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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