
 

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This won't look like a
convolution equation, but do

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you remember that it was -u''
plus a squared u

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equal some f(x)?
 

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00:10:34 --> 00:10:37
 
 

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So that's a constant,
it's certainly constant

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coefficient linear.
 

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Time invariant.
 

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Right, OK.
 

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And how did we solve that?
 

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We took Fourier transforms.
 

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So this was the second
derivative, the

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Fourier transform.
 

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What is the rule for
the Fourier transform

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of derivative?
 

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Every derivative brings down
an ik, in the transform.

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So we get ik twice.
 

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So it's k squared. i squared
cancels the minus one.

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So that's the
transform of -u''.

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This is the transform
of ordinary a squared

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u, just a squared.
 

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And this is f hat.
 

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So we've got into
frequency space.

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Where we are just seeing a
multiplication, K squared

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plus a squared, u hat.
 

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Of, this is u hat of k,
equals f hat of k, right?

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Oh well, sorry we were
- yeah, that's right.

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f hat of k, right.
 

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So we're in frequency
space, where we just

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see a multiplication.
 

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So again, this is now we just
demultiply, just divide.

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And then we have the answer,
but we have its transform.

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And then we have to
transform back.

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So we have to do the Fourier
transform to get to the,

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I'll say the inverse
Fourier transform.

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To get back to u(x).
 

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The answer.
 

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That's the model.
 

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That's the model.
 

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And that's maybe the one that
we've seen, now we're able

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to think about all
these four topics.

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Right.
 

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OK, so what was the key idea?
 

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Get into frequency space and
then it's just a, the equation

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is just a multiplication,
so the solution is

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just a division.
 

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00:12:46 --> 00:12:52
So can I do that now with these
four examples, just see.

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So this is like bring
the pieces together.

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OK, and deconvolution is
a very key thing to do.

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OK, so I'll take all four
of those and bring them

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00:13:01 --> 00:13:04
into frequency space.
 

223
00:13:04 --> 00:13:10
So this will be maybe, you'll
let me use k hat of, oh, k hat

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00:13:10 --> 00:13:14
of k, that's not too good.
 

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00:13:14 --> 00:13:27
Well, stuck with it.
 

226
00:13:27 --> 00:13:28
What am I doing here?
 

227
00:13:28 --> 00:13:31
In this 2pi periodic one?
 

228
00:13:31 --> 00:13:35
That's the one I started with,
but now I've got, I I'm

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using hats and so on.
 

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00:13:37 --> 00:13:42
I didn't do that
in Section 4.1.

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What the heck am I going to,
what notation am I going to do?

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And I really didn't
do convolution that

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much, for functions.
 

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So let me jump to here.
 

235
00:13:55 --> 00:13:57
I'll come back.
 

236
00:13:57 --> 00:13:59
It follows exactly
the same pattern.

237
00:13:59 --> 00:14:01
So let me jump to this one.
 

238
00:14:01 --> 00:14:05
OK, so I have a
convolution equation now.

239
00:14:05 --> 00:14:10
This is one where you
could do this one.

240
00:14:10 --> 00:14:16
This could appear on the quiz
because I can do all of it.

241
00:14:16 --> 00:14:18
So what is this convolution?
 

242
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OK.
 

243
00:14:19 --> 00:14:21
I've got n equations,
n unknowns.

244
00:14:21 --> 00:14:24
Let me write them in
matrix form, just so you

245
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see it that way too.
 

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00:14:26 --> 00:14:28
c_1, c_2, c_3>.
 

247
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I'll make n=4.
 

248
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And then these are, this
convolution has c_0, c_2, c_1.

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00:14:39 --> 00:14:43
c_3, c_2, c_1, c_0.
 

250
00:14:44 --> 00:14:51
This'll be c_3, c_3, c_3, I'm
writing down all the right

251
00:14:51 --> 00:14:53
numbers in the right places.
 

252
00:14:53 --> 00:14:56
So that when I do that
multiplication with the

253
00:14:56 --> 00:15:08
unknown, ,
I get the right-hand, the

254
00:15:08 --> 00:15:10
known right-hand side.
 

255
00:15:10 --> 00:15:12
Maybe b would be
a little better.

256
00:15:12 --> 00:15:15
Because we're more used
to b as as a known.

257
00:15:15 --> 00:15:20
It's just an Ax=b problem,
or an Au=b problem.

258
00:15:20 --> 00:15:25
But it looks like a convolution
but now it's just a

259
00:15:25 --> 00:15:26
matrix multiplication.
 

260
00:15:26 --> 00:15:29
So this is just <b_0,
b_1, b_2, b_3>.

261
00:15:30 --> 00:15:32
OK.
 

262
00:15:32 --> 00:15:34
That's our equation.
 

263
00:15:34 --> 00:15:36
Special type of matrix.
 

264
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Circulant matrix.
 

265
00:15:38 --> 00:15:43
So this is just literally
the same as c circularly

266
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convolved with u equals b.
 

267
00:15:46 --> 00:15:50
I just wrote it out
in matrix language.

268
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So you could call MATLAB with
that matrix, and so one way to

269
00:15:59 --> 00:16:04
answer it would be get the
inverse of the matrix.

270
00:16:04 --> 00:16:13
But if it was large, a better
way would be switch over

271
00:16:13 --> 00:16:16
to frequency space.
 

272
00:16:16 --> 00:16:16
Think, now.
 

273
00:16:16 --> 00:16:22
What happens when I
switch these vectors

274
00:16:22 --> 00:16:23
to frequency space?
 

275
00:16:23 --> 00:16:25
It becomes a multiplication.
 

276
00:16:25 --> 00:16:29
So this becomes a
multiplication.

277
00:16:29 --> 00:16:38
Now so c, I want the Fourier,
from the c's, what am I going

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00:16:38 --> 00:16:46
to - so these are all in the
space where it's a convolution.

279
00:16:46 --> 00:16:49
What am I going to call it
where it's in the space where

280
00:16:49 --> 00:16:51
it's a multiplication?
 

281
00:16:51 --> 00:16:54
I just need three new names.
 

282
00:16:54 --> 00:17:01
Maybe I'll use c hat, u hat,
and b hat just because there's

283
00:17:01 --> 00:17:05
no doubt in anybody's mind that
when you see that hat, you've

284
00:17:05 --> 00:17:07
gone into frequency space.
 

285
00:17:07 --> 00:17:11
Now, what's the equation
in frequency space?

286
00:17:11 --> 00:17:15
And then I'll do an example.
 

287
00:17:15 --> 00:17:21
It's a multiplication, but
I don't usually see a

288
00:17:21 --> 00:17:27
vector and nothing there.
 

289
00:17:27 --> 00:17:31
What's the multiplication
in frequency space?

290
00:17:31 --> 00:17:37
It's component by
component. c_0*u_0=b_0.

291
00:17:42 --> 00:17:50
c hat 1 u hat 1 equals b
hat 1. c hat 2, u hat

292
00:17:50 --> 00:17:53
2, equals b hat 2.
 

293
00:17:53 --> 00:18:00
And finally, c hat 3, u
hat 3, equals b hat 3.

294
00:18:00 --> 00:18:04
And there might be, I don't
swear that there isn't,

295
00:18:04 --> 00:18:06
a 1/4 somewhere.
 

296
00:18:06 --> 00:18:07
Right?
 

297
00:18:07 --> 00:18:12
But the point is, we're
in frequency space now.

298
00:18:12 --> 00:18:19
We just have a component by
component, each component of c

299
00:18:19 --> 00:18:22
hat times each component of u
hat gives us a component of b

300
00:18:22 --> 00:18:26
hat; now we're ready for a
deconvolution; just divide.

301
00:18:26 --> 00:18:31
So now u hat, obviously I don't
have to write all these,

302
00:18:31 --> 00:18:34
b hat 0 over c hat 0.
 

303
00:18:34 --> 00:18:35
Right?
 

304
00:18:35 --> 00:18:36
I just do a division.
 

305
00:18:36 --> 00:18:44
So on down to u hat 3 is b hat,
is the third component of b,

306
00:18:44 --> 00:18:47
divided by the third
component of c.

307
00:18:47 --> 00:18:50
OK, now don't forget here.
 

308
00:18:50 --> 00:18:54
That in going from here to
here, I had to figure out

309
00:18:54 --> 00:18:57
what the c hats were, right?
 

310
00:18:57 --> 00:19:03
I had to do the Fourier matrix,
or the inverse Fourier matrix

311
00:19:03 --> 00:19:08
to go from c to c hat, from b
to b hat, so everything

312
00:19:08 --> 00:19:10
got Fourier transforms.
 

313
00:19:10 --> 00:19:17
But the object was to
make the equation easy.

314
00:19:17 --> 00:19:21
And of course, now we've got
four trivial equations that

315
00:19:21 --> 00:19:24
we just solved that way.
 

316
00:19:24 --> 00:19:27
Alright, let me see if I
can just pull this down

317
00:19:27 --> 00:19:32
with some questions.
 

318
00:19:32 --> 00:19:34
Here's a good question.
 

319
00:19:34 --> 00:19:39
When is a circulant
matrix invertible.

320
00:19:39 --> 00:19:42
When will this method work?
 

321
00:19:42 --> 00:19:45
The circulant matrix could
fail to be invertible.

322
00:19:45 --> 00:19:50
How would I know that?
 

323
00:19:50 --> 00:19:54
If it's singular, and how would
I, if I proceed this way,

324
00:19:54 --> 00:19:57
here I've got an answer.
 

325
00:19:57 --> 00:19:59
But if it's singular I'm
not really expecting

326
00:19:59 --> 00:20:00
to get an answer.
 

327
00:20:00 --> 00:20:03
Let me left the board a little.
 

328
00:20:03 --> 00:20:09
So where would I get, oops.
 

329
00:20:09 --> 00:20:14
Have to stop this method.
 

330
00:20:14 --> 00:20:16
In solving those
four equations.

331
00:20:16 --> 00:20:20
Where would I learn
that it's is singular?

332
00:20:20 --> 00:20:22
What could go wrong in this?
 

333
00:20:22 --> 00:20:23
Yes.
 

334
00:20:23 --> 00:20:25
AUDIENCE: [INAUDIBLE]
 

335
00:20:25 --> 00:20:26
PROFESSOR STRANG: That's right.
 

336
00:20:26 --> 00:20:32
Always in math, the question
is are you dividing by zero.

337
00:20:32 --> 00:20:36
So the question of whether the
matrix is singular, is the same

338
00:20:36 --> 00:20:43
as the question of whether c_0
hat, c_01, c_02, and c_03, -

339
00:20:43 --> 00:20:48
sorry, c_0 hat, c_1 hat, c_2
hat, and c_3 hat,

340
00:20:48 --> 00:20:49
can't be zero.
 

341
00:20:49 --> 00:20:55
That's, in fact, even better
those four numbers, those four

342
00:20:55 --> 00:21:00
c hats, are actually the
eigenvalues of the matrix.

343
00:21:00 --> 00:21:04
We've switched, what the
Fourier transform did, was

344
00:21:04 --> 00:21:08
switch over to the eigenvalues
and eigenvectors.

345
00:21:08 --> 00:21:12
And there, that's the whole
message of those guys is, you

346
00:21:12 --> 00:21:14
follow each one separately.
 

347
00:21:14 --> 00:21:15
Just the way we're doing here.
 

348
00:21:15 --> 00:21:19
So this is the component
of the b in the for

349
00:21:19 --> 00:21:21
eigenvector directions.
 

350
00:21:21 --> 00:21:26
Those are the four eigenvalues,
and I have to divide by them.

351
00:21:26 --> 00:21:30
You see, the idea is, like,
we've diagonalized the matrix.

352
00:21:30 --> 00:21:35
We've had that matrix,
which is full.

353
00:21:35 --> 00:21:41
And we take by taking the
Fourier transforms, that's the

354
00:21:41 --> 00:21:47
same thing as as putting in the
eigenvectors, switching the

355
00:21:47 --> 00:21:51
matrix to this diagonal
matrix, right?

356
00:21:51 --> 00:21:58
Our problem has become like the
diagonalized form is c_0 hat

357
00:21:58 --> 00:22:01
down to c_3 hat, sitting
on the diagonal.

358
00:22:01 --> 00:22:03
All zeroes elsewhere.
 

359
00:22:03 --> 00:22:06
That's when we switched, when
we did Fourier transform we

360
00:22:06 --> 00:22:08
were switching to eigenvectors.
 

361
00:22:08 --> 00:22:10
OK, so that's the message.
 

362
00:22:10 --> 00:22:17
That the test for singularity
is the Fourier, the

363
00:22:17 --> 00:22:21
transform of c hits zero.
 

364
00:22:21 --> 00:22:22
Then we're in trouble.
 

365
00:22:22 --> 00:22:25
Let me do an example you know.
 

366
00:22:25 --> 00:22:27
Let me do an example you know.
 

367
00:22:27 --> 00:22:33
OK here's, so finally now we
get a numerical example.

368
00:22:33 --> 00:22:38
The example we really
know is this one, right?

369
00:22:38 --> 00:22:41
As I start writing that,
you may say in your

370
00:22:41 --> 00:22:43
mind, oh no not again.
 

371
00:22:43 --> 00:22:47
But give it to me, one more
week with these matrices.

372
00:22:47 --> 00:22:54
But it'll be the C matrix, so
it's going to be the circulant.

373
00:22:54 --> 00:22:57
Recognize this?
 

374
00:22:57 --> 00:23:02
And it's got those minus
ones in the corners, too.

375
00:23:02 --> 00:23:05
OK, let's go back to day one.
 

376
00:23:05 --> 00:23:09
Is that matrix invertible?
 

377
00:23:09 --> 00:23:12
Yes or no.
 

378
00:23:12 --> 00:23:14
Please, no.
 

379
00:23:14 --> 00:23:17
Everybody knows that
matrix is not invertible.

380
00:23:17 --> 00:23:21
And do you remember what's
in the null space?

381
00:23:21 --> 00:23:25
Yes, what's the vector in the
null space of that matrix?

382
00:23:25 --> 00:23:27
All ones.
 

383
00:23:27 --> 00:23:30
Now, when I take,
just think now.

384
00:23:30 --> 00:23:35
When I take Fourier transform,
that all ones is going

385
00:23:35 --> 00:23:38
to transform to what?
 

386
00:23:38 --> 00:23:40
It's going to transform
to the delta.

387
00:23:40 --> 00:23:45
It'll transform to the one
that is like, .

388
00:23:45 --> 00:23:47
Or maybe it's .
 

389
00:23:47 --> 00:23:56
But it's that, well, OK,
now I'm ready to take.

390
00:23:56 --> 00:24:01
So here's my c.
 

391
00:24:01 --> 00:24:02
So what's my method now?
 

392
00:24:02 --> 00:24:07
I'm going to do this method,
and I'm going to run into, this

393
00:24:07 --> 00:24:09
thing is going to be zero.
 

394
00:24:09 --> 00:24:13
Because that's the eigenvalue
that goes with the <1, 1, 1, 1,

395
00:24:13 --> 00:24:18
1> column, the constant, the
zero frequency in

396
00:24:18 --> 00:24:19
frequency space.
 

397
00:24:19 --> 00:24:20
You'll see it happen.
 

398
00:24:20 --> 00:24:24
So let's take the Fourier
transform of that.

399
00:24:24 --> 00:24:27
And then we would have to take
the Fourier transform of the

400
00:24:27 --> 00:24:29
right-hand side, b, whatever
that happened to be.

401
00:24:29 --> 00:24:31
But it's always the left side.
 

402
00:24:31 --> 00:24:33
The singular or not matrix.
 

403
00:24:33 --> 00:24:35
I believe we'll be
singular here.

404
00:24:35 --> 00:24:42
So, OK, just remind me how do I
take transforms of this guy?

405
00:24:42 --> 00:24:44
Gosh, we have to be
able to do that.

406
00:24:44 --> 00:24:48
That's Section 4.-
well, 4.3 isn't, yeah.

407
00:24:48 --> 00:24:53
The DFT of that vector.
 

408
00:24:53 --> 00:24:56
What do I get?
 

409
00:24:56 --> 00:24:58
Yes.
 

410
00:24:58 --> 00:25:02
How do I take the
DFT of a vector?

411
00:25:02 --> 00:25:05
I multiply by the
Fourier matrix, right?

412
00:25:05 --> 00:25:06
Yes.
 

413
00:25:06 --> 00:25:10
So I have to multiply that
thing by the Fourier matrix.

414
00:25:10 --> 00:25:17
So to get c hat, this was big C
for the matrix, little c for

415
00:25:17 --> 00:25:21
the vector that goes into
it, into column zero.

416
00:25:21 --> 00:25:24
And c hat for its transform.
 

417
00:25:24 --> 00:25:29
OK, so now here comes the
Fourier matrix that we know, 1,

418
00:25:29 --> 00:25:34
i, i^2, i^3, 1, i^2, i^4, i^6.
 

419
00:25:35 --> 00:25:38
1, i^3, i^6, and i^9.
 

420
00:25:39 --> 00:25:46
So I want to transform that c
to get, to find out c hat.

421
00:25:46 --> 00:25:52
OK, and what do I get up there?
 

422
00:25:52 --> 00:25:58
What's the first component, the
zeroth component I should say,

423
00:25:58 --> 00:26:03
when I take this guy, this
four, this vector with four

424
00:26:03 --> 00:26:07
components and I get back four
components, the frequency

425
00:26:07 --> 00:26:10
components, what's
the first one?

426
00:26:10 --> 00:26:13
Ones times this,
what am I getting?

427
00:26:13 --> 00:26:14
Zero.
 

428
00:26:14 --> 00:26:16
That's what I expected.
 

429
00:26:16 --> 00:26:20
That tells me the matrix is
not going to be invertible.

430
00:26:20 --> 00:26:27
Because in a different
language, I'm finding

431
00:26:27 --> 00:26:30
the eigenvalues and
that's one of them.

432
00:26:30 --> 00:26:34
And if an eigenvalue is zero,
that means the eigenvector is

433
00:26:34 --> 00:26:36
getting knocked out completely.
 

434
00:26:36 --> 00:26:40
And there's snow way a c
inverse could recover when

435
00:26:40 --> 00:26:42
that eigenvector is gone.
 

436
00:26:42 --> 00:26:43
OK, let's do the other ones.
 

437
00:26:43 --> 00:26:49
Two minus i, nothing.
 

438
00:26:49 --> 00:26:51
What's the other one here?
 

439
00:26:51 --> 00:26:56
Two, this is minus i, and
that's plus i, I think.

440
00:26:56 --> 00:26:59
So I think it's just two.
 

441
00:26:59 --> 00:27:02
Alright, this is 2 i squared,
can I write in some of these

442
00:27:02 --> 00:27:07
just so I have a little, i
squared is minus one, and i^4

443
00:27:07 --> 00:27:10
is one, and that's minus one.
 

444
00:27:10 --> 00:27:15
So that's two plus one,
plus one, I think is four.

445
00:27:15 --> 00:27:20
And then i^3 is the
same as minus i.

446
00:27:20 --> 00:27:22
And i^9 is the same as plus i.
 

447
00:27:22 --> 00:27:34
So I think I'm getting two plus
i, nothing, minus i, two.

448
00:27:34 --> 00:27:36
So what's my claim?
 

449
00:27:36 --> 00:27:39
My claim is that these are
the four eigenvalues that

450
00:27:39 --> 00:27:43
the Fourier - Fourier
diagonalizes these problems.

451
00:27:43 --> 00:27:44
That's what it comes to.
 

452
00:27:44 --> 00:27:49
Fourier diagonalizes all
constant coefficients, shift

453
00:27:49 --> 00:27:51
invariant, linear problems.
 

454
00:27:51 --> 00:27:55
And tells us here
are eigenvalues.

455
00:27:55 --> 00:27:57
So .
 

456
00:27:57 --> 00:28:00
Would you like to, how do
I check the eigenvalues

457
00:28:00 --> 00:28:00
of a matrix?
 

458
00:28:00 --> 00:28:02
Let's just remember.
 

459
00:28:02 --> 00:28:05
If I give you four numbers
and I say those are the

460
00:28:05 --> 00:28:09
eigenvalues, and you look
at that matrix, what quick

461
00:28:09 --> 00:28:12
check does everybody do?
 

462
00:28:12 --> 00:28:15
Compute the - the trace.
 

463
00:28:15 --> 00:28:19
Add up the diagonal of the
matrix, add up the proposed

464
00:28:19 --> 00:28:21
eigenvalues, they had
better be the same.

465
00:28:21 --> 00:28:22
And they are.
 

466
00:28:22 --> 00:28:23
I get eight both ways.
 

467
00:28:23 --> 00:28:26
That doesn't mean, of course,
that these four numbers are

468
00:28:26 --> 00:28:28
right, but I think they are.
 

469
00:28:28 --> 00:28:29
Yeah, yeah.
 

470
00:28:29 --> 00:28:31
So those added up to
eight, those numbers

471
00:28:31 --> 00:28:33
added up to eight.
 

472
00:28:33 --> 00:28:36
And yep.
 

473
00:28:36 --> 00:28:38
And these are real.
 

474
00:28:38 --> 00:28:40
They came out real, and
how did I know that would

475
00:28:40 --> 00:28:44
happen from the matrix?
 

476
00:28:44 --> 00:28:51
What matrices am I certain to
get real eigenvalues for?

477
00:28:51 --> 00:28:51
Symmetric.
 

478
00:28:51 --> 00:28:52
Right.
 

479
00:28:52 --> 00:28:55
Now, what about - I heard
the word positive.

480
00:28:55 --> 00:28:58
Of course, that's the other
question I have to ask.

481
00:28:58 --> 00:29:03
Is this matrix
positive definite?

482
00:29:03 --> 00:29:05
OK, everybody this is
the language we've

483
00:29:05 --> 00:29:06
learned in 18.085.
 

484
00:29:06 --> 00:29:09
Is that matrix positive
definite, yes or no?

485
00:29:09 --> 00:29:11
No.
 

486
00:29:11 --> 00:29:12
What is it?
 

487
00:29:12 --> 00:29:15
It's positive semi-definite.
 

488
00:29:15 --> 00:29:18
What does that tell me
about eigenvalues?

489
00:29:18 --> 00:29:22
Their one is zero, that's why
it's not positive definite.

490
00:29:22 --> 00:29:24
But the others are positive.
 

491
00:29:24 --> 00:29:28
So that sure enough, in other
words, what we've done here,

492
00:29:28 --> 00:29:32
for that matrix that came
on day one and now we're

493
00:29:32 --> 00:29:38
seeing it on day N-1 here.
 

494
00:29:38 --> 00:29:42
We're we're seeing sort of in a
new way, because at that time

495
00:29:42 --> 00:29:47
we didn't know these four were
the eigenvectors

496
00:29:47 --> 00:29:49
of that matrix.
 

497
00:29:49 --> 00:29:50
But they are.
 

498
00:29:50 --> 00:29:56
And we're coming to the
same conclusion we

499
00:29:56 --> 00:29:58
came to on day one.
 

500
00:29:58 --> 00:30:03
That the matrix is positive
semi-definite and that

501
00:30:03 --> 00:30:05
we know its eigenvalues.
 

502
00:30:05 --> 00:30:09
And we can, actually, let
me even take it one more

503
00:30:09 --> 00:30:13
step, just because this
example is so perfect.

504
00:30:13 --> 00:30:17
Some right-hand sides we
could solve for, right?

505
00:30:17 --> 00:30:22
If I have a matrix that's
singular, way, way back.

506
00:30:22 --> 00:30:25
Even, I think it was like a
worked example in Section 1.1,

507
00:30:25 --> 00:30:32
I could ask the question
when is Cx=b solvable.

508
00:30:32 --> 00:30:34
Because there are some
right-hand sides that'll work.

509
00:30:34 --> 00:30:37
Because if I just take an x and
multiply by C, I'll get a

510
00:30:37 --> 00:30:40
right-hand side that works.
 

511
00:30:40 --> 00:30:48
But for which vectors
right-hand sides, b,

512
00:30:48 --> 00:30:52
well my method work?
 

513
00:30:52 --> 00:30:54
The ones that have...?
 

514
00:30:54 --> 00:30:56
Yeah, the ones that have which?
 

515
00:30:56 --> 00:31:02
What do I need with these c's,
c_0, c_1, c_2, and c_3, for

516
00:31:02 --> 00:31:06
my solution to be possible.
 

517
00:31:06 --> 00:31:09
I need b_0 hat.
 

518
00:31:09 --> 00:31:11
Equals zero.
 

519
00:31:11 --> 00:31:13
I need b_0 hat equals zero.
 

520
00:31:13 --> 00:31:15
And then what does that say?
 

521
00:31:15 --> 00:31:20
That means that the b, the
vector b, has no constant

522
00:31:20 --> 00:31:22
term in the Fourier series.
 

523
00:31:22 --> 00:31:26
It means that the vector b
is orthogonal to the <1,

524
00:31:26 --> 00:31:29
1, 1, 1>, eigenvector.
 

525
00:31:29 --> 00:31:34
So this is like a subtle point
but just driving home the point

526
00:31:34 --> 00:31:39
that what Fourier does is
diagonalize everything.

527
00:31:39 --> 00:31:43
It diagonalizes all the
important problems of, all

528
00:31:43 --> 00:31:47
the simplest problems, of
differential equations.

529
00:31:47 --> 00:31:52
You know, I mean this is
like 18.03 looked at from

530
00:31:52 --> 00:31:53
Fourier's point of view.
 

531
00:31:53 --> 00:31:57
OK, what more could I
do with that equation?

532
00:31:57 --> 00:32:00
I think you really are seeing
all the good stuff here.

533
00:32:00 --> 00:32:02
You're seeing the matrix.
 

534
00:32:02 --> 00:32:05
We're recognizing
it as a circulant.

535
00:32:05 --> 00:32:08
We're realizing that we could
take its Fourier transform.

536
00:32:08 --> 00:32:11
We get the eigenvalues.
 

537
00:32:11 --> 00:32:15
We're diagonalizing the matrix,
the convolution becomes a

538
00:32:15 --> 00:32:19
multiplication, and the
solution becomes, inversion

539
00:32:19 --> 00:32:22
becomes division.
 

540
00:32:22 --> 00:32:26
I hope you see that.
 

541
00:32:26 --> 00:32:30
That's really a model
problem for this course.

542
00:32:30 --> 00:32:33
OK. yeah.
 

543
00:32:33 --> 00:32:33
Questions.
 

544
00:32:33 --> 00:32:41
Good.
 

545
00:32:41 --> 00:32:42
AUDIENCE: [INAUDIBLE]
 

546
00:32:42 --> 00:32:44
PROFESSOR STRANG: Would I give
you a six by six Fourier

547
00:32:44 --> 00:32:45
matrix on a test?
 

548
00:32:45 --> 00:32:46
Probably not.
 

549
00:32:46 --> 00:32:48
No.
 

550
00:32:48 --> 00:32:49
I just about could.
 

551
00:32:49 --> 00:32:55
I mean, it's, six by six, those
are pretty decent numbers.

552
00:32:55 --> 00:32:56
Right.
 

553
00:32:56 --> 00:32:59
Those six roots of unit
e, but not quite.

554
00:32:59 --> 00:33:00
Right, yeah.
 

555
00:33:00 --> 00:33:01
Yeah.
 

556
00:33:01 --> 00:33:01
Yeah.
 

557
00:33:01 --> 00:33:04
So four by four is,
five by five would not

558
00:33:04 --> 00:33:05
be nice, certainly.
 

559
00:33:05 --> 00:33:09
Who knows the cosine
of 72 degrees?

560
00:33:09 --> 00:33:10
Crazy.
 

561
00:33:10 --> 00:33:13
But, at 60 degrees we could do.
 

562
00:33:13 --> 00:33:17
So the Fourier matrix would
be full of square roots of

563
00:33:17 --> 00:33:20
three over two, and one over
two, an i's, and so on.

564
00:33:20 --> 00:33:24
But it wouldn't be, so
really four by four

565
00:33:24 --> 00:33:26
is sort of the model.
 

566
00:33:26 --> 00:33:28
Yeah, yeah.
 

567
00:33:28 --> 00:33:31
So four by four is that model.
 

568
00:33:31 --> 00:33:33
Other questions?
 

569
00:33:33 --> 00:33:35
Because this is really
a key example.

570
00:33:35 --> 00:33:37
Yeah.
 

571
00:33:37 --> 00:33:41
When I calculated the
eigenvalues, yeah.

572
00:33:41 --> 00:33:42
Ah.
 

573
00:33:42 --> 00:33:46
Because this matrix, I know
everything about that matrix

574
00:33:46 --> 00:33:47
when I know its first vector.
 

575
00:33:47 --> 00:33:49
AUDIENCE: [INAUDIBLE]
 

576
00:33:49 --> 00:33:51
PROFESSOR STRANG: Yeah,
it's because it's a

577
00:33:51 --> 00:33:52
circulant matrix.
 

578
00:33:52 --> 00:33:57
It's because that matrix is
expressing convolution with

579
00:33:57 --> 00:33:59
this vector. .
 

580
00:33:59 --> 00:34:03
 
 

581
00:34:03 --> 00:34:07
That circulant matrix
essentially is built from

582
00:34:07 --> 00:34:09
four numbers, right.
 

583
00:34:09 --> 00:34:09
Yeah.
 

584
00:34:09 --> 00:34:11
Yeah, and they go in
the zeroth column.

585
00:34:11 --> 00:34:12
Right, yeah.
 

586
00:34:12 --> 00:34:14
Yeah.
 

587
00:34:14 --> 00:34:20
Right, so there is an
example where we could

588
00:34:20 --> 00:34:21
like do everything.
 

589
00:34:21 --> 00:34:27
Now, and let me just remember
that with this example,

590
00:34:27 --> 00:34:29
we could do everything.
 

591
00:34:29 --> 00:34:37
So this is an example of, you
could say this type of problem.

592
00:34:37 --> 00:34:42
But with a very special kernel
there, so it turned out to be,

593
00:34:42 --> 00:34:45
it looks like an integral
equation here but if that

594
00:34:45 --> 00:34:50
kernel involves delta functions
and so on then it can be just

595
00:34:50 --> 00:34:51
a differential equation.
 

596
00:34:51 --> 00:34:53
And then that's
what we got there.

597
00:34:53 --> 00:34:59
So we took all the same steps
we did here, we did here.

598
00:34:59 --> 00:35:06
We took the Fourier transform,
and I emphasize there, just to

599
00:35:06 --> 00:35:09
remember Wednesday, this
was a delta function.

600
00:35:09 --> 00:35:13
When I took the Fourier
transform I got a one, so

601
00:35:13 --> 00:35:16
this was a one over this.
 

602
00:35:16 --> 00:35:21
And I did the inverse transform
and I got back to the

603
00:35:21 --> 00:35:24
function that I drew.
 

604
00:35:24 --> 00:35:28
Which was e^(-ax) over 2a.
 

605
00:35:29 --> 00:35:31
And even.
 

606
00:35:31 --> 00:35:36
So, yeah. this was
the answer u(x).

607
00:35:38 --> 00:35:41
So I was able to do that, I
mean this step was easy,

608
00:35:41 --> 00:35:43
that step is easy.
 

609
00:35:43 --> 00:35:45
That step is easy, the
division is easy.

610
00:35:45 --> 00:35:55
And then I just recognize this
as the transform of this one,

611
00:35:55 --> 00:35:57
this example that we had done.
 

612
00:35:57 --> 00:35:58
Once I divided by 2a.
 

613
00:35:59 --> 00:36:01
So you should be
able to do this.

614
00:36:01 --> 00:36:05
So those are two that you
should really be able to do.

615
00:36:05 --> 00:36:08
I'm not going to, obviously I'm
not going to, ask you a 2-D

616
00:36:08 --> 00:36:13
problem on the exam or
even on a homework.

617
00:36:13 --> 00:36:17
But now if you'll allow me,
I'd like to spend a few

618
00:36:17 --> 00:36:20
minutes to get into 2-D.
 

619
00:36:20 --> 00:36:25
Because really, you've got
the main thoughts here.

620
00:36:25 --> 00:36:28
That Fourier is the same
as finding eigenvectors

621
00:36:28 --> 00:36:30
and eigenvalues.
 

622
00:36:30 --> 00:36:35
That's the main thought
for these LTI problems.

623
00:36:35 --> 00:36:40
OK, now suppose I have, let's
just get the formalities

624
00:36:40 --> 00:36:41
straight here.
 

625
00:36:41 --> 00:36:45
Suppose I have a
function of x and y.

626
00:36:45 --> 00:36:48
2pi periodic in x, and in y.
 

627
00:36:48 --> 00:36:55
So if I bump x by 2pi, or if I
bump y by 2pi - oh, I'm using

628
00:36:55 --> 00:36:59
capital F for the
periodic guys.

629
00:36:59 --> 00:37:01
So let me stay with
capital F(x,y+2pi).

630
00:37:01 --> 00:37:04
 
 

631
00:37:04 --> 00:37:08
OK, so I have a function.
 

632
00:37:08 --> 00:37:10
This is given.
 

633
00:37:10 --> 00:37:13
This is, it's in 2-D now.
 

634
00:37:13 --> 00:37:16
And I want to write
its Fourier series.

635
00:37:16 --> 00:37:19
So I'm just asking the question
what does the Fourier series

636
00:37:19 --> 00:37:25
look like for a function
of two variables.

637
00:37:25 --> 00:37:28
The point is, it's going
to be a nice answer.

638
00:37:28 --> 00:37:33
And so everything, what
you know how to do in

639
00:37:33 --> 00:37:35
1-D you can do in 2-D.
 

640
00:37:35 --> 00:37:41
So let me write the complex
form, the e^(ik) stuff.

641
00:37:41 --> 00:37:44
So what would I write,
how would I write this?

642
00:37:44 --> 00:37:49
I would write that as a sum,
but it'll have, I'll make it a

643
00:37:49 --> 00:37:53
double sum, I'll write two
sigmas just to emphasize that

644
00:37:53 --> 00:37:58
we're summing from k equal
minus infinity, to infinity,

645
00:37:58 --> 00:38:02
and from l equal minus
infinity to infinity.

646
00:38:02 --> 00:38:03
We have coefficients c_kl.
 

647
00:38:03 --> 00:38:07
 
 

648
00:38:07 --> 00:38:11
They depend on two indices,
this is the pattern to know.

649
00:38:11 --> 00:38:15
Multiplying our e^(ikx),
and our e^(ily).

650
00:38:15 --> 00:38:19
 
 

651
00:38:19 --> 00:38:23
Right, good.
 

652
00:38:23 --> 00:38:25
So, alright.
 

653
00:38:25 --> 00:38:27
Let me ask you.
 

654
00:38:27 --> 00:38:29
How would I find c_23?
 

655
00:38:30 --> 00:38:35
Just to know that - we could
find all these coefficients,

656
00:38:35 --> 00:38:36
find formulas for them.
 

657
00:38:36 --> 00:38:39
We could do examples.
 

658
00:38:39 --> 00:38:40
How would I find c_23?
 

659
00:38:41 --> 00:38:45
So this is my F.
 

660
00:38:45 --> 00:38:47
I know F.
 

661
00:38:47 --> 00:38:48
I want to find c_23.
 

662
00:38:49 --> 00:38:52
What's the magic trick?
 

663
00:38:52 --> 00:38:55
Then and a 2pi periodic,
so all integrals.

664
00:38:55 --> 00:38:58
All the integrals, and I'm
giving you a hint, of course.

665
00:38:58 --> 00:39:01
I'm going to integrate.
 

666
00:39:01 --> 00:39:04
And the integrals will
all go from minus pi

667
00:39:04 --> 00:39:06
to pi in x, and in y.
 

668
00:39:06 --> 00:39:09
They'll integrate over
the period square.

669
00:39:09 --> 00:39:14
Here's the period square
from, there's the center. x

670
00:39:14 --> 00:39:21
direction, y direction, goes
out to pi and goes up to pi.

671
00:39:21 --> 00:39:24
So all integrals
will be over dxdy.

672
00:39:26 --> 00:39:28
But what do I integrate?
 

673
00:39:28 --> 00:39:29
To find c_23.
 

674
00:39:29 --> 00:39:33
 
 

675
00:39:33 --> 00:39:36
Well, these guys
are orthogonal.

676
00:39:36 --> 00:39:39
That's what's making everything
work, they're orthogonal

677
00:39:39 --> 00:39:40
and very special.
 

678
00:39:40 --> 00:39:43
So that by u's orthogonality,
what do I do?

679
00:39:43 --> 00:39:46
I multiply by?
 

680
00:39:46 --> 00:39:50
Just tell me what
to multiply by.

681
00:39:50 --> 00:39:57
By this and integrate.
 

682
00:39:57 --> 00:40:04
OK, what is it that I multiply
by if I'm shooting for c_23,

683
00:40:04 --> 00:40:13
for example? e^(i2x),
is it e^(i2x)?

684
00:40:15 --> 00:40:17
Minus, right.
 

685
00:40:17 --> 00:40:27
I multiply by e^(-i2x),
e^(-i3y), and integrate.

686
00:40:27 --> 00:40:28
Yeah.
 

687
00:40:28 --> 00:40:32
So when I multiply by that and
integrate, everything will

688
00:40:32 --> 00:40:35
go except the c_23 term.
 

689
00:40:35 --> 00:40:38
Which will be
multiplied by what?

690
00:40:38 --> 00:40:43
So I'll just have c_23 times
probably 2pi squared.

691
00:40:43 --> 00:40:46
I guess 2pi will come in
from both integrals, so

692
00:40:46 --> 00:40:48
the formula will be c_kl.
 

693
00:40:49 --> 00:40:52
c_kl will be, do you want
me to write this formula?

694
00:40:52 --> 00:40:56
I'll write it here and then
forget it right away. c_kl

695
00:40:56 --> 00:41:00
will be one over 2pi squared.
 

696
00:41:00 --> 00:41:04
The integral of my function.
 

697
00:41:04 --> 00:41:07
Times my e^(-ikx),
times my e^(-ily)dxdy.

698
00:41:07 --> 00:41:15
 
 

699
00:41:15 --> 00:41:17
So that just makes the point.
 

700
00:41:17 --> 00:41:22
That there's nothing new here,
it's just up a dimension.

701
00:41:22 --> 00:41:28
But the formulas all look the
same, and if f was a- well, if

702
00:41:28 --> 00:41:34
F is a delta function, if f
is now a 2-D delta function.

703
00:41:34 --> 00:41:38
We haven't done delta functions
in 2-D, why don't we?

704
00:41:38 --> 00:41:45
Suppose F is the delta
function in 2-D.

705
00:41:45 --> 00:41:49
Then what are the coefficients?
 

706
00:41:49 --> 00:41:51
What do you think you
this means, this delta

707
00:41:51 --> 00:41:53
function in 2-D?
 

708
00:41:53 --> 00:41:57
So if I put in the delta
here, and I integrate.

709
00:41:57 --> 00:42:01
And what do I get then?
 

710
00:42:01 --> 00:42:06
So if this guy is a delta, a
two-dimensional delta function,

711
00:42:06 --> 00:42:12
the rule is that when I
integrate over a region that

712
00:42:12 --> 00:42:17
includes the spike, so it's a
spike sitting up above a plane

713
00:42:17 --> 00:42:20
now, instead of sitting above a
line it's sitting

714
00:42:20 --> 00:42:21
above a plane.
 

715
00:42:21 --> 00:42:24
Then I get the value,
so this is the delta

716
00:42:24 --> 00:42:27
function at the origin.
 

717
00:42:27 --> 00:42:29
So I get the value of
this at the origin.

718
00:42:29 --> 00:42:32
So what answer do I get?
 

719
00:42:32 --> 00:42:34
I get one out of the
integral and then I just

720
00:42:34 --> 00:42:35
have this constant.
 

721
00:42:35 --> 00:42:38
So it's constant again.
 

722
00:42:38 --> 00:42:41
And it's just one.
 

723
00:42:41 --> 00:42:45
So the Fourier coefficients
of the delta function

724
00:42:45 --> 00:42:46
are constant.
 

725
00:42:46 --> 00:42:49
All frequencies
there are the same.

726
00:42:49 --> 00:42:53
What about a line of
delta functions?

727
00:42:53 --> 00:42:54
And what does that mean?
 

728
00:42:54 --> 00:43:00
What about, yeah let me
try to draw delta(x).

729
00:43:01 --> 00:43:07
Suppose I have a function of
x and y - it's just worth

730
00:43:07 --> 00:43:12
imagining a line of
delta functions.

731
00:43:12 --> 00:43:21
So I'm in the x,y, let me
look again at this thing.

732
00:43:21 --> 00:43:25
I have delta functions
all along this line.

733
00:43:25 --> 00:43:26
Now.
 

734
00:43:26 --> 00:43:29
Here is a crazy example,
just to say well there

735
00:43:29 --> 00:43:32
is something new in 2-D.
 

736
00:43:32 --> 00:43:36
So previously my delta function
was just at that point.

737
00:43:36 --> 00:43:38
And the all integrals just
picked out the value

738
00:43:38 --> 00:43:40
at that point.
 

739
00:43:40 --> 00:43:44
But now think of a delta
function's a sort

740
00:43:44 --> 00:43:46
of line of spikes.
 

741
00:43:46 --> 00:43:48
Going up here, and then
of course it's periodic.

742
00:43:48 --> 00:43:52
Everything's periodic so that
line continues and this line

743
00:43:52 --> 00:43:55
appears here, and this
line appears here.

744
00:43:55 --> 00:44:02
But I only have to focus
on one period square.

745
00:44:02 --> 00:44:04
What's my answer now?
 

746
00:44:04 --> 00:44:10
If this function suddenly
changes from a one point

747
00:44:10 --> 00:44:13
delta function to a line
of delta functions?

748
00:44:13 --> 00:44:17
Now tell me what the
coefficients are.

749
00:44:17 --> 00:44:21
What are the Fourier
coefficients in 2-D for a

750
00:44:21 --> 00:44:22
line of delta functions?
 

751
00:44:22 --> 00:44:28
A straight line of delta
functions going up the y axis?

752
00:44:28 --> 00:44:32
It'll be - let's see.
 

753
00:44:32 --> 00:44:34
What do I do?
 

754
00:44:34 --> 00:44:39
I'm go to integrate -
oh yeah, what is it?

755
00:44:39 --> 00:44:40
Good question.
 

756
00:44:40 --> 00:44:42
OK.
 

757
00:44:42 --> 00:44:47
So what is the - when do
I get zero and when do I

758
00:44:47 --> 00:44:49
not get zero out of this?
 

759
00:44:49 --> 00:44:54
Yeah, tell me first when do I
get zero out of this integral?

760
00:44:54 --> 00:44:56
And when do I not?
 

761
00:44:56 --> 00:45:00
What am I doing here?
 

762
00:45:00 --> 00:45:01
Help me.
 

763
00:45:01 --> 00:45:06
I said 2-D was easy and I've
got in over my head here.

764
00:45:06 --> 00:45:10
So look.
 

765
00:45:10 --> 00:45:12
I can do the x integral, right?
 

766
00:45:12 --> 00:45:15
We all know how to
do the x integral.

767
00:45:15 --> 00:45:18
Yes, is that right?
 

768
00:45:18 --> 00:45:22
If I integrate with respect
to x, what do I get?

769
00:45:22 --> 00:45:24
Let's see, I'll keep that
one over 2pi squared.

770
00:45:24 --> 00:45:28
Now I'm trying to
do this integral.

771
00:45:28 --> 00:45:34
Do I get a one?
 

772
00:45:34 --> 00:45:37
If I get a one from
the x integral.

773
00:45:37 --> 00:45:41
So then I'm down to one
integral, just the y integral

774
00:45:41 --> 00:45:47
is left. e^(-ily)dy, right?
 

775
00:45:47 --> 00:45:51
I did the x part, which
said, OK, take the value

776
00:45:51 --> 00:45:53
at x=0, which was one.
 

777
00:45:53 --> 00:45:57
So the x integral
was one, good.

778
00:45:57 --> 00:45:59
And now I've got
down to this part.

779
00:45:59 --> 00:46:03
Now what is that integral?
 

780
00:46:03 --> 00:46:10
It's two - wait a minute.
 

781
00:46:10 --> 00:46:16
Depends on l, doesn't it?
 

782
00:46:16 --> 00:46:18
When l - yeah.
 

783
00:46:18 --> 00:46:23
So it's going to depend on
whether l is zero or not.

784
00:46:23 --> 00:46:26
Is that right?
 

785
00:46:26 --> 00:46:28
Yeah, that's sort
of interesting.

786
00:46:28 --> 00:46:32
If l is zero, then I'm getting
- then this is a 2pi.

787
00:46:33 --> 00:46:40
So the answer, so I'm getting
c_k0, when l is zero I'm

788
00:46:40 --> 00:46:42
getting a 2pi out of that.
 

789
00:46:42 --> 00:46:45
If l is zero, I'm integrating
one, I get a 2pi

790
00:46:45 --> 00:46:46
cancels one of those.
 

791
00:46:46 --> 00:46:48
I get a one over 2pi.
 

792
00:46:48 --> 00:46:51
 
 

793
00:46:51 --> 00:46:59
And otherwise the other c_kl's,
when l is not zero, are what?

794
00:46:59 --> 00:47:00
Just zero, I think.
 

795
00:47:00 --> 00:47:03
The integral of this thing,
this is a periodic guy, if I

796
00:47:03 --> 00:47:07
integrate it from minus
pi to pi it's zero.

797
00:47:07 --> 00:47:12
What am I - I'm making a big
deal out of something that

798
00:47:12 --> 00:47:14
shouldn't be a big deal.
 

799
00:47:14 --> 00:47:20
The delta(x) function, this
is just, its Fourier series

800
00:47:20 --> 00:47:23
is just the one we know.
 

801
00:47:23 --> 00:47:25
Sum of e^(ikx)'s.
 

802
00:47:25 --> 00:47:31
 
 

803
00:47:31 --> 00:47:33
Do you see what's
happened here?

804
00:47:33 --> 00:47:36
It was supposed to
be a double sum.

805
00:47:36 --> 00:47:41
But the ones, when l
wasn't zero aren't there.

806
00:47:41 --> 00:47:42
The only ones - yeah.
 

807
00:47:42 --> 00:47:48
So I'm back to the, for a line
of spikes, a line of deltas,

808
00:47:48 --> 00:47:51
I'm back to - it so it only
depended on x, so the

809
00:47:51 --> 00:47:54
Fourier series is just
the one I already know.

810
00:47:54 --> 00:47:56
All ones.
 

811
00:47:56 --> 00:48:01
When there's no - when l is
zero, all ones or 1/2 2pi's,

812
00:48:01 --> 00:48:08
all constants when l is
zero, but there's no y.

813
00:48:08 --> 00:48:11
There's no oscillation
in the y direction.

814
00:48:11 --> 00:48:18
OK, I don't know why I got into
that example, because the

815
00:48:18 --> 00:48:22
conclusion was just it's the
Fourier series that we already

816
00:48:22 --> 00:48:26
know and it doesn't depend on
l, because the function

817
00:48:26 --> 00:48:28
didn't depend on y.
 

818
00:48:28 --> 00:48:33
OK, then we could imagine delta
functions in other positions,

819
00:48:33 --> 00:48:36
or a general function.
 

820
00:48:36 --> 00:48:39
OK, so that's 2-D.
 

821
00:48:39 --> 00:48:44
Would I want to
tackle a 2-D - ha.

822
00:48:44 --> 00:48:48
We've got two minutes. that's
one dimension a minute.

823
00:48:48 --> 00:48:50
Right, OK.
 

824
00:48:50 --> 00:48:57
What happens, what's a 2-D
discrete convolution?

825
00:48:57 --> 00:49:00
What's a 2-D discrete
convolution?

826
00:49:00 --> 00:49:02
Now, you might say OK,
why is Professor Strang

827
00:49:02 --> 00:49:04
inventing these problems?
 

828
00:49:04 --> 00:49:07
Because a 2-D discrete
convolution is the core

829
00:49:07 --> 00:49:10
idea of image processing.
 

830
00:49:10 --> 00:49:14
If I have an image, what
does image processing do?

831
00:49:14 --> 00:49:19
Image processing takes my
image, it separates it into

832
00:49:19 --> 00:49:23
pixels, right, that's all the
image is, bunch of pixels.

833
00:49:23 --> 00:49:30
Then many 2-D image processing
algorithms, we'll jpeg, all

834
00:49:30 --> 00:49:35
jpeg for example, would take
an eight by eight, eight by

835
00:49:35 --> 00:49:37
eight 2-D, in other words.
 

836
00:49:37 --> 00:49:41
Eight by eight, 2-D
is the main point.

837
00:49:41 --> 00:49:42
Set of pixels.
 

838
00:49:42 --> 00:49:43
And transform it.
 

839
00:49:43 --> 00:49:46
Do a 2-D transform.
 

840
00:49:46 --> 00:49:48
So what is a 2-D transform?
 

841
00:49:48 --> 00:49:56
What would be the 2-D transform
that would correspond to this?

842
00:49:56 --> 00:49:59
First of all how
big's the matrix?

843
00:49:59 --> 00:50:01
Just so we get an idea.
 

844
00:50:01 --> 00:50:04
I probably won't get to
the end of this example.

845
00:50:04 --> 00:50:10
But just, so in 1-D, my
matrix was four by four.

846
00:50:10 --> 00:50:15
Now I've got, so that was
for four points on a line.

847
00:50:15 --> 00:50:22
Now I've got a
square of points.

848
00:50:22 --> 00:50:25
So how big is my matrix?
 

849
00:50:25 --> 00:50:26
16, right?
 

850
00:50:26 --> 00:50:32
16 by 16, because it's
operating on 16 pixels.

851
00:50:32 --> 00:50:37
It's operating on 16 pixels,
where in 1-D it only

852
00:50:37 --> 00:50:38
had four to act on.
 

853
00:50:38 --> 00:50:42
So I'm going to end up with
a 16 by 16 matrix here.

854
00:50:42 --> 00:50:48
And I think - let me
see, what do I need?

855
00:50:48 --> 00:50:49
Oh, wait a minute.
 

856
00:50:49 --> 00:50:51
Uh-oh.
 

857
00:50:51 --> 00:50:54
Yeah, I think the
time's up here.

858
00:50:54 --> 00:50:59
Yeah, because my C, has my C
got to have 16 components?

859
00:50:59 --> 00:51:00
Yes.
 

860
00:51:00 --> 00:51:03
My u has to have 16, my
right-hand side has got

861
00:51:03 --> 00:51:05
these 16 components.
 

862
00:51:05 --> 00:51:05
Yeah.
 

863
00:51:05 --> 00:51:09
So I'm up to 16 but a
very special circulant

864
00:51:09 --> 00:51:11
of a circulant.
 

865
00:51:11 --> 00:51:13
It'll be a circulant of
a circulant somehow.

866
00:51:13 --> 00:51:16
OK, enough for 2-D.
 

867
00:51:16 --> 00:51:18
I'll see you Wednesday and
we're back to reality.

868
00:51:18 --> 00:51:19
OK.
 

869
00:51:19 --> 00:51:20