New answers tagged backpropagation
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While this may not be the answer you were looking for, I hope this explanation will help you to understand applying backpropagation to a CNN. Fundamentally, convolutional layers are no different than dense layers, however there are restrictions. The key one is weight-sharing which allows a CNN to be much more efficient than a regular dense layer (as well as ...
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The quick answer is that you want to use an activation function on the output layer that does not compress values to $(0,1)$. Depending on your software, this might be called "linear" or "identity". It looks like Keras just wants you to leave off the activation function: model.add(Dense(1)).
The typical way of thinking of a neural network as a classifier (...
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Lets mock some data up.
"100 numbers, each one is a parameter, they together define a number X(also given)"
# i.e. size of X_train -> [n x d]
# i.e. size of X_train -> [??? x 100] , when d = 100
# "I have 20000 instances for training"
# i.e. size of X_train -> [20000 x 100], when n = 20000
import torch
import numpy as np
X_train = torch.rand(...
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for regression, you can use a hidden layer with sigmoid, then a LINEAR output layer, where the weighted sum goes straight through, without modification.
this way your output is not restricted to 0-1
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The product rule of partial derivative:
$\frac{\partial}{\partial x} f g = g \frac{\partial}{\partial x} f + f \frac{\partial}{\partial x} g$
According to this: $\frac{\partial}{\partial W^{x}} W^{h}W^{x}X^{1} = W^{h}X^{1}$, because derivative of other term with respect to $W^{x}$ is zero. (I am not considering the transpose notation as it depends on how ...
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If we write $ H^2 = W^{h}H1 + W^{x}X^{2} $ then it will be better to understand the backward propagation step.
Now,
$\frac{\partial}{\partial W^{x}} W^{h}W^{x}X^{1}$ can be written as:
$\frac{\partial H^2}{\partial H^1}\frac{\partial H^1}{\partial W^{x}} $
$\frac{\partial H^2}{\partial H^1} = (W^h)^T$ and
$\frac{\partial H^1}{\partial W^{x}} = (X^{1})^T ...
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I have found that you should compute derivatives $\frac{\partial L}{\partial A}, \frac{\partial L}{\partial Z}$ in Flatten layer and then reshape Conv2D input shape.
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There is no benefit to using ReLU as the output activation of a neural network. As you said, the network will ignore training labels below zero and it will train on labels above zero as if no output activation were present. However, the problem you're describing can also occur for individual units of hidden layers, where ReLU activations are common. This is ...
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However, the chain rule tells us how the first variable influences the second variable, and so on. Following that logic, we should only update the weights of the first hidden layer.
I don't see how the second statement follows from the first.
Each weight $w_i$ (not just the ones in the first layer) affects the loss $\mathcal{L}$ according to the partial ...
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