# n-Ball Volume and surface with $n \rightarrow \infty$

The volume of the n-ball is given by $V_n(r) = \frac{\pi^{n/2}}{\Gamma (\frac n 2 + 1)}r^n$ and its surface area is $S_n(r) = \frac{\pi^{n/2}}{\Gamma (\frac n 2 + 1)}r^{n-1}n$. So far so good. Now there is this statement:

Because the volume is dependent exponentially from n, the majority of points within the ball are very close to the surface. In fact, for $n \rightarrow \infty$, the point sets 'surface' and 'volume' become identical.

Is there any way to explain this statement more precisely? My point is: Taking the fraction $\lim_{n \rightarrow \infty} \frac V S = \lim_{n \rightarrow \infty} \frac r n = 0$. Thus I would assume that the surface becomes larger than the volume.

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I think that "larger" is not the correct term, because $V_n(r)$ and $S_n(r)$ are two values that cannot be directly compared (e.g. you cannot compare square meters with cubic meter). But you can say that fixed a radious, $S_n(r)$ grows faster than $V_n(r)$ – Ikki Nov 28 '13 at 17:02
I know about that problem. But after all, both $V$ and $S$ are pointsets $V_n(r) = \{x \in \mathbb R^n| |x| \leq r \}$ (and $S_n$ likewise). How would one compare them otherwise? Anyway: The original problem is still why the volume becomes the surface (to state it a bit clumsy) – disaster Nov 28 '13 at 17:06
I think that what is meant is that if $X_n$ is a point drawn uniformly from the ball $\Bbb B_n(0,1)$, then it will be arbitrarily close to the sphere $\Bbb S_n(0,1)$ with overwhelming probability as $n\to+\infty$. More precisely, $\Pr\{d(X_n,\Bbb S_n(0,1)) \leq \varepsilon\} = 1-\varepsilon^n$. – Siméon Nov 28 '13 at 17:17
What exactly does the quotient $V/S$ tell you? You should realise that it is a length. But what does it mean? – Fly by Night Nov 28 '13 at 17:28
It's the length in the remaining dimension between the $n-1$ dimensional surface and the $n$ dimensional volume, ie., the 'difference' so to speak in between the two in the new dimension, isn't it? – disaster Nov 29 '13 at 8:29

Yes, this is what happens in high dimensions. But one does not need the formulas with $\Gamma$-function to see this. Consider a ball $B_1$ of radius $1$, and inside of it draw a concentric ball $B_r$ of radius $r<1$. Then the fraction of volume of $B_1$ contained in $B_r$ is simply $r^n$. And this is how the function $r^n$ looks for $n=1,2,4,8,\dots,512$:
As you can see, in 64 dimensions there is practically nothing in $B_{0.9}$, in terms of volume. In 128 dimensions there is almost nothing in $B_{0.95}$.
In fact, for $n\to\infty$, the point sets 'surface' and 'volume' become identical.
I don't like this statement at all. In whatever sense we take $n\to\infty$, the Euclidean norm ball ought to have some convexity, which its sphere does not. (I understand the idea of imagining, e.g., the $L^1$ ball as something nonconvex, but here we consider balls for an inner product metric).