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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,

beginequation
x=frac11+m g^-N x
endequation

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
$$f(x)=x left(1+m g^-n xright)-1$$ 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 $x_1=frac11+m$ and just continue until convergence using
$$x_k+1=fracg^n x_k-m n x_k^2 log (g)g^n x_k-m n x_k log (g)+m$$

Edit

Assuming that $x$ could be small, we could try to write
$$y=x left(1+m g^-n xright)$$ Expand as a Taylor series built at $x=0$ to get
$$y=(m+1) x-a m x^2+fraca^2 m2 x^3 +Oleft(x^4right)$$ where $a=n log(g)$ and use series reversion to get
$$x=fracym+1+fraca m (m+1)^3y^2+fraca^2m left(3 m-1right)2
(m+1)^5y^3+Oleft(y^4right)$$ and set $y=1$ to get
$$xsimeqfrac1m+1+fraca m (m+1)^3+fraca^2m left(3 m-1right)2
(m+1)^5$$

Edit

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