The volume of a $d$ dimensional hypersphere of radius $r$ is given by:
$$V(r,d)=\frac{(\pi r^2)^{d/2}}{\Gamma\left(\frac{d}{2}+1\right)}$$
What intrigues me about this, is that $V\to 0$ as $d\to\infty$ for any fixed $r$. How can this be? For fixed $r$, I would have thought adding a dimension would make the volume bigger, but apparently it does not. Anyone got a good explanation?
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Let $X_j$ be a sequence of independent random variables with uniform distribution in the interval $[-r,r]$ (i.e. you are picking an infinite sequence of random numbers from the interval). The probability that $(X_1, \ldots, X_d)$ lies in your hypersphere, i.e. that $R_d = X_1^2 + \ldots + X_d^2 \le r^2$, is $V(r,d)/(2r)^d$ (which by scaling doesn't depend on $r$, so let's call it $P(d)$). Of course $P(d) \to 0$ as $d \to \infty$, but the point is that it goes to 0 faster than an exponential in $d$. Indeed, by the theory of large deviations, for any $t > 0$ we should have $P(R_d \leq t d) \approx e^{-d I(t)}$ as $d \to \infty$ for some function $I(t)$, where $I(t) \to \infty$ as $t \to 0+$.
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The reason is because the length of the diagonal cube goes to infinity.
The cube in some sense does exactly what we expect. If it's side lengths are $1$, it will have the same volume in any dimension. So lets take a cube centered at the origin with side lengths $r$. Then what is the smallest sphere which contains this cube? It would need to have radius $r\sqrt{d}$, so that radius of the sphere required goes to infinity.
Perhaps this gives some intuition.
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This was thoroughly discussed on MO. I quote from the top answer:
The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is of size $\frac{1}{\sqrt{n}}\ll 1$. This can be turned into a simple geometric argument (as suggested by fedja) using the fact that an $n$-element set has $2^n$ subsets:
At least $n/2$ of the coordinates of a point in the unit ball are at most $\sqrt{\frac{2}{n}}$ in absolute value, and the rest are at most $1$ in absolute value. Thus, the unit ball can be covered by at most $2^n$ bricks (right-angled parallelepipeds) of volume $$\left(2\sqrt{\frac{2}{n}}\right)^{n/2},$$ each corresponding to a subset for the small coordinates. Hence, the volume of the unit ball is at most $$2^n \cdot \left(2\sqrt{\frac{2}{n}}\right)^{n/2} = \left(\frac{128}{n}\right)^{n/4}\rightarrow0.$$ In fact, the argument shows that the volume of the unit ball decreases faster than any exponential, so the volume of the ball of any fixed radius also goes to $0$.
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Consider the distance from the center of the sphere of radius $R$ in $\mathbb{R}^n$ to the corner of its enclosing cube: $R\sqrt{n}$. This is like covering the surface of the sphere with a layer of tall corners.
For $x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$, define $$ Q(x)=x\frac{|x|}{\max_i|x_i|}\tag{1} $$ $Q$ maps a sphere to its enclosing cube. As described above, $\frac{|Q(x)|}{|x|}$ reaches a maximum of $\sqrt{n}$ in the corners. To be exact, $$ \frac{|Q(x)|^2}{|x|^2}=\sum_j\frac{|x_j|^2}{\max_i|x_i|^2}\tag{2} $$ By considering each $x_i$ to be normally distributed and using homogeneity of $(2)$, we get that the mean of $\frac{|Q(x)|^2}{|x|^2}$ over the sphere is $\frac{n+2}{3}$. Thus, the rms average of $\frac{|Q(x)|}{|x|}$ is $\sqrt{\frac{n+2}{3}}$. In this way, the volume of the cube should be approximated by $\sqrt{\frac{n+2}{3}}^{\;n}$ times the volume of the sphere. This grows faster than any $R^n$.
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Imagine you have n circles fitting perfectly in a square. Pick a point at random inside each of these squares. What's the probability that the randomly chosen point is inside each of the circles? It doesn't matter. As long as the area of the target (a circle) is less than the area of the square the probability of hitting all the targets will approach 0. This is (roughly) analogous to taking a 2n dimensional sphere and projecting it onto n planes. A point in the 2n sphere will fit into all n of the projected circles. This shows that the volume of the n-sphere relative to the box that contains it approaches 0.
Forget volume or area or anything with a physical interpretation. You are picking n positive numbers according to some probability distribution (in our case its distance from a fixed point). The larger n is the less likely their sum is to be less than any fixed constant.
I suppose you could say that adding a dimension "makes the volume bigger" for the hypersphere, but it does so even more for the unit you measure the volume with, namely the unit cube. So the numerical value of the volume does go towards zero.
Really, of course, it is apples to oranges because volumes of different dimensions are not commensurable -- it makes no sense to compare the area of the unit disk with the volume of the unit sphere.
All we can say is that in higher dimensions, a hypersphere is a successively worse approximation to a hypercube (of side length twice the radius). They coincide in dimension one, and it goes downward from there.