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I'm currently studying the implicitization of bezier curves (that is, finding a function that f(x, y) = 0 for any x and y pairs of a curve p(t)) as part of an algorithm for curve intersection. The approach I'm taking involves taking the resultant of two functions known as "moving lines." For a cubic curve, the two functions required are one quadratic and one linear:

a(t) = (a*x + b*y + c)*(1 - t)2 + (d*x + e*y + f)*(1 - t)*t + (g*x + h*y + i)*t2
b(t) = (k*x + l*y + m)*(1 - t) + (n*x + o*y + p)*t

In the paper I'm reading, a process for taking the resultant of two functions of similar degree, which I assume to be Bezout resultant, is presented, but not for functions of differing degree. From what I can gather, I need a resultant process that eliminates two variables, t and (1 - t), in functions of differing degree. As previously stated, the Bezout resultant is already out due to the differing degrees of my functions. Additionally, the Dixon approach seems to be inapplicable as it only works with a system of n+1 polynomials of degree n. I read that the Sylvester resultant could be used iteratively to eliminate two variables, but I couldn't find any details on that process.

What I'd like to ask is if anyone can provide details or a source on using Sylvester's resultants to iteratively remove more than one variable. However, if you know of another method to obtain the resultant of two polynomials of degree n and n - 1 in two variables, I'd be just as grateful. I apologize if this is not a suitable question for this site.

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up vote 3 down vote accepted

I don't really understand the question. You seem to want to eliminate the variable $t,$ which you can do with a single application of the Sylvester resultant (that is, the determinant of the Sylvester matrix). The determinant of the two polynomials (thought of as polynomials in $t$) will give you a single polynomial $f(x, y)$, the zeros of which you seek. Unfortunately for you, you are only seeking its real zeros (or so I assume).

If your question on how to eliminate more than one variable is more academic than practical, there is no trick to using the Sylvester determinant multiple times: if your polynomials are $f(x, y, z),$ $g(x, y, z)$, $h(x, y, z)$ you first compute the the resultants of $f$ and $g$ (as polynomials in $z$) and then of $g$ and $h$ (as polynomials in $z$), then compute the resultant of the two resulting resultants, you will have a single polynomial in your single favorite variable. However, this is, in general, horribly inefficient (the degrees blow up), so mankind has developed many improvements (I am not sure what the latest word is, googling for "Collins sub resultant" will point you in the right direction), which are used in Mathematica, Maple, and whichever other computer algebra system you might use.

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Thank you for your answer. I had assumed t and (1 - t) were to be thought of as two separate variables, but it looks like that was incorrect. – Lucas McCarthy Apr 9 '12 at 18:56

Resultants and sub-resultants in one variable are extensively studied in "Elimination : Résultants et Sous-résultants, le cas d'une variable", by François Apéry and Jean-Pierre Jouanolou (Hermann, edit. 2006), 477 pages. It includes many exercises with their solutions, e.g. a conjecture by Helou and Tarjanian on Wendt determinants (Exercice 59, p.221).

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