Are there any equation that could produce a monotonic smooth step function by parameter

Question

I want to write a mathematic formular that, given any number of monotonic arbitrary point, it will produce a monotonic smooth step function

Such as a figure below, I give it 2 point (the intersect point of red line and black curve). It will plot a graph as a function curve that pass all given points with derivative 0, so between each point will be like S curve

I want a function that was not being step, being a single function that could inverse and differentiate/integrate. So it could be use in forward and reverse calculation

Are there any function that fit this requirement?

Answer 1

I have found a simple solution. Which is just keep adding logistic function and shifting offset of the x. The result is nearly smooth step as I wish

This need more tweak to have each term join together at the same tangent point and manipulate variable to have each step start and end at the point I expect. But for now I think this is what satisfied me the most

https://www.desmos.com/calculator/hzynrk3tu1

Answer 2

One way to accomplish this would to begin with a basic transition graph such as

$$ T(x)=\frac{1-\cos(\pi x)}{2}\text{ for }0\le x<1 $$

which transitions between the points $(0,0)$ and $(1,1)$.

Next, a graph for just the portion between two points $(x_k,y_k)$ and $(x_{k+1},y_{k+1})$ can be written in terms of the transition graph

$$ f_k(x)=\begin{cases}(y_{k+1}-y_k)T\left(\frac{x-x_k}{x_{k+1}-x_k}\right)+y_k&\text{ for }x_k\le x<x_{k+1}\\ 0&\text{ otherwise.}\end{cases} $$

Then the entire graph can be defined as

$$ F(x)=\begin{cases}y_1&\text{ for }x<x_1\\ \sum_{k=1}^{n-1}f_k(x)&\text{ for }x_1\le x<x_n\\ y_n&\text{ for }x\ge x_n\end{cases} $$

You can also experiment with different S-like curves $T(x)$.

ADDENDUM: For example, if we use my suggestion in the comment section to use $T(x)=x^2(3-2x)$ we can define the function $F(x)$ on the interval $[x_1,x_n)$ which satisfies all your requirements.

$$ F(x)=\sum_{k=1}^{n-1}(y_{k+1}-y_k)\left[\left(\frac{x-x_k}{x_{k+1}-x_k}\right)^2\left(3-2\left(\frac{x-x_k}{x_{k+1}-x_k}\right)\right)\right]\cdot\left[U(x_{k})-U(x_{k+1})\right] $$

where $U(x)$ is the unit step function.