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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 '09 at 22:09
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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 '09 at 22:16
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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 '09 at 2:12
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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 '10 at 22:22
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@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 '10 at 14:20

15 Answers 15

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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(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 '09 at 0:23
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A very nice argument! –  Greg Kuperberg Dec 9 '09 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 '09 at 4:57
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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 '10 at 22:43

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.

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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.

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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 '09 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 '09 at 17:10
    
Right. Although so did Fedja. –  Greg Kuperberg Dec 10 '09 at 17:27

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 $[-1/\sqrt{n},1/\sqrt{n}]^n$.

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More precisely, the constant c is 2e. –  Michael Lugo Jan 20 '10 at 16:07
    
One can get the exact result with school calculus, so surely the challenge is to avoid using even that and it is a pity that you refer to "the unit ball of $\ell_1^n$". We could equivalently consider the principal $2^n$-ant of the ball, whose tangent hyperplane normal to the body diagonal has equation $\sum x_i=\sqrt{n}$. This meets each axis at distance $\sqrt{n}$ and has volume $n^{n/2}/n ! $. –  Paul Taylor Dec 7 '13 at 12:06

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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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 '09 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 '09 at 17:04

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.

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What does "$C^2$ isomorphic to a Hilbert space" mean? –  L Spice Dec 25 '10 at 17:25
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There is an isomorphism $T$ to a Hilbert space s.t. $\|T\|\cdot \|T^{-1}\| \le C^2$. –  Bill Johnson Dec 25 '10 at 18:51

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 '09 at 19:15

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 '09 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 '09 at 23:54
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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 '09 at 0:39

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.

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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.

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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.

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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.

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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.

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Permit me to suggest another "intuitive" approach, which hardly uses any calculations, just some basic Arithmetic (that means: Combinatorics)

Well, it is a fact that even for big numbers of n (dimension) the unit n-sphere inscribed in the unit n-cube is still the LARGEST possible. Its diameter is always equal to the "side length" of the cube, and the sphere's surface touches every "face" of the cube (as "Face" for an n-cube, is to be considered of course an (n-1)-cube).

So, why the "shrinkage" of the sphere for higher and higher dimensions? It is simple a matter of Arithmetic. The sphere always occupies the "central area"/center of the n-cube, but there is not much "center" left, as n goes to infinity. Most of the cube's volume "escapes" (centrifugally, sort of) towards the corners/"vertices". For example a 100-cube has 200 "faces" but $2^{100}$ vertices. Let's consider that we produce the "diameters" of the cube, namely all the straights that pass through the Center of the cube.

There are two main types, as far as "length" is concerned. The smallest, let's call them "good" or "short" which start from the center of a face, pass through center C of the cube and end on the center of the opposite face. For the usual 3-dimensional cube that prescribes a unit sphere (r=1), this length of the short diameter is 2 (since 2 is both the diameter of the sphere and the edge length of the cube). That shows that a "good/short" diameter is spreading entirely (its whole length) INSIDE the sphere. A "bad/long" one from the other side, are these diameters of course which start from a vertex---trough center C---to the opposite vertex. For an n-cube with side length=2 , the long "bad" diameter has length = $2\sqrt n$. Thus, for example for a 100-cube the long/bad diameter has length equal to 2*10=20. We notice that just 1/10th of this length lies in this case inside the sphere!

Moreover, there are 100 "good/short"/full" diameters, but $2^{99}$ long ones! ("empty/bad"). The above argument "by induction", may be perhaps not so mathematically strict, but it is logically very powerful, I think.

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