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5DUQ3-Y_gX4 | you need to multiply
by a row vector. |
5DUQ3-Y_gX4 | Another way of thinking
about it is you |
5DUQ3-Y_gX4 | need to take the dot product-- |
5DUQ3-Y_gX4 | ALAN EDELMAN: The vector
and the scalar out. |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: The vector
and scalar out-- sorry. |
5DUQ3-Y_gX4 | You need to multiply
it by a one row thing. |
5DUQ3-Y_gX4 | Another way of
thinking about it is |
5DUQ3-Y_gX4 | that, if you have a
linear operation that |
5DUQ3-Y_gX4 | takes a vector in and
gives you a scalar |
5DUQ3-Y_gX4 | out, the only type of thing
that does that is a dot product. |
5DUQ3-Y_gX4 | If you take a dot product of
the vector, you get a scalar. |
5DUQ3-Y_gX4 | And that's the only
linear operation |
5DUQ3-Y_gX4 | that gives you a scalar
from a vector in some sense. |
5DUQ3-Y_gX4 | And so this is a dot
product with some vector. |
5DUQ3-Y_gX4 | That vector must
be pretty special. |
5DUQ3-Y_gX4 | And so we'll give it a name. |
5DUQ3-Y_gX4 | And we'll call
that the gradient. |
5DUQ3-Y_gX4 | So I think this is going
to be the gradient of f. |
5DUQ3-Y_gX4 | And this is the thing we
take the dot product with |
5DUQ3-Y_gX4 | to get our scalar df. |
5DUQ3-Y_gX4 | So you can think of this
in linear algebra terms. |
5DUQ3-Y_gX4 | So this dot product
is the same thing |
5DUQ3-Y_gX4 | as multiplying by a transpose. |
5DUQ3-Y_gX4 | So this is the same thing as-- |
5DUQ3-Y_gX4 | this is saying that f prime
is really grad f transpose. |
5DUQ3-Y_gX4 | Or equivalently, f prime
dx is the operation |
5DUQ3-Y_gX4 | of a dot product
with a gradient. |
5DUQ3-Y_gX4 | This is going to be really
powerful pretty soon |
5DUQ3-Y_gX4 | because it's going to also
allow us to generalize gradients |
5DUQ3-Y_gX4 | to other kinds of vector spaces. |
5DUQ3-Y_gX4 | As long as we have a dot
product and a scalar function, |
5DUQ3-Y_gX4 | you'll be able to
define a gradient. |
5DUQ3-Y_gX4 | So if you have a scalar function
of-- if this is something that |
5DUQ3-Y_gX4 | takes a matrix in and a scalar
out, like a determinant, |
5DUQ3-Y_gX4 | pretty soon we're going
to be able to take |
5DUQ3-Y_gX4 | the gradient of a determinant. |
5DUQ3-Y_gX4 | We'll be able to
define what that means. |
5DUQ3-Y_gX4 | ALAN EDELMAN: Just
to be clear, what |
5DUQ3-Y_gX4 | is on the other side
of that equals sign |
5DUQ3-Y_gX4 | that you just wrote? |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON:
So, yes, the f-- |
5DUQ3-Y_gX4 | yeah, I should write that. |
5DUQ3-Y_gX4 | That's the f prime. |
5DUQ3-Y_gX4 | Yeah. |
5DUQ3-Y_gX4 | So this is-- yeah. |
5DUQ3-Y_gX4 | I need to-- my equals-- |
5DUQ3-Y_gX4 | let's see. |
5DUQ3-Y_gX4 | The gradient of f is the
thing we take a dot product. |
5DUQ3-Y_gX4 | And here it's f
prime of x is this. |
5DUQ3-Y_gX4 | ALAN EDELMAN: Exactly. |
5DUQ3-Y_gX4 | It's f prime of x. |
5DUQ3-Y_gX4 | And it's not df. |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: Yeah. |
5DUQ3-Y_gX4 | It's not-- no dx. |
5DUQ3-Y_gX4 | Just the f prime by
itself is a row vector. |
5DUQ3-Y_gX4 | That's the transpose
of the gradient. |
5DUQ3-Y_gX4 | So now, that's the
definition of the gradient. |
5DUQ3-Y_gX4 | It's only something
we're usually |
5DUQ3-Y_gX4 | going to define for scalar
functions of vectors. |
5DUQ3-Y_gX4 | And pretty soon, we'll be
able to generalize that |
5DUQ3-Y_gX4 | to other kinds of vectors,
but will still be a scalar. |
5DUQ3-Y_gX4 | And it's the thing you
take the dot product |
5DUQ3-Y_gX4 | with of dx, of the differential,
and the change in the input |
5DUQ3-Y_gX4 | with to get the change
in the output, OK? |
5DUQ3-Y_gX4 | Yeah. |
5DUQ3-Y_gX4 | So we did-- Alan did the
example of x transpose x. |
5DUQ3-Y_gX4 | Let's do another
example just for fun. |
5DUQ3-Y_gX4 | So suppose f of x. |
5DUQ3-Y_gX4 | So x is going to be a vector. |
5DUQ3-Y_gX4 | Suppose that x transpose
Ax where-- so this is x. |
5DUQ3-Y_gX4 | x here is going to
have m components. |
5DUQ3-Y_gX4 | And so we're going to let
A be an m by m matrix. |
5DUQ3-Y_gX4 | ALAN EDELMAN: And are you
assuming asymmetric or-- |
5DUQ3-Y_gX4 | STEVEN G. JOHNSON: No, I'm not. |
5DUQ3-Y_gX4 | I won't make it symmetric
just for generality, OK? |
5DUQ3-Y_gX4 | And so this is going to be-- |
5DUQ3-Y_gX4 | A is not going to be an input. |
5DUQ3-Y_gX4 | This is just going to
be a constant matrix. |
5DUQ3-Y_gX4 | So when I do my d's, when I
change x, A does not change. |
5DUQ3-Y_gX4 | OK. |
5DUQ3-Y_gX4 | So let me do it
the long way first, |
5DUQ3-Y_gX4 | and then we'll try
and derive some rules. |
5DUQ3-Y_gX4 | So let's do the long way-- |
5DUQ3-Y_gX4 | long way, but still
faster than doing it |
5DUQ3-Y_gX4 | component by component, faster
than the 18.02 component |
5DUQ3-Y_gX4 | by component. |
5DUQ3-Y_gX4 | Because I could take the
derivative of this with respect |
5DUQ3-Y_gX4 | to x1, with respect to x2, with
respect to all the components, |
5DUQ3-Y_gX4 | and then build up the gradient. |
5DUQ3-Y_gX4 | It starts to become really
awkward really quickly. |
5DUQ3-Y_gX4 | I know it can be tempting when
you're faced with new problems |
5DUQ3-Y_gX4 | to fall back on what
you know, right? |
5DUQ3-Y_gX4 | And that's not a bad strategy. |
5DUQ3-Y_gX4 | But we really want
to encourage you to-- |
5DUQ3-Y_gX4 | even though you know
calculus really, |
5DUQ3-Y_gX4 | really well, you know how to
take derivatives really, really |
5DUQ3-Y_gX4 | well, 18.02 and
18.01 style, we're |
5DUQ3-Y_gX4 | going to try and learn
something new here, |
5DUQ3-Y_gX4 | a new way that can be really
a lot more powerful besides |
5DUQ3-Y_gX4 | [INAUDIBLE]. |
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