## What’s a nice argument that shows the volume of the unit n-ball in R^n approaches 0?

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.

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Maybe "Can we prove without calculus that the volume of the unit n-sphere approaches 0 as n goes to infinity?" – Reid Barton Dec 8 2009 at 22:09
I'm pretty sure that we want the volume of the unit n-ball, since blah blah sphere is the boundary. – Harry Gindi Dec 8 2009 at 22:16
I would like to thank everyone who responded/revised/commented on this thread! You came up with very beautiful arguments, and quickly enough for me to put something together for tomorrow morning. – B. Bischof Dec 9 2009 at 2:12
What the heck kind of calculus 1 class is this?!? I didn't even think about this problem until vector calculus i.e. calculus 3! SIGH.I want my money back.......... – Andrew L Jun 9 2010 at 22:22
@Andrew All the machinery to do the cross-section limit proof is there in calc one automatically. I tend to deviate a little from the standard material on a fairly regular basis for my own entertainment. It is unclear the usefulness of these deviations. – B. Bischof Jun 10 2010 at 14:20

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

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Related to this point, compare the diameter of the unit n-ball and the diagonal of the unit n-cube. – Qiaochu Yuan Dec 8 2009 at 22:32
It is better to say that most of points in a ball lie near hyperplane comming through its center. – Anton Petrunin Dec 8 2009 at 23:54
And so the volume of the n-ball will be a small number times the volume of the n−1-ball pluss something negligible, and we expect exponential decay as n grows. This seems to be the intuitive content of some other answers, including Greg Kuperberg's and Agol's. – Harald Hanche-Olsen Dec 9 2009 at 0:39

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.

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But the volume of the unit ball is increasing from n=2 to n=5 which would seem to contradict your last line – alex Dec 9 2009 at 8:50
Not if the radius is 1/2. I interpreted the original question as about the case r=1. – Reid Barton Dec 9 2009 at 17:04

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 $n \to \infty$.

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.

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 can you please enclose your formulas using $signs? – psihodelia Dec 9 2009 at 19:15 Edit: As Matthias pointed out, the following argument only works for the ball with radius 1/2. To measure volume, we need to agree on a unit of volume [1]. The traditional way of doing this is to set the volume of the unit cube to one. Now, think about the$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$grows; it doesn't show that the volume goes to zero. I'm hopeful, however, that a more sophisticated version of the same argument might do the trick! Edit: A more sophisticated version of the same argument does do the trick, and Matthias posted it while I was writing my post! Hooray! [1] To be more sophisticated about it: the differential n-form in$\mathbb{R}^n$is only unique up to multiplication by a constant, so we need to settle on a constant. - Let$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. - 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. - 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. - 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$0$. - (128/n)^{-n/4} actually tends to infinity. I think you should change all your "-n/4" to "n/4" to fix it up. – Jason DeVito Dec 9 2009 at 0:23 A very nice argument! – Greg Kuperberg Dec 9 2009 at 2:33 Unfortunately, I do not see the last part of your argument, perhaps I am being dense. Could you explain how you conclude the volume is bounded by (128/n)^(n/4)? – B. Bischof Dec 9 2009 at 4:57 BTW, the software now makes it look like it was my argument. But it wasn't; it was contributed by fedya. – Greg Kuperberg Jun 9 2010 at 22:43 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. -  Because I like optimizing constants: h = 1/3 is not the best possible choice. However, for h = 1/3 we get that 2h/(1-(1-h^2)^(d/2)) < 1 for d > 18.654. The best possible d is around 18.295, which we get for h near 0.375. (No, not h = 3/8.) But no choice of h brings this critical d as low as 18. – Michael Lugo Dec 9 2009 at 13:57 In other words, you showed that the statement is true for n-spheres of any fixed radius, not just radius 1. – Reid Barton Dec 10 2009 at 17:10 Right. Although so did Fedja. – Greg Kuperberg Dec 10 2009 at 17:27 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. - 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$[-1/\sqrt{n},1/\sqrt{n}]^n$. - More precisely, the constant c is 2e. – Michael Lugo Jan 20 2010 at 16:07 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. - What does "$C^2$isomorphic to a Hilbert space" mean? – L Spice Dec 25 2010 at 17:25 There is an isomorphism$T$to a Hilbert space s.t. $\|T\|\cdot \|T^{-1}\| \le C^2$. – Bill Johnson Dec 25 2010 at 18:51 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. - 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.

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