most recent 30 from http://mathoverflow.net 2013-05-19T08:51:34Z http://mathoverflow.net/feeds/user/24233 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98860/how-to-make-an-approximation-of-path-with-polynom-px-y0 David 2012-06-05T11:02:24Z 2013-04-24T09:21:58Z

<p>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?</p> <p>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.</p> <p>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</p>

http://mathoverflow.net/questions/104609/algebraic-curve-approximation David 2012-08-13T12:01:44Z 2012-08-14T01:03:18Z

<p>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)?</p>

http://mathoverflow.net/questions/104609/algebraic-curve-approximation/104659#104659 David 2012-08-14T21:05:32Z 2012-08-14T21:05:32Z

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 - &quot;normal&quot; can we aproximate it with &quot;normal&quot; algebraic curve? Let say that &quot;normal&quot; is something that our intuition suggests. I don't know, let say &quot;normal&quot; curve is a curve that is diffeomorphic to interval. Or may be it should be some another definition that kills these biffurcation points.

http://mathoverflow.net/questions/104609/algebraic-curve-approximation/104614#104614 David 2012-08-13T13:13:17Z 2012-08-13T13:13:17Z

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?

http://mathoverflow.net/questions/98860/how-to-make-an-approximation-of-path-with-polynom-px-y0 David 2012-06-11T08:30:28Z 2012-06-11T08:30:28Z

@Douglas Zare. not necessarily

http://mathoverflow.net/questions/98860/how-to-make-an-approximation-of-path-with-polynom-px-y0 David 2012-06-11T08:28:02Z 2012-06-11T08:28:02Z

@Misha Thank you very much for your response. I looked trough your paper. Actually it's great! I will try to understand it.

http://mathoverflow.net/questions/98860/how-to-make-an-approximation-of-path-with-polynom-px-y0 David 2012-06-06T12:08:16Z 2012-06-06T12:08:16Z

Stanislav, because I am writing web application I write everything on Javascript and use JS library &quot;numeric&quot;. Thank you for your comment. It seems that it exists much faster algorithms that in my library. Then I'll try to write my own eig() function.

http://mathoverflow.net/questions/98860/how-to-make-an-approximation-of-path-with-polynom-px-y0/98871#98871 David 2012-06-05T14:26:43Z 2012-06-05T14:26:43Z

Thank you for your answer. But the problem is to find the polynom that approximate the given path but not find the path from the polynom.

http://mathoverflow.net/questions/98860/how-to-make-an-approximation-of-path-with-polynom-px-y0 David 2012-06-05T14:18:44Z 2012-06-05T14:18:44Z

I want to write following a programm. User draw any path (let say his signature) and programm costruct the mechanical linkage that can draw this path. Because of the &quot;Kempe linkage theorem&quot; I know that it is poossible for any algebraic path P(x,y)=0.