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The space between a unit 'sphere' in n dimensions within an n-dimensional cube

  1. Aug 7, 2012 #1
    If we have a unit circle within a square s.t. the square touches the circle in 4 places then the biggest gap we can find is just √2 - 1.

    Doing a similar thing with a sphere in a cube we get √3 - 1

    I've heard the n-dimensional analogue is √n - 1. Which is crazy as it means the gap is bigger than the radius of the sphere! (When n is greater than 4).

    Anyway, how is such a thing proven?
     
  2. jcsd
  3. Aug 7, 2012 #2

    mathwonk

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    pythagoras?
     
  4. Aug 7, 2012 #3
    Yeh the result can be proven by Pythagorus in 2 and 3 dimensions but what about n?
     
  5. Aug 8, 2012 #4

    mathwonk

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    induction?
     
  6. Aug 8, 2012 #5
    Higher-dimensional spheres have some surprising properties. The (hyper-) volume and (hyper-) surface area of a unit sphere increase for a while as the dimension goes up, but they reach their maximum values pretty soon, then decrease forever after that.

    Consider an n-dimensional "cube" that just contains the n-dimensional sphere of radius 1: the midpoints of the sides of the cube are at distance 1 from the origin, but as you noted, the corners are farther away. For n=2 we have a square, whose corners are at distance [itex]\sqrt2[/itex] from the origin, and when n=3, we have an ordinary cube whose corners are at distance [itex]\sqrt3[/itex]. This pattern continues; when n=4, the distance is 2, when n=9 the distance is 3, when n=100 the distance is 10, and so on. Obviously there's no limit to how far away those corners get as the dimension becomes very large. But since every point on the sphere, of any dimension, is at distance 1 exactly, you can see why the sphere takes up less and less of the cube as the dimension increases.
     
  7. Aug 8, 2012 #6
    My question though is, how do we know that this pattern continues for n greater than 3? If it's just defined to be that then I can kind of see why but is there any 'proof'? This result surely cannot be proven by induction?
     
  8. Aug 8, 2012 #7

    mathwonk

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    the face of an n cube is an n-1 cube. The diagonal of an n cube thus is the hypotenuse of a right triangle whose other two legs are: 1) an edge of the n cube, 2) a diagonal of a face, i.e. a diagonal of an n-1 cube.

    thus usual pythagoras supplies the inductive step to prove the diagonal of an n cube of edge length one, has length sqrt(n) = sqrt{1^2 + sqrt(n-1)^2}.
     
  9. Aug 9, 2012 #8
    what about the ∞-dimensional case...?
    In such a case the volume of the gap between the ∞-dimensional unit sphere and the ∞-dimensional unit cube is +∞.

    Moreover the ratio of the volumes of the unit sphere and unit cube is given by [tex]\lim_{n\to +\infty} \frac{\pi^{n/2}}{2^n \Gamma(1+n/2)}[/tex]
    and in this case the above quantity tends to 0.
    Shall we conclude that in a ∞-dimensional space the unit-sphere behaves likes a point ?
     
    Last edited: Aug 9, 2012
  10. Aug 9, 2012 #9

    micromass

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    No, you should conclude that volume in an infinite-dimensional space is not a well-defined concept. Furthermore, there are multiple kinds of infinite-dimensional spaces.
     
  11. Aug 9, 2012 #10
    Could you just make an example on how to make an ∞-dimensional space out of ℝn that would not be the trivial method of "increasing the number of dimensions" up to infinity?
     
  12. Aug 9, 2012 #11

    micromass

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    I'm not sure what you mean with with "make an infinite-dimensional space out of [itex]\mathbb{R}^n[/itex]".

    But anyway, the easiest infinite-dimensional space to work with is probably the set of all sequence [itex](x_n)_n[/itex] which are eventually 0. So an example of an element in that set is (2,5,2,3,0,0,0,0,0,...).

    But this is not a very interesting infinite-dimensional space because it has quite some deficiencies. One such deficiency is that it is not "complete". This means that it has gaps (just like [itex]\mathbb{Q}[/itex] has gaps compared to [itex]\mathbb{R}[/itex]).
    There are various ways to make this space complete, and all these ways are useful in some way.
     
  13. Aug 11, 2012 #12

    HallsofIvy

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    Proof of what? You seem to think that these concepts of "volume" and "area" have some kind of existence beyond the basic definitions. That's not true (not even for two and three dimensions).
     
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