If you save numerator_t and denominator_t from a previous computation, then h_t = f( (numerator_(t-1) + z + a) / (denominator_(t-1) + a) ). So it should be linear with t?
The model I used is: h_t = f( (numerator_(t-1) + z x e^a) / (denominator_(t-1) + e^a) )
Equations (7) in the arxiv link (here it is again: https://arxiv.org/pdf/1703.01253.pdf) provide the update equations. When implementing the equations, you sometimes have to scale both the numerator and denominator back by a constant factor to avoid an overflow error
eximius|9 years ago
(where z and a are z(x_i, h_(i-1)), etc)
jostmey|9 years ago
The model I used is: h_t = f( (numerator_(t-1) + z x e^a) / (denominator_(t-1) + e^a) )
Equations (7) in the arxiv link (here it is again: https://arxiv.org/pdf/1703.01253.pdf) provide the update equations. When implementing the equations, you sometimes have to scale both the numerator and denominator back by a constant factor to avoid an overflow error