fix point solution or approximation available? logistic regression? please, is there a simple closed-form or approximation to the following fixed-point problem in $x$? $x$ is the value searched for.  $m$, $g$ and $N$ are real parameters, all greater than 0,
\begin{equation}
x=\frac{1}{1+m g^{-N x}}
\end{equation}
any insights on properties of $x$ are very welcome!  In particular, comments about how $x$ evolves towards convergence, in case we use a fixed point algorithm to find $x$, are welcome as well. This problem looks a bit like logistic regression, but I'm not sure about how to further relate the two. thanks!
 A: I am afraid that there is no closed form of the solution.
Without any information about $x$, let us consider the function
$$f(x)=x \left(1+m g^{-n x}\right)-1$$ What we have is $f(0)=-1$  and $f(1)=m g^{-n} >0$.
We also have to notice that
$$f(0)=-1 \qquad f'(0)=1+m \qquad f''(0)=-2\, m\, n \log (g)$$ So, if $g>1$, $f(0) \times f''(0) >0$ and by Darboux theorem , starting with $x_0=0$, Newton method would converge without any overshoot of the solution. As a first iterate, Newton method will give $x_1=\frac{1}{1+m}$ and just continue until convergence using
$$x_{k+1}=\frac{g^{n x_k}-m n x_k^2 \log (g)}{g^{n x_k}-m n x_k \log (g)+m}$$
Edit
Assuming that $x$ could be small, we could try to write
$$y=x \left(1+m g^{-n x}\right)$$ Expand as a Taylor series built at $x=0$ to get
$$y=(m+1) x-a m x^2+\frac{a^2 m}{2}  x^3 +O\left(x^4\right)$$ where $a=n \log(g)$ and use series reversion to get
$$x=\frac{y}{m+1}+\frac{a m }{(m+1)^3}y^2+\frac{a^2m \left(3 m-1\right)}{2
   (m+1)^5}y^3+O\left(y^4\right)$$ and set $y=1$ to get
$$x\simeq\frac{1}{m+1}+\frac{a m }{(m+1)^3}+\frac{a^2m \left(3 m-1\right)}{2
   (m+1)^5}$$
Edit
Let $t=\frac 1{m+1}$ and get
$$x=t+a t^2+\frac{a (3 a-2)}{2!}  t^3+\frac{a^2 (16 a-21)}{3!}  t^4+\frac{a^2(125 a^2-244 a+48 )}{4!}  t^5+\frac{a^3 \left(1296 a^2-3355 a+1500\right) } {5!} t^6+O\left(t^7\right)$$
