A # ball (mathematical)

Mathematicians, unlike the rest of the human race, draw a sharp distinction between a sphere and a ball. A sphere (in mathematics) is only a surface, whereas a ball is everything inside, and possibly including, that surface – the filling of the sphere. An open ball consists of all the points that are less than a given distance (the radius) away from a given point (the center); a closed ball, consists of all the points that are less than or equal to the radius.

Mathematical balls can also exist in any number of dimensions. A one-dimensional ball of radius r is just a line segment. It consists of all the points on a line between -r and r, or, in the case of a one-dimensional unit ball (a ball with a radius of 1), between -1 and 1. A 1-d unit ball thus has a length, or "1-d volume," of 2. A 2-d unit ball, which is the filling of a unit circle, has an area, or 2-d volume, of π. The volume of a unit ball in 3-d is 4/3 π. In 4-d it is π2/2. Apparently, as the number of dimensions increases, so does the volume of the unit ball. What does this volume tend to as the dimension tends to infinity? Intuitively, it might seem that in higher and higher dimensions there's more and more "room" in the unit ball, allowing its volume to become larger and larger. Does the volume become infinite, or does it approach a sufficiently large constant as the dimension increases? The answer is surprising and shows how our intuition is often misleading. Using a technique called multivariable calculus the volume of the unit ball in n dimensions, V(n), can be shown to be πn/2 / Γ(n/2 + 1), where Γ is the gamma function that generalizes the factorial function (i.e., Γ(z + 1) = z!). For n even, say n = 2k, the volume of the unit ball is thus given by V(n) = πk / k!.