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Question on Boundary Points

  1. Sep 15, 2009 #1
    Question on Boundary Points!!

    Determine all boundary points:

    S={(x,y); 0 < x< 1 and y= sin(1/(1-x)}


    I'm really confused! I dont understand how I can actually find all the points.. can anyone get me started??

    The help is appreciated!
     
  2. jcsd
  3. Sep 15, 2009 #2

    Office_Shredder

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    Re: Question on Boundary Points!!

    S is going to be a curve, so it looks like it's asking you to find what's the boundary at x=0 and at x=1. At x=1 of course you're going to run into a snag because 1/(1-x) doesn't act so nice there, so start by finding the boundary point at x=0
     
  4. Sep 15, 2009 #3

    Hurkyl

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    Re: Question on Boundary Points!!

    Maybe start somewhere eaiser -- can you find any boundary points at all? Can you find all boundary points that lie in some easily-described subset of the plane?
     
  5. Sep 15, 2009 #4
    Re: Question on Boundary Points!!

    Hey! Thats funny I have the same question in my practice questions too!
    Anyone know how to do it? My prof gives no examples!
     
  6. Sep 15, 2009 #5

    Hurkyl

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    Re: Question on Boundary Points!!

    You could try starting with an easier problem -- can you find any boundary points at all? Can you find all boundary points that lie in some easily-described subset of the plane?
     
  7. Sep 15, 2009 #6
    Re: Question on Boundary Points!!

    well sin(x) would probably be easier but how to find the actual points. if u take x=0 then sin (0)=0 and sin(1)=π/2
     
  8. Sep 15, 2009 #7
    Re: Question on Boundary Points!!

    haha yah dont feel bad im pretty lost too!
     
  9. Sep 15, 2009 #8
    Re: Question on Boundary Points!!

    Above ur taking the new function of y to be sin(x) right? .. this seems wrong lol no offence
     
  10. Sep 16, 2009 #9
    Re: Question on Boundary Points!!

    k me and tara123 clearly dont know what were talking about can anyone help us???
     
  11. Sep 16, 2009 #10

    Office_Shredder

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    Re: Question on Boundary Points!!

    So for example, if you had S={(x,y)|y=x2+1 0<x<1} you can see that the boundary points of that curve are going to be exactly at (0,1) and (1,2). Now why is that. For any e>0, you can find a point on the curve S (x,y) such that |(x,y)-(0,1)|<e and also for any e>0, you can find a point on the curve (x,y) such that |(x,y)-(1,2)|<e. It's pretty easy here because y=x2+1 is a continuous function, so if you're told all the points x>0 are included, then the value at x=0 must be included.

    So for the set S that you have, it's continuous away from x=1. So by the same process, we conclude that it has a boundary point at x=0. You should be able to find that point by yourself.

    The tricky part is for x=1. It's not continuous there. What's the behavior of sin(1/(1-x)) as x approaches 1? What values of (1,y) might be part of the boundary?
     
  12. Sep 17, 2009 #11

    HallsofIvy

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    Re: Question on Boundary Points!!

    First, have you drawn a sketch of the set?
    S={(x,y); 0 < x< 1 and y= sin(1/(1-x)} is essentially the graph of y= sin(1/(1-x)) for 0< x< 1. (One thing I would be inclined to do is let u= 1-s so this is y= sin(1/u) for 0< 1< u.) Look closely at what happens around u= 0 (x= 1).
     
  13. Sep 18, 2009 #12
    Re: Question on Boundary Points!!

    360px-Sin1over_x.svg.png
    I found an image for when y=sin(1/u), but when you look at u=0, there is never a defined value for it as the function doesn't exist at u=0. I don't see how that would apply to our function.
     
  14. Sep 21, 2009 #13
    Re: Question on Boundary Points!!

    yah so for x=0 y=sin(1)=pi
    and for x=1 i just look at what the graph does as it approaches 0?
     
  15. Sep 22, 2009 #14
    Re: Question on Boundary Points!!

    sorry i meant sin(1)=PI/2

    but just one other thing, ive bin thinking and y=sin(1/1-x) o<x<1 0 isnt even contained in the set...so how is it a boundary point if 0 is strictly less than x???
    Help!
     
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