Does the expression $$\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n,$$ which gives the volume of an $n$-dimensional ball of radius $R$ when $n$ is a nonnegative integer, have any known significance when $n$ is an odd negative integer?
$\begingroup$
$\endgroup$
7
asked Mar 4, 2020 at 1:12
-
1$\begingroup$ You should explain the context why you are asking this question to make it more relevant. $\endgroup$– Piotr HajlaszCommented Mar 4, 2020 at 14:22
-
12$\begingroup$ Pure curiosity. $\endgroup$– James ProppCommented Mar 4, 2020 at 15:28
-
1$\begingroup$ It can occur in the dimensional regularization of Feynman diagram integrals in quantum field theory, but I don't think it has too much significance there. $\endgroup$– gmvhCommented Mar 6, 2020 at 10:12
-
1$\begingroup$ Here is a funny formula: with $\nu !=\Gamma (\nu+1)$, we have $$ \sum_{\nu\in \mathbb N\cup (\mathbb N+\frac12)}\frac{π^\nu}{\nu !} R^{2\nu}=e^{π R^2} +\sum_{k\in \mathbb N}\frac{π^{k+\frac12}}{(k+\frac12)!} R^{2k+1} =\sum_{\nu\in \mathbb N\cup (\mathbb N+\frac12)}\vert \mathbb B^{2\nu}\vert R^{2\nu} =\sum_{n\in \mathbb N} \vert \mathbb B^{n}\vert R^{n}. $$ $\endgroup$– BazinCommented Mar 6, 2020 at 20:46
-
1$\begingroup$ Maybe en.wikipedia.org/wiki/Reflection_formula is relevant here? $\endgroup$– Sam HopkinsCommented Mar 7, 2020 at 2:43
|
Show 2 more comments
2 Answers
$\begingroup$
$\endgroup$
7
I will answer regarding the dimension $-1$.
An example of such space is a set of periodic lattices on a real line.
Indeed, you can see that the Hausdorff dimension of a periodic lattice is $-1$: If we scale down the lattice twice, it would be able to include two original lattices. Since scaling the fractal down twice makes it twice as big, its dimension is $\frac{\ln 2}{\ln (1/2)}=-1$.
The formula for the volume of a ball gives $\frac1{\pi R}$ for $n=-1$. This means the unit ball ($R=1$) is the $\pi$-periodic one.
If we reduce the radius (the step of the lattice divided by $\pi$), its volume increases: a lattice with step $1/2$ can be represented as two lattices of step $1$. Thus the volume of a lattice of step $1/2$ consists of two $1$-periodic lattices, and so has volume twice the volume of $1$-periodic lattice.
So, the volume of the $-1$-ball is the density of the lattice.
The ball is also a solid cube with side equal to the step of the lattice. Thus a unit cube is a lattice with step 1, its volume is 1. Thus, in dimensions 0,1 and -1 a ball is also a cube.
A ball with radius $1$ is a cube with side $\pi$ and a cube with side $1$ is a ball with radius $\frac1\pi$.
Interesting observation: in positive-dimentional space the ball becomes a point with reduction of its radius to zero. In zero-dimensional Euclidean space the ball is always a point, disregarding the "radius" and in this $-1$-dimensional space the ball becomes a zero-volume point when we increase the radius infinitely.
You can find some further ideas here.
-
1$\begingroup$ This proposed value of the “volume” of the (-1)-ball does not agree with the formula given in the original post. $\endgroup$ Commented Apr 19, 2020 at 16:31
-
$\begingroup$ @JamesPropp oh yes, i missed a mistake $\endgroup$– AnixxCommented Apr 19, 2020 at 16:33
-
2$\begingroup$ I do not see why “the unit sphere is a pi-periodic lattice” (and I doubt that it is a “known fact”). Also, Anixx’s style of answer is unlikely to discriminate between even and odd (negative) dimensionalities, so I think it’s on the wrong track. $\endgroup$ Commented Apr 19, 2020 at 17:03
-
$\begingroup$ @JamesPropp This answer is specifically about the dimension -1, and the generalization to other dimensions may require efforts. Particularly, because in dimension -1 the periodic lattice is both a sphere and a cube. $\endgroup$– AnixxCommented Apr 19, 2020 at 17:10
-
$\begingroup$ @JamesPropp And, in general, when we are talking about negative dimensions, the bodies in them are littices, spectrums and other (infinite but countable) collections of dots, but their volumes are determined based on their overall density (or the infinite part of numerocity if you prefer), not on the number of dots they include. $\endgroup$– AnixxCommented Apr 19, 2020 at 17:25
$\begingroup$
$\endgroup$
I want to make another answer to this question, although it is is not really an answer, and I do not know whether it is relevant or not, but it may be useful.
Let's take
$$V(n,R)=\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n$$
Also, using the reflection formula for the Zeta function we can see:
$$n\zeta(1-n)\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}=(1-n)\zeta(n)\frac{\pi^{\frac{1-n}{2}}}{\Gamma(\frac{1-n}{2}+1)}$$
Interestingly enough, in the algebra of divergent integrals described here, there is a rule $\operatorname{reg}\omega_+^n=-n\zeta(1-n)$, (where $\omega_+=\int_{-1/2}^\infty dx$ is an infinite divergent integral and $\operatorname{reg}$ denotes regularization), the overall formula becomes
$$\operatorname{reg} V(n, \omega_+)=\operatorname{reg} V(1-n, \omega_+)$$
This is a very nice-looking relation but its deep meaning is unclear. Particularly because it is not evident what meaning should have the balls with infinite radius, and especially the balls of infinite radius in negative dimension :-). Still formally this relation looks beautiful.
See also this question: What is the connection between the Riemann Xi-function and n-sphere?