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?

Thanks in advance.

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.