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.
>
>>
>
>>
>
>

```

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