LOG#235. Hyperballs.

Posted on by October 13, 2019

Hi, everyone. The saddest thing about job (working with teenagers and other people) is that it delays other stuff, like blogging! So, you should be patient about getting more of “my stuff”.

What is going on today? Hyperballs. Or hyperspheres and some cool variations. I have written about hyperspheres here before…Even provided formulae for volume (and area) from the. You know, from higher dimensions, as you can get hyperparalletope or cross-polytope with hypervolume V_n=\prod_i X_i, for the known sphere S^2 you get V=\dfrac{4}{3}\pi R^3 and the more general formula for the nd hypersphere or usual hyperball volume

(1)   \begin{equation*}V_n=\dfrac{\Gamma(1/2)^n}{\Gamma(\dfrac{n}{2}+1)}R^n\end{equation*}

where \Gamma(1/2)=\sqrt{\pi} and \Gamma(n)=(n-1)!. The hyperarea can be obtained from the recurrence in nd from lowering dimensions and a useful derivative gadget tool:

(2)   \begin{equation*}A_{n-1}=\dfrac{dV_n}{dR}=\dfrac{n\pi^{n/2}R^{n-1}}{\Gamma(n/2+1)}\end{equation*}

For any 3d ellipsoid you also can derive that, given a,b,c: V=\dfrac{4}{3}\pi abc, and similar formulae in higher dimensional hyperellipsoids, following the pattern V(HE)=V_D(1)\prod R_i, where V_D(1) is the hypervolume of the unit hypersphere and R_i are the hyperellipsoid (HY) semiaxes.

By the other hand, you do not need to keep things so simple, you can even change the norm in \mathbb{R}. Thus, having a vector (x_1,\cdots,x_n) in L_p with norm

    \[\vert x\vert_p=\left(\sum_{i=1}^n\vert x_i\vert^p\right)^{1/p}\]

then the so-called p-normed hyperball volume in nd follows:

    \[V_n(p)=\dfrac{\left(2\Gamma(\dfrac{1}{p}+1)R\right)^n}{\Gamma(\dfrac{n}{p}+1)}\]

In particular, you get V^1_n=\dfrac{2^n}{n!}R^n y V^\infty_n=(2R)^n, and those match the expressions for the cross-polytope and the n-cube. Other possible generalization is the next one. For any real positive numbers you can even define the balls:

    \[B_ {p_1, \ldots, p_n} = \left\{ x = (x_1, \ldots, x_n) \in \mathbf{R}^n : \vert x_1 \vert^{p_1} + \cdots + \vert x_n \vert^{p_n} \le 1 \right\}\]

Since Dirichlet times, mathematicians know the general formula for these hyperballs/hyperspheres:

    \[ V(B_{p_1, \ldots, p_n}) = 2^n \frac{\Gamma\left(1 + \frac{1}{p_1}\right) \cdots \Gamma\left(1 + \frac{1}{p_n}\right)}{\Gamma\left(1 + \frac{1}{p_1} + \cdots + \frac{1}{p_n}\right)}\]

Enough balls today? Not yet! I wish! I am showing you in a moment why calculus rocks. And not a usual calculus indeed only. Fractional calculus is a variation from common calculus where you can get non-integer derivatives, even irrational, complex or more complicated definitions! Before that, let me remember you as caution that \Gamma(\nu)=(\nu-1)!. And know define the Riemann-Liouville operator (fractional derivative):

(3)   \begin{equation*} D^{-\nu}f\equiv \dfrac{1}{\Gamma(\nu)}\int_0^\sigma \left(\sigma-y\right)^{\nu-1}f(y)dy\end{equation*}

Take now f(y)=1. Wow. Then,

(4)   \begin{equation*} D^{-\nu}(1)\equiv \dfrac{1}{\Gamma(\nu)}\int_0^\sigma \left(\sigma-y\right)^{\nu-1}dy=\dfrac{\sigma^\nu}{\Gamma(1+\nu)}\end{equation*}

and then, you obtain the partial result

    \[D^{-\nu}(1)=\dfrac{\sigma^\nu}{\Gamma(1+\nu)}\]

Now, insert \sqrt{\pi^N} with \nu=N/2 and \sigma=R^2, then

    \[\sqrt{\pi^N} D^{-N/2}(1)=\dfrac{\sqrt{\pi^N}}{\Gamma\left(\dfrac{N}{2}+1\right)}\left(\sqrt{\sigma}\right)^N\]

so, you finally deduce that

(5)   \begin{equation*}\boxed{V_N(R=\sqrt{\sigma})=\dfrac{\Gamma(1/2)^NR^N}{\Gamma(N/2+1)}=\Gamma\left(\dfrac{1}{2}\right)^ND^{-N/2}(1)=\left(-\dfrac{1}{2}\right)!^ND^{-N/2}(1)}\end{equation*}

or equivalently

(6)   \begin{equation*}\boxed{V_N=\dfrac{\Gamma(1/2)^NR^N}{\Gamma(N/2+1)}=\Gamma\left(\dfrac{1}{2}\right)^ND^{-N/2}(1)=\dfrac{D^{-N/2}(1)}{\pi^{-N/2}}}\end{equation*}

The fractional recurrence

    \[V_N(\sigma)=\left(\dfrac{1}{\pi}\dfrac{\partial}{\partial \sigma}\right)^{-1/2} V_{N-1}(\sigma)\]

with \sigma=R^2 holds and note that the general Riemann-Liouville fractional derivative

    \[_a D^{-\nu}_x f(x)\equiv \dfrac{1}{\Gamma(\nu)}\int_a^x \left(x-y\right)^{\nu-1}f(y)dy\]

has gaps or poles, in principle, at values \nu=0,-1,-2,\ldots since \Gamma functions have singularities at negative integers, including zero.

See you in another wonderful blog post!

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