A visual proof that neural nets can compute any function


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One of the most striking facts about neural networks is that they can
compute any function at all. That is, suppose someone hands you some
complicated, wiggly function, $f(x)$:

No matter what the
function, there is guaranteed to be a neural network so that for every
possible input, $x$, the value $f(x)$ (or some close approximation) is
output from the network, e.g.:

This result holds even if the function has many inputs, $f = f(x_1,
ldots, x_m)$, and many outputs. For instance, here’s a network
computing a function with $m = 3$ inputs and $n = 2$ outputs:

This result tells us that neural networks have a kind of
universality. No matter what function we want to compute, we
know that there is a neural network which can do the job.

What’s more, this universality theorem holds even if we restrict our
networks to have just a single layer intermediate between the input
and the output neurons – a so-called single hidden layer. So even
very simple network architectures can be extremely powerful.

The universality theorem is well known by people who use neural
networks. But why it’s true is not so widely understood. Most of the
explanations available are quite technical. For instance, one of the
original papers proving the
by superpositions of a sigmoidal function
, by George Cybenko
(1989). The result was very much in the air at the time, and
several groups proved closely related results. Cybenko’s paper
contains a useful discussion of much of that work. Another
important early paper is
feedforward networks are universal approximators
, by Kurt Hornik,
Maxwell Stinchcombe, and Halbert White (1989). This paper uses the
Stone-Weierstrass theorem to arrive at similar results.
did so
using the Hahn-Banach theorem, the Riesz Representation theorem, and
some Fourier analysis. If you’re a mathematician the argument is not
difficult to follow, but it’s not so easy for most people. That’s a
pity, since the underlying reasons for universality are simple and

In this chapter I give a simple and mostly visual explanation of the
universality theorem. We’ll go step by step through the underlying
ideas. You’ll understand why it’s true that neural networks can
compute any function. You’ll understand some of the limitations of
the result. And you’ll understand how the result relates to deep
neural networks.

To follow the material in the chapter, you do not need to have read
earlier chapters in this book. Instead, the chapter is structured to
be enjoyable as a self-contained essay. Provided you have just a
little basic familiarity with neural networks, you should be able to
follow the explanation. I will, however, provide occasional links to
earlier material, to help fill in any gaps in your knowledge.

Universality theorems are a commonplace in computer science, so much
so that we sometimes forget how astonishing they are. But it’s worth
reminding ourselves: the ability to compute an arbitrary function is
truly remarkable. Almost any process you can imagine can be thought
of as function computation. Consider the problem of naming a piece of
music based on a short sample of the piece. That can be thought of as
computing a function. Or consider the problem of translating a
Chinese text into English. Again, that can be thought of as computing
a function*
*Actually, computing one of many functions, since
there are often many acceptable translations of a given piece of
. Or consider the problem of taking an mp4 movie file and
generating a description of the plot of the movie, and a discussion of
the quality of the acting. Again, that can be thought of as a kind of
function computation*
*Ditto the remark about translation and
there being many possible functions.
. Universality means that, in
principle, neural networks can do all these things and many more.

Of course, just because we know a neural network exists that can (say)
translate Chinese text into English, that doesn’t mean we have good
techniques for constructing or even recognizing such a network. This
limitation applies also to traditional universality theorems for
models such as Boolean circuits. But, as we’ve seen earlier in the
book, neural networks have powerful algorithms for learning functions.
That combination of learning algorithms + universality is an
attractive mix. Up to now, the book has focused on the learning
algorithms. In this chapter, we focus on universality, and what it

Two caveats

Before explaining why the universality theorem is true, I want to
mention two caveats to the informal statement “a neural network can
compute any function”.

First, this doesn’t mean that a network can be used to exactly
compute any function. Rather, we can get an approximation
that is as good as we want. By increasing the number of hidden
neurons we can improve the approximation. For instance,
earlier I illustrated a network
computing some function $f(x)$ using three hidden neurons. For most
functions only a low-quality approximation will be possible using
three hidden neurons. By increasing the number of hidden neurons
(say, to five) we can typically get a better approximation:

And we can do still better by further increasing the number of hidden

To make this statement more precise, suppose we’re given a function
$f(x)$ which we’d like to compute to within some desired accuracy
$epsilon > 0$. The guarantee is that by using enough hidden neurons
we can always find a neural network whose output $g(x)$ satisfies
$|g(x) – f(x)|

The second caveat is that the class of functions which can be
approximated in the way described are the continuous functions.
If a function is discontinuous, i.e., makes sudden, sharp jumps, then
it won’t in general be possible to approximate using a neural net.
This is not surprising, since our neural networks compute continuous
functions of their input. However, even if the function we’d really
like to compute is discontinuous, it’s often the case that a
continuous approximation is good enough. If that’s so, then we can
use a neural network. In practice, this is not usually an important

Summing up, a more precise statement of the universality theorem is
that neural networks with a single hidden layer can be used to
approximate any continuous function to any desired precision. In this
chapter we’ll actually prove a slightly weaker version of this result,
using two hidden layers instead of one. In the problems I’ll briefly
outline how the explanation can, with a few tweaks, be adapted to give
a proof which uses only a single hidden layer.

Universality with one input and one output

To understand why the universality theorem is true, let’s start by
understanding how to construct a neural network which approximates a
function with just one input and one output:

It turns out that this is the core of the problem of universality.
Once we’ve understood this special case it’s actually pretty easy to
extend to functions with many inputs and many outputs.

To build insight into how to construct a network to compute $f$, let’s
start with a network containing just a single hidden layer, with two
hidden neurons, and an output layer containing a single output neuron:

To get a feel for how components in the network work, let’s focus on
the top hidden neuron. In the diagram below, click on the weight,
$w$, and drag the mouse a little ways to the right to increase $w$.
You can immediately see how the function computed by the top hidden
neuron changes:

As we learnt earlier in the book,
what’s being computed by the hidden neuron is $sigma(wx + b)$, where
$sigma(z) equiv 1/(1+e^{-z})$ is the sigmoid function. Up to now,
we’ve made frequent use of this algebraic form. But for the proof of
universality we will obtain more insight by ignoring the algebra
entirely, and instead manipulating and observing the shape shown in
the graph. This won’t just give us a better feel for what’s going on,
it will also give us a proof*
*Strictly speaking, the visual
approach I’m taking isn’t what’s traditionally thought of as a
proof. But I believe the visual approach gives more insight into
why the result is true than a traditional proof. And, of course,
that kind of insight is the real purpose behind a proof.
Occasionally, there will be small gaps in the reasoning I present:
places where I make a visual argument that is plausible, but not
quite rigorous. If this bothers you, then consider it a challenge
to fill in the missing steps. But don’t lose sight of the real
purpose: to understand why the universality theorem is true.
universality that applies to activation functions other than the
sigmoid function.

To get started on this proof, try clicking on the bias, $b$, in the
diagram above, and dragging to the right to increase it. You’ll see
that as the bias increases the graph moves to the left, but its shape
doesn’t change.

Next, click and drag to the left in order to decrease the bias.
You’ll see that as the bias decreases the graph moves to the right,
but, again, its shape doesn’t change.

Next, decrease the weight to around $2$ or $3$. You’ll see that as
you decrease the weight, the curve broadens out. You might need to
change the bias as well, in order to keep the curve in-frame.

Finally, increase the weight up past $w = 100$. As you do, the curve
gets steeper, until eventually it begins to look like a step function.
Try to adjust the bias so the step occurs near $x = 0.3$. The
following short clip shows what your result should look like. Click
on the play button to play (or replay) the video:

We can simplify our analysis quite a bit by increasing the weight so
much that the output really is a step function, to a very good
approximation. Below I’ve plotted the output from the top hidden
neuron when the weight is $w = 999$. Note that this plot is static,
and you can’t change parameters such as the weight.

It’s actually quite a bit easier to work with step functions than
general sigmoid functions. The reason is that in the output layer we
add up contributions from all the hidden neurons. It’s easy to
analyze the sum of a bunch of step functions, but rather more
difficult to reason about what happens when you add up a bunch of
sigmoid shaped curves. And so it makes things much easier to assume
that our hidden neurons are outputting step functions. More
concretely, we do this by fixing the weight $w$ to be some very large
value, and then setting the position of the step by modifying the
bias. Of course, treating the output as a step function is an
approximation, but it’s a very good approximation, and for now we’ll
treat it as exact. I’ll come back later to discuss the impact of
deviations from this approximation.

At what value of $x$ does the step occur? Put another way, how does
the position of the step depend upon the weight and bias?

To answer this question, try modifying the weight and bias in the
diagram above (you may need to scroll back a bit). Can you figure out
how the position of the step depends on $w$ and $b$? With a little
work you should be able to convince yourself that the position of the
step is proportional to $b$, and inversely proportional
to $w$.

In fact, the step is at position $s = -b/w$, as you can see by
modifying the weight and bias in the following diagram:

It will greatly simplify our lives to describe hidden neurons using
just a single parameter, $s$, which is the step position, $s = -b/w$.
Try modifying $s$ in the following diagram, in order to get used to
the new parameterization:

As noted above, we’ve implicitly set the weight $w$ on the input to be
some large value – big enough that the step function is a very good
approximation. We can easily convert a neuron parameterized in this
way back into the conventional model, by choosing the bias $b = -w s$.

Up to now we’ve been focusing on the output from just the top hidden
neuron. Let’s take a look at the behavior of the entire network. In
particular, we’ll suppose the hidden neurons are computing step
functions parameterized by step points $s_1$ (top neuron) and $s_2$
(bottom neuron). And they’ll have respective output weights $w_1$ and
$w_2$. Here’s the network:

What’s being plotted on the right is the weighted output $w_1
a_1 + w_2 a_2$ from the hidden layer. Here, $a_1$ and $a_2$ are the
outputs from the top and bottom hidden neurons,
*Note, by the way, that the output from the whole
network is $sigma(w_1 a_1+w_2 a_2 + b)$, where $b$ is the bias on
the output neuron. Obviously, this isn’t the same as the weighted
output from the hidden layer, which is what we’re plotting here.
We’re going to focus on the weighted output from the hidden layer
right now, and only later will we think about how that relates to
the output from the whole network.
. These outputs are denoted with
$a$s because they’re often known as the neurons’ activations.

Try increasing and decreasing the step point $s_1$ of the top hidden
neuron. Get a feel for how this changes the weighted output from the
hidden layer.

It’s particularly worth understanding what happens when $s_1$ goes
past $s_2$. You’ll see that the graph changes shape when this
happens, since we have moved from a situation where the top hidden
neuron is the first to be activated to a situation where the bottom
hidden neuron is the first to be activated.

Similarly, try manipulating the step point $s_2$ of the bottom hidden
neuron, and get a feel for how this changes the combined output from
the hidden neurons.

Try increasing and decreasing each of the output weights. Notice how
this rescales the contribution from the respective hidden neurons.
What happens when one of the weights is zero?

Finally, try setting $w_1$ to be $0.8$ and $w_2$ to be $-0.8$. You
get a “bump” function, which starts at point $s_1$, ends at point
$s_2$, and has height $0.8$. For instance, the weighted output might
look like this:

Of course, we can rescale the bump to have any height at all. Let’s
use a single parameter, $h$, to denote the height. To reduce clutter
I’ll also remove the “$s_1 = ldots$” and “$w_1 = ldots$” notations.

Try changing the value of $h$ up and down, to see how the height of
the bump changes. Try changing the height so it’s negative, and
observe what happens. And try changing the step points to see how
that changes the shape of the bump.

You’ll notice, by the way, that we’re using our neurons in a way that
can be thought of not just in graphical terms, but in more
conventional programming terms, as a kind of if-then-else
statement, e.g.:

    if input >= step point:
        add 1 to the weighted output
        add 0 to the weighted output

For the most part I’m going to stick with the graphical point of view.
But in what follows you may sometimes find it helpful to switch points
of view, and think about things in terms of if-then-else.

We can use our bump-making trick to get two bumps, by gluing two pairs
of hidden neurons together into the same network:

I’ve suppressed the weights here, simply writing the $h$ values for
each pair of hidden neurons. Try increasing and decreasing both $h$
values, and observe how it changes the graph. Move the bumps around
by changing the step points.

More generally, we can use this idea to get as many peaks as we want,
of any height. In particular, we can divide the interval $[0, 1]$ up
into a large number, $N$, of subintervals, and use $N$ pairs of hidden
neurons to set up peaks of any desired height. Let’s see how this
works for $N = 5$. That’s quite a few neurons, so I’m going to pack
things in a bit. Apologies for the complexity of the diagram: I could
hide the complexity by abstracting away further, but I think it’s
worth putting up with a little complexity, for the sake of getting a
more concrete feel for how these networks work.

You can see that there are five pairs of hidden neurons. The step
points for the respective pairs of neurons are $0, 1/5$, then $1/5,
2/5$, and so on, out to $4/5, 5/5$. These values are fixed – they
make it so we get five evenly spaced bumps on the graph.

Each pair of neurons has a value of $h$ associated to it. Remember,
the connections output from the neurons have weights $h$ and $-h$ (not
marked). Click on one of the $h$ values, and drag the mouse to the
right or left to change the value. As you do so, watch the function
change. By changing the output weights we’re actually
designing the function!

Contrariwise, try clicking on the graph, and dragging up or down to
change the height of any of the bump functions. As you change the
heights, you can see the corresponding change in $h$ values. And,
although it’s not shown, there is also a change in the corresponding
output weights, which are $+h$ and $-h$.

In other words, we can directly manipulate the function appearing in
the graph on the right, and see that reflected in the $h$ values on
the left. A fun thing to do is to hold the mouse button down and drag
the mouse from one side of the graph to the other. As you do this you
draw out a function, and get to watch the parameters in the neural
network adapt.

Time for a challenge.

Let’s think back to the function I plotted at the beginning of the

I didn’t say it at the time, but what I plotted is actually the
f(x) = 0.2+0.4 x^2+0.3x sin(15 x) + 0.05 cos(50 x),
plotted over $x$ from $0$ to $1$, and with the $y$ axis taking
values from $0$ to $1$.

That’s obviously not a trivial function.

You’re going to figure out how to compute it using a neural network.

In our networks above we’ve been analyzing the weighted combination
$sum_j w_j a_j$ output from the hidden neurons. We now know how to
get a lot of control over this quantity. But, as I noted earlier,
this quantity is not what’s output from the network. What’s output
from the network is $sigma(sum_j w_j a_j + b)$ where $b$ is the bias
on the output neuron. Is there some way we can achieve control over
the actual output from the network?

The solution is to design a neural network whose hidden layer has a
weighted output given by $sigma^{-1} circ f(x)$, where $sigma^{-1}$
is just the inverse of the $sigma$ function. That is, we want the
weighted output from the hidden layer to be:

If we can do this, then the output from the network as a whole will be
a good approximation to $f(x)$*
*Note that I have set the bias
on the output neuron to $0$.

Your challenge, then, is to design a neural network to approximate the
goal function shown just above. To learn as much as possible, I want
you to solve the problem twice. The first time, please click on the
graph, directly adjusting the heights of the different bump functions.
You should find it fairly easy to get a good match to the goal
function. How well you’re doing is measured by the average
between the goal function and the function the network is
actually computing. Your challenge is to drive the average deviation
as low as possible. You complete the challenge when you drive
the average deviation to $0.40$ or below.

Once you’ve done that, click on “Reset” to randomly re-initialize
the bumps. The second time you solve the problem, resist the urge to
click on the graph. Instead, modify the $h$ values on the left-hand
side, and again attempt to drive the average deviation to $0.40$ or

You’ve now figured out all the elements necessary for the network to
approximately compute the function $f(x)$! It’s only a coarse
approximation, but we could easily do much better, merely by
increasing the number of pairs of hidden neurons, allowing more bumps.

In particular, it’s easy to convert all the data we have found back
into the standard parameterization used for neural networks. Let me
just recap quickly how that works.

The first layer of weights all have some large, constant value, say $w
= 1000$.

The biases on the hidden neurons are just $b = -w s$. So, for
instance, for the second hidden neuron $s = 0.2$ becomes $b = -1000
times 0.2 = -200$.

The final layer of weights are determined by the $h$ values. So, for
instance, the value you’ve chosen above for the first $h$, $h = $
, means that
the output weights from the top two hidden neurons are and , respectively. And
so on, for the entire layer of output weights.

Finally, the bias on the output neuron is $0$.

That’s everything: we now have a complete description of a neural
network which does a pretty good job computing our original goal
function. And we understand how to improve the quality of the
approximation by improving the number of hidden neurons.

What’s more, there was nothing special about our original goal
function, $f(x) = 0.2+0.4 x^2+0.3 sin(15 x) + 0.05 cos(50 x)$. We
could have used this procedure for any continuous function from $[0,
1]$ to $[0, 1]$. In essence, we’re using our single-layer neural
networks to build a lookup table for the function. And we’ll be able
to build on this idea to provide a general proof of universality.

Many input variables

Let’s extend our results to the case of many input variables. This
sounds complicated, but all the ideas we need can be understood in the
case of just two inputs. So let’s address the two-input case.

We’ll start by considering what happens when we have two inputs to a

Here, we have inputs $x$ and $y$, with corresponding weights $w_1$ and
$w_2$, and a bias $b$ on the neuron. Let’s set the weight $w_2$ to
$0$, and then play around with the first weight, $w_1$, and the bias,
$b$, to see how they affect the output from the neuron:

As you can see, with $w_2 = 0$ the input $y$ makes no difference to
the output from the neuron. It’s as though $x$ is the only input.

Given this, what do you think happens when we increase the weight
$w_1$ to $w_1 = 100$, with $w_2$ remaining $0$? If you don’t
immediately see the answer, ponder the question for a bit, and see if
you can figure out what happens. Then try it out and see if you’re
right. I’ve shown what happens in the following movie:

Just as in our earlier discussion, as the input weight gets larger the
output approaches a step function. The difference is that now the
step function is in three dimensions. Also as before, we can move the
location of the step point around by modifying the bias. The actual
location of the step point is $s_x equiv -b / w_1$.

Let’s redo the above using the position of the step as the parameter:

Here, we assume the weight on the $x$ input has some large value
– I’ve used $w_1 = 1000$ – and the weight $w_2 = 0$. The
number on the neuron is the step point, and the little $x$ above the
number reminds us that the step is in the $x$ direction.

Of course, it’s also possible to get a step function in the $y$
direction, by making the weight on the $y$ input very large (say, $w_2
= 1000$), and the weight on the $x$ equal to $0$, i.e., $w_1 = 0$:

The number on the neuron is again the step point, and in this case the
little $y$ above the number reminds us that the step is in the $y$
direction. I could have explicitly marked the weights on the $x$ and
$y$ inputs, but decided not to, since it would make the diagram rather
cluttered. But do keep in mind that the little $y$ marker implicitly
tells us that the $y$ weight is large, and the $x$ weight is $0$.

We can use the step functions we’ve just constructed to compute a
three-dimensional bump function. To do this, we use two neurons, each
computing a step function in the $x$ direction. Then we combine those
step functions with weight $h$ and $-h$, respectively, where $h$ is
the desired height of the bump. It’s all illustrated in the following

Try changing the value of the height, $h$. Observe how it relates to
the weights in the network. And see how it changes the height of the
bump function on the right.

Also, try changing the step point $0.30$ associated to the top hidden
neuron. Witness how it changes the shape of the bump. What happens
when you move it past the step point $0.70$ associated to the bottom
hidden neuron?

We’ve figured out how to make a bump function in the $x$ direction.
Of course, we can easily make a bump function in the $y$ direction, by
using two step functions in the $y$ direction. Recall that we do this
by making the weight large on the $y$ input, and the weight $0$ on the
$x$ input. Here’s the result:

This looks nearly identical to the earlier network! The only thing
explicitly shown as changing is that there’s now little $y$ markers on
our hidden neurons. That reminds us that they’re producing $y$ step
functions, not $x$ step functions, and so the weight is very large on
the $y$ input, and zero on the $x$ input, not vice versa. As before,
I decided not to show this explicitly, in order to avoid clutter.

Let’s consider what happens when we add up two bump functions, one in
the $x$ direction, the other in the $y$ direction, both of height $h$:

To simplify the diagram I’ve dropped the connections with zero weight.
For now, I’ve left in the little $x$ and $y$ markers on the hidden
neurons, to remind you in what directions the bump functions are being
computed. We’ll drop even those markers later, since they’re implied
by the input variable.

Try varying the parameter $h$. As you can see, this causes the output
weights to change, and also the heights of both the $x$ and $y$ bump

What we’ve built looks a little like a tower function:

If we could build such tower functions, then we could use them to
approximate arbitrary functions, just by adding up many towers of
different heights, and in different locations:

Of course, we haven’t yet figured out how to build a tower function.
What we have constructed looks like a central tower, of height $2h$,
with a surrounding plateau, of height $h$.

But we can make a tower function. Remember that earlier we saw
neurons can be used to implement a type of if-then-else

    if input >= threshold: 
        output 1
        output 0

That was for a neuron with just a single input. What we want is to
apply a similar idea to the combined output from the hidden neurons:

    if combined output from hidden neurons >= threshold:
        output 1
        output 0

If we choose the threshold appropriately – say, a value of
$3h/2$, which is sandwiched between the height of the plateau and the
height of the central tower – we could squash the plateau down to
zero, and leave just the tower standing.

Can you see how to do this? Try experimenting with the following
network to figure it out. Note that we’re now plotting the output
from the entire network, not just the weighted output from the hidden
layer. This means we add a bias term to the weighted output from the
hidden layer, and apply the sigma function. Can you find values for
$h$ and $b$ which produce a tower? This is a bit tricky, so if you
think about this for a while and remain stuck, here’s two hints: (1)
To get the output neuron to show the right kind of if-then-else
behaviour, we need the input weights (all $h$ or $-h$) to be large;
and (2) the value of $b$ determines the scale of the
if-then-else threshold.

With our initial parameters, the output looks like a flattened version
of the earlier diagram, with its tower and plateau. To get the
desired behaviour, we increase the parameter $h$ until it becomes
large. That gives the if-then-else thresholding
behaviour. Second, to get the threshold right, we’ll choose $b
approx -3h/2$. Try it, and see how it works!

Here’s what it looks like, when we use $h = 10$:

Even for this relatively modest value of $h$, we get a pretty good
tower function. And, of course, we can make it as good as we want by
increasing $h$ still further, and keeping the bias as $b = -3h/2$.

Let’s try gluing two such networks together, in order to compute two
different tower functions. To make the respective roles of the two
sub-networks clear I’ve put them in separate boxes, below: each box
computes a tower function, using the technique described above. The
graph on the right shows the weighted output from the second
hidden layer, that is, it’s a weighted combination of tower functions.

In particular, you can see that by modifying the weights in the final
layer you can change the height of the output towers.

The same idea can be used to compute as many towers as we like. We
can also make them as thin as we like, and whatever height we like.
As a result, we can ensure that the weighted output from the second
hidden layer approximates any desired function of two variables:

In particular, by making the weighted output from the second hidden
layer a good approximation to $sigma^{-1} circ f$, we ensure the
output from our network will be a good approximation to any desired
function, $f$.

What about functions of more than two variables?

Let’s try three variables $x_1, x_2, x_3$. The following network can
be used to compute a tower function in four dimensions:

Here, the $x_1, x_2, x_3$ denote inputs to the network. The $s_1,
t_1$ and so on are step points for neurons – that is, all the
weights in the first layer are large, and the biases are set to give
the step points $s_1, t_1, s_2, ldots$. The weights in the second
layer alternate $+h, -h$, where $h$ is some very large number. And
the output bias is $-5h/2$.

This network computes a function which is $1$ provided three
conditions are met: $x_1$ is between $s_1$ and $t_1$; $x_2$ is between
$s_2$ and $t_2$; and $x_3$ is between $s_3$ and $t_3$. The network is
$0$ everywhere else. That is, it’s a kind of tower which is $1$ in a
little region of input space, and $0$ everywhere else.

By gluing together many such networks we can get as many towers as we
want, and so approximate an arbitrary function of three variables.
Exactly the same idea works in $m$ dimensions. The only change needed
is to make the output bias $(-m+1/2)h$, in order to get the right kind
of sandwiching behavior to level the plateau.

Okay, so we now know how to use neural networks to approximate a
real-valued function of many variables. What about vector-valued
functions $f(x_1, ldots, x_m) in R^n$? Of course, such a function
can be regarded as just $n$ separate real-valued functions, $f^1(x_1,
ldots, x_m), f^2(x_1, ldots, x_m)$, and so on. So we create a
network approximating $f^1$, another network for $f^2$, and so on.
And then we simply glue all the networks together. So that’s also
easy to cope with.


  • We’ve seen how to use networks with two hidden layers to
    approximate an arbitrary function. Can you find a proof showing
    that it’s possible with just a single hidden layer? As a hint, try
    working in the case of just two input variables, and showing that:
    (a) it’s possible to get step functions not just in the $x$ or $y$
    directions, but in an arbitrary direction; (b) by adding up many of
    the constructions from part (a) it’s possible to approximate a tower
    function which is circular in shape, rather than rectangular; (c)
    using these circular towers, it’s possible to approximate an
    arbitrary function. To do part (c) it may help to use ideas from a
    bit later in this

Extension beyond sigmoid neurons

We’ve proved that networks made up of sigmoid neurons can compute any
function. Recall that in a sigmoid neuron the inputs $x_1, x_2,
ldots$ result in the output $sigma(sum_j w_j x_j + b)$, where $w_j$
are the weights, $b$ is the bias, and $sigma$ is the sigmoid

What if we consider a different type of neuron, one using some other
activation function, $s(z)$:

That is, we’ll assume that if our neurons has inputs $x_1, x_2,
ldots$, weights $w_1, w_2, ldots$ and bias $b$, then the output is
$s(sum_j w_j x_j + b)$.

We can use this activation function to get a step function, just as we
did with the sigmoid. Try ramping up the weight in the following, say
to $w = 100$:

Just as with the sigmoid, this causes the activation function to
contract, and ultimately it becomes a very good approximation to a
step function. Try changing the bias, and you’ll see that we can set
the position of the step to be wherever we choose. And so we can use
all the same tricks as before to compute any desired function.

What properties does $s(z)$ need to satisfy in order for this to work?
We do need to assume that $s(z)$ is well-defined as $z rightarrow
-infty$ and $z rightarrow infty$. These two limits are the two
values taken on by our step function. We also need to assume that
these limits are different from one another. If they weren’t, there’d
be no step, simply a flat graph! But provided the activation function
$s(z)$ satisfies these properties, neurons based on such an activation
function are universal for computation.


  • Earlier in the book we met another type of neuron known as a rectified linear
    . Explain why such neurons don’t satisfy the conditions
    just given for universality. Find a proof of universality showing
    that rectified linear units are universal for computation.
  • Suppose we consider linear neurons, i.e., neurons with the
    activation function $s(z) = z$. Explain why linear neurons don’t
    satisfy the conditions just given for universality. Show that such
    neurons can’t be used to do universal computation.

Fixing up the step functions

Up to now, we’ve been assuming that our neurons can produce step
functions exactly. That’s a pretty good approximation, but it is only
an approximation. In fact, there will be a narrow window of failure,
illustrated in the following graph, in which the function behaves very
differently from a step function:

In these windows of failure the explanation I’ve given for
universality will fail.

Now, it’s not a terrible failure. By making the weights input to the
neurons big enough we can make these windows of failure as small as we
like. Certainly, we can make the window much narrower than I’ve shown
above – narrower, indeed, than our eye could see. So perhaps we
might not worry too much about this problem.

Nonetheless, it’d be nice to have some way of addressing the problem.

In fact, the problem turns out to be easy to fix. Let’s look at the
fix for neural networks computing functions with just one input and
one output. The same ideas work also to address the problem when
there are more inputs and outputs.

In particular, suppose we want our network to compute some function,
$f$. As before, we do this by trying to design our network so that
the weighted output from our hidden layer of neurons is $sigma^{-1}
circ f(x)$:

If we were to do this using the technique described earlier, we’d use
the hidden neurons to produce a sequence of bump functions:

Again, I’ve exaggerated the size of the windows of failure, in order
to make them easier to see. It should be pretty clear that if we add
all these bump functions up we’ll end up with a reasonable
approximation to $sigma^{-1} circ f(x)$, except within the windows
of failure.

Suppose that instead of using the approximation just described, we use
a set of hidden neurons to compute an approximation to half our
original goal function, i.e., to $sigma^{-1} circ f(x) / 2$. Of
course, this looks just like a scaled down version of the last graph:

And suppose we use another set of hidden neurons to compute an
approximation to $sigma^{-1} circ f(x)/ 2$, but with the bases of
the bumps shifted by half the width of a bump:

Now we have two different approximations to $sigma^{-1} circ f(x) /
2$. If we add up the two approximations we’ll get an overall
approximation to $sigma^{-1} circ f(x)$. That overall approximation
will still have failures in small windows. But the problem will be
much less than before. The reason is that points in a failure window
for one approximation won’t be in a failure window for the other. And
so the approximation will be a factor roughly $2$ better in those

We could do even better by adding up a large number, $M$, of
overlapping approximations to the function $sigma^{-1} circ f(x) /
M$. Provided the windows of failure are narrow enough, a point will
only ever be in one window of failure. And provided we’re using a
large enough number $M$ of overlapping approximations, the result will
be an excellent overall approximation.


The explanation for universality we’ve discussed is certainly not a
practical prescription for how to compute using neural networks! In
this, it’s much like proofs of universality for NAND gates and
the like. For this reason, I’ve focused mostly on trying to make the
construction clear and easy to follow, and not on optimizing the
details of the construction. However, you may find it a fun and
instructive exercise to see if you can improve the construction.

Although the result isn’t directly useful in constructing networks,
it’s important because it takes off the table the question of whether
any particular function is computable using a neural network. The
answer to that question is always “yes”. So the right question to
ask is not whether any particular function is computable, but rather
what’s a good way to compute the function.

The universality construction we’ve developed uses just two hidden
layers to compute an arbitrary function. Furthermore, as we’ve
discussed, it’s possible to get the same result with just a single
hidden layer. Given this, you might wonder why we would ever be
interested in deep networks, i.e., networks with many hidden layers.
Can’t we simply replace those networks with shallow, single hidden
layer networks?

Thanks to Jen Dodd and Chris Olah for many
discussions about universality in neural networks. My thanks, in
particular, to Chris for suggesting the use of a lookup table to
prove universality. The interactive visual form of the chapter is
inspired by the work of people such as Mike Bostock, Amit
, Bret Victor, and Steven Wittens.

While in principle that’s possible, there are good practical reasons
to use deep networks. As argued in
Chapter 1, deep networks
have a hierarchical structure which makes them particularly well
adapted to learn the hierarchies of knowledge that seem to be useful
in solving real-world problems. Put more concretely, when attacking
problems such as image recognition, it helps to use a system that
understands not just individual pixels, but also increasingly more
complex concepts: from edges to simple geometric shapes, all the way
up through complex, multi-object scenes. In later chapters, we’ll see
evidence suggesting that deep networks do a better job than shallow
networks at learning such hierarchies of knowledge. To sum up:
universality tells us that neural networks can compute any function;
and empirical evidence suggests that deep networks are the networks
best adapted to learn the functions useful in solving many real-world

. .

Next Post

The eyes of mammals reveal a dark past

Sat Apr 20 , 2019

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