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N-sphere math problem

  1. Jul 24, 2005 #1
    [tex]\sum_{i=1}^{n+1} (x_i-c_i)^2{\leq}r^2[/tex] is not the set of all points at a certain distance from the center, but it is a "solid ball", so is it still an n-sphere?
     
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  3. Jul 24, 2005 #2

    Gokul43201

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    No, the n-sphere is the surface defined by the equality.
     
  4. Jul 24, 2005 #3
    who what is the inequality?
     
  5. Jul 24, 2005 #4

    Gokul43201

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  6. Jul 24, 2005 #5

    Gokul43201

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    All points on [itex]\sum_{i=1}^{n+1} (x_i-c_i)^2 = r^2[/itex] belong in an n-sphere (not points inside).
     
  7. Jul 24, 2005 #6
    i know that, i was asking what the points inside(the inequality) is called, since it isn't a sphere.
     
  8. Jul 24, 2005 #7

    Gokul43201

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    I guess you'd simply call them "points inside the n-sphere" (?), though I'm not sure if 'inside' is well defined. In this case, though, you could define 'inside' as being on the same side of the surface as the center.
     
    Last edited: Jul 24, 2005
  9. Jul 24, 2005 #8
    the volume which said n-sphere encapsulates?
     
  10. Jul 25, 2005 #9

    robphy

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    How about "interior of the (n+1)-ball"?
     
  11. Jul 25, 2005 #10
    The convention in topology as I learned it is

    [tex]\sum_{i=1}^{n} (x_i-c_i)^2{\leq}r^2[/tex]
    is a closed n-ball or radius r (or just "closed ball" when you're in R^n Euclidean space)

    [tex]\sum_{i=1}^{n} (x_i-c_i)^2< r^2[/tex]
    is an open n-ball or radius r or "open ball" in R^n.

    Note that in R^n, a ball is n-dimensional and a sphere is (n-1)-dimensional; i.e., a topologist's n-sphere is the boundary of an (n+1)-ball.
     
  12. Jul 25, 2005 #11

    matt grime

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    The interiot is commonly called a ball or a disc.
     
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