| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1609.1 | | 3D::ROTH | Geometry is the real life! | Wed May 13 1992 03:13 | 27 |
| If you consider your function f(x,y) as a plane algebraic curve,
then a rational parameterization only exists if the curve has
genus zero. This is the case for the conic sections (circles,
hyperbolas, parabolas), for cubics with a self intersection
(or double point) or for quartics with a triple point.
In implicit form you have a quartic:
f(x,y) = x^2*y^2 - (x^2+y^2) - 4*E*x*y + 1 = 0
Is there a point where this curve intersects itself twice? If so,
a line drawn thru that point will only meet the curve at one other
point and the slope of that line will be a rational parameterization.
I may have to look this up at home since I don't remember the
exact details, but it's in Semple and Roth's (no relation) book on
algebraic geometry.
[genus zero means the curve as a Riemann surface is topologically a
sphere; the most general functions on the sphere are the rational
functions. A general cubic is genus one, or a torus and elliptic
functions are required for parameterization. It rapidly gets
hairy for higher degrees...]
later...
- Jim
|
| 1609.2 | | 3D::ROTH | Geometry is the real life! | Wed May 13 1992 04:39 | 5 |
| I sketched some examples of this curve for various values of E in the
range [-2,2] and it looks like you're out of luck for a rational
parameterization. (It makes a nice symmetric design, actually...)
- Jim
|
| 1609.3 | | TRACE::GILBERT | Ownership Obligates | Wed May 13 1992 17:55 | 25 |
| 1609.4 | Left as a PhD thesis for the reader | CIV009::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Wed May 13 1992 18:41 | 18 |
| > (BTW, is there some 'easy' way of recognizing an expression as the sum
> of two squares? Will MAPLE do this?)
I think the answers are {No, No}. FWIW, every *integer* is the sum of 4
integer squares, and they can be found by a simple backtrack scheme,
largest first. But I don't see any way of dealing formally with something
like
2 2 2 2 2 2
5 x y - 12 x y - 6 x y + 9 x + 9 y
(original form appears after the spoiler!)
2 2
(3 x - 2 x y) + (x y - 3 y)
|
| 1609.5 | | TRACE::GILBERT | Ownership Obligates | Thu May 14 1992 14:55 | 7 |
| 1609.6 | 'Here's something that doesn't fit' | CIV009::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Fri May 15 1992 17:32 | 3 |
| re .-1: Yes, but that's like the moral equivalent of saying "For any
object, one can find a set of objects of the same type of which the given
object is a non-member." Not a lot of help in attacking the problem!
|
| 1609.7 | | GUESS::DERAMO | Dan D'Eramo, zfc::deramo | Mon May 25 1992 21:27 | 18 |
| re .3,
> (BTW, is there some 'easy' way of recognizing an expression as the sum
> of two squares? Will MAPLE do this?)
The first order theory of real closed fields in the
language of fields {0,1,+,*} is complete. So it can be
proven whether or not a polynomial in x,y,z,... with
rational coefficients ever takes on a negative value for
real x,y,z,.... If it does, then it cannot be a sum of
squares. If it never takes on negative values, then I
think it may be possible to write as a sum of squares of
rational functions (quotients of polynomials), but I
haven't been able to track down where I read that to see
a precise statement of the theorem (such as, does it
apply here).
Dan
|