I''m going to render the volume in 3d. I have an input ray to a bounded volume (an octree node) and I know the formula for the cubic function inside the volume of the node, and I want to find where the line intersects the surface that you get wherever the function equals the cutoff point. So, I figure, for each subtree of the node, if there are any points inside where the value of the function exceeds the cutoff, trace the ray inside that node. Otherwise, skip it, and do the same for the next subnode that the ray hits.

FallingFrog

Newton''s Method is pretty cool (and it works for pretty much anything). It uses an iterative approach, so you can be as accurate or inaccurate as you want.

NewX = OldX - (f(OldX) / f''(oldX))

Provided that f''(OldX) isn''t 0.

What it actually does, is starting from a point, the tangent to the curve at that point is found, then at the point where that tangent hits the x axis, is the new x coordinate. This will eventually reach (or converge on) a solution most of the time. Problems come when you can''t get the solution you are looking for, in which case, your best bet is to find the turning points and other useful points which are often found near roots (points of inflexion) and use them for initial values of x.

The other possible problem can come when f''(oldX) is a really small number for a few values of x which totally puts off the search because the tangent is almost horizontal and intercepts the x-axis a long way away.

And, if you don''t know calculus, then this will be a bit harder. But here is the crash course:

Polynomial f(x) = ax^n + bx^m + cx^o + dx...

Derivative f''(x) = anx^(n - 1) + bmx^(m - 1) + cox^(o - 1) + d

(so you take the power, multiply it by the coefficient and then subtract 1 from the power)

Trying is the first step towards failure.

Enigma- that would be great
Incidentally, I thought of another way of solving my problem-
if(f(x + q) > v || f''(x + q) > v || f''''(x + q) > v)
then it is possible that there is a value above v somewhere between x and x + q. But this method will give me a lot of false positives. Because the cubic will usually be non-flat, if you know what I mean. Yours will be faster, I think. Say, you wouldn''t want to help me implement reflection and shadows, would you?

FallingFrog

PS. I just invented the number of times it has to run in a second. I multiplied 320 * 200 * 20fps * 5 recursions, and rounded it down a little.

PPS. This is just an idea, in case you haven''t done it yet: Since I only need to know if at least one max exists, you can return as soon as you find one- you don''t have to find them all.