Reversing the coefficients of the continued fraction expansion of a rational number

Question

On this Wikipedia page about the Markov constant, there's a surprising theorem. Basically if $\alpha$ is a real number with continued fraction expansion $[a_0;a_1,a_2,...]$, the claim is that the Markov constant $M(\alpha)$ is equal to

$$ M(\alpha) = \limsup_{k\to\infty} ([a_k;a_{k+1},a_{k+2},...] + [0;a_{k-1},a_{k-2},...,a_1, a_0]) $$

This is very surprising to me. So we take, for each position $k$, the tail of the continued fraction starting at $a_k$ onward, and treat it as a real number $\alpha_k$. Then we add to it the rational number one gets by taking the initial segment of the continued fraction up to $a_{k-1}$, reversing the coefficients, and prepending a 0 to it.

It is amazing to me that this yields anything coherent at all. What, in general, does it mean to reverse the continued fraction sequence of a rational number? And if we add the tail to the reversed initial segment, what interpretation does the resulting real number have, and why does any of this yield the Markov constant?

Answer 1

"What, in general, does it mean to reverse the continued fraction sequence of a rational number?" This is the only part I can help to answer.

Suppose [0; k₀, k₁....k_n] = $\frac{A}{B}$ , A and B being coprime natural numbers,

The same fraction, truncated by omitting the final term is

[0; k₀, k₁....k_n-1] = $\frac{C}{D}$.

Now AD - BC = |1|. This is because calculating the continued fraction that has been truncated of its final term is equivalent to applying the extended Euclidean algorithm to the original continued fraction. C and D represent co-efficients of a Bezout identity for B and A.

The reverse continued fraction is:

[0; k_n, k_n-1....k₀] = $\frac{D}{B}$

The reverse truncated continued fraction is:

[0; k_n-1, k_n-2....k₀] = $\frac{C}{A}$

The reason for this neat symmetry can be seen more clearly by successively truncating each fraction and reversed fraction. I have posted more details on this blog: https://simplyfractions.blogspot.com/2022/10/continued-fraction-triangle.html

A relevant observation is that we can swap the numerator of $\frac{A}{B}$ with the denominator of $\frac{C}{D}$ without altering the cross-multiplication result.

Now this may be a start for your "in general" question. However, the equation you cite is not the reversal of a simple continued fraction, as you point out.