I am wondering wether it exists a theorem that any continuous path on the plane one can approximate with algebraic curve $P(x,y)=0$ ($P$- is a polynom)?
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$\begingroup$ See en.wikipedia.org/wiki/Polynomial_interpolation or en.wikipedia.org/wiki/Bernstein_polynomial. $\endgroup$– Lalit JainCommented Aug 13, 2012 at 13:09
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3 Answers
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Given any compact set $K$ in the plane (in particular the image of a compact interval under a continuous function) and $\epsilon > 0$, there is a finite set $\{(x_j, y_j)\}_{j=1}^n \subseteq K$ such that $K$ is contained in the union of the disks of radius $\epsilon$ centred at $(x_j, y_j)$. Then $K$ is within distance $\epsilon$ in the Hausdorff metric of the real algebraic curve $P(x,y) = 0$, where $P(x,y) = \prod_{j=1}^n ((x-x_j)^2 + (y-y_j)^2 - \epsilon^2)$.
answered Aug 14, 2012 at 1:03
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$\begingroup$ thank you very much for your answer. This solution is very nice! But what is bed that solution contains a lot of bifurcation points. I am wonderig if the initial path is more or less - "normal" can we aproximate it with "normal" algebraic curve? Let say that "normal" is something that our intuition suggests. I don't know, let say "normal" curve is a curve that is diffeomorphic to interval. Or may be it should be some another definition that kills these biffurcation points. $\endgroup$– DavidCommented Aug 14, 2012 at 21:05
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$\begingroup$ If by bifurcations you mean critical points (intersections of the circles), those are easy to get rid of by modifying the definition. With a bit more work, if $K$ is diffeomorphic to an interval we should be able to get a curve diffeomorphic to a circle. $\endgroup$ Commented Aug 15, 2012 at 17:50
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There might be problems with the Peano curve which is a continuous map $[0,1]\to\mathbb{R}^2$ whose image is the square $[0,1]\times [0,1]$.
answered Aug 13, 2012 at 13:09
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2$\begingroup$ That square can be approximated (in the sense of the Hausdorff metric) by the curve $$P(x,y) = \prod_{j=0}^N \prod_{k=0}^N \left((x-j/N)^2 + (y-k/N)^2 - 1/N^2\right) = 0$$ $\endgroup$ Commented Aug 13, 2012 at 16:00
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It's hard to tell what this question means exactly.
If a "continuous path" is a continuous image of the unit interval, then any continuous path can be uniformly approximated by polynomial paths; this is the Weierstrass approximation theorem. Any of those polynomial paths can be extended to a polynomial image of the entire real line, which (because you are in the plane) is a set-theoretic complete intersection (i.e. the locus of $P(x,y)=0$ for some $P$).
But --- the continuous path consisting of a line segment in the plane is of course not the vanishing set of any polynomial, nor does there seem to be any reasonable sense in which it could be approximated by such.
On the other hand, if a "continuous path" is a continuous image of the real line, then it can be uniformly approximated on compact sets by polynomial paths, each of which is the locus of vanishing of some $P$. Whether this satisfies your needs depends on what you mean by approximating the path.
answered Aug 13, 2012 at 13:07
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$\begingroup$ Sorry but I didn't understand. Using Weierstrass theorem you can approximate your continuous path with polinoms $x(t)$ and $y(t)$. How from this polynoms one can сonstruct an algebraic curve $P(x,y)$=0? $\endgroup$– DavidCommented Aug 13, 2012 at 13:13
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$\begingroup$ If $X(t)$ and $Y(t)$ are polynomials, then the resultant of $x-X(t)$ and $y - Y(t)$ is a polynomial in $x$ and $y$ that is $0$ iff there is $t$ (not necessarily real) such that $x = X(t)$ and $y = Y(t)$. $\endgroup$ Commented Aug 13, 2012 at 15:48
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$\begingroup$ A line segment in the plane can be reasonably approximated by ellipses. $\endgroup$ Commented Aug 13, 2012 at 15:53
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$\begingroup$ Robert Israel: Good point regarding the ellipses. $\endgroup$ Commented Aug 13, 2012 at 17:35