Comments | nolacoaster: Question about function... optimization?

nolacoaster

Question about function... optimization?

Feb 05, 2010 14:16

Computer science people, I am looking to be pointed in the right direction. I am looking for a solution to a problem, and I think the answer might lie in an area of Computer Science/Statistics/O.R. that I know very little about. Here's the deal:

I have two mathematical functions, f_1 and f_2. Both functions intersect with the Y axis at exactly two ( Read more... )

nerds

Comments 5

simrob February 5 2010, 22:33:09 UTC

SIGBOVIK surprise or I'm-totally-gonna-get-best-OOPSLA-paper surprise or I AM YOUR NEW ROBOT OVERLORD surprise?

I don't have any idea, beyond vague notions like hill-climbing that only work if there aren't a lot of global maximums. If you want me to explain my vague notions on what hill-climbing does I can.

nolacoaster February 6 2010, 18:19:24 UTC

Now that you reminded me, definitely a SIGBOVIK submission!

lordspaz February 6 2010, 05:19:22 UTC

I actually know a decent amount about optimization, but I think I need more details about your problem to do anything with this. Do you mean "intersect the x axis at two points"? Is f_2 continuous, and does it have a closed form? How is f_2 specified, if not?

This sounds like a convex optimization problem, in general, but I don't know enough to be sure. Convex optimization problems are "easy."

nolacoaster February 6 2010, 17:18:20 UTC

Okay, awesome! So let's see:

f_1 is easy. It's just a quadratic equation. When I say it intersects the x-axis in two points, I mean that it looks like the quadratic equation in this image:
http://www.freemathhelp.com/images/lessons/graph14.gif
and I really don't care what happens to it below y=0. In fact, I'll assume x=0 whenever y<=0. So I am just trying to say that it creates an enclosed area above the x axis. Maybe that doesn't matter ( ... )

lordspaz February 6 2010, 18:25:19 UTC

Oh, this actually sounds easy. So f1 is a parabola. f2 is a piecewise linear function of 3 variables. But the 3 variables are of two different kinds - x and the shift/scaling factors. I'll call the last two "parameters."

So f1 = ax^2 + bx + c. Easy enough.

f2 = m12 * (x - \beta * (x1 + \alpha)) + y1 when x1 < x <= x2,
m23 * (x - \beta * (x1 + \alpha)) + y2 when x2 < x <= x3 ( ... )