Obviously the volume of a unit hypercube is always $1$.

I know that the volume of a unit hypersphere is given by: $$ \dfrac{\pi^\frac{d}{2}}{\Gamma\Big(\dfrac{d}{2} + 1\Big)} $$ which rises sharply until about $d=5$, then falls off a cliff.

Why should it be this way? I can't get my head around why volume should increase with dimension only to a point.

  • $\begingroup$ Tag edits or suggestions very much appreciated. $\endgroup$ – OJFord Nov 1 '16 at 1:00
  • $\begingroup$ It is more meaningful to look at the volume of the hypersphere inscribed in the unit hypercube. $\endgroup$ – dxiv Nov 1 '16 at 1:24

This question has a thorough answer here

For your convenience, Let me quote top answers. ( I advise you to read the full discussion on MathOverflow).

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 $0$.

Another one says,

The reason is because the length of the diagonal cube goes to infinity.

The cube in some sense does exactly what we expect. If it's side lengths are $1$, it will have the same volume in any dimension. So lets take a cube centered at the origin with side lengths $r$. Then what is the smallest sphere which contains this cube? It would need to have radius $r\sqrt{d}$, so that radius of the sphere required goes to infinity.

Perhaps this gives some intuition.

Hope this helps.

  • $\begingroup$ Thanks for that, the second quote in particular was exactly what I needed. :) $\endgroup$ – OJFord Nov 1 '16 at 15:25

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