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