Re: Re: Finding the periphery of a region

*To*: mathgroup at smc.vnet.net*Subject*: [mg72042] Re: [mg72005] Re: Finding the periphery of a region*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sat, 9 Dec 2006 06:09:51 -0500 (EST)*References*: <el8ufm$st3$1@smc.vnet.net> <200612081117.GAA20172@smc.vnet.net>

It's well known fact in real algebraic geometry that this does not work in general. Here is a well known example example (also included in my response to the OP): x^3 - x^2 - y^2 > 0 && x < 10 The boundary is not x^3 - x^2 - y^2 >= 0 && x <= 10 This can be seen also on a picture: <<Graphics`InequalityGraphics` InequalityPlot[x^3-x^2-y^2>0&&x<10,{x},{y}] You can see that the point {0,0} which lies on x^3 - x^2 - y^2 == 0 is not on the boundary of the region. Andrzej Kozlowski On 8 Dec 2006, at 20:17, dh wrote: > > > Hi, > > try replacing inequalities by equalities. This should work fine as > long > > as you do not have intersecting regions. E.g.: > > x^2+y^2<100 ==> x^2+y^2=100, obviously a circle > > (5<=x<=25 and -10<=y<=10) ===>( x==5&&-10<=y<=10) || > > (x==25&&-10<=y<=10) || (5<=x<=25&&y==-10) || (5<=x<=25&&y==10), four > > line segements. > > Daniel > > > > Bonny Banerjee wrote: > >> I have a region specified by a logical combination of equatlities and > >> inequalities. For example, r(x,y) is a region defined as follows: > >> > >> r(x,y) = x^2+y^2<100 or (5<=x<=25 and -10<=y<=10) > >> > >> How do I obtain the periphery of r(x,y)? I am only interested in >> finite > >> regions i.e. x or y never extends to infinity. > >> > >> Thanks, > >> Bonny. > >> > >> > >

**References**:**Re: Finding the periphery of a region***From:*dh <dh@metrohm.ch>