Do there exist smooth sigmoid functions similar to atan but with different upper and lower bounds? This function must go through origin.
Something like on the picture: Function
If not, why? Where can I get more info about this?
Sigmoid functions similar to atan but with different upper and lower bound
functionssmooth-functions
Related Solutions
We can start with a simple soft staircase function:
and then feed it into itself:
$$ y(x) = f(f(x)) $$
then again:
$$ y(x) = f(f(f(x))) $$
and again:
$$ y(x) = f^4(x) $$
As you can see, each iteration makes the "flat" part of the step longer, and the rise steeper.
The period and the height of each step is $ 2 \pi $, so multiply $x$ by $2 \pi / w$ and $y$ by $h / 2 \pi$ to reach your desired scale.
In reality, the curve is only truly flat (zero derivative) at the centre of each step — at every $ 2 \pi k $ — and only close to flat on either side of that point.
Configurability is limited: The softness of the step can only be specified in integer amounts (the number of times we reapply $f$ to itself), and it requires many/infinite applications to make the step really sharp.
I was just working on finding a similar function. In case anyone else lands here in the future, here is a sigmoidal function which satisfies:
$$f(0)=0$$ $$f(1)=1$$
and has a single parameter, $k$, controlling steepness:
$$f(x) = 1-\frac{1}{1+(\frac{1}{x}-1)^{-k}} $$
Note that, by symmetry, these will also always be true: $f(1/2)=1/2$, and $\int_0^1 f(x)dx = 1/2$.
Method: Since $\frac{1}{1+e^x}$ is a bijection from $\mathbb{R} \to [0,1]$; we can invert it to get a function mapping $[0,1] \to \mathbb{R}$, which is $h(x) = \ln(\frac{1}{x}-1)$. Then we can put $h$ into a standard (increasing) sigmoid which maps the reals back to $[0,1]$. The form of the sigmoid is:
$$ g(y) = 1-\frac{1}{1+e^{-ky}}$$
$f(x)$ is what you get when you simplify $g(h(x))$. Note that this $f$ is defined outside $x \in [0,1]$ and it does not asymptote out to 0 or 1, but it does achieve the desired behavior within that interval.
One additional note, in the special case where $k=1$, this function reduces to $f(x)=x$.
Best Answer
Welcome to math.stackexhange. You can transform any standard sigmoidal function $f$ such as $f = \arctan$ or the hyperbolic tangent $f = \tanh$ by
Combining these, new functions $g(x) = c(f(x + b) - f(b))$ can be made that are increasing, have desired limits at $\pm \infty$, and satisfy $g(0) = 0$.
For example, $g(x) = 4(\arctan(x + 1) - \arctan(1))$ has limits $- 3\pi$ at $- \infty$, $\pi$ at $\infty$, and $g(0) = 0$. See the graph.