How to make an approximation of path with polynom P(x,y)=0? Hi. Imagine that a user draws on the canvas any path. I want to approximate this path with a path $P(x,y)=0$ where $P(x,y)$ - is unknown polynom. May be somebody can suggest an appropriate algorithm?
1) I tried to use the method of least squares to find this polynom. Just choosed on the path a lot of points $(x_i,y_i)$ . And minimized unknown $\sum\limits_i P(x_i,y_i)^2$ among all polinoms of degree n. This problem of minimization is a problem of finding an eigen vector of huge matrix that grows as $n^2\times n^2$. Already for $n=7$ this matrix has a size $36\times36$ it's hard work for PC   to find the solution. And for $n=7$ it doesn't give appropriate result.
2) Spline doesn't work for me. Because spline - is a union of curves $P_i(x,y)=0$. To each spline  of course we can correspond $P(x,y)=\prod\limits_i P_i(x,y) $. But this union  $P(x,y)=0$ will have a lot of bifurcation points on the curve. And for my project it is very bad
 A: This is a very hard problem to do right. The simplest solution is to sample your polynomial on a grid, and for every pair of adjacent grid points where the sign changes, find a zero (exactly or approximately) on the relevant edge, then connect nearby zeros by segments. This is easily modified to be adaptive (that is, if there is a lot of action in a region, you refine the grid), but will still miss singularities. Splinification is bad, because it introduces smoothness where it is not warranted.
A: Your reference to eigenvalues suggests that you are minimizing $\sum_i P(x_i, y_i)^2$ subject to a constraint like $\sum_{a,b} P_{ab}^2=1$, where $P(x,y) = \sum_{a,b} P_{ab} x^a y^b$. Why not instead minimize $\sum_i P(x_i, y_i)^2$ subject to a linear constraint like $P_{00}=1$? 
This is a standard approach when fitting a conic through a cloud of points (see, for example, Paul L. Rosin, A note on the least squares fitting of ellipses, Pattern Recognition Letters, Volume 14, Issue 10, October 1993, Pages 799–808) and I don't see why it wouldn't be a good idea for higher degree.
This puts your problem in the form "minimize $\vec{x}^T A \vec{x}$, subject to $\vec{b} \cdot \vec{x}=1$", to which the solution is $\vec{x} = \frac{A^{-1} \vec{b}}{\vec{b}^T A^{-1} \vec{b}}$; no eigenvalues needed. 
I suspect Misha's comment will be more useful than this answer, though.
I also second everyone who said that $36 \times 36$ is not large. It looks like this thread might be able to help you find a better linear algebra library.
A: New method for solving underdetermined systems of nonlinear equations. The first link describes (in Russian) calculations of linkages. The second gives an example from another area on mapleprimes (with the text, however, in Maple (answer from myself, "one man")). 
http://www.exponenta.ru/educat/systemat/selitskiy-ivanov/index.asp 
http://www.mapleprimes.com/questions/205606-Find-Root-Method 
