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5DUQ3-Y_gX4 | This somehow is a source
of endless confusion. |
5DUQ3-Y_gX4 | Alan already talked
about it when |
5DUQ3-Y_gX4 | he said, for
example, dx transpose |
5DUQ3-Y_gX4 | x equals x transpose dx. |
5DUQ3-Y_gX4 | That's a little easier to
understand because it's just |
5DUQ3-Y_gX4 | a dot product. |
5DUQ3-Y_gX4 | You can swap things,
a scalar product. |
5DUQ3-Y_gX4 | So this is also an
instance of the same rule. |
5DUQ3-Y_gX4 | But you know, it's a little
bit more complicated looking. |
5DUQ3-Y_gX4 | It's the same idea. |
5DUQ3-Y_gX4 | Because this is a number,
I can transpose it. |
5DUQ3-Y_gX4 | And that swaps
everything around. |
5DUQ3-Y_gX4 | And the transpose of a product-- |
5DUQ3-Y_gX4 | hopefully, you remember
from linear algebra |
5DUQ3-Y_gX4 | that the transpose
of a product is |
5DUQ3-Y_gX4 | the product of the
transpose in reverse order. |
5DUQ3-Y_gX4 | So this term here,
this whole term, |
5DUQ3-Y_gX4 | equals x transpose
A transpose dx. |
5DUQ3-Y_gX4 | And what that allows
me to do is it allows |
5DUQ3-Y_gX4 | me to combine the two terms. |
5DUQ3-Y_gX4 | Now, this term and
this term, they |
5DUQ3-Y_gX4 | both have a dx on the right. |
5DUQ3-Y_gX4 | And so I can put that
over in the right. |
5DUQ3-Y_gX4 | And I can put parentheses. |
5DUQ3-Y_gX4 | And I have two terms,
and both of them |
5DUQ3-Y_gX4 | have an x transpose on the left. |
5DUQ3-Y_gX4 | And the first term,
one of the terms, |
5DUQ3-Y_gX4 | has an A. The other
term has an A transpose. |
5DUQ3-Y_gX4 | So that means this thing here
is our derivative, f prime. |
5DUQ3-Y_gX4 | Again, don't get that confused
with the differential. |
5DUQ3-Y_gX4 | f prime times dx is df. |
5DUQ3-Y_gX4 | That's the change in the output. |
5DUQ3-Y_gX4 | That's the little
change in the output. |
5DUQ3-Y_gX4 | f prime is the rate of change. |
5DUQ3-Y_gX4 | It's the thing
you operate on dx. |
5DUQ3-Y_gX4 | This is a row vector. |
5DUQ3-Y_gX4 | Notice that this
is a row vector. |
5DUQ3-Y_gX4 | ALAN EDELMAN: Could
I ask the class? |
5DUQ3-Y_gX4 | Then quickly tell me what is
the gradient of x transpose dx. |
5DUQ3-Y_gX4 | Anybody want to shout it out? |
5DUQ3-Y_gX4 | What is the gradient? |
5DUQ3-Y_gX4 | AUDIENCE: You [INAUDIBLE]? |
5DUQ3-Y_gX4 | ALAN EDELMAN: Right. |
5DUQ3-Y_gX4 | Do you want to say
it in its full glory? |
5DUQ3-Y_gX4 | AUDIENCE: OK. |
5DUQ3-Y_gX4 | A plus-- A plus A
transpose times x. |
5DUQ3-Y_gX4 | ALAN EDELMAN: Good, yup. |
5DUQ3-Y_gX4 | A plus A transpose times x
is the gradient, exactly. |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: Right. |
5DUQ3-Y_gX4 | It's just the transpose
of this thing. |
5DUQ3-Y_gX4 | So we transpose this thing.
dx goes over on the right. |
5DUQ3-Y_gX4 | This thing gets transposed. |
5DUQ3-Y_gX4 | But this is symmetric,
so it equals itself. |
5DUQ3-Y_gX4 | ALAN EDELMAN: So,
Steven, on the clock |
5DUQ3-Y_gX4 | here, we're already at 12:56. |
5DUQ3-Y_gX4 | So you might want to kind of-- |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: OK. |
5DUQ3-Y_gX4 | ALAN EDELMAN: --come
to a conclusion. |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: Yeah. |
5DUQ3-Y_gX4 | So this is just revisiting
the notion of a gradient. |
5DUQ3-Y_gX4 | And so next time, we're going
to continue so that next time, |
5DUQ3-Y_gX4 | basically, we're going to
do 18.06 revisited part two. |
5DUQ3-Y_gX4 | And we're going to
have f is now going |
5DUQ3-Y_gX4 | to be a vector function that
takes a vector of outputs |
5DUQ3-Y_gX4 | and also takes a
vector of inputs. |
5DUQ3-Y_gX4 | And so we have outputs in,
say, Rm inputs in, say, Rn. |
5DUQ3-Y_gX4 | I probably should have used n
just to be consistent before. |
5DUQ3-Y_gX4 | And then what we're going
to find is that then-- |
5DUQ3-Y_gX4 | you can almost do it right now. |
5DUQ3-Y_gX4 | df has to be a
linear operator that |
5DUQ3-Y_gX4 | takes a small
change in the input |
5DUQ3-Y_gX4 | and gives you a small
change in the output. |
5DUQ3-Y_gX4 | And so this has to have m
outputs, m components here. |
5DUQ3-Y_gX4 | It has to have n
components there. |
5DUQ3-Y_gX4 | The only way you can get
a linear operator that |
5DUQ3-Y_gX4 | takes n inputs and m
outputs is that this |
5DUQ3-Y_gX4 | has to be an m by n matrix. |
5DUQ3-Y_gX4 | ALAN EDELMAN: It has to
be expressible as an m |
5DUQ3-Y_gX4 | by n matrix. |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: Exactly. |
5DUQ3-Y_gX4 | ALAN EDELMAN: But you don't
have to write down the matrix. |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON:
Yes, that's right. |
5DUQ3-Y_gX4 | So this is our f prime
of x linear operator. |
5DUQ3-Y_gX4 | And this, if we write
it down as a matrix, |
5DUQ3-Y_gX4 | we call it the Jacobian. |
5DUQ3-Y_gX4 | ALAN EDELMAN: I
think what I'll do |
5DUQ3-Y_gX4 | next time is show my
favorite nonlinear |
5DUQ3-Y_gX4 | operator on two-dimensional
space, which is |
5DUQ3-Y_gX4 | hyperbolic operators on corgis. |
5DUQ3-Y_gX4 | So you'll see hyperbolic
corgis on Friday if you come. |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: So probably
on Friday maybe we'll switch. |
5DUQ3-Y_gX4 | And I'll start with the first
half and then finish this up. |
5DUQ3-Y_gX4 | And then Alan can
do the second half. |
5DUQ3-Y_gX4 | ALAN EDELMAN: OK, we
could do it that way. |
5DUQ3-Y_gX4 | OK. |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: Thanks, all. |
5DUQ3-Y_gX4 | Any questions at this point? |
5DUQ3-Y_gX4 | But let's-- |
5DUQ3-Y_gX4 | ALAN EDELMAN: I'll
also stick around. |
5DUQ3-Y_gX4 | You can ask Steven
or, you know-- |
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