N-sphere math problem

  • #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?
 

Answers and Replies

  • #2
Gokul43201
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No, the n-sphere is the surface defined by the equality.
 
  • #3
who what is the inequality?
 
  • #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).
 
  • #6
i know that, i was asking what the points inside(the inequality) is called, since it isn't a sphere.
 
  • #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.
 
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  • #9
robphy
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How about "interior of the (n+1)-ball"?
 
  • #10
rachmaninoff
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.
 
  • #11
matt grime
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The interiot is commonly called a ball or a disc.
 

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