8
$\begingroup$

Why does the volume of an $n$-dimensional cube approach $\infty$ as $n \to \infty$, but the volume of an $n$-dimensional sphere approach $0$ as $n \to \infty$?

To be more precise,

$$\frac{V_n(\text{Sphere})}{V_n (\text{Cube})} \to 0, \,\text{as}\; n \to \infty$$

I understand how this is evident using the expression for the volume of an $n$-dimensional sphere and cube.

Volume of an $n$-dim sphere (radius = $1$) $$V_n=\frac{\pi^{n/2}}{\Gamma(n/2+1)}$$

Volume of an $n$-dim cube (side length = $2$) $$V_n = 2^n$$

I am looking for a geometric explanation. In what way is a n-dim sphere different from a n-dim cube? Are there more general families which exhibit these properties?

Edit: For the sake of keeping the problem relatively simple, I am referring to a sphere of unit radius and a cube with side length equal to the diameter of the sphere.

$\endgroup$
  • 1
    $\begingroup$ What size are your cubes and balls? The size of an $n$ dimensional cube with side $a$ is $a^n$, which tends to zero, one or infinity depending on $a$. $\endgroup$ – Joonas Ilmavirta Aug 11 '14 at 19:32
  • 1
    $\begingroup$ Ahh sorry about that. I am asking about a sphere of unit radius and a cube of side equal to the diameter of the sphere. $\endgroup$ – saq7 Aug 11 '14 at 19:38
  • $\begingroup$ "the volume of an n-dimensional cube approach infinity..:" Well, the volume of a cube of edge length 10cm tends to infinity, buy the volume of a cube of edge length 0.1m tends to zero. Uh, wait... math.stackexchange.com/questions/67039/… math.stackexchange.com/questions/15656/… $\endgroup$ – leonbloy Aug 11 '14 at 19:49
  • 1
    $\begingroup$ "I'm looking for a geometric explanation". Geometrically, it makes absolutely no sense to compare the volume of cubes (or spheres) of different dimensions. If you don't believe this, tell me: which is larger, a cube of 1cm$^3$ or a circle of 1m$^2$? $\endgroup$ – leonbloy Aug 11 '14 at 19:53
  • 2
    $\begingroup$ @saq7 When you say, in your question, that some volume "tends to infity" or "approaches zero" when increasing the dimension, you are comparing them. That's wrong. The only way of making sense of this is to compare the (dimensionless) ratios, as does Joonas in his answer - or in the linked ones. $\endgroup$ – leonbloy Aug 11 '14 at 20:03
12
$\begingroup$

In very high dimensional space, the opposite corners of a cube are very far from one another. The unit cube is a huge object as compared to the unit sphere, where every point is 2 units or less from every other point.

$\endgroup$
5
$\begingroup$

A first remark: For a cube of side $a$ in $n$ dimensions, the volume is $a^n$. Thus the limit depends on whether $a\in(0,1)$, $a=1$ or $a>1$.

You are comparing the volumes $B_n$ of the ball of radius one and $C_n$ of the cube of side two. The reason why $C_n/B_n\to\infty$ as $n\to\infty$ is that, loosely speaking, much of the volume of the cube is near its corners which the ball doesn't reach, and this phenomenon becomes stronger and stronger as dimension grows. In other words, a ball of radius one cannot go very far in every direction at once, but the cube of side two can.

$\endgroup$
  • $\begingroup$ Thanks for the answer! I tried and convinced myself of the fact that much of the volume of a cube is near its corners. Since the ratio of the distance from the centre of a hypercube to its corner to the perpendicular distance from the centre of the hypercube to its side, approaches infinity as n->infinity. Does this imply that if you had a n-dim pentagon it would be less 'cornery' than a cube but more so than a sphere? Are there rules that govern this? What should I search for? $\endgroup$ – saq7 Aug 11 '14 at 20:06

protected by Community Mar 22 at 18:17

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.