## Some questions about the volume of the n-ball.

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- Hackfleischkannibale
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### Some questions about the volume of the n-ball.

We just had the formula for the volume of the n-dimensional ball of radius r, which is [imath]B_{2n}(r)=\frac{r^{2n}}{n!}\cdot\pi^n[/imath] for even numbers and [imath]B_{2n+1}(r)=\frac{r^{2n+1}2^{n+1}}{(2n+1)!!}\cdot\pi^n[/imath] for odd ones (with !! denoting the double factorial, i.e. the product of all odd numbers up to that point). And this is just confusing.

Why does the power of pi increase by one when you go up from an odd n to an even one, but not from even to odd? And what's the intuition for the fact that [imath]lim_{n\rightarrow\infty}B_n(r)=0[/imath]? Even our prof admitted to being confused by that.

Also, the factor by which you have to multiply the formula if you want to increase the dimension by one looks rather weird and is different for even and odd n, as hinted at above, but the factor for increasing the dimension by two, from n to n+2, say, is the surprisingly simple [imath]\frac{2\pi r^2}{n+2}[/imath], as if that was the natural thing to do.

Why does the power of pi increase by one when you go up from an odd n to an even one, but not from even to odd? And what's the intuition for the fact that [imath]lim_{n\rightarrow\infty}B_n(r)=0[/imath]? Even our prof admitted to being confused by that.

Also, the factor by which you have to multiply the formula if you want to increase the dimension by one looks rather weird and is different for even and odd n, as hinted at above, but the factor for increasing the dimension by two, from n to n+2, say, is the surprisingly simple [imath]\frac{2\pi r^2}{n+2}[/imath], as if that was the natural thing to do.

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- jestingrabbit
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### Re: Some questions about the volume of the n-ball.

Regarding the parity issue, its relatively straightforward if we accept a simple assumption, namely that [imath]B_n(r) = C_n r^n.[/imath]

Imagine an n-ball, and slice it into parallel sheets. The n-volume of the ball is the integral of the (n-1)-volume of each of those sheets, and each sheet is an (n-1)-ball. Putting this into an equation,

[math]B_n(r) = \int_{-r}^r B_{n-1} \left( \sqrt{r^2 - t^2} \right)\ dt[/math]

which you can prove with algebra too, but I'm content with a handwave and a gesture to lower dimensional cases.

Now, if n is odd, the square root will disappear when we substitute [imath]C_{n-1} r^{n-1}[/imath], but if n is even we'll still have something under the square root to worry about, and getting out from under it will introduce another pi into C_n.

As for why the volume goes to zero as r increases, I can only suggest that for large n, most of the volume of an n-cube is in the corners. That this is counter intuitive is a lesson: your intuition is useless in ridiculously high dimensions.

Imagine an n-ball, and slice it into parallel sheets. The n-volume of the ball is the integral of the (n-1)-volume of each of those sheets, and each sheet is an (n-1)-ball. Putting this into an equation,

[math]B_n(r) = \int_{-r}^r B_{n-1} \left( \sqrt{r^2 - t^2} \right)\ dt[/math]

which you can prove with algebra too, but I'm content with a handwave and a gesture to lower dimensional cases.

Now, if n is odd, the square root will disappear when we substitute [imath]C_{n-1} r^{n-1}[/imath], but if n is even we'll still have something under the square root to worry about, and getting out from under it will introduce another pi into C_n.

As for why the volume goes to zero as r increases, I can only suggest that for large n, most of the volume of an n-cube is in the corners. That this is counter intuitive is a lesson: your intuition is useless in ridiculously high dimensions.

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### Re: Some questions about the volume of the n-ball.

jestingrabbit wrote:As for why the volume goes to zero as r increases, I can only suggest that for large n, most of the volume of an n-cube is in the corners. That this is counter intuitive is a lesson: your intuition is useless in ridiculously high dimensions.

Seems perfectly intuitive to me. The long diagonals of a unit n-cube have length 2√n. As n grows without bound, the unit n-cube becomes arbitrarily large along the diagonals. But the longest chord in a unit n-ball has length 2, as it’s a diameter. Sure, that doesn’t prove the volume fraction gets small (maybe the corners of the cube get really narrow as n grows—they don’t, but this argument doesn’t prove it) but it certainly appeals to my intuition.

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### Re: Some questions about the volume of the n-ball.

Hackfleischkannibale wrote:And what's the intuition for the fact that [imath]lim_{n\rightarrow\infty}B_n(r)=0[/imath]? Even our prof admitted to being confused by that.

Note, other people: Hack here was asking about the fact that the volume decreases to 0, not that the ratio of the n-ball and the n-cube approaches 0. The latter is pretty easy to understand; the former is a little more confusing.

To fix your intuition, recall that the volume of an n-ball is in n-dimensional units, too. It only shrinks in the higher-dimension units. An n-ball has a higher surface area (in units

^{n-1}) than than (n-1) ball has volume, but it's n-volume is smaller (in units

^{n}).

Thinking of it in terms of taking an n-cube and "shaving off" the bits that aren't in the n-ball doesn't help our intuition much, because there's no good intuitive connection between an n-cube and an (n-1)-ball. However, there *is* such a connection if we instead use an n-cylinder. The volume of an n-cylinder is simply the volume (in units

^{n-1}) of an (n-1)-ball times the height. If we're talking about a unit (n-1)-ball with radius 1, then the appropriate n-cylinder has height 2, and thus is twice the volume (in units

^{n}) as the (n-1)-ball (in units

^{n-1}). We can then shave off the parts of the n-cylinder that lie outside the n-ball. If we shave off more than we keep, then the factor-of-2 increase in volume is negated, and the n-ball is smaller than the (n-1)-ball. By a similar argument to the n-cube one, as you increase the dimension, more and more of the n-cylinder's volume is contained in its "edges", which are the parts you shave off. The tipping point happens to be at n=6.

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

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### Re: Some questions about the volume of the n-ball.

But they amount to exactly the same thing, since n-volume is defined in terms of the size of the n-cube...Xanthir wrote:Hack here was asking about the fact that the volume decreases to 0, not that the ratio of the n-ball and the n-cube approaches 0. The latter is pretty easy to understand; the former is a little more confusing.

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### Re: Some questions about the volume of the n-ball.

Yes, of course; I was merely arguing that's it's not as useful from an intuition perspective.

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

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### Re: Some questions about the volume of the n-ball.

But if you can intuitively grasp why it takes up an ever-smaller portion of the n-cube, and then simply remember that size relative to the n-cube is how n-volume is defined, I don't see how the intuitiveness of the two facts is any different.

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### Re: Some questions about the volume of the n-ball.

gmalivuk wrote:But if you can intuitively grasp why it takes up an ever-smaller portion of the n-cube, and then simply remember that size relative to the n-cube is how n-volume is defined, I don't see how the intuitiveness of the two facts is any different.

I think we can all intuitively accept that an n-dimensional ball inscribed in a cube is takes up a smaller and smaller fraction of the cube. But here we're talking about a cube of side length 1 relative to a sphere of radius 1, so it is much less obvious that this ratio should go to zero. In fact, our intuition from small dimensional cases says the opposite, since for small dimensions, the measure of the n-ball goes up as n goes from 1 to 2 to 3 (and even to 4 and 5).

The fact that the volume of the n-sphere goes to zero is equivalent to the ratio of an n-sphere inscribed in an n-cube going to zero faster than 2

^{-n}, which I don't find at all intuitive.

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### Re: Some questions about the volume of the n-ball.

I like the Xanthir's cylinder formulation. Another way to phrase it is that if you take an n-sphere and deform it into a cylinder of the same radius, the height of the cylinder goes to 0 as n goes to infinity. This at least seems plausible, but it doesn't directly involve comparing an n-sphere to an n-cube.

The key point, I think, is that the inscribed cube gives you a much better approximation for the sphere's volume than the circumscribing cube. If I didn't make any mistakes, the side length of the cube inscribed in a unit sphere is 2/n

For instance, is there an easy way to see that a sphere circumscribing a cube has less volume than the cube with three times the side length?

The key point, I think, is that the inscribed cube gives you a much better approximation for the sphere's volume than the circumscribing cube. If I didn't make any mistakes, the side length of the cube inscribed in a unit sphere is 2/n

^{1/2}, and the side length of the cube that has the same volume as the sphere is (2 e pi)^{1/2}/ n^{1/2}(1 + o(1)). There may be a purely geometric way to see that these are roughly proportional.For instance, is there an easy way to see that a sphere circumscribing a cube has less volume than the cube with three times the side length?

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- Hackfleischkannibale
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### Re: Some questions about the volume of the n-ball.

Another idea: What's easy to see is that the volume of the n-ball of the [imath]||\cdot||_1[/imath]-norm goes to 0, because its volume is [imath]\frac{(2r)^n}{n!}[/imath] (which I think is rather easy to see), and since (I think) the volume of the 2-norm-n-ball behaves roughly like the square root of the 1-norm-n-ball, there might be a way to the original problem, too.

Error- correcting edit: The volume of the 1-norm-n-ball/Cross-Polytope doesn't behave like the square root of the n-ball, but the double factorial behaves roughly like square root of the factorial, and that's the important part.

Error- correcting edit: The volume of the 1-norm-n-ball/Cross-Polytope doesn't behave like the square root of the n-ball, but the double factorial behaves roughly like square root of the factorial, and that's the important part.

Last edited by Hackfleischkannibale on Fri Dec 09, 2011 2:04 pm UTC, edited 2 times in total.

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### Re: Some questions about the volume of the n-ball.

That answers my question. Consider a unit cube, inscribed in a sphere, which is inscribed in a cross-polytope. The cube has volume 1, the cross-polytope has volume n

The intuition is, then, if an infinity-ball (cube) is inscribed in a 1-ball (cross-polytope), their volumes are pretty close, and hence so is the volume of the 2-ball (ball) which can be stuck between them. This is the only such arrangement where the volumes are close.

This suggests that for any p > q, the p-ball inscribed in the q-ball and the p-ball with the same volume as the q-ball have roughly the same radius, and in fact that as n goes to infinity, the ratio of these radii approaches a fixed constant. If so, what is it?

For p = infinity, q = 1, it is e.

For p = infinity, q = 2, it is (e pi / 2)

For p = infinity, q = infinity, it is 1.

It also suggests this more philosophical point: if you want to compare p-balls and q-balls, "the right place" to look is at the uniform vector (1,1,1, ..., 1), where the p-ball is tangent to the circumscribing q-ball when p > q, and not at the concentrated vector (1,0,0,...,0), where they are tangent when p < q. Vectors that look uniform or near it are more common than vectors that are concentrated.

^{n}/n!, which is at most e^n. That is, the cube that has the same volume as the sphere has side length at most e times that of the cube inscribed in the sphere.The intuition is, then, if an infinity-ball (cube) is inscribed in a 1-ball (cross-polytope), their volumes are pretty close, and hence so is the volume of the 2-ball (ball) which can be stuck between them. This is the only such arrangement where the volumes are close.

This suggests that for any p > q, the p-ball inscribed in the q-ball and the p-ball with the same volume as the q-ball have roughly the same radius, and in fact that as n goes to infinity, the ratio of these radii approaches a fixed constant. If so, what is it?

For p = infinity, q = 1, it is e.

For p = infinity, q = 2, it is (e pi / 2)

^{1/2}.For p = infinity, q = infinity, it is 1.

It also suggests this more philosophical point: if you want to compare p-balls and q-balls, "the right place" to look is at the uniform vector (1,1,1, ..., 1), where the p-ball is tangent to the circumscribing q-ball when p > q, and not at the concentrated vector (1,0,0,...,0), where they are tangent when p < q. Vectors that look uniform or near it are more common than vectors that are concentrated.

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### Re: Some questions about the volume of the n-ball.

Ah, right. I was picturing the sphere of diameter 1, inscribed within said unit cube.skeptical scientist wrote:But here we're talking about a cube of side length 1 relative to a sphere of radius 1

### Re: Some questions about the volume of the n-ball.

gmalivuk wrote:Ah, right. I was picturing the sphere of diameter 1, inscribed within said unit cube.skeptical scientist wrote:But here we're talking about a cube of side length 1 relative to a sphere of radius 1

But why wouldn't we just phrase it that way? That makes it much more obvious that the n-volume drops to 0. The n-volume of a unit n-hypercube is always 1, the n-hypersphere always fits in the n-hypercube, and the sphere must make cuts with every additional dimension to remain a n dimensional sphere as opposed to a cylinder.

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### Re: Some questions about the volume of the n-ball.

WarDaft wrote:The n-volume of a unit n-hypercube is always 1, the n-hypersphere always fits in the n-hypercube, and the sphere must make cuts with every additional dimension to remain a n dimensional sphere as opposed to a cylinder.

But the n-sphere has radius 1, not diameter 1, and so doesn't fit inside the unit hypercube.

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### Re: Some questions about the volume of the n-ball.

Right. As skeptical scientist pointed out, an n-cube that fits around a unit n-sphere has a volume of 2

^{n}. Which means that the ratio of sphere volume to cube volume could go to zero as long as the sphere's volume grows more slowly than 2^{n}. But in reality it doesn't simply grow more slowly than that, it actually shrinks. Meaning that the portion of the cube an inscribed sphere takes up goes to zero faster than 2^{-n}, which isn't terribly intuitive.### Re: Some questions about the volume of the n-ball.

Try doing it with just a unit cube and a unit sphere, so the sphere is bigger than the cube in the axis directions.

Pick a random direction to go in the sphere, and go that way until you exit the cube. For large n, 'most' of the time you'll leave the sphere really really early in your journey. So what if in a few we only make it half way out? (Eg take directions where all components are equal, negatives, or 0, then you only exit the cube if all but 4 or less are 0)

So you have these piddly little 4d bulges outside of the unit cube, and you take up next to no space inside the cube itself.

Pick a random direction to go in the sphere, and go that way until you exit the cube. For large n, 'most' of the time you'll leave the sphere really really early in your journey. So what if in a few we only make it half way out? (Eg take directions where all components are equal, negatives, or 0, then you only exit the cube if all but 4 or less are 0)

So you have these piddly little 4d bulges outside of the unit cube, and you take up next to no space inside the cube itself.

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### Re: Some questions about the volume of the n-ball.

quokka wrote:But the n-sphere has radius 1, not diameter 1, and so doesn't fit inside the unit hypercube.

A unit n-sphere yes, but the original statement was every radius goes to 0 volume zero. Radius 0.5 makes it by far the most readily apparent... and that's what we were going for, isn't it?

It decidedly does not segue into a proof, but it doesn't have to, it's for intuition.

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### Re: Some questions about the volume of the n-ball.

I see no reason off hand why something true for one radius should be true for all, for example it's trivial that the '0.5-unit' cubes go to 0 volume, but of course the unit cube does not. It does happen to be true for spheres, but I don't know of an intuitive way to see that without addressing what skep's concern was, that it goes to 0 faster than an exponent of the dimension.

Last edited by mike-l on Sat Dec 10, 2011 2:30 am UTC, edited 1 time in total.

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### Re: Some questions about the volume of the n-ball.

I feel like all these handwavy versions are kinda weak guys. There's a much more direct approach. Its called algebra.

So again, let [imath]B_n(r) = C_n r^n.[/imath] We need a decent formula for C_n that isn't "presolved" like Hack put in the OP ie a formula for the volume of the unit ball that is a lot more intuitively built. So, as before, we have

[math]\begin{align*}

C_n = B_n(1) &= \int B_{n-1}\left(\sqrt{1 - t^2}\right)\ dt \\

&= C_{n-1} \int_{-1}^1 (1 - t^2)^{(n-1)/2}\ dt\\

&= \prod_{i=1}^n \int_{-1}^1 (1 - t^2)^{(i-1)/2}\ dt\\

&= \prod_{i=1}^n \int_{-\pi/2}^{\pi/2} \cos^i(\theta)\ d\theta\\

&= \prod_{i=1}^n \int_{0}^{\pi} \sin^i(\theta)\ d\theta\end{align*}[/math]

which is a pretty nice place to stop. So, we're taking the positive segment of a period of sin or cos, taking successively higher powers and integrating. Now, perhaps its not obvious why the terms in that product are less any epsilon after some n, but they're at least something that you can start to think about properly instead of being all "I will flail my arms around in the direction of the results and something magical will occur as a result".

Incidentally, that particular formula for C_n is a good place to realise that there isn't a great deal of numeric difference in changing from odd to even cases. In fact, the Wallis product is a good thing to read about here.

This is a great question. More thoughts later.

@mke-I: I had a concern?

So again, let [imath]B_n(r) = C_n r^n.[/imath] We need a decent formula for C_n that isn't "presolved" like Hack put in the OP ie a formula for the volume of the unit ball that is a lot more intuitively built. So, as before, we have

[math]\begin{align*}

C_n = B_n(1) &= \int B_{n-1}\left(\sqrt{1 - t^2}\right)\ dt \\

&= C_{n-1} \int_{-1}^1 (1 - t^2)^{(n-1)/2}\ dt\\

&= \prod_{i=1}^n \int_{-1}^1 (1 - t^2)^{(i-1)/2}\ dt\\

&= \prod_{i=1}^n \int_{-\pi/2}^{\pi/2} \cos^i(\theta)\ d\theta\\

&= \prod_{i=1}^n \int_{0}^{\pi} \sin^i(\theta)\ d\theta\end{align*}[/math]

which is a pretty nice place to stop. So, we're taking the positive segment of a period of sin or cos, taking successively higher powers and integrating. Now, perhaps its not obvious why the terms in that product are less any epsilon after some n, but they're at least something that you can start to think about properly instead of being all "I will flail my arms around in the direction of the results and something magical will occur as a result".

Incidentally, that particular formula for C_n is a good place to realise that there isn't a great deal of numeric difference in changing from odd to even cases. In fact, the Wallis product is a good thing to read about here.

antonfire wrote:This suggests that for any p > q, the p-ball inscribed in the q-ball and the p-ball with the same volume as the q-ball have roughly the same radius, and in fact that as n goes to infinity, the ratio of these radii approaches a fixed constant. If so, what is it?

This is a great question. More thoughts later.

@mke-I: I had a concern?

Last edited by jestingrabbit on Sat Dec 10, 2011 2:31 am UTC, edited 1 time in total.

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### Re: Some questions about the volume of the n-ball.

Sorry, Skeptical's concerns, not JR.

Assuming your integrals are correct, it shouldn't be hard to show that the integral of sufficiently high powers of sin is less than 1 (or epsilon for that matter, the curve approaches a pointmass at pi/2), so it's clear from that that the volume must go to 0. It's also interesting that the half integer values of the gamma function can be expressed as a product of integrals of sin, that's neat.

I, at least, though, don't get any intuition from that particular algebra. Of course, you may be right that intuition is worthless in high dimensions, but something along the lines of looking at the various diagonals to get a notion of size satiates my intuition.

If you wanted this geometric approach to actually do something instead of hand waving, I'd suggest seeing if you can put a lower bound on what portion of the volume lies within an inscribed cube, and since the inscribed cube must go to size 0, so must the whole thing.

Assuming your integrals are correct, it shouldn't be hard to show that the integral of sufficiently high powers of sin is less than 1 (or epsilon for that matter, the curve approaches a pointmass at pi/2), so it's clear from that that the volume must go to 0. It's also interesting that the half integer values of the gamma function can be expressed as a product of integrals of sin, that's neat.

I, at least, though, don't get any intuition from that particular algebra. Of course, you may be right that intuition is worthless in high dimensions, but something along the lines of looking at the various diagonals to get a notion of size satiates my intuition.

If you wanted this geometric approach to actually do something instead of hand waving, I'd suggest seeing if you can put a lower bound on what portion of the volume lies within an inscribed cube, and since the inscribed cube must go to size 0, so must the whole thing.

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### Re: Some questions about the volume of the n-ball.

That's... pretty much what I did above?mike-l wrote:If you wanted this geometric approach to actually do something instead of hand waving, I'd suggest seeing if you can put a lower bound on what portion of the volume lies within an inscribed cube, and since the inscribed cube must go to size 0, so must the whole thing.

The cube inscribed in a unit sphere has side length 2nantonfire wrote:The cube that has the same volume as the sphere has side length at most e times that of the cube inscribed in the sphere.

^{-1/2}, so if we fix s, when n > (2e/s)

^{2}, the unit sphere is smaller in volume than the cube of side length s.

Last edited by antonfire on Sat Dec 10, 2011 3:59 am UTC, edited 2 times in total.

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### Re: Some questions about the volume of the n-ball.

Hrm, I somehow didn't see your posts

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### Re: Some questions about the volume of the n-ball.

jestingrabbit wrote:I feel like all these handwavy versions are kinda weak guys. There's a much more direct approach. Its called algebra.

So again, let [imath]B_n(r) = C_n r^n.[/imath] We need a decent formula for C_n that isn't "presolved" like Hack put in the OP ie a formula for the volume of the unit ball that is a lot more intuitively built. So, as before, we have

[math]\begin{align*}

C_n = B_n(1) = \prod_{i=1}^n \int_{0}^{\pi} \sin^i(\theta)\ d\theta\end{align*}[/math]

which is a pretty nice place to stop. So, we're taking the positive segment of a period of sin or cos, taking successively higher powers and integrating. Now, perhaps its not obvious why the terms in that product are less any epsilon after some n, but they're at least something that you can start to think about properly instead of being all "I will flail my arms around in the direction of the results and something magical will occur as a result".

Yes, this does look helpful. In particular, this looks somewhat similar to the volume of the n-cross-polytope:

[math]\beta_n(1) = \prod_{i=1}^n \int_{-1}^1|x|^i dx[/math]

Products (or sums) make it most visible what happens in the next dimension, and I thinks it's rather obvious that [imath]lim_{n\rightarrow \infty}\int_0^{\pi}sin^n(x) dx = 0[/imath], so this is a good intuition to use.

Also I think the terms needn't actually approach 0, just any number smaller than 1/2, and that's even easier to see.

And I second an interest in the question raised by antonfire, even though I guess there won't be nice formulas (formulae?) for most of the ratios. Oh, one could still be interesting: what's the ratio of the volumes of the n-ball to the circumscribing n-cross-polytope?

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### Re: Some questions about the volume of the n-ball.

That one follows from the ones we already have; it's (2e/pi)

^{-1/2}. Be careful though, this isn't the (approximate) ratio of the n-ball to the cross-polytope it's inscribed in; it's the n'th root of that.-
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### Re: Some questions about the volume of the n-ball.

Here's a fun fact, first pointed out to me in a column by Martin Gardner: if you sum the volumes of all the unit balls in spaces of even dimension, you get [math]e^\pi[/math] I think that's kind of a beautiful variation on the Euler formula for exponentiating imaginary numbers.

### Re: Some questions about the volume of the n-ball.

Well, I finally got around to doing this computation and the result is pretty nice, so here we go.

First, normalize the p-norm so that (1,1,...,1) has norm 1. This corresponds to making it so that (||x||

In particular, for p = infinity, 2, and 1, we have s = 0, 1/2, and 1 respectively, and plugging that into this expression tells us that p-balls of unit volume have diameters 1, (e pi / 2)

This can be worked out with the usual recursive formula for volumes of p-balls in n dimensions. I suspect there's a cleaner argument of some sort, maybe a probabilistic one, but I don't know one.

First, normalize the p-norm so that (1,1,...,1) has norm 1. This corresponds to making it so that (||x||

_{p})^{p}is the mean of the numbers |x_{1}|^{p}, ..., |x_{n}|^{p}rather than their sum. This leaves the infinity-norm unchanged. Then with respect to this normalization, in large dimensions the p-ball of unit volume has diameter approximately e^{-s}s^{s}/ s!, where s = 1/p. (So with respect to the usual normalization it has diameter approximately n^{s}e^{-s}s^{s}/ s!.)In particular, for p = infinity, 2, and 1, we have s = 0, 1/2, and 1 respectively, and plugging that into this expression tells us that p-balls of unit volume have diameters 1, (e pi / 2)

^{-1/2}, and e^{-1}with respect to this normalization, as before.This can be worked out with the usual recursive formula for volumes of p-balls in n dimensions. I suspect there's a cleaner argument of some sort, maybe a probabilistic one, but I don't know one.

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