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!


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 $$x_{k+1}=\frac{g^{n x_k}-m n x_k^2 \log (g)}{g^{n x_k}-m n x_k \log (g)+m}$$


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 $$x\simeq\frac{1}{m+1}+\frac{a m }{(m+1)^3}+\frac{a^2m \left(3 m-1\right)}{2 (m+1)^5}$$


Let $t=\frac 1{m+1}$ and get $$x=t+a t^2+\frac{a (3 a-2)}{2!} t^3+\frac{a^2 (16 a-21)}{3!} t^4+\frac{a^2(125 a^2-244 a+48 )}{4!} t^5+\frac{a^3 \left(1296 a^2-3355 a+1500\right) } {5!} t^6+O\left(t^7\right)$$

  • $\begingroup$ thanks a lot, @ClaudeLeibovici Please, do you have a reference to Darboux-Fourier theorem? Also, does your solution possibly lead to an approximation of the root? $\endgroup$ – Daniel S. Dec 20 '18 at 15:20
  • $\begingroup$ @DanielS. Here is a link numdam.org/article/NAM_1869_2_8__17_0.pdf $\endgroup$ – Claude Leibovici Dec 20 '18 at 15:38
  • $\begingroup$ thanks a lot for the reference! so, $x_1$, $x_2$, and so on will be refined approximations of the root, right? maybe closed-form expressions for those quantities are good approximations for the root? $\endgroup$ – Daniel S. Dec 20 '18 at 16:35
  • $\begingroup$ thanks again, @ClaudeLeibovici $\endgroup$ – Daniel S. Dec 21 '18 at 11:25
  • $\begingroup$ @DanielS. Let me know how my last stuff works for approximation (run a few cases). Thanks & cheers. :-) $\endgroup$ – Claude Leibovici Dec 21 '18 at 11:27

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