# Volume Of An N-ball

 Posts 114
Casual Member
Pinbacker
Casual Member
Joined: Feb 7 2014, 03:32 PM
Hello.

Iâ€™ve tried to express V(n) / V(n-1) where V(n) is â€œthe n-dimensional volume of a Euclidean ball of radius R in n-dimensional Euclidean spaceâ€.

Iâ€™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 Iâ€™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 â€œ!!â€ 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. Itâ€™s like even telling us that the fundamental circle constant is not pi, nor 2pi, but sqrt(2pi).

But the problem is that it doesnâ€™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?

 Posts 817
Dozens Disciple
m1n1f1g
Dozens Disciple
Joined: Feb 20 2011, 10:15 AM
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 â€˜modâ€™ 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 â€˜modâ€™ 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 (&#8242;). 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 11&#400;7-03-1X, and a whole range of similar radix points. I will often use X and &#400; for and .

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.

wendy.krieger
wendy.krieger
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.

 Posts 114
Casual Member
Pinbacker
Casual Member
Joined: Feb 7 2014, 03:32 PM
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 pagesâ€¦

Iâ€™m still waiting for someone to point me where did I make an error.

And also Iâ€™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 $k \pi$

 Posts 817
Dozens Disciple
m1n1f1g
Dozens Disciple
Joined: Feb 20 2011, 10:15 AM
For a bit of anachronism, I've edited my original reply.
A few little conventions:
- Dozenal integers suffixed with prime (&#8242;). 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 11&#400;7-03-1X, and a whole range of similar radix points. I will often use X and &#400; for and .

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.

wendy.krieger
wendy.krieger
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.

 Posts 114
Casual Member
Pinbacker
Casual Member
Joined: Feb 7 2014, 03:32 PM
@m1n1f1g: Yes interesting, but this doesnâ€™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 â€œnâ€ and your â€œdâ€ represent? And what are these â€œcrind-unitsâ€? (Iâ€™ve tried to Google it, but there was no result)

wendy.krieger
wendy.krieger
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.

 Posts 817
Dozens Disciple
m1n1f1g
Dozens Disciple
Joined: Feb 20 2011, 10:15 AM
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 (&#8242;). 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 11&#400;7-03-1X, and a whole range of similar radix points. I will often use X and &#400; for and .

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.

 Posts 186
Regular
arbiteroftruth
Regular
Joined: Feb 17 2013, 05:51 AM
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.