18

Almost all of the functionalities provided by the non-linear activation functions are given by other answers. Let me sum them up: First, what does non-linearity mean? It means something (a function in this case) which is not linear with respect to a given variable/variables i.e. $f(c1.x1 + c2.x2...cn.xn + b) != c1.f(x1) + c2.f(x2) ... cn.f(xn) + f(b).$ NOTE:...


12

It seems to me that you already understand the shortcomings of ReLUs and sigmoids (like dead neurons in the case of plain ReLU). I would suggest looking at ELU (exponential linear units) and SELU (self-normalising version of ELU). Under some mild assumptions, the latter have the nice property of self-normalisation, which mitigates the problem of vanishing ...


10

A neuron is said activated when its output is more than a threshold, generally 0. For examples : \begin{equation} y = Relu(a) > 0 \end{equation} when \begin{equation} a = w^Tx+b > 0 \end{equation} Same goes for sigmoid or other activation functions.


9

In a neural network (NN), a neuron can act as a linear operator, but it usually acts as a non-linear one. The usual equation of a neuron $i$ in layer $l$ of an NN is $$o_i^l = \sigma(\mathbf{x}_i^l \cdot \mathbf{w}_i^l + b_i^l),$$ where $\sigma$ is a so-called activation function, which is usually a non-linearity, but it can also be the identity ...


8

If you only had linear layers in a neural network, all the layers would essentially collapse to one linear layer, and, therefore, a "deep" neural network architecture effectively wouldn't be deep anymore but just a linear classifier. $$y = f(W_1 W_2 W_3x) = f(Wx)$$ where $W$ corresponds to the matrix that represents the network weights and biases for one ...


7

Combining ReLU, the hyper-parameterized1 leaky variant, and variant with dynamic parametrization during learning confuses two distinct things: The comparison between ReLU with the leaky variant is closely related to whether there is a need, in the particular ML case at hand, to avoid saturation — Saturation is thee loss of signal to either zero ...


6

If what you are asking is what is the intuition for using the derivative in backpropagation learning, instead of an in-depth mathematical explanation: Recall that the derivative tells you a function's sensitivity to change with respect to a change in its input. A high (absolute) value for the derivative at a certain point means that the function is very ...


5

Let's first talk about linearity. Linearity means the map (a function), $f: V \rightarrow W$, used is a linear map, that is, it satisfies the following two conditions $f(x + y) = f(x) + f(y), \; x, y \in V$ $f(c x) = cf(x), \; c \in \mathbb{R}$ You should be familiar with this definition if you have studied linear algebra in the past. However, it's more ...


5

Consider a dataset $\mathcal{D}=\{x^{(i)},y^{(i)}:i=1,2,\ldots,N\}$ where $x^{(i)}\in\mathbb{R}^3$ and $y^{(i)}\in\mathbb{R}$ $\forall i$ The goal is to fit a function that best explains our dataset.We can fit a simple function, as we do in linear regression. But that's different about neural networks, where we fit a complex function, say: $\begin{align}h(...


5

Let us suppose we have a network without any functions in between. Each layer consists of a linear function. i.e layer_output = Weights.layer_input + bias Consider a 2 layer neural network, the outputs from layer one will be: x2 = W1*x1 + b1 Now we pass the same input to the second layer, which will be x3 = W2x*2 + b2 Also x2 = W1*x1 + b1 Substituting ...


5

There is no strict definition of suitability of an activation function for neural networks. Instead there are a number of desirable traits, and functions that don't meet them or come close enough may perform badly in general (but those functions may still work in specific cases) If you are using gradient descent as a training method, then differentiability ...


4

Consider a very simple neural network, with just 2 layers, where the first has 2 neurons and the last 1 neuron, and the input size is 2. The inputs are $x_1$ and $x_1$. The weights of the first layer are $w_{11}, w_{12}, w_{21}$ and $w_{22}$. We do not have activations, so the outputs of the neurons in the first layer are \begin{align} o_1 = w_{11}x_1 + w_{...


4

First Degree Linear Polynomials Non-linearity is not the correct mathematical term. Those that use it probably intend to refer to a first degree polynomial relationship between input and output, the kind of relationship that would be graphed as a straight line, a flat plane, or a higher degree surface with no curvature. To model relations more complex ...


4

Nonlinear relations between input and output can be achieved by using a nonlinear activation function on the value of each neuron, before it's passed on to the neurons in the next layer.


4

The term "activated" is mostly used when talking about activation functions which only outputs a value (except 0) when the input to the activation function is greater than a certain treshold. Especially when discussing ReLU the term "activated" may be used. ReLU will be "activated" when it's output is greater than 0, which is also when it's input is greater ...


4

I can't speak for individual researchers, but I can guess why the community as a whole hasn't adopted this activation function. ReLU is just so incredibly cheap. This benefit continues to grow as networks grow deeper. Also, they work reasonably well. As pointed out in Searching for Activation Functions, the performance improvements of the other ...


4

The basic (and usual) algorithm used to update the weights of the artificial neural network (ANN) is an iterative, numerical and optimization algorithm, called gradient descent, which is based on and requires the computation of the derivative of the function you want to find the minimum of. If the function you want to find the minimum of is multivariable, ...


4

Almost never. The sum of linear functions is another linear function, so if neurons were only linear transformations there would be basically no point to having more than one neuron per layer. Instead, every neuron applies some kind of nonlinear function to its input. There are lots of different variations, but in the end the combination of the nonlinear ...


4

It is definitley possible to make the links between neurons use more complex functions. Provided those functions are differentiable, backpropagation still works, and the resulting compound function might be able to learn something useful. The general name for such a thing is a computational graph and the standardised structures used in most neural networks ...


3

The main difference your change would have is to allow you to apply a loss function to a different part of the network. This may affect training. If you keep the same loss function (e.g. MSE), but apply it to the pre-transformed values, then you will have changed the objective of the network, perhaps significantly. Whether or not this is a good thing ...


3

There are a few traits that you want the activation function to have, and cube roots rate as OK-ish: Nonlinear – check. Continuously differentiable – no. There is a problem at $x=0$. Unlike other discontinuous functions like ReLU, although the gradient can be calculated near zero, it can be arbitrarily high as you approach $x=0$, because $\frac{d}{dx}x^{\...


3

The ReLu is a non-linear activation function. Check out this question for the intuition behind using ReLu's (also check out the comments). There is a very simple reason of why we do not use a linear activation function. Say you have a feature vector $x_0$ and weight vector $W_1$. Passing through a layer in a Neural Net will give the output as $W_1^T * x_0 ...


3

No, it is not necessary that an activation function is differentiable. In fact, one of the most popular activation functions, the rectifier, is non-differentiable at zero! This can create problems with learning, as numerical gradients calculated near a non-differentiable point can be incorrect. The "kinks" section in these lecture notes discuss the issue. ...


3

It is almost mandatory to have a differentiable activation function unless, of course, you have an alternative to training the network by back-propagating the error.


3

Indeed I haven't seen the term "logit probability" used in many places other than that specific paper. So, I cannot really comment on why they're using that term / where it comes from / if anyone else uses it, but I can confirm that what they mean by "logit probability" is basically the same thing that is more commonly referred to simply as "logits": they ...


3

Taking the question from comments on nbro's answer. Am I wrong to see a clear relationship between how we are currently training networks and the classic function that defines a line? You are right about it. This is an intuitive way to understand neural networks. You can create a neural network that only does simple linear regression, by using linear ...


3

Let's first recapitulate why the function that calculates the maximum between two or more numbers, $z=\operatorname{max}(x_1, x_2)$, is not a linear function. A linear function is defined as $y=f(x) = ax + b$, so $y$ linearly increases with $x$. Visually, $f$ corresponds to a straight line (or hyperplane, in the case of 2 or more input variables). If $z$ ...


3

Hi and welcome to the community. It's important to understand these basic concepts very clearly. You have to first understand the basic unit of a neural network, a single node/neuron/perceptron. Let us forget all about Neural Networks for a bit, and talk about something far simpler. Linear Regression In the above figure, we clearly have one independent ...


2

Training While "running" a neural network can be done with any activation functions, we usually want to train it - i.e., adjust its parameters so that the result would be closer to what we desire. This is commonly done by backpropagation and variations of gradient descent, which requires the existence of a gradient - i.e., requires activation function to ...


2

Derivative gives the rate of change in $y$ for a small change in $x$ or the slope of a function at point $x$. In the above function, y = x for x >= 0, i.e. y/x = 1 y = x/20 for x < 0, i.e. y/x = 1/20 The following function returns the derivative of leaky ReLU as explained private double leaky_relu_derivative(double x) { if (x &...


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