## Curious Constants (part 2 of 4)

This continues my series of notes for my talk on some mathematical constants. I’d like to make a small diversion from talking about these constants to provide some background on continued fractions. What follows is a brief, abridged summary of Khinchin‘s short book, “Continued Fractions,” which I highly recommend.

## Continued Fractions

A (simple, regular) continued fraction is an expression (possibly finite)

$a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\ddots}}}$

Which I abbreviate $[a_0;a_1,\ldots,a_n,\ldots]$. Here the values $a_i$ are natural numbers (allowing $a_0$ to be 0, but all others positive), and we call them the elements of the continued fraction. If the continued fraction is finite, ending at $a_k$, we will say that it has order $k$.

Finite continued fractions clearly correspond to rational numbers. We can convert any number to a continued fraction, by repeatedly splitting off integer parts, and then inverting. For example

$\frac{13}{9}=1+\frac{4}{9}=1+\frac{1}{9/4}=1+\cfrac{1}{2+\frac{1}{4}}$

The same process works for real numbers, we just iterate without end.

Fix a continued fraction $[a_0;a_1,\ldots,a_n,\ldots]$, corresponding to the real number $\alpha$. Let $s_k=[a_0;a_1,\ldots,a_k]$, which we call the $k$th segment. This is a fraction that is already in lowest terms, which we typically denote $p_k/q_k$, and call the $k$th convergent. We highlight, now, a few equations and inequalities concerning convergents:

$\displaystyle \frac{p_{k-1}}{q_{k-1}}-\frac{p_k}{q_k} =\frac{(-1)^k}{q_kq_{k+1}}$

$q_k \geq 2^{\frac{k-1}{2}}$

$\displaystyle \frac{1}{q_k(q_{k+1}+q_k)}<\left| \alpha-\frac{p_k}{q_k}\right|<\frac{1}{q_kq_{k+1}}$

The convergents of even order form an increasing sequence, and those of odd order form a decreasing sequence, with the limit, $\alpha$ in the middle.

One of the most interesting things about continued fractions is their relation to best rational approximations. Loosely, every convergent is a best rational approximation, and vice-versa. For more specifics, see Khinchin’s book. Along the same lines, if $a/b$ is an irreducible fraction with $|\alpha-a/b|<1/(2b^2)$, then $a/b$ is a convergent of the continued fraction for $\alpha$.

For every irrational $\alpha$ with bounded elements, and small enough value for $c$, $|\alpha-p/q| has no integer solutions for $p$ and $q$. However, if the elements are unbounded, then for any $c$, there are infinitely many solutions. More generally, Liouville showed that if $\alpha$ is an irrational with degree $n$ (i.e., it is algebraic with minimal polynomial $n$), then there is a $c>0$ such that for any $p,q$, $|\alpha-p/q|>c/q^n$. Thus, if for every positive $c$, and every positive integer $n$, there are integers $p,q$ such that $|\alpha-p/q|\leq c/q^n$, then $\alpha$ is transcendental. In a sense, this is saying that transcendental numbers have a rapidly converging sequence of rational approximations. This led to the first proofs that chosen numbers were transcendental. To construct a transcendental, you can pick any collection of positive integers $a_0,\ldots,a_k$, and then form the continued fraction $[a_0;a_1,\ldots,a_k]$, find it’s denominator $q_k$, and choose any $a_{k+1}>q_k^{k+1}$. Repeating this process will generate an infinite continued fraction with unbounded denominators that is provably transcendental. Considerations like the above, concerning just how good rational approximations can be, earned Klaus Roth a Fields Medal in 1958.

Other posts in this series

1. The First Batch
2. Continued Fractions
3. The Second Batch
4. Resources

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