On Sat, 20 Jun 2009 04:00:47 -0500, yogesh_gharote wrote:

>>Now hopefully I have some momentum: The four dimensional analog of a
>>sphere (commonly called a hypersphere) is a four dimensional figure
>>enclosed by the locus of all points that are a distance of r away from
>>the center point. It contains an amount of space (not a volume,
>>certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
>>volume of 2 pi^2 r^3.
>>
>>I've done the math in a way that seems obvious to me (integrate the
>>volume of the sphere that forms the surface of the hypersphere as one of
>
>>the dimensions varies from -r to r), and that pi^2 just _belongs_ there,
>
>>no matter how much it surprises me.
>
>
> Both the formulae for S.A. & volume for a 4D sphere are correct but if
> ur logic of integration is followed then the pi squared term cant be
> accounted for (if u integrate wrt only r ) .
> Generally the hypervolume & the hyper surface area of a N
> dimensional sphere( of which urs is a N=4 case) is not calculated by the
> method mentioned by u but the result rather arrives thru principle of
> mathematical induction & the result involves beta & gamma functions
> which is why u get to see the powers of pi (e.g. gamma(1/2) = sqrt(pi)
> and the general formula involves powers of gamma(1/2))
> Remarkably for a n-sphere as it is generally called , u get
> hyper
> surface area = derivative of hypervolume , again theory of beta n gamma
> functions is involved in this , so once u know the volume of a
> hypersphere , u differentiate it wrt r to get hypersurface area.
> The formula for a n-sphere hypervolume is
> Vn(r) = (gamma(1/2))^n * r^n/ gamma(1/2*n +1)
> &
> Sn(r) = d/dr(Vn(r))
>
> Note : The above discussion applies to N dimensional Euclidean
> space.

My choice of oddball integrals was intentional, as I want to go on to
calculating various moments for the probability distributions of the
surface of the hypersphere when 3D probability distributions are mapped
onto it. Clearly if I map a tight Gaussian distribution onto the
hypersphere with a standard deviation that's much smaller than the
hypersphere radius the resulting probability distribution will be easy;
it's figuring out what happens as that probability distribution opens up
that's making my brain cramp.


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