Generalized Markov constant and continued fractions · Introduction Lagrange spectrum Properties of S( )Approximation of values of ThesetL 0 p1 5 p 1 8 p 5 221 p 13 1571 1 3 largest

$Page 1: Generalized Markov constant and continued fractions · Introduction Lagrange spectrum Properties of S( )Approximation of values of ThesetL 0 p1 5 p 1 8 p 5 221 p 13 1571 1 3 largest$
Generalized Markov constant and continuedfractions

Štepán Starosta

Faculty of Information TechnologyCzech Technical University in Prague

1 / 24

Introduction Lagrange spectrum Properties of S(α) Approximation of values of S(α)

Motivation

When analyzing the spectrum of the operator

�f (x , y) = λf (x , y) , � =∂2

∂x2 −∂2

∂y2 , f |∂R = 0,

acting on R with R being a rectangle with sides a and b:

b

aR

the following set comes up

S(α) = set of all accumulation points of

{m2(

km− α

): k ,m ∈ Z

},

where α = ab ∈ R.

2 / 24


Motivation

When analyzing the spectrum of the operator

�f (x , y) = λf (x , y) , � =∂2

∂x2 −∂2

∂y2 , f |∂R = 0,

acting on R with R being a rectangle with sides a and b:

b

aR

the following set comes up


{m2(

km− α

): k ,m ∈ Z

},

where α = ab ∈ R.

2 / 24


Questions


{m2(

km− α

): k ,m ∈ Z

}

Questions:

1 Is the set empty? When is it empty?

2 How does it depend on α?

3 Do we know an element from the set?

4 Is the set finite or infinite?

5 How does the set look like?

6 Can we obtain all/some elements of the set efficiently?

7 ...

3 / 24


Questions


{m2(

km− α

): k ,m ∈ Z

}Questions:

1 Is the set empty? When is it empty?

2 How does it depend on α?

3 Do we know an element from the set?

4 Is the set finite or infinite?

5 How does the set look like?

6 Can we obtain all/some elements of the set efficiently?

7 ...

3 / 24


Markov constant

S(α) = set of all accumulation points of{

m2(

km − α

): k ,m ∈ Z

}Let ‖x‖ denote the distance to the nearest integer:‖x‖ = min{|x − n| : n ∈ Z}.

The Markov constant of α is

µ(α) = lim infm→+∞

m ‖mα‖

=

if α 6∈ Q

min |S(α)|

L = {µ(α) : α ∈ R}

(L (or 1L ) is sometimes called the Lagrange spectrum)

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Markov constant


m2(

km − α

): k ,m ∈ Z




m ‖mα‖ =

if α 6∈ Q

min |S(α)|

L = {µ(α) : α ∈ R}


4 / 24


Markov constant


m2(

km − α

): k ,m ∈ Z




m ‖mα‖ =

if α 6∈ Q

min |S(α)|

L = {µ(α) : α ∈ R}


4 / 24


Hurwitz’s theorem


m2(

km − α

): k ,m ∈ Z

}

Hurwitz’s theorem (1891): there exists infinitely many p, q ∈ Z suchthat ∣∣∣∣α− p

q

∣∣∣∣ < 1√5q2

(also follows from an earlier result of Markov)

thus, µ(α) ≤ 1√5

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Hurwitz’s theorem


m2(

km − α

): k ,m ∈ Z

}

Hurwitz’s theorem (1891): there exists infinitely many p, q ∈ Z suchthat ∣∣∣∣α− p

q

∣∣∣∣ < 1√5q2

(also follows from an earlier result of Markov)

thus, µ(α) ≤ 1√5

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The set L

0

1√5

1√8

5√221

13√1571

13

largest accumulation point

discrete part

1√12

partial coefficients in {1, 2}

1√13

at least one partial coefficient larger than 2

F

F = 555391024−70937√

4622507812168 ≈ 0.22

(Freiman’s constant, 1975)

continuous part (Hall’s ray)

explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


The set L

0 1√5

1√8

5√221

13√1571

13


discrete part

1√12


1√13


F

F = 555391024−70937√

4622507812168 ≈ 0.22



explored jungle

largest gap

6 / 24


Calculation of Markov constants


m ‖mα‖ =

if α 6∈ Q

min |S(α)| ,

How do we calculate Markov constants?

If α ∈ Q, µ(α) = 0.

If α 6∈ Q, we can use the continued fraction expansion of α:

(Legendre) If α is irrational and∣∣∣α− p

q

∣∣∣ < 12q2 , then p

q is a convergent

of α.

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m ‖mα‖ =

if α 6∈ Q

min |S(α)| ,


If α ∈ Q, µ(α) = 0.



q

∣∣∣ < 12q2 , then p

q is a convergent

of α.

7 / 24




m ‖mα‖ =

if α 6∈ Q

min |S(α)| ,


If α ∈ Q, µ(α) = 0.



q

∣∣∣ < 12q2 , then p

q is a convergent

of α.

7 / 24


Calculation of µ(α) for irrational α

Let pNqN

be the convergents of α = [a0, a1, a2, . . .], then

q2N

(α− pN

qN

)= (−1)N 1

[aN+1, aN+2, . . .] + [0, aN , . . . , a1].

µ(α) =

µ(α) = lim infm→+∞m ‖mα‖

lim infN→+∞

1[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

= min |M(α)

from the exercise

| .

µ

(1 +√

52

)= µ([1]) =

11+√

52 + 1+

√5

2 − 1=

1√5

µ(

1 +√

2)

= µ([2]) =1

1 +√

2 + 1 +√

2− 2=

1

2√

2=

1√8

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Let pNqN


q2N

(α− pN

qN

)= (−1)N 1

[aN+1, aN+2, . . .] + [0, aN , . . . , a1].

µ(α) =


lim infN→+∞

1[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

= min |M(α)

from the exercise

| .

µ

(1 +√

52

)= µ([1]) =

11+√

52 + 1+

√5

2 − 1=

1√5

µ(

1 +√

2)

= µ([2]) =1

1 +√

2 + 1 +√

2− 2=

1

2√

2=

1√8

8 / 24



Let pNqN


q2N

(α− pN

qN

)= (−1)N 1

[aN+1, aN+2, . . .] + [0, aN , . . . , a1].

µ(α) =


lim infN→+∞

1[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

= min |M(α)

from the exercise

| .

µ

(1 +√

52

)= µ([1]) =

11+√

52 + 1+

√5

2 − 1=

1√5

µ(

1 +√

2)

= µ([2]) =1

1 +√

2 + 1 +√

2− 2=

1

2√

2=

1√8

8 / 24



Let pNqN


q2N

(α− pN

qN

)= (−1)N 1

[aN+1, aN+2, . . .] + [0, aN , . . . , a1].

µ(α) =


lim infN→+∞

1[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

= min |M(α)

from the exercise

| .

µ

(1 +√

52

)= µ([1]) =

11+√

52 + 1+

√5

2 − 1=

1√5

µ(

1 +√

2)

= µ([2]) =1

1 +√

2 + 1 +√

2− 2=

1

2√

2=

1√8

8 / 24


Easy cases

Set

M(α) = acc. pts of

{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}

M(α) ⊂ S(α)

S(α) ∩(−1

2,

12

)= M(α) ∩

(−1

2,

12

)If α is a quadratic irrational number, its continued fraction is eventuallyperiodic, and we may find M(α) exactly (given as the exercise).

If ` is the period of the continued fraction expansion of α, then

#M(α) ≤

{` if ` is even;

2` otherwise.

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Easy cases

Set

M(α) = acc. pts of

{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}

M(α) ⊂ S(α)

S(α) ∩(−1

2,

12

)= M(α) ∩

(−1

2,

12

)If α is a quadratic irrational number, its continued fraction is eventuallyperiodic, and we may find M(α) exactly (given as the exercise).If ` is the period of the continued fraction expansion of α, then

#M(α) ≤

{` if ` is even;

2` otherwise.

9 / 24


Easy cases - examples

M

(√5 + 12

)= M([1]) =

{− 1√

5,

1√5

}

M

(√37 + 4

7

)= M([1, 2, 3]) ={

237

√37,− 2

37

√37,− 3

74

√37,− 7

74

√37,

374

√37,

774

√37

}

M(√

6 + 2)

= M([4, 2]) =

{− 1

12

√6,

16

√6

}

10 / 24



M

(√5 + 12

)= M([1]) =

{− 1√

5,

1√5

}

M

(√37 + 4

7

)= M([1, 2, 3]) ={

237

√37,− 2

37

√37,− 3

74

√37,− 7

74

√37,

374

√37,

774

√37

}

M(√

6 + 2)

= M([4, 2]) =

{− 1

12

√6,

16

√6

}

10 / 24



M

(√5 + 12

)= M([1]) =

{− 1√

5,

1√5

}

M

(√37 + 4

7

)= M([1, 2, 3]) ={

237

√37,− 2

37

√37,− 3

74

√37,− 7

74

√37,

374

√37,

774

√37

}

M(√

6 + 2)

= M([4, 2]) =

{− 1

12

√6,

16

√6

}

10 / 24


From µ(α) to S(α)


m2(

km − α

): k ,m ∈ Z

}• S(α) is a closed set, closed under multiplication by squares

• α ∈ Q⇒ S(α) = ∅• α /∈ Q⇒ min |S(α)| = µ(α) and S(α) is not empty

• If a, b, c, d ∈ Z such that ad − bc = ∆ 6= 0, then

∆ · S(α) ⊂ S(

aα + bcα + d

).

• S([an+k , an+1+k , an+2+k , . . .]) =(−1)kS([an, an+1, an+2, . . .]) for any k , n ∈ N

• If α /∈ Q and I is an interval, then there exists β ∈ I such thatS(α) = S(β).

11 / 24




m2(

km − α

): k ,m ∈ Z


• α ∈ Q⇒ S(α) = ∅

• α /∈ Q⇒ min |S(α)| = µ(α) and S(α) is not empty


∆ · S(α) ⊂ S(

aα + bcα + d

).



11 / 24




m2(

km − α

): k ,m ∈ Z




∆ · S(α) ⊂ S(

aα + bcα + d

).



11 / 24




m2(

km − α

): k ,m ∈ Z




∆ · S(α) ⊂ S(

aα + bcα + d

).



11 / 24




m2(

km − α

): k ,m ∈ Z




∆ · S(α) ⊂ S(

aα + bcα + d

).



11 / 24




m2(

km − α

): k ,m ∈ Z




∆ · S(α) ⊂ S(

aα + bcα + d

).



11 / 24


α quadratic irrational outside M(α)

a different approach leads to the following

S(α) = S(α′) = C · {NQ(α)/Q(k + mα) : k ,m ∈ Z}

for some constant C depending on α(joint work with V. Kála, work in progress)

12 / 24


Well-approximable numbers

α is well approximable it it has unbounded partial coefficients

α is well approximableα/∈Q⇔ µ(α) = 0⇔ 0 ∈ S(α)

Example: for e = [2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, . . .] we have

(−12 ,

12 ) ∩ S(e) = {0} and {a + 1

2 : a ∈ Z} ⊂ S(e).

Let α be an irrational well approximable number. For any n ∈ N theinterval [n, n + 1] or the interval [−n − 1,−n] has a non-emptyintersection with S(α).

13 / 24




α is well approximableα/∈Q⇔ µ(α) = 0

⇔ 0 ∈ S(α)


(−12 ,

12 ) ∩ S(e) = {0} and {a + 1

2 : a ∈ Z} ⊂ S(e).


13 / 24






(−12 ,

12 ) ∩ S(e) = {0} and {a + 1

2 : a ∈ Z} ⊂ S(e).


13 / 24






(−12 ,

12 ) ∩ S(e) = {0} and {a + 1

2 : a ∈ Z} ⊂ S(e).


13 / 24






(−12 ,

12 ) ∩ S(e) = {0} and {a + 1

2 : a ∈ Z} ⊂ S(e).


13 / 24


Badly approximable numbers

If an irrational number is not well approximable, it is badlyapproximable.

Recall

S(α) ⊃ M(α) = acc. pts of

{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}

Set F(m) = {[t, a1, a2, a3, . . .] : t ∈ Z, 1 ≤ ai ≤ m }F(4) + F(4) = R [Hall]F(3) + F(4) = R = F(2) + F(5)F(3) + F(3) 6= R 6= F(2) + F(4)

14 / 24




Recall


{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}


14 / 24




Recall


{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}

Set F(m) = {[t, a1, a2, a3, . . .] : t ∈ Z, 1 ≤ ai ≤ m }F(4) + F(4) = R [Hall]

F(3) + F(4) = R = F(2) + F(5)F(3) + F(3) 6= R 6= F(2) + F(4)

14 / 24




Recall


{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}

Set F(m) = {[t, a1, a2, a3, . . .] : t ∈ Z, 1 ≤ ai ≤ m }F(4) + F(4) = R [Hall]F(3) + F(4) = R = F(2) + F(5)

F(3) + F(3) 6= R 6= F(2) + F(4)

14 / 24




Recall


{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}


14 / 24


Shape of F(2)

0 1

[0, . . .]

[0, 1, . . .][0, 2, . . .]

[0, 1, 1, . . .]

[0, 1, 2, . . .]

[0, 2, 1, . . .]

[0, 2, 2, . . .]

[0, 1, 1, 2 . . .]

[0, 1, 1, 1 . . .]

[0, 1, 2, 2 . . .]

[0, 1, 2, 1 . . .]

[0, 2, 1, 2 . . .]

[0, 2, 1, 1 . . .]

[0, 2, 2, 2 . . .]

[0, 2, 2, 1 . . .]

15 / 24


Shape of F(2)

0 1

[0, . . .]

[0, 1, . . .][0, 2, . . .]

[0, 1, 1, . . .]

[0, 1, 2, . . .]

[0, 2, 1, . . .]

[0, 2, 2, . . .]

[0, 1, 1, 2 . . .]

[0, 1, 1, 1 . . .]

[0, 1, 2, 2 . . .]

[0, 1, 2, 1 . . .]

[0, 2, 1, 2 . . .]

[0, 2, 1, 1 . . .]

[0, 2, 2, 2 . . .]

[0, 2, 2, 1 . . .]

15 / 24


Shape of F(2)

0 1

[0, . . .]

[0, 1, . . .][0, 2, . . .]

[0, 1, 1, . . .]

[0, 1, 2, . . .]

[0, 2, 1, . . .]

[0, 2, 2, . . .]

[0, 1, 1, 2 . . .]

[0, 1, 1, 1 . . .]

[0, 1, 2, 2 . . .]

[0, 1, 2, 1 . . .]

[0, 2, 1, 2 . . .]

[0, 2, 1, 1 . . .]

[0, 2, 2, 2 . . .]

[0, 2, 2, 1 . . .]

15 / 24


Shape of F(2)

0 1

[0, . . .]

[0, 1, . . .][0, 2, . . .]

[0, 1, 1, . . .]

[0, 1, 2, . . .]

[0, 2, 1, . . .]

[0, 2, 2, . . .]

[0, 1, 1, 2 . . .]

[0, 1, 1, 1 . . .]

[0, 1, 2, 2 . . .]

[0, 1, 2, 1 . . .]

[0, 2, 1, 2 . . .]

[0, 2, 1, 1 . . .]

[0, 2, 2, 2 . . .]

[0, 2, 2, 1 . . .]

15 / 24


Badly approximable numbers II

There exist badly approximable irrational numbers α1, α2 and α3 suchthat:

1 S(α1) = R;

2 S(α2) = (−∞,−ε] ∪ [ε,+∞), where ε =√

28 ≈ 0.18;

3 the Hausdorff dimension of S(α3) ∩(−1

2 ,12

)is positive but less

than 1.

[Pelantová, S., Znojil]

16 / 24


General case

What can be done for α irrational not quadratic?

M(α) = acc. pts of

{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}

To find approximations of values of M(α), we need subsequences Ni

such that

limi→+∞

(−1)Ni

[aNi+1 , aNi+2 , . . .] + [0, aNi , . . . , a1]

exists

17 / 24


General case

What can be done for α irrational not quadratic?

M(α) = acc. pts of

{(−1)N

[aN+1, aN+2, . . .] + [0, aN , . . . , a1]

}

To find approximations of values of M(α), we need subsequences Ni

such that

limi→+∞

(−1)Ni

[aNi+1 , aNi+2 , . . .] + [0, aNi , . . . , a1]

exists

17 / 24


General case — fixed points of morphisms

Let f be the Fibonacci word over {1, 2}: α = [f] ≈ 1.3879545879671(transcendental [Allouche])

The word f is fixed by ϕ : 1 7→ 12, 2 7→ 1.

Its suitable prefixes may be obtained by using the mappingΦ : w 7→ ϕ(w)1.

1Φ3

7−→ w1Φ3

7−→ w2 . . .

w1 = Ψ3(1) = 12112121121.

w2 = Ψ3(w1) =121121211211212112121 12112121121︸︷︷︸

w1

121211212112112121121.

To obtain the words aNi+1 , aNi+2 . . . and aNi · · · a1, we cut each wi atthe same position and reverse the prefix. (We need to deal with thesign also...)

18 / 24






1Φ3

7−→ w1Φ3

7−→ w2 . . .

w1 = Ψ3(1) = 12112121121.

w2 = Ψ3(w1) =121121211211212112121 12112121121︸︷︷︸

w1

121211212112112121121.


18 / 24






1Φ3

7−→ w1Φ3

7−→ w2 . . .

w1 = Ψ3(1) = 12112121121.

w2 = Ψ3(w1) =121121211211212112121 12112121121︸︷︷︸

w1

121211212112112121121.


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1Φ3

7−→ w1Φ3

7−→ w2 . . .

w1 = Ψ3(1) = 12112121121.

w2 = Ψ3(w1) =121121211211212112121 12112121121︸︷︷︸

w1

121211212112112121121.


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1Φ3

7−→ w1Φ3

7−→ w2 . . .

w1 = Ψ3(1) = 12112121121.

w2 = Ψ3(w1) =121121211211212112121 12112121121︸︷︷︸

w1

121211212112112121121.


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1Φ3

7−→ w1Φ3

7−→ w2 . . .

w1 = Ψ3(1) = 12112121121.

w2 = Ψ3(w1) =121121211211212112121 12112121121︸︷︷︸

w1

121211212112112121121.


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Fixed points - another example

Let ϕ : 1 7→ 122, 2 7→ 112.

Another technique: produce factors starting with the factor 21:

2|1 ϕ7−→ 112|122ϕ7−→ 122122112|122112112

and cutting at the same position relatively to |.

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Fixed points - another example

Let ϕ : 1 7→ 122, 2 7→ 112.

Another technique: produce factors starting with the factor 21:

2|1 ϕ7−→ 112|122ϕ7−→ 122122112|122112112

and cutting at the same position relatively to |.

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Fixed points — systematic approach

Every factor resides inside a unique bispecial factor of minimal length.(A factor of a word is bispecial if it can be extended to the right and tothe left in more than one way.)

To enumerate all factors, we may enumerate all bispecial factors andkeep cutting at the same positions.

For instance for ϕ : 1 7→ 122, 2 7→ 112

we set

Φ : w 7→ lcs

longest common suffix

(ϕ(1), ϕ(2))ϕ(w)lcp

longest common prefix

(ϕ(1), ϕ(2)) = 2ϕ(w)1

all bispecial factors are{

Φk (w) : k ∈ N,w ∈ {1, 2, 12, 21}}

([Klouda]in general for circular morphisms)

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we set

Φ : w 7→ lcs


(ϕ(1), ϕ(2))ϕ(w)lcp


(ϕ(1), ϕ(2)) = 2ϕ(w)1


Φk (w) : k ∈ N,w ∈ {1, 2, 12, 21}}


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we set

Φ : w 7→ lcs


(ϕ(1), ϕ(2))ϕ(w)lcp


(ϕ(1), ϕ(2)) = 2ϕ(w)1


Φk (w) : k ∈ N,w ∈ {1, 2, 12, 21}}


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Algebraic operations on continued fractions

Is there a better way to sum [aNi+1 , aNi+2 , . . .] + [0, aNi , . . . , a1]?

Denote

Fa0a1a2···an = {x ∈ R : x = [a0, a1, a2, . . . , an, . . .]}

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Is there a better way to sum [aNi+1 , aNi+2 , . . .] + [0, aNi , . . . , a1]?

Denote

Fa0a1a2···an = {x ∈ R : x = [a0, a1, a2, . . . , an, . . .]}

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In general, any operation

(x , y) 7→ p(x , y)

q(x , y)with p(x , y), q(x , y) polynomials over Q

can be performed in a similar manner [Raney; Gosper; Vuillemin;Liarder & Stambul . . . ]

However

Theorem (Kurka, Vávra)

If G is an analytic function computed in a sofic Möbius number systemusing a finite state transducer, then G is a Möbius transformation (thatis, a mapping of type x 7→ ax+b

cx+d ).

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Questions and problems/exercises

1 Find values of M(α) for α quadratic and exhibit various possiblebehaviors (see the exercise);

2 Find approximate values [with given precision] of M(α) based oncombinatorial properties of the continued fraction expansion ofα = [u]:

1 u is a fixed point of morphism;2 ?

Sample code located at http://goo.gl/KxknZ8

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http://goo.gl/KxknZ8

Thank you

Documents

Generalized Markov constant and continued fractions · Introduction Lagrange spectrum Properties of S( )Approximation of values of ThesetL 0 p1 5 p 1 8 p 5 221 p 13 1571 1 3 largest