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It is well known that the volume of an $n$-ball of radius $R$ is

$$V_{n}(R)=\frac{\pi^{n/2}}{\Gamma(1+n/2)}R^n.$$

Graphically, the volume as a function of dimension looks like

enter image description here

We see how the volume increases and then decreases in the dimension. Differentiating gives the volume-maximizing dimension:

$$\frac{\partial}{\partial n} \log V_n(R)=0\implies n^*=2(\psi^{-1} (\log \pi R^2)-1) \qquad (1) $$

where $\psi^{-1}$ is the inverse digamma function (this answer rounded accordingly to restrict to integer dimensions).

Now let's consider each of the terms in $V_n(R)$. The $R^n$ term is fairly intuitive; it captures the effect of the size of the ball on its volume. The interesting terms are $\pi^{n/2}$ and $\Gamma(1+n/2)$, which are purely dimensional effects on the volume, the former being an increasing effect and the latter being a decreasing effect. As the dimension grows large, the decreasing effect wins.

To control for the size effect, take a unit ball ($R=1$). Then $(1)$ evaluates to $\approx 5.257,$ and we can verify dimension $n=5$ (among positive integers) maximizes the volume (seen in the plot above).

My questions are

  1. Is there a simple geometric intuition behind the increasing effect ($\pi^{n/2}$) and decreasing effect ($\Gamma(1+n/2)$) of dimension on volume?
  2. What intuitively makes dimension 5 "optimal" for the unit ball?

Preliminary thoughts thus far on question 1:

First, the notion of comparing volumes across dimensions may seem like an apples and oranges comparison, so it helps to keep in mind what we are really comparing. We are simply comparing the amount of unit hypercubes that make up the unit hypersphere.

Now there has already been discussion here on the intuition for the long-term decreasing effect of dimension on volume. The simplest explanation I found is to use the cube, this being our yardstick to measure volume after all. The smallest ball that inscribes a cube of side length $2k$ has radius $R=k\sqrt n.$ So a bigger sized ball is required to contain a cube of fixed size as the dimension increases. This suggests why we may expect a unit ball to be made up of a smaller amount of unit cubes as the dimension grows.

Nonetheless, this intuition seems incomplete as it does not explain the initial increasing effect of dimension on volume. I guess understanding the increasing effect can then help with question 2.

Golden_Ratio
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    To me, it's more natural to compare the volume of an $n$-ball of radius $r$ with the volume of an $n$-cube of radius $r$ (i.e., with the $n$-ball inscribed). That ratio is non-increasing, as One Might Naively Expect. <> Using cubes of side length $r$ arguably does compare apples to oranges: We're using a ball in each dimension, but subdividing the cube in a dimension-dependent way. The subdivision is hidden because each cubelet is a scaled copy to the whole, but that fact doesn't make the subdivision geometrically natural. – Andrew D. Hwang Mar 06 '22 at 14:25
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    As you said,$$V(B_{n,r})=\frac{\pi^{n/2}}{\Gamma(1+n/2)}r^n=:x_n r^n$$ tells us the volume of the $n$-ball with radius $r$ as a multiple of the volume of the $n$-cube with radius $r$, i.e. we can fit $x_n$ $n$-cubes inside an $n$-ball (independetly of $r$). You have plotted the graph of $$n\mapsto x_n$$ and you are surprised that $n=5$ is a maximum. I doubt that there is an "intuitive" explanation, but before we think about higher dimensions, let's think about the cases $n=1,2,3$ which we can visualize. Do you have an "intuitive" explanation for $$x_1=2<x_2=\pi<x_3=\frac{4}{3}\pi$$? – Filippo Mar 06 '22 at 15:15
  • Related: https://math.stackexchange.com/questions/2644700/whats-new-in-higher-dimensions/2644740#2644740 – Ethan Bolker Mar 06 '22 at 16:20
  • "We see how the volume increases and then decreases in the dimension" $;-;$ This is just a play on numbers, to which it's hard to assign a meaning or look for an intuition. Along the same line, I suppose you would say that the volume of the unit hypercube is "constant" across dimensions. But I don't see the intuition behind saying that the length of a unit segment in 1D is the same as the area of a unit square in 2D and the volume of a unit cube in 3D. – dxiv Mar 07 '22 at 03:40
  • @dxiv, why is it hard to assign meaning to the number of unit hypercubes that can fit into an object, which is what volume is? – Golden_Ratio Mar 07 '22 at 03:43
  • @Golden_Ratio Because those hypercubes have different dimensions in each case. Stating it in terms of a count of unit hypercubes does not change the fact that, in the end, you are comparing volumes across different dimensions. And, at the risk of repeating some of my previous comment, I don't see the point of a statement like "the volume of the unit hypercube is constant across dimensions". Not saying that the statement is technically wrong, just that I don't find it particularly meaningful or intuitive. – dxiv Mar 07 '22 at 03:52
  • @dxiv Stating volume as a count of unit hypercubes removes the units attached to volume and makes a fair comparison across dimensions. I agree that "the volume of the unit hypercube is constant across dimensions" is completely useless; that just tells us one hypercube is made up of one hypercube. The scenario I am looking at is not as trivial and has meaning. – Golden_Ratio Mar 07 '22 at 03:58
  • @dxiv Also if you read my whole post, I specifically addressed this notion of an apples and oranges comparison in anticipation of your reservation – Golden_Ratio Mar 07 '22 at 04:12
  • @Golden_Ratio Calling a hypervolume a count of unit hypervolumes, instead, does not change the fact that those unit hypercubes have different dimensions. Again, I am not arguing the numbers or the math behind. But the question is asking about "geometric intuition", and I find that to be elusive here. Maybe someone could tweak Archimedes' proof into an "intuition" that a circle of radius $6$ and a sphere of radius $3$ have the same hypervolume, though it's not obvious, and not obvious how/that it carries over to higher dimensions. – dxiv Mar 07 '22 at 04:30
  • @dxiv I think I finally understand what you mean and I think the problem is that the terminology is misleading: If we consider up to three dimensions, then we have different names for the objects in consideration: Line segments in 1D, squares and circles in 2D, cubes and spheres in 3D. As you said, one might argue that comparing the amount of squares that fit into a circle with the amount of cubes fitting into a sphere is an apples and oranges comparison. But talking about $n$-balls and $n$-cubes suggests that there is a relation, which as I said, might be misleading. – Filippo Mar 10 '22 at 07:38

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