Saturday, April 23, 2016

Wallis sieve, and lp n-balls II

This post is a continuation of this one. In it I talked about this video by Matt Parker and some
of its consequences for two dimensions. In the video, he also asserted that, for three dimensions, the approach of  cutting off pieces of a cube in tha same way that is done in the Wallis sieve but in three dimensions, would give back the volume of a sphere (see the pretty drawings in the post by Evelyn Lamb).
This opened the  question of whether this would happen in higher dimensions. Someone in twitter (thanks!) pointed me to this post in the xkcd forum, where they proof this fact. I am going to transcript, dissect and explain this proof. The original ideas are all from the post, and the mistakes all mine.
First, remember the basic Wallis product formula,

$$\frac{\pi}{4}=\prod_{n=1}^{\infty}\frac{4n(n+1)}{(2n+1)^2}=lim_{n\to\infty} \frac{4^n n! (n+1)!}{(2n+1)!!^2}.$$

Remember that the doble factorial is the product $n!! = n(n-2)(n-4)\cdots~$ which stops when the terms would cease to be positive, i.e. with $⌈n/2⌉$ terms. An important property of the  double factorial is that it can be written in terms of Euler gamma function,
$$\Gamma\Big(n+\frac{1}{2}\Big) = \frac{(2n-1)!!\sqrt{\pi}}{2^n}.$$

Remember also that the volume of an $l_p$ hyperball is
$$V_d^p(R) = \frac{(2\Gamma(\frac{1}{p} + 1)R)^d}{\Gamma(\frac{d}{p} + 1)}.$$
So the, taking into account $\Gamma\big(\frac{3}{2}\big)=\frac{\sqrt{\pi}}{2}$, the volume of an d-hypersphere, which is the $l_2$ hyperball is
$$V_d^2(R) = \frac{R^d\pi^{d/2}}{\Gamma(\frac{d}{2} + 1)}.$$

The Wallis product can be written,
$$\frac{\pi}{4}=\prod_{n=1}^{\infty}\Bigg[1-\frac{1}{(2n+1)^2}\Bigg],$$
which is convergent, and we can move around terms (if we are careful), because the series
$$\sum_{n=0}^{\infty}a_n=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2},$$
is absolutely convergent and we can apply the test for product convergence.

The general product for d dimensions for the Wallis sieve turns out to be
$$\prod_{n=1}^\infty\frac{2^dn^{d-1}(n+\frac{d}{2})}{(2n+1)^d}.$$

For even $d$ we can write this product as
$$lim_{n\to\infty} \frac{2^{nd}(n!)^{d-1}(n+\frac{d}{2})!}{(\frac{d}{2})!((2n+1)!!)^d}.$$
What we have done here is that the factors coming from $n+\frac{d}{2}$ have been expanded into
$\frac{(n+\frac{d}{2})!}{(\frac{d}{2})!}$. This is the tricky part where we have assumed $
d$ to be even. For odd $d$ we can either rewrite in terms of $k$, i.e. $d=2k+1$, and follow a similar approach or be more general and use the gamma function.

 The trick to calculate the limit is to separate it into the product of two parts, one of which can be identified as a Wallis product, between square brackets,
$$lim_{n\to\infty}  A_n^d = lim_{n\to\infty}  \Bigg[\frac{4^{n-1} n! (n-1)!}{(2n-1)!!}\Bigg]^{d/2}\Bigg(\frac{2^dn^{d/2}(n+\frac{d}{2})!}{(\frac{d}{2})!n!(2n+1)^d}\Bigg),$$
$$lim_{n\to\infty}  A_n^d =\Big(\frac{\pi}{2}\Big)^{d/2}\Bigg(\frac{2^dn^{d/2}(n+\frac{d}{2})!}{(\frac{d}{2})!n!(2n+1)^d}\Bigg).$$

The last part is to show that the right term converges to $\Big(\frac{d}{2}\Big)!$.
We show it by parts, first, when $n \to \infty$
$$\frac{\Big(n+\frac{d}{2}\Big)!}{n!}=(n+1)(n+2)\cdots(n+\frac{d}{2}) \sim n^{d/2},$$
 so the limit can be rewritten as
$$lim_{n\to\infty}  A_n^d =\Big(\frac{\pi}{2}\Big)^{d/2}\Bigg(\frac{(2n)^d}{(\frac{d}{2})!(2n+1)^d}\Bigg).$$.

 Finally, for big enough $n$, we can use approximate the second term as,
$$\frac{(2n)^d}{(\frac{d}{2})!(2n+1)^d}\sim\frac{1}{(\frac{d}{2})!}=\frac{1}{\Gamma(\frac{d}{2}+1)},$$
so
$$lim_{n\to\infty}  A_n^d = \frac{(\frac{\pi}{4})^{d/2}}{\Gamma(\frac{d}{2} + 1)}=V_d^2\Big(\frac{1}{2}\Big).$$

So, now that we have the general formula for $d$ dimensions, the question still stands, can we interpret $A_n^d$ in geometrical terms as we did for $d=2$? What about $l_p$ for $p\neq2$? Stay tuned.






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