Markov constant

{{short description|Property of an irrational number}}

{{About|a property of an irrational number|integer solutions to the Markov Diophantine equation|Markov number|a set of real numbers|Markov spectrum}}

{{Infobox mathematics function|image={{Markov constant chart}}|name=Markov constant of a number|parity=even|domain=Irrational numbers|codomain=Lagrange spectrum with \infty|period=1|max=+\infty|min= {{math|{{sqrt|5}}}} |vr1=\phi|f1= {{math|{{sqrt|5}}}} |vr2= {{math|{{sqrt|2}}}} |f2={{math|2{{sqrt|2}}}}|notes=This function is undefined on rationals; hence, it is not continuous.}}

In number theory, specifically in Diophantine approximation theory, the Markov constant M(\alpha) of an irrational number \alpha is the factor for which Dirichlet's approximation theorem can be improved for \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 \pi\approx\frac{22}{7} is the best rational approximation among rational numbers with denominator up to 56.{{Cite web|url=https://oeis.org/A063673|title=A063673 (Denominators of sequence {3/1, 13/4, 16/5, 19/6, 22/7, 179/57, 201/64, 223/71, 245/78, 267/85, 289/92, 311/99, 333/106, ... } of approximations to Pi with increasing denominators, where each approximation is an improvement on its predecessors.) |last=Fernando|first=Suren L.|date=27 July 2001|website=The On-Line Encyclopedia of Integer Sequences|access-date=2 December 2019}} Also, some numbers can be approximated more readily than others. Dirichlet proved in 1840 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. Specifically, he proved that for any number \alpha there are infinitely many pairs of relatively prime numbers (p,q) such that \left|\alpha - \frac{p}{q}\right| < \frac{1}{q^2} if and only if \alpha is irrational.

51 years later, Hurwitz further improved Dirichlet's approximation theorem by a factor of {{radic|5}},{{cite journal|last=Hurwitz|first=A.|author-link=Adolf Hurwitz|year=1891|title=Ueber die angenäherte Darstellung der Irrationalzahlen durch rationale Brüche (On the approximate representation of irrational numbers by rational fractions)|journal=Mathematische Annalen|language=de|url=https://gdz.sub.uni-goettingen.de/id/PPN235181684_0039 |volume=39|issue=2|pages=279–284|doi=10.1007/BF01206656|jfm=23.0222.02|s2cid=119535189 }} contains the actual proof in German. improving the right-hand side from 1/q^2 to 1/\sqrt{5}q^2 for irrational numbers:

: \left|\alpha - \frac{p}{q}\right| < \frac{1}{\sqrt{5}q^2}.

The above result is best possible since the golden ratio \phi is irrational but if we replace {{radic|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 \alpha=\phi.

Furthermore, he showed that among the irrational numbers, the least readily approximable numbers are those of the form \frac{a\phi+b}{c\phi+d} where \phi is the golden ratio, a,b,c,d\in\Z and ad-bc=\pm1.{{Cite web|url=http://mathworld.wolfram.com/HurwitzsIrrationalNumberTheorem.html|title=Hurwitz's Irrational Number Theorem|last=Weisstein|first=Eric W.|date=25 November 2019|website=Wolfram Mathworld|access-date=2 December 2019}} (These numbers are said to be equivalent to \phi.) If we omit these numbers, just as we omitted the rational numbers in Dirichlet's theorem, then we can increase the number {{radic|5}} to 2{{radic|2}}. Again this new bound is best possible in the new setting, but this time the number square root of 2, and numbers equivalent to it, limits the bound. If we don't allow those numbers then we can again increase the number on the right hand side of the inequality from 2{{radic|2}} to {{radic|221}}/5, for which the numbers equivalent to \frac{1+\sqrt{221}}{10} limit the bound. The numbers generated show how well these numbers can be approximated; this 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 \phi, {{radic|2}} or \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 \alpha is defined as M(\alpha)=\sup \left\{\lambda\in\R : \left\vert \alpha - \frac{p}{q} \right\vert<\frac{1}{\lambda q^2} \text{ has infinitely many solutions for }p,q\in\N \right\}.{{Cite book|title=Fundamentals of Number Theory|last=LeVeque|first=William|publisher=Addison-Wesley Publishing Company, Inc.|year=1977|isbn=0-201-04287-8|pages=251–254}} If the set does not have an upper bound we define M(\alpha)=\infty.

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

Properties and results

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

If \alpha = [a_0; a_1, a_2, ...] is its continued fraction expansion then M(\alpha)=\limsup_{k\to\infty}{([a_{k+1}; a_{k+2}, a_{k+3}, ...] + [0; a_{k}, a_{k-1}, ...,a_2,a_1])}.

From the above, if p=\limsup_{k\to\infty}{a_k} then p. This implies that M(\alpha)=\infty if and only if (a_k) is not bounded. In particular, M(\alpha)<\infty if \alpha is a quadratic irrational number. In fact, the lower bound for M(\alpha) can be strengthened to M(\alpha)\ge\sqrt{p^2+4}, the tightest possible.{{Cite journal|last=Hancl|first=Jaroslav|date=January 2016|title=Second basic theorem of Hurwitz|journal=Lithuanian Mathematical Journal|volume=56|pages=72–76|doi=10.1007/s10986-016-9305-4|s2cid=124639896 }}

The values of \alpha for which M(\alpha)<3 are families of quadratic irrationalities having the same period (but at different offsets), and the values of M(\alpha) for these \alpha are limited to Lagrange numbers. There are uncountably many numbers for which M(\alpha)=3, no two of which have the same ending; for instance, for each number \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 r_1, M(\alpha)=3.

If \beta=\frac{p\alpha+q}{r\alpha+s} where p,q,r,s\in\Z then M(\beta)\ge\frac{M(\alpha)}{\left\vert ps-rq \right\vert}.{{Cite journal|arxiv=1510.02407|title=Markov constant and quantum instabilities|journal=Journal of Physics A: Mathematical and Theoretical|volume=49|issue=15|pages=155201|doi=10.1088/1751-8113/49/15/155201|year=2016|last1=Pelantová|first1=Edita|last2=Starosta|first2=Štěpán|last3=Znojil|first3=Miloslav|bibcode=2016JPhA...49o5201P|s2cid=119161523 }} In particular if \left\vert ps-rq \right\vert=1 then M(\beta)=M(\alpha) .{{Cite book|title=Encyclopaedia of Mathematics|last=Hazewinkel|first=Michiel|publisher=Springer Science & Business Media|year=1990|isbn=9781556080050|pages=106}}

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

= Numbers having a Markov constant less than 3 =

Burger et al. (2002){{Cite journal|last1=Burger|first1=Edward B.|last2=Folsom|first2=Amanda|last3=Pekker|first3=Alexander|last4=Roengpitya|first4=Rungporn|last5=Snyder|first5=Julia|date=2002|title=On a quantitative refinement of the Lagrange spectrum|journal=Acta Arithmetica|volume=102|issue=1|pages=59–60|bibcode=2002AcAri.102...55B|doi=10.4064/aa102-1-5|doi-access=free}} provides a formula for which the quadratic irrationality \alpha_n

whose Markov constant is the nth Lagrange number:

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

Nicholls (1978){{Cite journal|last=Nicholls|first=Peter|date=1978|title=Diophantine Approximation via the Modular Group|journal=Journal of the London Mathematical Society |series=Second Series|volume=17|pages=11–17|doi=10.1112/jlms/s2-17.1.11}} provides a geometric proof of this (based on circles tangent to each other), providing a method that these numbers can be systematically found.

Examples

{{Image frame

|caption=A demonstration that {{math|{{sqrt|10}}/2}} has Markov constant {{math|{{sqrt|10}}}}, as stated in the example below. This plot graphs {{math|y(k)}} = {{math|{{sfrac|k2{{!}}α-{{sfrac|f(αk)|k}}{{!}}}}}} against {{math|log(k)}} (the natural log of k) where {{math|f(x)}} is the nearest integer to {{math|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 {{math|{{sqrt|10}}}} is approached.

|max-width=30%|content={{Markov constant chart}}

|width=360

|align=right}}

= Markov constant of two numbers =

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

: \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

e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, \ldots, 1, 2n, 1, \ldots], M(e)=\infty

because the E (mathematical constant)#Representations is unbounded.

= Numbers &alpha;<sub>n</sub> having Markov constant less than 3 =

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

: \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}

See also

References