if we have a sigmoid function $1/(1+e^a)$, where $a$ is in the form of $(x-b)/c$ and the sigmoid equals function a value, called $k$. This $k$ is in the form of $1/x^2$. So something like: $1/x^2= 1/(1+e^a)$ ,where $a = (x-b)/c$
Is it possible to find a unique value for x algebraicelly?? And if not, why not?
It can be an approximation, but not a numerical one.
Edit: I had to use a numerical one :) Steven
On
It sounds like you are seeking an analytic solution to the equation $$ \frac{1}{1+\exp((x-b)/c)} = 0, $$ but the LHS is always positive for any $x$ since exponent is always positive, so both numerator and denominator are always positive.
However, as $x \to \infty$ if $c>0$ you will get the desired effect.
If we have that $$f(x)=\frac1{1+e^{\frac{x-b}{c}}}=k$$ for some $k\in\mathbb{C}$ then we can rearrange this to get $$1+e^{\frac{x-b}{c}}=\frac1k$$ $$e^{\frac{x-b}{c}}=\frac1k-1$$ $$\frac{x-b}{c}=\ln{\left(\frac1k-1\right)}$$ $$x-b=c\ln{\left(\frac1k-1\right)}$$ $$x=b+c\ln{\left(\frac1k-1\right)}$$ Which is undefined for $k=0$ but otherwise allows a unique value of $x$ to be found.