I heard about the Wallis sieve the first time in this video by Matt Parker, which is fascinating. Instantly I recognized the pattern. I thought the relation between the Wallis sieve and the formula for the volume of an lp n-ball would be trivial and well-known, it turns out it is neither. Later, I read the blog post in scientific american by Evelyn Lamb and I still thought it would be easy to relate both. Finally, I sat down and did the work and found that the result is not only surprising, but (at least to me), completely non-obvious and has the potential to be very interesting.

The volume of an $l_p$ n-ball, a generalized ball is of radius $R$ is, (for more details and the calculation and history of the formula, see Xiafu Wang's paper),

$$V_d^p(R) = \frac{(2\Gamma(\frac{1}{p} + 1)R)^d}{\Gamma(\frac{d}{p} + 1)}.$$

Note that the ball for $l_2$ is an hypersphere of dimension $d$.

This explains the relation between the Euler gamma function and $\pi$, one of my favorites,

$$\Gamma\Big(\frac{3}{2}\Big) = \frac{\sqrt{\pi}}{2}.$$

At the same time, the gamma function is a generalization of the factorial and satisfies all sorts of recursive formulas similar to the Wallis sieve.

I will refer you to Evelyn Lamb's post for a detailed introduction, but the Wallis sieve can be easily written as a limit using the gamma formula,

$$\frac{\pi}{2} = \prod_{n=1}^{\infty}\Bigg[\frac{(2n)^2}{(2n-1)(2n+1)}\Bigg] = \frac{2\cdot 2\cdot 4\cdot 6\cdot 6\ldots}{1\cdot 2\cdot 2\cdot 5\cdot 5\cdot 7\ldots},$$

which can be rewritten as a limit,

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

We can then apply the gamma duplication formula,

$$\Gamma(z)\Gamma(z+\frac{1}{2})= 2^{1-2z}\sqrt{\pi}\Gamma(2z),$$

and the functional relation,

$$\Gamma(z+1) = z\Gamma(z),$$

to rewrite again the limit,

$$\pi = \lim_{n\to\infty} n\Bigg[\frac{\Gamma(n)^2}{\Gamma(2n)2^{1-2n}}\Bigg]^2 = \lim_{n\to\infty} \frac{1}{n}\Bigg[\frac{\Gamma(1+n)^2 2^{2n}}{\Gamma(2n+1)}\Bigg]^2,$$

so

$$\pi = \lim_{n\to\infty}\frac{V_2^{\frac{1}{n}}(2^n)^2}{4n}.$$

This is, to say the least, surprising. Instead of hyperspheres and a trivial relationship, we get something which looks like an astroid (the image comes from Wikipedia).

So it is the limit of the square of the volume of this figure as it collapses upon itself, its inner radius getting smaller while the outer radio grows. This result is bizarre and not at all trivial.

The formula can be generalized (I will play with this the next time I have some free time) to higher dimensions. The video talks about this, but I have not written their formula down. Also, it will be interesting if fat Cantor sets can be written in terms of hyperballs too.

I have the conjecture which it will be related with the taxicab measure astroid ball, whatever that is.

Edit: fixed a missing n in the denominator in the limit.

Nice post!

ReplyDeleteTypo in the title :/

Fixed, thanks!!!

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