Question

I have a little math problem. I would like to have a function with these properties:

1. for x much bigger than 0: lim f(x) = x
2. for x much smaller than 0: lim f(x) = 0
3. and f(0) = 1 (sorry, I had here f(1)=1 which was wrong!)
4. f(x) should be monotonically increasing

So the function should look somewhat like this:

        ^         |   /         |  /         | /    ___.-+´ --´-----+------>         | 

The best I got so far is x/(1 + e^(-x)) but then I recognized that it drops below 0 and is not monotonically increasing.

A great help for playing around with these function is GraphFunc Online.

Also, it would be helpful if the function is fast to calculate as I need to execute it very often.

EDIT: I am using this in a program to limit values. I have an optimization algorithm, that uses curve fitting with a Levenberg-Marquardt algorithm. But this algorithm does not allow constraints, and optimizes over the full range of real values. So I need a function like this so that I can add an artificial constraint so that the function is bigger than 0. A simple approach would be to use f(x) = x² but then the function is not monotonically increasing and it has two minimas.

The Levenberg-Marquardt approximates derivatives, so I think it would be best when the function is smooth too. But I am not sure if this is absolutely necessary.

Answer 1

Except for a discontinuty at 0, x/(1 - e^(-x)) works. So define f(0) to be 1, and you’re set.

#define E 2.71828183 double SimpleFunc(double x) {    if (x == 0)       return 1;    return x / (1 - pow(E, (-x))); } 

Likely faster:

double SimpleFunc2(double x) {    if (x < 0)      return 1/(1 - x);    return x+1; } 

Both are continuous in the first derivative, but the second one has a jump at 1 in the second derivative)

If you really don’t want to do the piece-wise function, try this: (x^2+.1)^.5 / ((1 - e^(-x))^2+.1)^.5