$$z = \frac{y}{1+C \exp(-Wx)}$$
Is it possible to simplify this equation by a sigmoid function or something similar?
You could do this: Let $\displaystyle C = e^{-d}$ then
$\displaystyle z = \frac{y}{1 + C e^{-Wx}} = y \cdot \frac{1}{1 + e^{-(Wx + d)}}$
Then let A = Wx + d which gives
$\displaystyle z = y \cdot \frac{1}{1 + e^{-A}}$, which is your sigmoid function.
This looks nice but there is a catch: Setting $\displaystyle C = e^{-d}$ restricts the value of C to be positive, which would not be true in general.
-Dan
I am just wondering what is meant by "simplify". You could convert the equation to spherical or cylindrical coordinates. Would that be "simpler"? You can solve for $y$ as a function of $C,x,W,z$. That would eliminate the fraction. You could solve for $x$ in terms of $C,W,y,z$. That would eliminate the exponential (but introduce a logarithm). Is that "simpler"?
Take your pick.
$ \begin{align*} z& = \dfrac{y}{{1 + C{e^{ - wx}}}} \\ z &= \dfrac{{{e^{wx}}y}}{{{e^{wx}} + C}} \\z{e^{wx}} + zC& = {e^{wx}}y \\ z{e^{wx}} - y{e^{wx}}& = - Cz \\ {e^{wx}}& = \frac{{Cz}}{{y - z}} \\ x& = \dfrac{1}{w}\left( {\log (C) + \log (z) - \log (y - z)} \right) \end{align*}$