Volumes of n-balls: what is so special about n=5? - MathOverflow

Volumes of n-balls: what is so special about n=5?

2011-01-24T20:27:02Z

I am reposting this question from math.stackexchange where it has not yet generated an answer I had been looking for.

The volume of an $n$-dimensional ball of radius $R$ is given by the classical formula $$V_n(R)=\frac{\pi^{n/2}R^n}{\Gamma(n/2+1)}.$$ It is not difficult to see that the "dimensionless" ratio $V_n(R)/R^n$ attains its maximal value when $n=5$.
The "dimensionless" ratio $S_n(R)/R^n$ where $S_n(R)$ is the $n$-dimensional volume of an $n$-sphere attains its maximum when $n=7$.

Question. Is there a purely geometric explanation of why the maximal values in each case are attained at these particular values of the dimension?

[EDIT. Thanks to all for the answers and comments.]

Answer by Marty for Volumes of n-balls: what is so special about n=5?

2011-01-24T21:18:49Z

In my opinion, nothing is special about $n = 5$.

The "dimensionless ratio" $V_n(R) / R^n$ is the ratio of the volume of the $n$-ball of radius $1$ to the volume of the $n$-cube of side-length $1$. So this is maximized at $n =5$, but bluntly, so what?

More interesting geometrically might be the equally dimensionless ratio $V_n(R) / (2R)^n$, which is the ratio of the volume of the $n$-ball to the volume of the smallest $n$-cube containing it. This is monotonic decreasing (for $n \geq 1$), showing that balls decrease in volume relative to their smallest cubical container, as the dimension increases. This has more geometric content, since there is a simple geometric relationship between the sphere and cube here.

One could consider many similar problems, involving inscribing a cube inside a sphere (instead of the other way around), or using an orthoplex or polycylinder or other figure instead of a cube. All of these have some geometric content, and are expressed as a sequence of dimensionless ratios.

Answer by Igor Rivin for Volumes of n-balls: what is so special about n=5?

2011-01-24T21:25:17Z

A very celebrated result on a related subject is the following, which was a very major advance in the Busemann-Petty problem (don't worry, the math review has all you need to know). EDIT by popular demand, the review can be seen here: http://dl.dropbox.com/u/5188175/BallReview.pdf

@incollection {MR950983,

AUTHOR = {Ball, Keith},

 TITLE = {Some remarks on the geometry of convex sets},

BOOKTITLE = {Geometric aspects of functional analysis (1986/87)},

SERIES = {Lecture Notes in Math.},

VOLUME = {1317},

 PAGES = {224--231},

PUBLISHER = {Springer},

ADDRESS = {Berlin},

  YEAR = {1988},

MRCLASS = {52A40},

MRNUMBER = {950983 (89h:52009)},

MRREVIEWER = {G. D. Chakerian},

   DOI = {10.1007/BFb0081743},

   URL = {http://dx.doi.org/10.1007/BFb0081743},

}

Answer by Bill Thurston for Volumes of n-balls: what is so special about n=5?

2011-01-24T21:27:24Z

There are many "dimensionless" ratios: choosing $R$ as the linear measurement is arbitrary. For instance, the ratio of the volume of the sphere to the volume of the circumscribed cube has a maximum at $n=1$. The ratio to the volume of the inscribed cube never attains a maximum. There are intermediate geometrically-related "midscribed" cubes, where all faces of some dimension are tangent to the unit sphere. Here is the graph for the ratio of volumes when the codimension 2 faces of a cube are tangent. It attains the maximum for $n = 12$ (just barely more than for $n=11$). There are many other reasonable dimensionless comparisons, for instance comparing to a simplex, etc. etc. Since the Gamma function grows super-exponentially, these simple geometric variations tend to shift the maximum --- there's nothing special about 5 or 7.

Answer by I. J. Kennedy for Volumes of n-balls: what is so special about n=5?

2011-10-22T17:22:40Z

Brian Hayes has very nice article about the volume of the n-ball in the current issue of American Scientist (Nov 2011). In particular, he discusses the surprising fact that the maximum volume occurs at $n=5$.