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Pinbacker 
Posted: Apr 7 2014, 04:21 PM

Casual Member Group: Members Posts: 117 Member No.: 795 Joined: 7February 14 
Hello.
I’ve tried to express V(n) / V(n1) where V(n) is “the ndimensional volume of a Euclidean ball of radius R in ndimensional Euclidean space”. I’ve found on http://en.wikipedia.org/wiki/Volume_of_an_nball#Recursions this equation: V(n) = V(n1) R sqrt(pi) G[(n+1) / 2] / G[(n/2) + 1] Where G(n) is the gamma function. I reorganized this a little to: (1/R) V(n) / V(n1) = sqrt(pi) G[(n+1) / 2] / G[(n+2) + 2] And I’ve found on http://mathworld.wolfram.com/GammaFunction.html this equation valid for all positive integers k: G(k/2) = sqrt(pi) [(k2)!!] 2^[(k1) / 2] Where “!!” is the double factorial function. By using that previous equation I get: (1/R) V(n) / V(n1) = sqrt(pi) [ [(n1)!!] 2^[(n+1) / 2] ] / [ (n!!) 2^(n/2) ] Which finally becomes this: (1/R) V(n) / V(n1) = sqrt(2pi) [(n1)!!] / (n!!) This looks like a beautiful equation. It’s like even telling us that the fundamental circle constant is not pi, nor 2pi, but sqrt(2pi). But the problem is that it doesn’t work when you test it for particular values of n. For example, for n=5: (1/R) V(5) / V(4) = sqrt(2pi) [(4)!!] / (5!!) (1/R) V(5) / V(4) = sqrt(2pi) (8/15) But the correct value was 16/15. Can someone tell me what I did wrong? Thanks in advance. 
m1n1f1g 
Posted: Apr 7 2014, 10:26 PM

Dozens Disciple Group: Members Posts: 826 Member No.: 591 Joined: 20February 11 
You know, we can use the lovely lovely LaTeX on here!
Anyway, from the Tau Manifesto (ironically), we get:\[V_n=\frac{2^n\eta^{\left\lfloor\frac{n}{2}\right\rfloor}r^{n1}}{n!!} \quad\text{where }\eta=\frac{\pi}{2}\]Then, I started working on something, then got tired. Here's what I've got (EDIT: I added the ‘mod’ line):\[\begin{aligned} \frac{V_n}{V_{n1}}&=\frac{2^n\eta^{\left\lfloor\frac{n}{2}\right\rfloor}r^{n1}(n1)!!}{2^{n1}\eta^{\left\lfloor\frac{n1}{2}\right\rfloor}r^{n2}n!!}\\ &=\frac{2\eta^{\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n1}{2}\right\rfloor}r(n1)!!}{n!!}\\ &=\frac{2\eta^{(n1)\ \bmod\ 2}r(n1)!!}{n!!}\\ \end{aligned}\]That's the nicest form I can get it in. I'll show how I got the ‘mod’ bit:\[\text{Consider }\left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n1}{2}\right\rfloor\\[24pt] \text{Even }n\text{:}\\ \text{Let }n=2m,\ m\in\mathbb{Z}\\ \begin{aligned} \left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n1}{2}\right\rfloor&=\left\lfloor\frac{2m}{2}\right\rfloor\left\lfloor\frac{2m1}{2}\right\rfloor\\ &=\lfloor m\rfloor\left\lfloor m\frac{1}{2}\right\rfloor\\ &=m(m1)\\ &=1\\ \end{aligned}\\[48pt] \text{Odd }n\text{:}\\ \text{Let }n=2m+1,\ m\in\mathbb{Z}\\ \begin{aligned} \left\lfloor\frac{n}{2}\right\rfloor\left\lfloor\frac{n1}{2}\right\rfloor&=\left\lfloor\frac{2m+1}{2}\right\rfloor\left\lfloor\frac{2m+11}{2}\right\rfloor\\ &=\left\lfloor m+\frac{1}{2}\right\rfloor\lfloor m\rfloor\\ &=mm\\ &=0 \end{aligned}\]How those floored terms reduce is nonobvious (unless you happen to be wellversed in discrete maths or whatever). But in any case, there are never half powers of eta or pi, so there are no square roots. 
wendy.krieger 
Posted: Apr 7 2014, 10:43 PM

Dozens Demigod Group: Members Posts: 2,432 Member No.: 655 Joined: 11July 12 
The volume of a unit sphere in n dimensions, is (\pi /2)^{n div2}/n!!.
Mathematicians suppose a "unit sphere" is actually what everyone else calls a size 2 sphere, because they use the radius. So you have to multiply yhe above by 2^n. 
Pinbacker 
Posted: Apr 12 2014, 05:41 PM

Casual Member Group: Members Posts: 117 Member No.: 795 Joined: 7February 14 
I installed the addon which makes me see LaTex (Greasemonkey). The only inconvenient is that I now have a stupid monkey face on the upper right section of every Firefox pages…
I’m still waiting for someone to point me where did I make an error. And also I’ve got another question: is it possible that the true most fundamental circle/sphere constant is actually of the form \[\sqrt(k \pi)\] instead of just \[k \pi\] ? (with k = 1, 2, 3, 4, 1/2, 1/3, 1/4, or something like that) Because clearly in the formulas giving the volume and surface areas of nballs it is \[\sqrt(k \pi)\] that appears and not just \[k \pi\] 
m1n1f1g 
Posted: Apr 12 2014, 07:50 PM

Dozens Disciple Group: Members Posts: 826 Member No.: 591 Joined: 20February 11 
For a bit of anachronism, I've edited my original reply.

wendy.krieger 
Posted: Apr 13 2014, 12:00 AM

Dozens Demigod Group: Members Posts: 2,432 Member No.: 655 Joined: 11July 12 
The volume of a sphere in crindunits is already \( d^n \). The bits with pi in it are just conversion factors.
d is the diameter, this is 1 for a unit sphere. n is the dimension. The crind product is one of the three known coherent products: you are better off looking at the pi vs tau thread, i mentioned it there there. 
Pinbacker 
Posted: Apr 14 2014, 09:37 AM

Casual Member Group: Members Posts: 117 Member No.: 795 Joined: 7February 14 
@m1n1f1g: Yes interesting, but this doesn’t answer my original question: where is the mistake in my calculations?
@wendy.krieger : I have absolutely no idea what you are talking about. What do your “n” and your “d” represent? And what are these “crindunits”? (I’ve tried to Google it, but there was no result) 
wendy.krieger 
Posted: Apr 16 2014, 12:03 AM

Dozens Demigod Group: Members Posts: 2,432 Member No.: 655 Joined: 11July 12 
Where Pinbacker makes the error is when he replaces (n/2)! with n!!/2^(n%2), and sheads the sqrt pi in the process. The volume of a sphere of radius r is pi^{n/2}/(n/2)! r^n.

m1n1f1g 
Posted: Apr 16 2014, 10:33 PM

Dozens Disciple Group: Members Posts: 826 Member No.: 591 Joined: 20February 11 
Sorry; I couldn't find any mistake. I derived the same formula as you just now, so I'm a bit confused.
Okay, the problem is that the identity \(\Gamma\left(\frac{n}{2}\right)=\frac{(n2)!!\sqrt\pi}{2^{(n1)/2}}\) only works for odd \(n\). An easy test: \(\Gamma(2)=1!=1\) but \(\frac{(42)!!\sqrt\pi}{2^{(41)/2}}\) has a factor of \(\sqrt\pi\). Incidentally, I prefer the Pi function, where \(\Pi(z)=\Gamma(z+1)\) and hence \(n!=\Pi(n)\). 
arbiteroftruth 
Posted: Apr 21 2014, 06:25 PM

Regular Group: Members Posts: 167 Member No.: 704 Joined: 17February 13 
The algebraically simplest formula for the volumn of an nball with diameter d is:
V=d^n*eta^floor(n/2)/(n!!) where eta=pi/2. This implies that V(n)/V(n1)=d*(n1)!!/(n!!)*eta^((1+(1)^n)/2). Where that mess in the exponent of eta just means to multiply by eta if n is even, and don't multiply by eta if n is odd. There's no way to simplify that ratio of doublefactorials though. Directly applying the formula for V(n) is simpler than trying to use the ratio. Although, if you don't mind jumping 2 dimensions at a time, you can use the ratio V(n)/V(n2)=d^2*eta/n. 