What simple math function f(x) has these properties?

Question

Hello, 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

Answer 2

f(x) = abs( x/2 ) + x/2

where abs(x) is absolute value of x

This simple function would obviously be fast to compute and meets all four criteria.

Answer 3

I don't know what exactly you're using it for; what's wrong with a piecewise function? If you're going to be executing it a lot, something like this is going to be faster than doing exponents:

f(x) = -1/x, x < -1
f(x) = 1, -1 <= x <= 1
f(x) = x, x > 1

EDIT: Fixed it so it actually works.

Answer 4

Just to give you ideas, this is a solution without the constraint f(1)=1 and not monotonically increasing.

Basically you want to blend between two functions: f1(x) = 0 for x < 0, and f2(x) = x for x > 0. You want to blend that smoothly. A simple step function with a constant limit at -inf and +inf is atan (limits are -pi/2 and +pi/2 respectively).

So combining an atan blend function with f1 and f2, you get:

blend(x) = atan(x)/pi + 0.5 f(x) = (1 - blend(x)) * f1(x) + blend(x) * f2(x)

Which gives:

f(x) = (atan(x)/pi + 0.5) * x

There are probably other blending functions that you can use instead of atan. Also note than with small negative values, f(x) will be negative.

If you want your curve to go through (1,1), you can use the fact that atan(0)=0.

Answer 5

1/2 * (x + ABS(x))

It is monotonic.

f(1) = 1.

When x is less than zero, f(x) = 0, otherwise it's equal to x.

Answer 6

Here's a smooth function that satisfies your requirements:

f(x) = (x + sqrt(x^2 + 4)) / 2

For x = 0, you can see that f(x) = 1. For very large positive x, sqrt(x^2 + 4) is approximately x, so f(x) ≈ x. For very large negative x, sqrt(x^2 + 4) is approximately -x, so f(x) ≈ 0.

The first derivative is

f'(x) = 1/2 + 1/2*x/sqrt(x^2 + 4)

For x > 0, x/sqrt(x^2 + 4) > 0, so f'(x) > 0. For x < 0,

0 < x^2/(x^2 + 4) < 1
0 < |x|/sqrt(x^2 + 4) < 1
-1 < x/sqrt(x^2 + 4) < 0
-1/2 < 1/2*x/sqrt(x^2 + 4) < 0
1/2 + 1/2*x/sqrt(x^2 + 4) > 0

Hence, f'(x) > 0 for all x, so f(x) is monotonically increasing as desired.