Re: [Phys-L] playing for keeps

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding:

> Good point, John M. (Are half of the people on this list named
> John?) So here's a more complete summary.
>
> Bounding square/circle:
>  perimeter 8r/2pi*r = 4/pi = 1.27
>  area 4r^2/pi*r^2 = 4/pi = 1.27 (odd that the ratios are the
>  same)
>
> Bounding cube/sphere
>  area 24r^2/4*pi*r^2 = 6/pi = 1.91
>  volume  8r^3/(4/3)pi*r^3 = 6/pi = 1.91
>
> Seems like it ought to be easy to see why these pairs work....??
>
> Bruce
>
>
> On Sat, Jun 29, 2013 at 6:09 PM, John Mallinckrodt
> <ajm@csupomona.edu>wrote:
>
> > Bruce,
> >
> > You left out the area of the bounding cube/area of sphere which
> > also equals 6/pi.  Bet we can find a "reason" for those two
> > apparent coincidences!
> >
> > John Mallinckrodt
> > Cal Poly Pomona

The generalization of this coincidence works on all dimensions.  The ratio of the d-dimensional hypervolume (d-volume) of the smallest hypercube (d-cube) containing a d-dimensional ball to the d-volume of the contained ball is 

((d/2)!)*(2/sqrt([pi])^d .

And the ratio of the total (d-1)-volume of the 2*d 'faces' of the (d-1)-cubes forming the boundary of the d-cube to the (d-1)-volume of the (d-1)-sphere that is the boundary of the d-ball is also

((d/2)!)*(2/sqrt([pi])^d .

The reason this coincidence works is as follows:

Suppose we work out the ratio of the interior d-volumes of the objects and then get the
formula for that ratio to be ((d/2)!)*(2/sqrt([pi])^d .  (We can prove this using some calculus.)

Next, we go down one dimension and look at the corresponding ratio of (d-1)-volumes of the *boundaries* of these objects  To go down from the d-volume (= L^d) of the d-cube to its boundary containing 2*d separate (d-1)-cubes gives a total (d-1)-volume of 2*d*L^(d-1).  This means the drop of one dimension changes the measure by a factor of 2*d/L.

OTOH, if we consider a d-ball of radius r its d-volume is (([pi]^(d/2))/(d/2)!)*r^ d .  To get the (d-1)-volume of its bounding (d-1)-sphere we take the d-ball's d-volume and differentiate it WRT r.  Because the formula for the d-volume is proportional to r^d its derivative is d/r times what it was before differentiating it.  Now the smallest bounding d-cube has L = 2*r  This means the (d-1)-volume of the (d-1)-sphere is 2*d/L times the d-volume of the d-ball.

Thus, going down in dimension from d to d-1 also changes the ball's hypervolume by the *same* ratio as going down one dimension does for the bounding d-cube.  Since going down in dimension changes both the d-ball's and the d-cube's measure by the same factor this means the ratio of the "cube" value to the "ball" value remains the same for the d-1 dimensional boundaries as it is for the d-dimensional interior measures.

Note: the value of (d/2)! when d is an odd number can be found using the gamma function to provide the analytic continuation of the factorial function to fractional arguments values.  In particular, when d is a positive odd number we have

(d/2)! = (d!!)*sqrt([pi]/2^(d+1))

where the double factorial function for odd positive integers is defined by

d!! = 1*3*5*7*...*d .

David Bowman