# The Lumber Room

"Consign them to dust and damp by way of preserving them"

## Some incredibly beautiful things

First, a totally completely absolutely incredible fact about convergents of continued fractions: the Khinchin-Levy theorem, seen here: For almost all real numbers $\alpha$, the denominators $q_n$ of the continued-fraction convergents of $\alpha$ satisfy $\displaystyle \lim_{n \to \infty} q_n^{1/n} = e^{\pi^2/(12 \log 2)}$

Why should such a thing be a true?! Why should the limit even exist, why should it be the same for almost all real numbers, and where do $e$, $\pi$, $\log 2$ and 12 come from?!!

Intrigued by these questions, and embarrassed by my near-complete ignorance of continued fractions (each time I had encountered them before, I had been intimidated by the notation and skipped pages), I started reading a couple of books on continued fractions (including Khinchin’s)… other things intervened and I had to give up for now, but I hope to return to their study soon. But from what I’ve already seen, continued fractions are awesome, and it is a real shame they seem to be out of fashion these days and not more widely taught and known. Neither of these books proves the above theorem actually, but an editor’s footnote in Khinchin’s book gives the reference: it was proved by Khinchin in 1935 that the limit exists and is the same for almost all real numbers, and Lévy found the constant soon after (published 1937).

I suspect (based on nothing at all) that it bears some relation to, among other things, (some generalization of) the following well-known fact, itself quite surprising (probably the most surprising fact I know a proof of): $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} = \prod_{p \text{ prime}}{\left(1-\frac1{p^2}\right)}$

This fact is known as the Basel problem, was proved by (who else?) Euler, and fourteen proofs of it are collected here. But let this not distract you from reading about continued fractions!

Second, pointed out at John Baez’s This Week’s Finds in Mathematical Physics (Week 230), and from this paper by Sondow, are these (quoting Prof. Baez directly, just typeset in LaTeX):

eerily similar formulas for some of our favorite math constants. First, one for e: $\displaystyle e = \left(\frac{2}{1}\right)^{1/1} \left(\frac{2^2}{1\times3}\right)^{1/2} \left(\frac{2^3\times4}{1\times3^3}\right)^{1/3} \left(\frac{2^4\times4^4}{1\times3^6\times5}\right)^{1/4} \dots$

Then, one for π/2: $\displaystyle \frac{\pi}{2} = \left(\frac{2}{1}\right)^{1/2} \left(\frac{2^2}{1\times3}\right)^{1/4} \left(\frac{2^3\times4}{1\times3^3}\right)^{1/8} \left(\frac{2^4\times4^4}{1\times3^6\times5}\right)^{1/16} \dots$

Then one for eγ, where γ is Euler’s constant: $\displaystyle e^\gamma = \left(\frac{2}{1}\right)^{1/2} \left(\frac{2^2}{1\times3}\right)^{1/3} \left(\frac{2^3\times4}{1\times3^3}\right)^{1/4} \left(\frac{2^4\times4^4}{1\times3^6\times5}\right)^{1/5} \dots$

He also points out Wallis’ product for π/2 and Pippenger’s for e: $\displaystyle \frac{\pi}{2} = \left(\frac{2}{1}\right)^{1/1} \left(\frac{2\times 4}{3\times3}\right)^{1/1} \left(\frac{4\times6\times6\times8}{5\times5\times7\times7}\right)^{1/1} \dots$ $\displaystyle e = \left(\frac{2}{1}\right)^{1/2} \left(\frac{2\times 4}{3\times3}\right)^{1/4} \left(\frac{4\times6\times6\times8}{5\times5\times7\times7}\right)^{1/8} \dots$

What does it all mean? I haven’t a clue! Another mystery thrown down to us by the math gods, like a bone from on high… we can merely choose to chew on it or not, as we wish.

Indeed!

Third, my favourite way to prove the Möbius inversion formula. Well, it’s actually the only way I know, besides induction, but I think it’s beautiful. (Ignore if you already know what Dirichlet convolutions are.) It’s from Herbert Wilf’s generatingfunctionology, one of my favourite books.

Let us consider a kind of generating function known as a Dirichlet generating function. (“A generating function is a clothesline on which we hang up a sequence of numbers for display.”) For a sequence $a_n$, the Dirichlet generating function is defined as the (formal) series (in s) $\displaystyle G_a(s) = \sum_{n=1}^{\infty} \frac{a_n}{n^s} = a_1 + \frac{a_2}{2^s} + \frac{a_3}{3^s} + \frac{a_4}{4^s} + \dots$.

(A formal series means that we can ignore issues of convergence etc.) Now if we have the generating functions $G_a(s)$ and $G_b(s)$ for two sequences $a_n$ and $b_n$, what is $G_a(s)G_b(s)$? Well, it’s easy to see that the coefficient of $1/n^s$ in $G_a(s)G_b(s)$, which comes exactly from the terms of the form $i^sj^s$ with $ij=n$, is $\displaystyle \sum_{ij=n} a_ib_j$, or equivalently $\displaystyle \sum_{d \vert n} a_db_{n/d}$. So the product is the generating function for a special kind of convolution, the sequence $c_n$ defined as $\displaystyle c_n = \sum_{d \vert n} a_db_{n/d}$

Let us call this fact “(*)” and put it aside for now; we will use it shortly. Instead, let us look at the generating function for a very special sequence, the sequence of all 1s. It is, by definition, $\displaystyle \sum_{n\ge 1} \frac{1}{n^s}$. While we could call this $G_1(s)$, it goes by the more familiar name $\zeta(s)$ — it is (when viewed as a function instead of a formal series) the Riemann zeta function. But we also know (this is the Euler product formula, and follows simply from the unique factorisation of integers): $\displaystyle \zeta(s) = \sum_{n\ge1}\frac{1}{n^s} = \prod_{p\text{ prime}}\left(1+\frac1{p^s} + \frac1{p^{2s}} + \frac1{p^{3s}} + \dots\right) = \prod_{p\text{ prime}}\frac1{\left(1-\frac1{p^s}\right)}$.

This means that $\displaystyle \frac1{\zeta(s)} = \prod_{p\text{ prime}}{\left(1-\frac1{p^s}\right)} = G_\mu(s)$.

Why did I write $G_\mu(s)$ on the right? Is this a generating function of anything? Why, yes! By expanding the product, you can see that it is the generating function for the sequence $\mu(n)$, defined as $\displaystyle \mu(s) = \begin{cases} 0 \quad \text{if }n\text{ is not squarefree,} \\ (-1)^k \quad \text{if }n = p_1p_2\dots p_k\text{ is the product of k primes}\end{cases}$.

(This function “mu” is known as the “Möbius function”.) With this background, the Möbius inversion formula falls out entirely naturally: if $G_f(s)$ is the generating function of a sequence $f(n)$, and the function $F(n)$ is defined as $\displaystyle F(n) = \sum_{d\vert n} f(d)$,

then its generating function $G_F(s)$ is, by (*), simply $\displaystyle G_F(s) = G_f(s)\zeta(s)$,

which means that $\displaystyle G_f(s) = G_F(s)\frac{1}{\zeta(s)} = G_F(s)G_\mu(s)$,

or in other words, again by (*), $\displaystyle f(n) = \sum_{d\vert n} F(d)\mu(n/d)$,

which is precisely the Möbius inversion formula. How cool is that?

Written by S

Fri, 2010-04-30 at 01:12:02

Posted in mathematics