# Markov constant

Markov constant of a number
Template:Markov constant chart
Basic features
Parityeven
DomainIrrational numbers
CodomainLagrange spectrum with $\displaystyle{ \infty }$
Period1

Specific values
Maxima$\displaystyle{ +\infty }$
Minima5
Value at $\displaystyle{ \phi }$5
Value at 222

This function is undefined on rationals; hence, it is not continuous.

In number theory, specifically in Diophantine approximation theory, the Markov constant $\displaystyle{ M(\alpha) }$ of an irrational number $\displaystyle{ \alpha }$ is the factor for which Dirichlet's approximation theorem can be improved for $\displaystyle{ \alpha }$.

## History and motivation

Certain numbers can be approximated well by certain rationals; specifically, the convergents of the continued fraction are the best approximations by rational numbers having denominators less than a certain bound. For example, the approximation $\displaystyle{ \pi\approx\frac{22}{7} }$ is the best rational approximation among rational numbers with denominator up to 56.[1] Also, some numbers can be approximated more readily than others. Dirichlet proved in 1840[2] that the least readily approximable numbers are the rational numbers, in the sense that for every irrational number there exists infinitely many rational numbers approximating it to a certain degree of accuracy that only finitely many such rational approximations exist for rational numbers[further explanation needed]. Specifically, he proved that for any number $\displaystyle{ \alpha }$ there are infinitely many pairs of relatively prime numbers $\displaystyle{ (p,q) }$ such that $\displaystyle{ \left|\alpha - \frac{p}{q}\right| \lt \frac{1}{q^2} }$ if and only if $\displaystyle{ \alpha }$ is irrational.

51 years later, Hurwitz further improved Dirichlet's approximation theorem by a factor of 5,[3] improving the right-hand side from $\displaystyle{ 1/q^2 }$ to $\displaystyle{ 1/\sqrt{5}q^2 }$ for irrational numbers:

$\displaystyle{ \left|\alpha - \frac{p}{q}\right| \lt \frac{1}{\sqrt{5}q^2}. }$

The above result is best possible since the golden ratio $\displaystyle{ \phi }$ is irrational but if we replace 5 by any larger number in the above expression then we will only be able to find finitely many rational numbers that satisfy the inequality for $\displaystyle{ \alpha=\phi }$.

Furthermore, he showed that among the irrational numbers, the least readily approximable numbers are those of the form $\displaystyle{ \frac{a\phi+b}{c\phi+d} }$ where $\displaystyle{ \phi }$ is the golden ratio, $\displaystyle{ a,b,c,d\in\Z }$ and $\displaystyle{ ad-bc=\pm1 }$.[4] (These numbers are said to be equivalent to $\displaystyle{ \phi }$.) If we omit these numbers, just as we omitted the rational numbers in Dirichlet's theorem, then we can improve the number 5 by 22. Again this new bound is best possible in the new setting, but this time the number 2, and numbers equivalent to it, limits the bound.[4] If we don't allow those numbers then we can again increase the number on the right hand side of the inequality from 22 to 221/5,[4] for which the numbers equivalent to $\displaystyle{ \frac{1+\sqrt{221}}{10} }$ limits the bound. The numbers generated show how well these numbers can be approximated, can can be seen as a property of the real numbers.

However, instead of considering Hurwitz's theorem (and the extensions mentioned above) as a property of the real numbers except certain special numbers, we can consider it as a property of each excluded number. Thus, the theorem can be interpreted as "numbers equivalent to $\displaystyle{ \phi }$, 2 or $\displaystyle{ \frac{1+\sqrt{221}}{10} }$ are among the least readily approximable irrational numbers." This leads us to consider how accurately each number can be approximated by rationals - specifically, by how much can the factor in Dirichlet's approximation theorem be increased to from 1 for that specific number.

## Definition

Mathematically, the Markov constant of irrational $\displaystyle{ \alpha }$ is defined as $\displaystyle{ M(\alpha)=\sup \{\lambda\in\R|\left\vert \alpha - \frac{p}{q} \right\vert\lt \frac{1}{\lambda q^2} \text{ has infinitely many solutions for }p,q\in\N \} }$.[5] If the set does not have an upper bound we define $\displaystyle{ M(\alpha)=\infty }$.

Alternatively, it can be defined as $\displaystyle{ \limsup_{k\to\infty}\frac{1}{k^2\left\vert \alpha-\frac{f(k)}{k} \right\vert} }$ where $\displaystyle{ f(k) }$ is defined as the closest integer to $\displaystyle{ \alpha k }$.

## Properties and results

Hurwitz's theorem implies that $\displaystyle{ M(\alpha)\ge\sqrt{5} }$ for all $\displaystyle{ \alpha\in\R }$.

If $\displaystyle{ \alpha = [a_0; a_1, a_2, ...] }$ is its continued fraction expansion then $\displaystyle{ M(\alpha)=\limsup_{k\to\infty}{([a_k; a_{k+1}, a_{k+2}, ...] + [0; a_{k-1}, a_{k-2}, ...,a_1,a_0])} }$.[5]

From the above, if $\displaystyle{ p=\limsup_{k\to\infty}{a_k} }$ then $\displaystyle{ p\lt M(\alpha)\lt p+2 }$. This implies that $\displaystyle{ M(\alpha)=\infty }$ if and only if $\displaystyle{ (a k) }$ is not bounded. In particular, $\displaystyle{ M(\alpha)\lt \infty }$ if $\displaystyle{ \alpha }$ is a quadratic irrationality. In fact, the lower bound for $\displaystyle{ M(\alpha) }$ can be strengthened to $\displaystyle{ M(\alpha)\ge\sqrt{p^2+4} }$, the tightest possible.[6]

The values of $\displaystyle{ \alpha }$ for which $\displaystyle{ M(\alpha)\lt 3 }$ are families of quadratic irrationalities having the same period (but at different offsets), and the values of $\displaystyle{ M(\alpha) }$ for these $\displaystyle{ \alpha }$ are limited to Lagrange numbers. There are uncountably many numbers for which $\displaystyle{ M(\alpha)=3 }$, no two of which have the same ending; for instance, for each number $\displaystyle{ \alpha = [\underbrace{1;1,...,1}_{r_1},2,2,\underbrace{1;1,...,1}_{r_2},2,2,\underbrace{1;1,...,1}_{r_3},2,2,...] }$ where $\displaystyle{ r_1\lt r_2\lt r_3\lt \cdots }$, $\displaystyle{ M(\alpha)=3 }$.[5]

If $\displaystyle{ \beta=\frac{p\alpha+q}{r\alpha+s} }$ where $\displaystyle{ p,q,r,s\in\Z }$ then $\displaystyle{ M(\beta)\ge\frac{M(\alpha)}{\left\vert ps-rq \right\vert} }$.[7] In particular if $\displaystyle{ \left\vert ps-rq \right\vert=1 }$ them $\displaystyle{ M(\beta)=M(\alpha) }$.[8]

The set $\displaystyle{ L=\{M(\alpha)|\alpha\in\R-\Q\} }$ forms the Lagrange spectrum. It contains the interval $\displaystyle{ [F,\infty] }$ where F is Freiman's constant.[8] Hence, if $\displaystyle{ m\gt F\approx4.52783 }$ then there exists irrational $\displaystyle{ \alpha }$ whose Markov constant is $\displaystyle{ m }$.

### Numbers having a Markov constant less than 3

Burger et al. (2002)[9] provides a formula for which the quadratic irrationality $\displaystyle{ \alpha_n }$ whose Markov constant is the nth Lagrange number:

$\displaystyle{ \alpha_n=\frac{2u-3m_n+\sqrt{9m_n^2-4}}{2m_n} }$ where $\displaystyle{ m_n }$ is the nth Markov number, and u is the smallest positive integer such that $\displaystyle{ m_n\mid u^2+1 }$.

Nicholls (1978)[10] provides a geometric proof of this (based on circles tangent to each other), providing a method that these numbers can be systematically found.

## Examples

A demonstration that 10/2 has Markov constant 10, as stated in the example below. This plot graphs y(k) = 1/k2|α-f(αk)/k| against log(k) (the natural log of k) where f(x) is the nearest integer to x. The dots at the top corresponding to an x-axis value of 0.7, 2.5, 4.3 and 6.1 (k=2,12,74,456) are the points for which the limit superior of 10 is approached.

### Markov constant of a number

Since $\displaystyle{ \frac{\sqrt{10}}{2}=[1;\overline{1,1,2}] }$,

\displaystyle{ \begin{align} M\left ( \frac{\sqrt{10}}{2} \right ) & = \max([1;\overline{2,1,1}]+[0;\overline{1,2,1}],[1;\overline{1,2,1}]+[0;\overline{2,1,1}],[2;\overline{1,1,2}]+[0;\overline{1,1,2}]) \\ & = \max\left ( \frac{2\sqrt{10}}{3},\frac{2\sqrt{10}}{3},\sqrt{10} \right ) \\ & = \sqrt{10}. \end{align} }

As $\displaystyle{ e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, \ldots, 1, 2n, 1, \ldots], M(e)=\infty }$ because the continued fraction representation of e is unbounded.

### Numbers $\displaystyle{ \alpha_n }$ having Markov constant less than 3

Consider $\displaystyle{ n=6 }$; Then $\displaystyle{ m_n=34 }$. By trial and error it can be found that $\displaystyle{ u=13 }$. Then

\displaystyle{ \begin{align} \alpha_6 & = \frac{2u-3m_6+\sqrt{9m_6^2-4}}{2m_6} \\[6pt] & = \frac{-76+\sqrt{10400}}{68} \\[6pt] &= \frac{-19+5\sqrt{26}}{17} \\[6pt] &=[0;\overline{2,1,1,1,1,1,1,2}]. \end{align} }