Question

Before you close for "homework problem", please note the tags.

Last week, I gave my calculus 1 class the assignment to calculate the $n$-volume of the $n$-ball. They had finished up talking about finding volume by integrating the area of the cross-sections. I asked them to calculate a formula for $4$ and $5$, and take the limit of the general formula to get 0.

Tomorrow I would like to give them a more geometric idea of why the volume goes to zero. Anyone have any ideas? :)

Comm wiki in case people want to add/modify this a bit.

Answer 1

Maybe the fact that most point of the sphere are very close to the equator (concentration of measure) gives some conceptual explanation.

Answer 2

For $r=\frac{1}{2}$, I have a geometric argument.

Let $I=[-\frac{1}{2},\frac{1}{2}]$.

Now $Vol(D^{n-1} \times I) = Vol(D^{n-1})$. Take an annulus $A$, say with outer radius $\frac{1}{2}$ and inner radius $0.9\frac{1}{2}$. Remove $A \times [0.9\frac{1}{2}, \frac{1}{2}]$ from the top of $D^{n-1} \times I$. This still contains $D^n$ and has volume at least 0.1% less than $Vol(D^{n-1} \times I) = Vol(D^{n-1})$. So $Vol(D^n) < 0.999 Vol(D^{n-1})$. Done.

Answer 3

By a strange coincidence I found myself thinking about almost this very question last week on a walk home early one morning (yes, that's correct). I wanted however only the weaker result that the ratio of the volume of the unit ball in the $l^2$ norm of $\mathbb{R}^n$ to that of the unit ball in the $l^{\infty}$ norm of $\mathbb{R}^n$ (i.e., $2^n$), goes to zero as $n \to \infty$. This is what I came up with:

Start with the volume of the 4-ball in $\mathbb{R}^4$. Notice that the region is entirely contained in that of the polydisk = $\{(x_1,x_2,x_3,x_4) | x_1^2 + x_2^2 \le 1, x_3^2 + x_4^2 \le 1\}$ and therefore the proportion of the volume of the 4-sphere to that of the unit ball in the $l^{\infty}$ norm is at most $(\pi/4)^2$. Repeating this argument shows that the corresponding proportion in $2n$ or $2n+1$ variables is at most $(\pi/4)^n$. This goes to 0 as $n \to \infty$.

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ADDENDUM: Having thought about it a little more, the above "polydisk" argument can be easily modified to answer the orginal question provided you exhibit one value of n for which the volume of the n-ball is less than one. The good news is that finding such an n is a straightforward calculus exercise involving repeated integration by parts. The bad news is that, if I've done the exercise correctly, the smallest such n is n = 13.

Answer 4

Edit: As Matthias pointed out, the following argument only works for the ball with radius 1/2.

To measure volume, we need to agree on a unit of volume [1]. The traditional way of doing this is to set the volume of the unit cube to one.

Now, think about the $n$-ball inscribed in the unit $n$-cube. As we increase $n$, the ball's diameter stays constant, but what happens to its volume? When $n = 1$, the ball takes up the whole unit cube, so its volume is one. When $n = 2$, the ball no longer takes up the whole unit cube, so its volume is less than one. When $n = 3$, the ball takes up even less of the unit cube, so its volume is even smaller.

There's an easy way to see that when you go from $\mathbb{R}^n$ to $\mathbb{R}^{n + 1}$, the fraction of the unit cube occupied by the inscribed ball goes down. Start with an $n$-ball inscribed in the unit $n$-cube, and extrude both objects into the $(n + 1)$st dimenion. Now you have an $(n + 1)$-cylinder inscribed in an $(n + 1)$-cube. The fraction of the $(n + 1)$-cube occupied by the $(n + 1)$-cylinder is clearly the same as the fraction of the $n$-cube occupied by the $n$-ball. It's easy to see, however, that the $(n + 1)$-ball inscribed in the $(n + 1)$-cube fits inside the inscribed $(n + 1)$-cylinder with room to spare.

This argument only shows that the volume of the unit-diameter $n$-ball decreases as $n$ grows; it doesn't show that the volume goes to zero. I'm hopeful, however, that a more sophisticated version of the same argument might do the trick!

Edit: A more sophisticated version of the same argument does do the trick, and Matthias posted it while I was writing my post! Hooray!

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[1] To be more sophisticated about it: the differential n-form in $\mathbb{R}^n$ is only unique up to multiplication by a constant, so we need to settle on a constant.

Answer 5

Let $f(n)$ be the n-volume of the n-sphere. Then the natural thing to ask about is not $f(n)$, but $f(n)/2^n$; this is the ratio of the volume of the n-sphere to the volume of the n-cube in which it is inscribed. It's natural that this should be very small by a concentration of measure argument. The "typical" distance of a point in the n-cube $[-1,1]^n$ from the origin is a constant times $n$. More rigorously, pick a point uniformly at random from the $n$-cube; the square of its distance from the origin is $X_1^2 + X_2^2 + \ldots + X_n^2$, where $X_i$ is the $i$th coordinate, a uniform[-1,1] random variable. Thus $X_i^2$ has mean 1/3 and variance 4/45 (*), so the squared distance of a random point from the origin is roughly normally distributed with mean $n/3$ and variance $4n/45$. But points in the sphere are just those which have distance at most 1 from the origin, and these are quite rare.

(*) I'm not actually sure of this "4/45". In any case, it's a positive constant.

Answer 6

Here's a geometric argument (still with a bit of calculus). The volume of the unit $n+1$-ball may be obtained from integrating the volumes of $n$-ball cross-sections from say south pole to north pole. We have $Vol(D^{n+1}) = Vol(D^{n}) \int_{-1}^1 \sqrt{1-z^2}^{n} dz$, since the volume of the $n$-ball of radius $r$ is the volume of the unit $n$-ball times $r^n$, and the radius of the $z$-cross-section is $\sqrt{1-z^2}$. Since for any $1 >\delta >0$, $(1-z^2)^{n/2}$ converges to $0$ uniformly on $[-1,-\delta] \cup [\delta, 1]$, it is not hard to see that these integrals converge to zero.

So the ratio $Vol(D^{n+1})/Vol(D^n)$ converges to zero, and therefore $Vol(D^n)\to 0$ as $n\to \infty$. As Gil Kalai says, this argument shows that the volume gets concentrated near the equator.

Answer 7

A variation on some of the previous arguments that gives some intuition without actually doing any calculation.

Consider $B_n$, the ball in $R^n$, and $C_n$, the cube $[-\frac{1}{2}, \frac{1}{2}]^n$. We make the following observations.

1. $C_n$ has volume $1$.

2. A typical point in $C_n$ will have about half its coordinates larger than $\frac{1}{4}$ in absolute value, so will be outside of $B_n$. In other words, almost none of the volume of $C_n$ is contained in $B_n$.

3. A typical point in $B_n$ will have no coordinates larger than $\frac{1}{2}$, since the sum of the squares of the coordinates is $1$ and this sum has to be divided among $n$ coordinates. (This is a weak version of the concentration of measure mentioned by Gil Kalai, and may be intuitively palatable).

Looking at these, we see that in going from $C_n$ to $B_n$ we start with a volume of $1$, throw almost all of it away, and add almost nothing back in.

Answer 8

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 $0$.

Answer 9

A calculus-free proof that the volume $V_n$ of the unit $n$-sphere goes to 0 faster than any exponential. Equivalently, the volume $r^nV_n$ of the sphere of radius $r$ goes to 0 for every $r$. It is inspired by the intuitive answers about concentration of measure.

Claim. For any $0 < h < 1$, $$V_n \le 2h V_{n-1} + (1-h^2)^{n/2} V_n.$$

Proof. Remove a slab from the middle of the $n$-ball with thickness $2h$, and bring together the remaining slices to make a lens shape. The radius of the equator of this lens is $\sqrt{1-h^2}$, and it clearly fits inside of an $n$-ball of that radius.

Proof of main result. Rearrange the claim as a volume relation between adjacent dimensions: $$V_n \le \frac{2h}{1-(1-h^2)^{n/2}} V_{n-1}.$$ For every $h$, the factor on the right is eventually close to $2h$, qed. In particular, if we take $h = 1/3$, then by the time that $n \ge 19$, the volume has turned around and is decreasing.

Answer 10

A sphere packing argument (and some kissing number construction), because having smaller n-balls is equivalent to be able to pack more of them in the unity n-cube.

edit: This needed to use r=1/2, not unity n-balls. Fixing the associated values. This hurts the usefulness of the argument, but still give some geometric insight on how the n-ball fills the n-cube and the shape of the space between them.

First, we take n=4 and show how to put two n-balls $B_n$ having a radius of 1/2 into the unity n-cube $C_n$: we place the first at the center of the n-cube, and use this as origin. Then we place the second at (1/2,1/2,1/2,1/2) and wrap it into the n-cube. These two $B_n$ are now into $C_n$ (even if one can be considered as split in parts) and are disjoints because the distance between them is $\sqrt{4(1/2)^2} = 1$.

This proves that the volume $V(B_4) < 1/2$.

Now, when n=4k, notice that we can place k other $B_n$ plus the one at center by using the same type of translation as before. For the first, we only set the 4 first coordinates as 1/2 and the rest as 0. For the second, only the 4 next ones, etc... All these additional $B_n$ are at distance 1 from the centered one, and at distance $\sqrt{2}>1$ from each other, making them all disjoint. Thus, we have

$V(B_{4k}) < 1/(k+1)$,

which goes to 0 when n=4k increases. Of course, when n=4k+p, the same trick still work, but only with k+1 n-balls, which is not a problem.

Answer 11

There is a simple argument by comparing to the unit ball of $\ell_1^n$.

Let $K$ be the unit ball of $\ell_1^n$, i.e. the set of points with sum of coordinates (in absolute value) bounded by $1$. Then $K$ is the disjoint union of $2^n$ simplices (one per octant), and each simplex has volume $1/n!$.

Now the Euclidean unit ball is contained in $\sqrt{n}K$, so its volume is at most $n^{n/2}2^n/n!$. This tends to $0$ and behaves like $(c/\sqrt{n})^n$ for some constant $c$.

The value is sharp up to the value of $c$, as shown by the dual argument : the unit ball contains the cube $[-1/\sqrt{n},1/\sqrt{n}]^n$.

Answer 12

Given an $n$ dimensional closed bounded convex symmetric body $E$, situate it in $R^n$ so that the Euclidean ball $B$ is the ellipsoid of maximal volume contained in $E$. In 1978 Szarek, building on work of Kashin, showed (much more than) that if $({{vol(E)}\over{vol(B)}})^{1/n}\le C$, then the Banach space that has $E$ for its unit ball contains a subspace of dimension $n/2$ which is $C^2$ isomorphic to a Hilbert space. However, it is easy to see that $\ell_\infty^n$ contains a subspace well isomorphic to Hilbert spaces only of dimension of order $\log n$.

This is the most complicated answer I can think of to the original question but does show why someone might care about computing volume ratios.

Answer 13

Replacing the characteristic function of the unit ball by a suitable normal distribution with spherical symmetry when computing the volume should give approximatively the correct answer. Since $$\frac{1}{(2\pi)^{n/2}}\int_{\mathbb R^n}(x_1^2+\dots+x_n^2)e^{-(x_1^2+\dots+x_n^2)/2}dx_1\cdots dx_n$$ is linear in $n$, one has to rescale by a factor of order $\frac{1}{\sqrt{n}}$ leading to a decay of order $(\lambda n)^{-n/2}$ for the volume of the unit sphere.

Answer 14

I've come a bit late to this particular party, but here's another argument. This one includes most of the sphere in a suitable cone.

Choose a small positive number x, to be optimized later. Then the volume of that part of the sphere with $0\leq x_n\leq x$ is at most $2^n x$ (the volume of the cube being $2^n$).

Now consider the plane $x_n=x$. This intersects the sphere in an (n-1)-dimensional subsphere. Let C be the smallest cone that contains everything in the sphere that lies above this plane. A simple argument using similar triangles shows that the height of this cone is at most 1/x. Therefore, its volume is at most $2^{n-1}/nx$.

Doubling all this to get both halves of the sphere, we get an upper bond of $2^n(2x+1/nx)$, and taking $x=n^{-1/2}$ we get an upper bound for the ratio of $3n^{-1/2}$.

Of course, this is a weak bound, but I was trying to make the argument as simple as possible. (It's simpler in my head than I've managed to make it written down.)

Apologies if this duplicates someone else's argument -- I checked, but could have missed something.