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While digging through the topic of neural networks and how to efficiently train them, I came across the method of using very simple activation functions, such as the rectified linear unit (ReLU), instead of the classic smooth sigmoids. The ReLU-function is not differentiable at the origin, so according to my understanding the backpropagation algorithm (BPA) is not suitable for training a neural network with ReLUs, since the chain rule of multivariable calculus refers to smooth functions only. However, none of the papers about using ReLUs that I read address this issue. ReLUs seem to be very effective and seem to be used virtually everywhere while not causing any unexpected behavior. Can somebody explain to me why ReLUs can be trained at all via the backpropagation algorithm?
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An ideal activation function is both nonlinear and differentiable. The nonlinear behavior of an activation function allows our neural network to learn nonlinear relationships in the data. Differentiability is important because it allows us to backpropagate the model's error when training to optimize the weights.
The backpropagation algorithm requires a differentiable activation function, and the most commonly used are tan-sigmoid, log-sigmoid, and, occasionally, linear. Feed-forward networks often have one or more hidden layers of sigmoid neurons followed by an output layer of linear neurons.
A requirement for backpropagation algorithm is a differentiable activation function. However, the Heaviside step function is non-differentiable at x = 0 and it has 0 derivative elsewhere. This means that gradient descent won't be able to make a progress in updating the weights.
ReLU Function Although it gives an impression of a linear function, ReLU has a derivative function and allows for backpropagation while simultaneously making it computationally efficient. The main catch here is that the ReLU function does not activate all the neurons at the same time.
To understand how backpropagation is even possible with functions like ReLU you need to understand what is the most important property of derivative that makes backpropagation algorithm works so well. This property is that :
f(x) ~ f(x0) + f'(x0)(x - x0) If you treat x0 as actual value of your parameter at the moment - you can tell (knowing value of a cost function and it's derivative) how the cost function will behave when you change your parameters a little bit. This is most crucial thing in backpropagation.
Because of the fact that computing cost function is crucial for a cost computation - you will need your cost function to satisfy the property stated above. It's easy to check that ReLU satisfy this property everywhere except a small neighbourhood of 0. And this is the only problem with ReLU - the fact that we cannot use this property when we are close to 0.
To overcome that you may choose the value of ReLU derivative in 0 to either 1 or 0. On the other hand most of researchers don't treat this problem as serious simply because of the fact, that being close to 0 during ReLU computations is relatively rare.
From the above - of course - from the pure mathematical point of view it's not plausible to use ReLU with backpropagation algorithm. On the other hand - in practice it usually doesn't make any difference that it has this weird behaviour around 0.
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