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> . For example, when we apply convolution filters on images the network consists of repetitive blocks of convolutional layers, and one linear output layer at t
by gugagore 5y ago
> . For example, when we apply convolution filters on images the network consists of repetitive blocks of convolutional layers, and one linear output layer at the very end. It's essentially f(f(f(...f(x))...)) where f is the neural network, and we call this "deep" because of the layers of composition. But what if we make this composition go to infinity?
This really does not jive with my understanding. Each layer of, e.g. VGG 16 [1] does not implement the same function. Each layer has its own weights. There are certain architectures that tie weights across layers, but not all of them.
[1] https://neurohive.io/en/popular-networks/vgg16/ https://neurohive.io/en/popular-networks/vgg16/
- taeric 5y agoI think this works if you consider subscripted fs. That is, it is always the same arity, and the output is the same. So, would be better it they said f_n, where n is the layer if they network. (I mean this as a question, but don't see an obvious place for a question mark...)
- ChrisRackauckas 5y agoCorrect, not all models have equivalent weights at each step. But it is an abstraction that needs to be done to send it to infinity.