The unit ball in n dimensions under the Lp norm has volume
I ran across this formula via A nice formula for the volume of an L_p ball. That post gives an even more general result that allows different values of p along each axis.
There have been several blog posts lately on the volume of balls in higher dimensions that correspond to the case p = 2. The formula above is valid for all p > 0.
Note that as p goes to ∞ the volume goes to 2n because the terms involving gamma functions go to 1. This is as we’d expect since the unit “ball” in the infinity norm is a cube, two units wide on each side.
Related post: Means and inequalities
For all p>0, not just p>=1? Cool!
Yes, p can be less than 1, leading to balls that are not convex.
Understanding the” shape” of a ball using an arbitrary norm-induced distance is central to many fields: approximation, theoretical CS, statistics, probability (esp. concentration of measure, with implication with random matrices and CLT), geometric and functional analysis. Most importantly, it’s an elementary topic (in terms of methods, not results) and revises our intuition, a bit like probability. The survey of Keith Ball is a great place to start (http://www.msri.org/publications/books/Book31/files/ball.pdf); the focus there is more on order of magnitude than on constant. Highly recommended.
There are lots of beautiful results in this area: sections of balls (or of symmetric convex bodies), “widths” of balls (in the Gelfand or Kolmogorov sense), etc. They are waiting for an elementary exposition.
Also note how the volume goes to zero while n->∞, due to faster growing of gamma function when compared to 2^n (also recall curse dimensionality).
Dear John,
thank you for your kindness to share this valuable information.
I need to calculate the volume of a space which defined as follow:
x1^p+….+xn^p<r 0<r<1.
I would be grateful if you help me.
thank you in advance
Regards,
Asghar
One of the pathological aspects of volume you can see in the n-dimensional ball case is that as n increases the volume gets pushed out to the boundary. Calculate the volume of a unit ball and subtract the volume of one with radius 1-epsilon. I.e., the volume of an epsilon-thin shell. They become equal as n increases. As a consequence if you pick a point inside the unit ball via a uniform-by-volume probability (a radial vector), then as n gets large all the selected radial vectors all have length 1 with variance 0. Thus you inject certainty into the problem that wasn’t there to start with.