Minimal distance between quadratic function and point I have a function or line $R\rightarrow R^n$
$$ y_i = f(x) = {-b_i \pm \sqrt{-4 \cdot a_i\cdot c_i + b_i^2+4\cdot c_i \cdot x}\over 2 \cdot c_i}$$
where the parameter vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are known. For any value x, I will get a n-dimensional point out. (The equation should look familiar, meaning x(y) are quadratic functions.)
I would like to find for an arbitrary point $\mathbf{p}$ the closest point on this line $f$, and its corresponding $x$.
What I did so far was to differentiate the euclidean norm $\Delta = \sum_i (y_i(x) - p_i)^2 $ and setting it to $0$, yielding, after dividing away non-zero terms:
$$2\cdot \sum_i {p_i + {b_i\over 2 \cdot c_i }\over \sqrt{4 \cdot c_i\cdot  x - 4 \cdot a_i\cdot c_i+ b_i^2}} - {1\over2\cdot c_i} \overset{!}{=} 0 $$
Now I'm stuck. Does an analytic solution for finding $x$ exist?
 A: I doubt this has any closed form solution. Here's why. The difficult part of the formula you have is simplifying an expression of the form $\frac1{\sqrt x} + \frac1{\sqrt y} + \frac1{\sqrt z}$, and some additional terms, so that there is a common denominator with no radical in it. Once you get there, you have an expression of the form $\sqrt a + \sqrt b + \sqrt c$, for which there is at least some hope.
However, to get there, you cannot really avoid converting to the expression $\frac{yz\sqrt{x}+xz\sqrt{y}+xy\sqrt{z}}{xyz}$. Similar things work for longer sums than three, but unfortunately, this is where the good luck stops. We multiply both sides by $xyz$ and what we find is that there are degree-three polynomials on each side of the equality sign, in addition to things in square roots. So the smallest polynomial you can hope to have here is degree six.
Furthermore, there is almost no relationship between the coefficients. Yes, it all looks nice and symmetric when we're doing the heuristics, but $x$, $y$, and $z$ are all nontrivially linear in the variable and there are also constant terms floating around out there. Therefore, Abel is going to come in and say that you probably cannot solve this polynomial algebraically.
Sorry about that :(
