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gradient descent for custom function

4 Ansichten (letzte 30 Tage)
L
L am 24 Apr. 2024
Bearbeitet: Torsten am 24 Apr. 2024
I have four equations:
1) e = m - y
2) y = W_3 * h
3) h = z + W_2 * z + f
4) f = W_1 * x
I want to update W_1, W_2 and W_3 in order to minimize a cost function J = (e^T e ) by using gradient descent.
x is an input, y is the output and m is the desired value for each sample in the dataset
I would like to do W_1 = W_1 - eta* grad(J)_w_1
W_2 = W_2 - eta* grad(J)_w_2
W_3 = W_3 - eta* grad(J)_w_3
Going through documentation I found out that you can train standard neural networks. But notice that I have some custom functions, so I guess it would be more of an optimization built in function to use.
Any ideas?
  2 Kommentare
Matt J
Matt J am 24 Apr. 2024
x is an input, y is the output and m is the desired value for each sample in the dataset
It looks like z is also an input. It is not given by any other equations.
L
L am 24 Apr. 2024
Yes, z is another input.

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Antworten (2)

Matt J
Matt J am 24 Apr. 2024
Bearbeitet: Matt J am 24 Apr. 2024
so I guess it would be more of an optimization built in function to use.
No, not necessarily. Your equations can be implemented with fullyConnectedLayers and additionLayers.
  3 Kommentare
Matt J
Matt J am 24 Apr. 2024
You would create a dlnetwork and then use trainnet.
L
L am 24 Apr. 2024
Bearbeitet: L am 24 Apr. 2024
Thanks @Matt J. I found the documentation hard to follow. I tried using the DL tool box, but I was only able to generate the topology. What should I do next ? how to train it on the x, z inputs? For each x,z input, I have a desired output m. For all matrices I don't wish to learn a bias term, so, in the tool box I set learnratebias =0

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Torsten
Torsten am 24 Apr. 2024
Verschoben: Torsten am 24 Apr. 2024
e = m - y = m - W_3*h = m - W_3*(z + W_2 * z + W_1 * x )
Now if you formulate this as
e = W1*z + W2*x - m
with
W1 = W_3 + W_2*W_3 and W2 = W_1*W_3
your problem is
min: || [z.',x.']*[W1;W2] - m ||_2
and you can use "lsqlin" to solve.
  10 Kommentare
L
L am 24 Apr. 2024
Well, I can prescribe some variables.
Torsten
Torsten am 24 Apr. 2024
Bearbeitet: Torsten am 24 Apr. 2024
As far as I understand, you have 10200 free parameters and 200 known values. I think you should re-consider your problem.

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