I have a question similar to this.
I want the sigmoid to have asymptotes to $+1$ and $0$ in specific points $\frac{1}{A}$ and $-\frac{1}{A}$, as in the Figure (where $\frac{1}{A}=2$ and $-\frac{1}{A}=-2$).
How can I modify the curve
$S(t) = \frac{1}{1+e^{-t}}$
so as to make this happen?
Thank you.
The original $S(t)$ has flexes at $t=\log(2\pm\sqrt 3)$. Now define $$ Z(t)=S(kt) $$ for some stretch scalar $k\in\mathbb R^+$, and convince yourself that $Z'''(t)=0$ iff $S'''(kt)\cdot k^3=0$ iff $S'''(kt)=0$. Hence $Z(t)$ has flexes at $$ t=\frac{\log(2\pm\sqrt 3)}{k}=\pm\frac{1}{A} $$ and for $A,k>0$ this leads to $$ k=A\log(2+\sqrt 3) $$ which makes $Z(t)=S(kt)$ solve your problem. Thus the final form matching requirements for the flexes is $$ Z(t)=\frac{1}{1+\text{e}^{-A\log(2+\sqrt 3)t}} $$
If in addition to that we want $W(\pm\frac 1A)=\pm 1$, then by symmetry around the point $(0,0.5)$ this simply means $$ W(t)=\frac{1}{Z(1/A)-0.5}(Z(t)-0.5) $$ here is a graph showing $S(t),Z(t)$ and $W(t)$ defined this way.
Here $S(t)$ is blue, $Z(t)$ is red, and $W(t)$ is purple.
EDIT: OK, I think that I know a function that could do the job, if you want it. In the theory of smooth manifolds, we use so-called bump functions based on the function $$ f(t)= \begin{cases} \text{e}^{-1/t}&\text{ for } t>0\\ 0&\text{ otherwise } \end{cases}$$ which can then be used to define a function with a graph similar to what you are suggesting here, namely $$h(t)=\frac{f(t/4+1/2)}{f(t/4+1/2)+f(1/2-t/4)}$$
having the following graph:
where $f(t)$ is the red curve and $h(t)$ is the blue curve. Note that $h(t)\in(0,1)$ for $t\in(-2,2)$ and $h(t)\in\{0,1\}$ elsewhere.