Volume Of An N-ball

Volume Of An N-ball

Pinbacker
Casual Member
Pinbacker
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Joined: Feb 7 2014, 03:32 PM

Apr 7 2014, 04:21 PM #1

Hello.

I’ve tried to express V(n) / V(n-1) where V(n) is “the n-dimensional volume of a Euclidean ball of radius R in n-dimensional Euclidean space”.

I’ve found on http://en.wikipedia.org/wiki/Volume_of_ ... Recursions this equation:

V(n) = V(n-1) 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(n-1) = 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) [(k-2)!!] 2^[(k-1) / 2]

Where “!!” is the double factorial function.

By using that previous equation I get:

(1/R) V(n) / V(n-1) = sqrt(pi) [ [(n-1)!!] 2^[(n+1) / 2] ] / [ (n!!) 2^(n/2) ]

Which finally becomes this:

(1/R) V(n) / V(n-1) = sqrt(2pi) [(n-1)!!] / (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.
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m1n1f1g
Dozens Disciple
m1n1f1g
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Joined: Feb 20 2011, 10:15 AM

Apr 7 2014, 10:26 PM #2

You know, we can use the lovely lovely LaTeX on here!

Anyway, from the Tau Manifesto (ironically), we get:\[V_n=\frac{2^n\eta^{\left\lfloor\frac{n}{2}\right\rfloor}r^{n-1}}{n!!} \quad\text{where }\eta=\frac{\pi}{2}\]Then, I started working on something, then got tired. Here's what I've got (EDIT: I added the ‘mod’ line):\[\begin{aligned}
\frac{V_n}{V_{n-1}}&=\frac{2^n\eta^{\left\lfloor\frac{n}{2}\right\rfloor}r^{n-1}(n-1)!!}{2^{n-1}\eta^{\left\lfloor\frac{n-1}{2}\right\rfloor}r^{n-2}n!!}\\
&=\frac{2\eta^{\left\lfloor\frac{n}{2}\right\rfloor-\left\lfloor\frac{n-1}{2}\right\rfloor}r(n-1)!!}{n!!}\\
&=\frac{2\eta^{(n-1)\ \bmod\ 2}r(n-1)!!}{n!!}\\
\end{aligned}\]That's the nicest form I can get it in. I'll show how I got the ‘mod’ bit:\[\text{Consider }\left\lfloor\frac{n}{2}\right\rfloor-\left\lfloor\frac{n-1}{2}\right\rfloor\\[24pt]
\text{Even }n\text{:}\\
\text{Let }n=2m,\ m\in\mathbb{Z}\\
\begin{aligned}
\left\lfloor\frac{n}{2}\right\rfloor-\left\lfloor\frac{n-1}{2}\right\rfloor&=\left\lfloor\frac{2m}{2}\right\rfloor-\left\lfloor\frac{2m-1}{2}\right\rfloor\\
&=\lfloor m\rfloor-\left\lfloor m-\frac{1}{2}\right\rfloor\\
&=m-(m-1)\\
&=1\\
\end{aligned}\\[48pt]
\text{Odd }n\text{:}\\
\text{Let }n=2m+1,\ m\in\mathbb{Z}\\
\begin{aligned}
\left\lfloor\frac{n}{2}\right\rfloor-\left\lfloor\frac{n-1}{2}\right\rfloor&=\left\lfloor\frac{2m+1}{2}\right\rfloor-\left\lfloor\frac{2m+1-1}{2}\right\rfloor\\
&=\left\lfloor m+\frac{1}{2}\right\rfloor-\lfloor m\rfloor\\
&=m-m\\
&=0
\end{aligned}\]How those floored terms reduce is non-obvious (unless you happen to be well-versed in discrete maths or whatever). But in any case, there are never half powers of eta or pi, so there are no square roots.
A few little conventions:
- Dozenal integers suffixed with prime (′). This is the uncial point.
- Decimal integers suffixed with middle dot (·). This is the decimal point.

You may see me use * prefix for messages before 11Ɛ7-03-1X, and a whole range of similar radix points. I will often use X and Ɛ for :A and :B.

Sometimes, I will imply that an integer is in dozenal, so I won't add any marks to it. You should be able to tell that "10 = 22 * 3" is in dozenal.
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wendy.krieger
wendy.krieger

Apr 7 2014, 10:43 PM #3

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.
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Pinbacker
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Pinbacker
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Joined: Feb 7 2014, 03:32 PM

Apr 12 2014, 05:41 PM #4

I installed the add-on 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 n-balls it is \[\sqrt(k \pi)\] that appears and not just \[k \pi\]
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m1n1f1g
Dozens Disciple
m1n1f1g
Dozens Disciple
Joined: Feb 20 2011, 10:15 AM

Apr 12 2014, 07:50 PM #5

For a bit of anachronism, I've edited my original reply.
A few little conventions:
- Dozenal integers suffixed with prime (′). This is the uncial point.
- Decimal integers suffixed with middle dot (·). This is the decimal point.

You may see me use * prefix for messages before 11Ɛ7-03-1X, and a whole range of similar radix points. I will often use X and Ɛ for :A and :B.

Sometimes, I will imply that an integer is in dozenal, so I won't add any marks to it. You should be able to tell that "10 = 22 * 3" is in dozenal.
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wendy.krieger
wendy.krieger

Apr 13 2014, 12:00 AM #6

The volume of a sphere in crind-units 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.
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Pinbacker
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Pinbacker
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Joined: Feb 7 2014, 03:32 PM

Apr 14 2014, 09:37 AM #7

@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 “crind-units”? (I’ve tried to Google it, but there was no result)
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wendy.krieger
wendy.krieger

Apr 16 2014, 12:03 AM #8

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.
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m1n1f1g
Dozens Disciple
m1n1f1g
Dozens Disciple
Joined: Feb 20 2011, 10:15 AM

Apr 16 2014, 10:33 PM #9

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 \(\Gamma\left(\frac{n}{2}\right)=\frac{(n-2)!!\sqrt\pi}{2^{(n-1)/2}}\) only works for odd \(n\). An easy test: \(\Gamma(2)=1!=1\) but \(\frac{(4-2)!!\sqrt\pi}{2^{(4-1)/2}}\) has a factor of \(\sqrt\pi\).

Incidentally, I prefer the Pi function, where \(\Pi(z)=\Gamma(z+1)\) and hence \(n!=\Pi(n)\).
A few little conventions:
- Dozenal integers suffixed with prime (′). This is the uncial point.
- Decimal integers suffixed with middle dot (·). This is the decimal point.

You may see me use * prefix for messages before 11Ɛ7-03-1X, and a whole range of similar radix points. I will often use X and Ɛ for :A and :B.

Sometimes, I will imply that an integer is in dozenal, so I won't add any marks to it. You should be able to tell that "10 = 22 * 3" is in dozenal.
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arbiteroftruth
Regular
Joined: Feb 17 2013, 05:51 AM

Apr 21 2014, 06:25 PM #10

The algebraically simplest formula for the volumn of an n-ball with diameter d is:

V=d^n*eta^floor(n/2)/(n!!)

where eta=pi/2.

This implies that V(n)/V(n-1)=d*(n-1)!!/(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 double-factorials 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(n-2)=d^2*eta/n.
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