Consider a real polynomial
$ f(x) := \sum_{k=0}^n c_k x^k $
of degree $n$, where $c_k \in \mathbb{R}$ are constant coefficients. Given $n+1$ samples $f_i$ of $f$ at distinct sample points $x_i$, one can recover the coefficients by solving the linear system
$ f(x_i) = f_i. $
Easy enough. However, the problem seems to get harder when $f$ has a special form. For instance, suppose that
$f(x) := (ax-b)^2$
where $a,b \in \mathbb{R}$ are now the unknown coefficients. This function describes a parabola whose minimum is zero -- in other words, it "sits" on the x-axis. Since this polynomial has fewer degrees of freedom, you would think you could recover its coefficients using fewer samples -- in particular, you would think two samples would suffice (one to recover the center, the other to recover the width). Surprisingly enough, however, actually computing these coefficients seems more challenging: we now have to solve a pair of simultaneous quadratic equations
$ (ax_1-b)^2 = f_1 $
$ (ax_2-b)^2 = f_2 $
In reality, I'm interested in the case where the domain is $\mathbb{R}^2$, not $\mathbb{R}$. In this case, I want to recover the three parameters $a$, $u$, and $v$ of the isotropic paraboloid
$ g(x,y) := (ax-u)^2 + (ay-v)^2 $
from three samples $(x_i,y_i)$. Again, it seems that three samples should suffice: two for the center and one for the "width." But so far I don't see any good way of computing a solution short of applying Newton's method (and keeping my fingers crossed). Even in the 1D case, asking Mathematica for a Gröbner basis results in a hairy mess!
Thanks. :-)