As a warm up with recurrent neural networks, I'm trying to predict a sine wave from another sine wave of another frequency.

My model is a simple RNN, its forward pass can be expressed as follow:

$$ \begin{aligned} r_t &= \sigma(W_{in} \cdot x_t + W_{rec} \cdot r_{t-1}))\\ z_t &= W_{out} \cdot r_t \end{aligned} $$ where $\sigma$ is the sigmoïd function.

When both input the input and expected output are two sine waves of the same frequency but with (possibly) a phase shift, the model is able to properly converge to a reasonable approximation.

However, in the following case, the model converge to a local minima and predicts zero all the time:

  • input: $x = sin(t)$
  • expected output: $y = sin(\frac{t}{2})$

Here's what the network predicts when given the full input sequence after 10 epochs of training, using mini-batches of size 16, a learning rate of 0.01, a sequence length of 16 and hidden layers of size 32:

Network prediction after 10 epochs, using mini-batches of size 16

Which leads me to think the network is unable to learn through time and relies only on the current input to make its prediction.

I tried to tune the learning rate, sequences length and hidden layers size without much success.

I'm having the exact same issue with an LSTM. I don't want to believe these architectures are that flawed, any hints on what am I doing wrong ?

I'm using an rnn package for Torch, the code is in a Gist.


Your data basically cannot be learned with an RNN trained that way. Your input is $\sin(t)$ is $2\pi$-periodic $\sin(t) = \sin(t+2\pi)$

but your target $\sin(t/2)$ is $4\pi$-periodic and $\sin(t/2) = -\sin(t+2\pi)$

Therefore, in your dataset you'll have pairs of identical inputs with opposite outputs. In terms of Mean Squared Error, it means that the optimal solution is a null function.

These are two slices of your plot where you can see identical inputs but opposite targets enter image description here

  • 1
    $\begingroup$ To elaborate on this answer, the issue came for using the same feedback initialization for different inputs. I solved this by doing (randomly) more forwards than backwards in order to learn the full sequence. $\endgroup$ – Simon Jul 25 '16 at 15:23

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