Creating a Variable Bend in a Function with Preserved Properties

[Ad VanderVen]

TL;DR Summary

With a given function how to make a variable bend.

I have the following function: $$y = s-1 + exp (-ln (-1 / (s-1)) x)$$ with 0 < s < 1. For $x = 0$ you have $y = s$ and for $x = 1$ you have $y = 0$. The function has a fixed bend, but now I want to make the bend variable, while the two properties, that for $x = 0$ you have $y = s$ and for $x = 1$, $y = 0$, should remain unchanged. How do I do that?

 

[pbuk]

Where did you get that equation from? Do you realize it can be simplified to $y = (1-s)^x + s - 1$? What is 'bend'? If you mean you want a similar function with an increased radius of curvature, try $y = (1-s)^{x^2} + s - 1$, although that has a very different behaviour for $x < 0$. How do you think you could modify this to get a curve that is similar for any value of $x$?

 

[Ad VanderVen]

I want a similar function with the same restrictions as described above with a parameter, say $b$, representing the curvature.

 

[pbuk]

I want a similar function with the same restrictions as described above with a parameter, say $b$, representing the curvature.

Have you looked at the function I suggested? How do you think you could incorporate the parameter $b$ into it?

 

[Ad VanderVen]

Yes, I looked at the function you suggested and the function $$s-1+(1-s)^{x^b}$$ with $0<b<1$ would do perfectly. I cannot say how grateful I am..

 

[Ad VanderVen]

What could you call such a function? Is it a special case of some well-known function?

 

[pbuk]

I'm not a big fan of naming things: some people (and some educational cultures) get over-obsessed with naming and categorization.

I suppose you could call it a sub-exponential function (and for b = 0.5 a half-exponential function) but I have only seen this term used for increasing functions (i.e. $s < -1$) so I think using that term in this context would not be helpful.

 

[Ad VanderVen]

@pbuk Anyway, thanks for the attempt.