Suppose we have four points, $x_1<x_2<x_3<x_4$
Is anyone able to provide me with an equation for a function that is
- nearly flat $x_1,x_2$ (i.e. $f(x_1)/f(x_2) \approx 1)$
- Steep between $x_3,x_4$ (specifically, I want $\frac{f(x_3)}{f(x_4)}<\frac{f(x_1)}{f(x_2)}$)
- strictly increasing, continuous, positive, and convex over the positive reals
For example, suppose $x_1 = 2.2,x_2=5,x_3=7.5,x_4=9$,
Is there a convex, strictly increasing, function where $\frac{f(x_3)}{f(x_4)}<\frac{f(x_1)}{f(x_2)}$
Some comments:
- Im okay with either a general answer or an answer for the specific points i have given
- convexity is not a requirement as long as the function is strictly increasing (I prefer if it is convex though
- I'd like $f(0)=0$ if possible (but not required)
I've been generating a bunch of points in mathematica and trying to generate a curve, but i haven't had luck.
I have also tried an exponential function (this is the steepest thing I could think of),but then $x_1,x_2,x_3,x_4$ need to satisfy $x_1+x_4 > x_3+x_2$, (which my example points don't satisfy)
This has been bugging me because its so easy to draw such a function, but I haven't been able to come up with an equation.
Best Answer
A simple answer is to define $x_c=\frac 12{x_2+x_3}$, then $$f(x)=e^{k(x-x_c)}$$ The side below $x_c$ will be pretty flat because it is like $e^x$ for $x \lt 0$ while the side above is like $x \gt 0$. Increase $k$ to make it more pronounced.