# Tag Info

## Hot answers tagged relu

3

By convention, the $\mathrm{ReLU}$ activation is treated as if it is differentiable at zero (e.g. in [1]). Therefore it makes sense for TensorFlow to adopt this convention for tf.nn.relu. As you've found, of course, it's not true in general that we treat the gradient of the absolute value function as zero in the same situation; it makes sense for it to be an ...

2

I suppose, the situation is as follows - PReLU increases the expressiveness of a model for a bit at a small cost, but the gain is almost negligible as well (according to this post). There is, indeed, a noticeable difference between ReLU and PReLU, since the former takes the same value for all $\mathbb{R}_{\leq 0}$. However, compared with a LeakyReLU, note ...

2

$F_l$ is the activation of the filter. They state in the paper that they base their method on VGG-Network, which uses ReLU as its activation function. In fact, VGG uses it in all of its hidden layers. ReLU is defined as $$f(x) = max(0,x)$$ Since ReLU is 0 for all x's below 0, the equation above holds; When x is non-positive, all terms in the loss function ...

1

I'm sure the biases are initially initialized to zero but I don't know how the weights are handled. Looking at the Dense layer docs: by default Dense layers biases are initialized with zeros (bias_initializer='zeros') and weights are initialized with Glorot uniform (kernel_initializer='glorot_uniform'). ... "unusual" element to point here; I've ...

1

Creating custom gradient for tf.abs may solve the problem: @tf.custom_gradient def abs_with_grad(x): y = tf.abs(x); def grad(div): # Derivation intermediate value g = 1; # Use 1 to make the chain rule just skip abs return div*g; return y,grad;

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