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En provenance du cours de Imperial College London

Mathematics for Machine Learning: Multivariate Calculus

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Imperial College London

Mathematics for Machine Learning: Multivariate Calculus

1128 notes

Cours 2 sur 3 dans Specialization Mathematics for Machine Learning

This course offers a brief introduction to the multivariate calculus required to build many common machine learning techniques. We start at the very beginning with a refresher on the “rise over run” formulation of a slope, before converting this to the formal definition of the gradient of a function. We then start to build up a set of tools for making calculus easier and faster. Next, we learn how to calculate vectors that point up hill on multidimensional surfaces and even put this into action using an interactive game. We take a look at how we can use calculus to build approximations to functions, as well as helping us to quantify how accurate we should expect those approximations to be. We also spend some time talking about where calculus comes up in the training of neural networks, before finally showing you how it is applied in linear regression models. This course is intended to offer an intuitive understanding of calculus, as well as the language necessary to look concepts up yourselves when you get stuck. Hopefully, without going into too much detail, you’ll still come away with the confidence to dive into some more focused machine learning courses in future.

À partir de la leçon

Multivariate chain rule and its applications

Having seen that multivariate calculus is really no more complicated than the univariate case, we now focus on applications of the chain rule. Neural networks are one of the most popular and successful conceptual structures in machine learning. They are build up from a connected web of neurons and inspired by the structure of biological brains. The behaviour of each neuron is influenced by a set of control parameters, each of which needs to be optimised to best fit the data. The multivariate chain rule can be used to calculate the influence of each parameter of the networks, allow them to be updated during training.

Welcome to Module 3!0:36

Multivariate chain rule2:38

More multivariate chain rule5:38

Rencontrer les enseignants

Samuel J. Cooper
Lecturer
Dyson School of Design Engineering
David Dye
Professor of Metallurgy
Department of Materials
A. Freddie Page
Strategic Teaching Fellow
Dyson School of Design Engineering

[MUSIC].

Last video, we finished up an expression for the multivariate chain rule.

In this video, we're going to start by picking up

one more little detail which you might have already spotted.

And then follow this up by adding another link to our chain.

If we have a function, f(x), where x is a vector,

in which each term in x depends on t, which we can compactly write like this.

We also have a compact form of the multivariate chain rule to go with it.

What I hope you might have noticed last time is that our vector of partial

derivatives, df by dx, is just the same as a Jacobian vector,

which we saw last module.

Except that we wrote as a column instead of a raw vector.

So from our knowledge of linear algebra,

we can say that df by dx must be the transverse of the Jacobian of f.

The last thing to realize is that taking the dot product of two column vectors

is the same operation as multiplying a row vector by a column vector.

So finally, we can see that our old friend, the Jacobian, offers us perhaps

the most convenient representation of the multivariate chain rule.

Next, we're going to see that the chain rule still works for more than two links.

To start us off, we'll work through a really quick univariate example

where we're going to add in another function of separating f from t.

So we can say f(x) = 5x.

And we can have x(u) = 1- u.

And u(t) = t squared.

So we've got three functions, and we're separating f from t now by an extra step.

Of course, we can just sub in each step into each other and

find an expression for f as a function directly of t where we say, well,

it's going to be 5 of whatever x is and x is 1- whatever u is.

And u is t squared, so

this thing goes to 5- 5t squared.

And of course, we can now directly differentiate

this thing, and say df by dt= -10t.

Or we can apply a two-step chain rule and say the following.

So we can have df by dt =

df by dx times dx by du and

finally du by dt, okay?

Now subbing in for each of our terms, we just get,

well, this thing's going to equal df by dt is, so df by dx is just 5.

We're going to multiply that by dx by du, which is just -1.

And finally, du by dt is just going to be 2t.

So once again, we're going to recover the same answer,

which is- 10t.

So we can see that this approach works for chains of univariate functions.

And we could extend it out to as many intermediary functions between f and

t as we would like.

But what about the multivariate case?

Well, the chain rule does work here, too, but

we do just have to pay attention to a few extra details.

Let's start by considering the function f(x(u(t))),

again, where the function f takes the vector x as an input, but

this time x is a vector valued function, which also takes a vector u as its input.

As the last video, the both symbols indicate vectors.

So u is again a vector valued function and it takes this scalar t as its input.

Ultimately, we are still relating the scalar input t to a scalar output f.

But we're doing this via two intermediary vector valued functions, x and u.

If, once again, we'd like to know the derivative of f with respect to t,

we can essentially write the same expression as the univariate case,

except that now seven of our terms are in bold.

We've already seen differentiating the scale of the valued function f,

with respect to its input vector x, gives us the Jacobian row vector.

We've also seen that differentiating a vector valued function

u with respect to the scalar variable t gives us a column vector of derivatives.

But what about the middle term dx by du?

Well, for the function x, we need to find the derivative of each of the two

output variables with respect to each of the two input variables.

So we end up with four terms in total,

which as we saw in the last module, can be conveniently arranged as a matrix.

We still refer to this object as a Jacobian, so

we can now say the derivative of f with respect to t, is the product

of the Jacobian of f, with the Jacobian of x and the derivative vector of u.

Notice that the dimensions of these vectors and matrices as shown here,

are such that this operation is possible.

And that they will return a scaler just as we expected.

That's it for this video.

I hope you're now starting to see the various threads of linear algebra and

multivariate calculus weaved together.

See you in the next one.

[MUSIC]

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