Posted by: matheuscmss | April 5, 2017

## New numbers in M-L

In 1973, Freiman found an interesting number ${\alpha_{\infty}}$ in the complement ${M\setminus L}$ of the Lagrange spectrum ${L}$ in the Markov spectrum ${M}$.

By carefully analyzing Freiman’s argument, Cusick and Flahive constructed in 1989 a sequence ${\alpha_n}$ converging to ${\alpha_{\infty}}$ as ${n\rightarrow\infty}$ such that ${\alpha_n\in M\setminus L}$ for all ${n\geq 4}$, and, as it turns out, ${\alpha_4}$ was the largest known element of ${M\setminus L}$.

In our recent preprint, Gugu and I described the structure of the complement ${M\setminus L}$ of the Lagrange spectrum ${L}$ in the Markov spectrum ${M}$ near ${\alpha_{\infty}}$, and this led us to wonder if our description could be used to find new numbers in ${M\setminus L}$ which are larger than ${\alpha_4}$.

As it turns out, Gugu and I succeeded in finding such numbers and we are currently working on the combinatorial arguments needed to extract the largest number given by our methods. (Of course, we plan to include a section on this matter in a forthcoming revised version of our preprint.)

In order to give a flavour on our construction of new numbers in ${M\setminus L}$, we will prove in this post that a certain number

$\displaystyle x=3.2930444288\dots > \alpha_4 = 3.2930442719\dots$

belongs to ${M\setminus L}$.

1. Preliminaries

Let

$\displaystyle [a_0; a_1, a_2,\dots] = a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots}}}$

be the usual continued fraction expansion.

We abbreviate periodic continued fractions by putting a bar over the period: for instance, ${[2; \overline{1,1,2,2,2,1,2}] = [2; 1,1,2,2,2,1,2,1,1,2,2,2,1,2,\dots]}$. Moreover, we use subscripts to indicate the multiplicity of a digit in a sequence: for example, ${[2; 1,1,2,2,2,1,2, \dots] = [2; 1_2,2_3,1,2,\dots]}$.

Given a bi-infinite sequence ${A=(a_n)_{n\in\mathbb{Z}}\in(\mathbb{N}^*)^{\mathbb{Z}}}$ and ${i\in\mathbb{Z}}$, let

$\displaystyle \lambda_i(A) := [a_i; a_{i+1}, a_{i+2}, \dots] + [0; a_{i-1}, a_{i-2}, \dots]$

In this context, recall that the classical Lagrange and Markov spectra ${L}$ and ${M}$ are the sets

$\displaystyle L=\{\ell(A)<\infty: A\in(\mathbb{N}^*)^{\mathbb{Z}}\} \quad \textrm{and} \quad M=\{m(A)<\infty: A\in(\mathbb{N}^*)^{\mathbb{Z}} \}$

where

$\displaystyle \ell(A)=\limsup\limits_{i\rightarrow\infty}\lambda_i(A) \quad \textrm{and} \quad m(A) = \sup\limits_{i\in\mathbb{Z}} \lambda_i(A)$

As we already mentioned, Freiman proved that

$\displaystyle \alpha_{\infty} = [2;\overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, \overline{2}] = 3.2930442654\dots\in M\setminus L,$

and Cusick and Flahive extended Freiman’s argument to show that the sequence

$\displaystyle \alpha_n := [2;\overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_n, \overline{1, 2, 1_2, 2_3}]$

accumulating on ${\alpha_{\infty}}$ has the property that ${\alpha_n\in M\setminus L}$ for all ${n\geq 4}$. In particular,

$\displaystyle \alpha_4 = 3.2930442719\dots$

was the largest known number in ${M\setminus L}$.

2. A new number in ${M\setminus L}$

In what follows, we will show that

Theorem 1

$\displaystyle x:=[2;\overline{1_2, 2_3, 1, 2}]+[0;1, 2_3, 1_2, 2, 1, 2_2,1,\overline{1_2,2}] = 3.2930444288\dots \in M\setminus L$

Remark 1 ${x}$ is a “good” variant of ${\alpha_2 = 3.2930444886\dots}$ in the sense that it falls in a certain interval ${I}$ which can be proved to avoid the Lagrange spectrum: see Proposition 5 below.

Remark 2 Note that ${\frac{x-\alpha_{\infty}}{\alpha_4-\alpha_{\infty}}=24.8321\dots}$, i.e., if we center our discussion at ${\alpha_{\infty}}$, then ${x}$ is almost 25 times bigger than ${\alpha_4}$.

Similarly to the arguments of Freiman and Cusick-Flahive, the proof of Theorem 1 starts by locating an appropriate interval ${I}$ centered at ${\alpha_{\infty}}$ such that ${I}$ does not intersect the Lagrange spectrum.

In this direction, one needs the following three lemmas:

Lemma 2 If ${B\in\{1,2\}^{\mathbb{Z}}}$ contains any of the subsequences

• (a) ${1^*}$
• (b) ${22^*}$
• (c) ${1_2 2^* 1_2}$
• (d) ${2_2 1 2^* 1_2 2 1}$
• (e) ${1 2_2 1 2^* 1_2 2}$
• (f) ${2_4 1 2^* 1_2 2_3}$

then ${\lambda_j(B)<\alpha_{\infty} - 10^{-5}}$ where ${j}$ indicates the position in asterisk.

Proof: See Lemma 2 in Chapter 3 of Cusick-Flahive’s book. $\Box$

Lemma 3 If ${B\in\{1,2\}^{\mathbb{Z}}}$ contains any of the subsequences:

• (i) ${212^*12}$
• (ii) ${212^*1_3}$
• (iii) ${1212^*1_2}$
• (iv) ${2_3 1 2^* 1_2 2_2 1}$
• (v) ${2 1 2_3 1 2^* 1_2 2_3}$
• (vi) ${1_2 2_3 1 2^* 1_2 2_4}$
• (vii) ${1_2 2_3 1 2^* 1_2 2_3 1_2}$
• (viii) ${1_3 2_3 1 2^* 1_2 2_3 1 2}$
• (ix) ${2 1_2 2_3 1 2^* 1_2 2_3 1 2_2}$
• (x) ${2_2 1_2 2_3 1 2^* 1_2 2_3 1 2 1}$
• (xi) ${1_2 2 1_2 2_3 1 2^* 1_2 2_3 1 2 1_2 2}$

then ${\lambda_j(B)>\alpha_{\infty}+10^{-6}}$ where ${j}$ indicates the position in asterisk.

Proof: See Lemma 1 in Chapter 3 of Cusick-Flahive’s book and also Lemma 3.2 of our preprint with Gugu. $\Box$

Lemma 4 If ${B\in\{1,2\}^{\mathbb{Z}}}$ contains the subsequence:

• (xii) ${2_2 1 2 1_2 2_3 1 2^* 1_2 2_3 1 2 1_2 2 1}$

then ${\lambda_j(B)>\alpha_{\infty}+1.968\times 10^{-7}}$ where ${j}$ indicates the position in asterisk.

Proof: In this situation,

$\displaystyle \begin{array}{rcl} \lambda_j(B) &=& [2;1_2,2_3,1,2,1_2,2,1,\dots] + [0;1,2_3,1_2,2,1,2_2,\dots] \\ &\geq& [2;1_2,2_3,1,2,1_2,2,1,\overline{1,2}] + [0;1,2_3,1_2,2,1,2_2,\overline{2,1}] \\ &=& 3.2930444624\dots \end{array}$

thanks to the standard fact that if

$\displaystyle \alpha=[a_0; a_1,\dots, a_n, a_{n+1},\dots]$

and

$\displaystyle \beta=[a_0; a_1,\dots, a_n, b_{n+1},\dots]$

with ${a_{n+1}\neq b_{n+1}}$, then ${\alpha>\beta}$ if and only if ${(-1)^{n+1}(a_{n+1}-b_{n+1})>0}$. $\Box$

As it is explained in Chapter 3 of Cusick-Flahive’s book and also in the proof of Proposition 3.7 of our preprint with Gugu, Lemmas 2, 3 and 4 allow to show that:

Proposition 5 The interval

$\displaystyle I=[\alpha_{\infty}-10^{-8}, \alpha_{\infty}+ 1.968\times 10^{-7})$

does not intersect the Lagrange spectrum ${L}$.

Of course, this proposition gives a natural strategy to exhibit new numbers in ${M\setminus L}$: it suffices to build elements of ${M\cap I}$ as close as possible to the right endpoint of ${I}$.

Remark 3 As the reader can guess from the statement of the previous proposition, the right endpoint of ${I}$ is intimately related to Lemma 4. In other terms, the natural limit of this method for producing the largest known numbers in ${M\setminus L}$ is given by how far we can push to the right the boundary of ${I}$.Here, Gugu and I are currently trying to optimize the choice of ${I}$ by exploiting the simple observation that the proof of Lemma 4 is certainly not sharp in our situation: indeed, we used the sequence

$\displaystyle B':=\overline{1,2},2_2,1,2,1_2,2_3,1,2;1_2,2_3,1,2,1_2,2,1,\overline{1,2}$

to bound ${\lambda_j(B)}$, but we could do better by noticing that this sequence provides a pessimistic bound because ${\overline{1,2}}$ contains a copy of the word ${21212}$ (and, thus, ${m(B')>\alpha_{\infty}+10^{-6}}$, i.e., ${B}$ can’t coincide with ${B'}$ on a large chunk when ${m(B)<\alpha_{\infty}+10^{-6}}$).

Anyhow, Proposition 5 ensures that

$\displaystyle x:=[2;\overline{1_2, 2_3, 1, 2}]+[0;1, 2_3, 1_2, 2, 1, 2_2,1,\overline{1_2,2}] = 3.2930444288\dots$

does not belong to the Lagrange spectrum ${L}$.

At this point, it remains only to check that ${x}$ belongs to the Markov spectrum.

For this sake, let us verify that

$\displaystyle x= [2;\overline{1_2, 2_3, 1, 2}]+[0;1, 2_3, 1_2, 2, 1, 2_2,1,\overline{1_2,2}]=:\lambda_0(A) =m(A)$

By items (a), (b), (c) and (e) of Lemma 2,

$\displaystyle \lambda_i(A)<\alpha_{\infty}-10^{-5}$

except possibly for ${i=7k}$ with ${k\geq 1}$. Since

$\displaystyle \begin{array}{rcl} \lambda_{7k}(A) &=& [2; \overline{1_2, 2_3, 1, 2}] + [0;\underbrace{1, 2_3, 1_2, 2, \dots, 1, 2_3, 1_2, 2}_{k \textrm{ times }}, 1, 2_3, 1_2, 2, 1, 2_2,\dots] \\ &<& [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_3, 1_2, 2, 1,\overline{2,1}] \\ &<& [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_3,\overline{2,1}] \\ &<& [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_2,1,\overline{1_2,2}] = \lambda_0(A), \end{array}$

we have

$\displaystyle \begin{array}{rcl} \lambda_0(A)-\lambda_{7k}(A)&>&[0; 1, 2_3, 1_2, 2, 1, 2_3,\overline{2,1}] - [0; 1, 2_3, 1_2, 2, 1, 2_3, 1_2, 2, 1,\overline{2,1}] \\ &>& 2.817153\dots\times 10^{-8} \end{array}$

for all ${k\geq 1}$. This proves that ${x=\lambda_0(A)=m(A)}$ belongs to the Markov spectrum.

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