http://www.xing.com/net/mathe/ Latest posts in XING group "Mathematics / Mathematik / Matemática" en-en Copyright XING AG 11 Jul 2014 16:00:07 GMT Fri, 11 Jul 2014 16:00:07 -0000 en-support@xing.com en-support@xing.com epublica http://www.xing.com/img/xing/xing_logo_small.gif http://www.xing.com/net/mathe/ 90 45 http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46721464#46721464 Ok, whatever :-| u need code? Sascha Oppermann <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46721464#46721464 Sat, 05 Jul 2014 14:03:20 GMT http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46709563#46709563 Hm, das verstehe ich widerum nicht, es geht doch hier um topologische Raeume bzw deren Zerlegung/Triangulation. Ohne Umgebungsfilter kann man doch nicht differenzieren? Sascha Oppermann <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46709563#46709563 Wed, 02 Jul 2014 09:24:51 GMT http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46707139#46707139 Verstehe ich nicht. Das ist ein analytisches Problem, kein topologisches. Dr. Thomas Müller <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46707139#46707139 Tue, 01 Jul 2014 15:09:10 GMT http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46706793#46706793 Okay, so you need a transition from differential forms to cochains? That possess' a differentiable - as a result of the choosen basis - surface representation with the source vertices embedded within. Or doesn't that make sense to u? Sascha Oppermann <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46706793#46706793 Tue, 01 Jul 2014 13:59:34 GMT http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46705897#46705897 If you start with any closed surface consisting of vertices which encloses the point set, I would like to use a sort of mean curvature flow in order to find a surface which is close enough in some sense, e.g. distance to point set. But a surface consisting of vertices is not differentiable, i.e. mean curvature flow makes no sense. Dr. Thomas Müller <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46705897#46705897 Tue, 01 Jul 2014 11:43:45 GMT http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46704633#46704633 Hi Thomas, what exactly is the prob? Finding a proper algorithm to decompose your desired vector bundle? Just askin' Sascha Oppermann <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46704633#46704633 Tue, 01 Jul 2014 08:48:07 GMT http://www.xing.com/app/forum?op=showarticles&id=46674172&articleid=46699805#46699805 Thanks for your answers. M.S. schrieb: ... Ah, yes, that would be an interesting information, wouldn't it? :) I am a graduated mathematician (Diplom-Mathematiker) and my (profesional) special interest is probability theory. I am also using statistical methods (R / R studio) in various areas: data verification, data analysis, market analysis. Most of this can be done with the very basics of statistics (estimator for the mean? take the sample mean... test if the simulated numbers are wrong? take ... Thomas Runge <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=46674172&articleid=46699805#46699805 Mon, 30 Jun 2014 08:37:08 GMT http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46699345#46699345 Hello Martin, did you actually solve your problem? I have a similar problem: I would like to construct a triangulated &quot;tubular&quot; neighbourhood of a finite point set. It should not be the convex hull, since this is to &quot;big&quot;. Regards, Thomas Dr. Thomas Müller <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46699345#46699345 Mon, 30 Jun 2014 07:32:54 GMT http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46699322#46699322 This is very likely the whole space, since if you have at least points which are linear independent, they are a basis of R^3. All linear combinations of them are the whole space. Dr. Thomas Müller <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46699322#46699322 Mon, 30 Jun 2014 07:29:54 GMT http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46698520#46698520 I think that we must consider all the linear combinations between the points of the initial ensemble, given in a finite number. Edoardo Angeloni <en-support@xing.com> http://www.xing.com/app/forum?op=showarticles&id=22619171&articleid=46698520#46698520 Sun, 29 Jun 2014 19:07:43 GMT