What is the most simple formula to achieve this pattern?

Question

Can the pattern in the attached diagram be captured by tuning a single parameter in a simple function? If so, what function is that?

Answer 1

If you scale and flip almost any Cumulative Distribution Function, you should get exactly what you're looking for. The simplest CDF is from the Kumaraswamy Distribution, which has CDF $1-(1-x^a)^b$.

Answer 2

$$ f_n(x)=\frac{\alpha\,}{1+\left(\frac{2x}{1-2x}\right)^n} $$ For $\alpha=1$ and $n=1,2,3,4,5$, we get

Answer 3

A simple way to construct the desired curve is to start with a sigmoid $$f(x)=\frac{1}{1+\exp(A(x-1/4))}$$ where $A$ is a positive parameter that influences the steepness.

Add a line through $(1/4,0)$ $$g(x)=\alpha f(x)+\beta (x-1/4)$$ with $\beta$ to be determined such that the curve passes through $(0,\alpha)$ and $(1/2,0)$: $$g(0)=\alpha \frac{1}{1+\exp(-A/4)}-\frac{\beta}{4}=\alpha$$ $$\implies 4 \alpha \left( \frac{1}{1+\exp(-A/4)}-1\right)=\beta$$

So $$\frac{\alpha}{1+\exp(A(x-1/4))}+4 \alpha \left( \frac{1}{1+\exp(-A/4)}-1\right)(x-1/4)$$ behaves as desired. (Note: at $A=0$ this function is exactly linear, in the $A\rightarrow \infty $ limit it becomes a step function)

It works just fine in Mathematica:

\[Alpha]/(1 + Exp[A*(x - 1/4)]) + 4*\[Alpha]*(1/(1 + Exp[-A/4]) - 1)*(x - 1/4)
{% /. {\[Alpha] -> 1, A -> 0}, % /. {\[Alpha] -> 1, A -> 20}, % /. {\[Alpha] -> 1, A -> 100}}
Plot[%, {x, 0, 1/2}]