convergence of integer continued fractions: a geometric approach · 2019-10-27 · convergence of...

Convergence of integer continued fractions: Ageometric approach

Mairi Walker

The Open Universitymairi.walker@open.ac.uk

The University of Manchester

25th June 2015

Mairi Walker (The Open University) Convergence of Integer Continued Fractions 25th June 2015 1 / 1

Contents

Introduction Integer continued fractions

Integer continued fractions

A finite integer continued fraction is a continued fraction of the form

[b1,b2, . . . ,bn] = b1 +1

b3 · · ·+1bn

where b1,b2, . . . ∈ Z.

An infinite simple continued fraction is defined to be the limit

[b1,b2, . . . ] = limi→∞

[b1,b2, . . . ,bi ] ∈ R ∪ {∞},

of its sequence of convergents Ci = [b1,b2, . . .bi ], if it exists.

Integer continued fractions

A finite integer continued fraction is a continued fraction of the form

[b1,b2, . . . ,bn] = b1 +1

b3 · · ·+1bn

where b1,b2, . . . ∈ Z.

An infinite simple continued fraction is defined to be the limit

[b1,b2, . . . ] = limi→∞

[b1,b2, . . . ,bi ] ∈ R ∪ {∞},

of its sequence of convergents Ci = [b1,b2, . . .bi ], if it exists.

Convergence of continued fractions

An integer continued fraction is simple if for all i > 1, bi > 0. Everyinfinite simple continued fraction converges.

What about general integer continued fractions?

[0,1,−1,1,−1, . . .] = 0 +1

−1 +1

1 + . . .

C1 = 0, C2 = 1, C3 =∞, C4 = 0, C5 = 1, C6 =∞, . . .

When does an integer continued fraction converge?

[0,1,−1,1,−1, . . .] = 0 +1

−1 +1

1 + . . .

C1 = 0, C2 = 1, C3 =∞, C4 = 0, C5 = 1, C6 =∞, . . .

[0,1,−1,1,−1, . . .] = 0 +1

−1 +1

1 + . . .

C1 = 0, C2 = 1, C3 =∞, C4 = 0, C5 = 1, C6 =∞, . . .

[0,1,−1,1,−1, . . .] = 0 +1

−1 +1

1 + . . .

C1 = 0, C2 = 1, C3 =∞, C4 = 0, C5 = 1, C6 =∞, . . .

Introduction Convergence of continued fractions

What’s known about convergence?

TheoremA continued fraction [b1,b2, . . . ], with bi ∈ R and bi > 0 for i > 1,converges if and only if

∑∞i=1 bi diverges.

Theorem (Sleszynski-Pringsheim)If ai ,bi ∈ R with |bi+1| > |ai |+ 1 for all i , then

b1 +a1

b2 +a2

b3 + . . .

converges.

Can we find a necessary and sufficient condition for convergence ofinteger continued fractions?

b1 +a1

b2 +a2

b3 + . . .

converges.

b1 +a1

b2 +a2

b3 + . . .

converges.

The hyperbolic geometry of continued fractions The Modular group and continued fractions

The theta graphThe Modular group, Γ, is the group generated by the Möbius maps

S(z) = z + 1 and T (z) = −1z.

The elements of Γ are the Möbius maps

z 7→ az + bcz + d

with a,b, c,d ∈ Z and ad − bc = 1. They are isometries of thehyperbolic upper half-plane H.

S(z) = z + 1 and T (z) = −1z.

z 7→ az + bcz + d

S(z) = z + 1 and T (z) = −1z.

z 7→ az + bcz + d

Hyperbolic geometryThe hyperbolic plane is the set H = {z = x + iy ∈ C | y > 0} equippedwith the infinitesimal metric

ds2 =dx2 + dy2

Geodesic lines in H are vertical lines and arcs of circles perpendicularto the boundary R.

Rewriting integer continued fractions

We rewrite an integer continued fraction in the form

[b1,b2,b3, . . . ] = b1 −1

b2 −1

b3 − . . .

= b1 +1

−b2 +1

b3 + . . .

The Modular group and continued fractions

An element of Γ can be written Sb1TSb2T . . .SbnTSbn+1 .

Sb1TSb2T . . .SbnTSbn+1(∞) = b1 −1

b2 −1

b3 − . . .1bn

Conversely, [b1,b2, . . . ,bn] can be associated the wordSb1TSb2T . . .TSbn .

There is a correspondence between elements of Γ and finite integercontinued fractions.

Sb1TSb2T . . .SbnTSbn+1(∞) = b1 −

b2 −1

b3 − . . .1bn

b2 −1

b3 − . . .1bn

b2 −

b3 − . . .1bn

b2 −1

b3 − . . .1bn

b2 −1

b3 −

. . .1bn

b2 −1

b3 − . . .1bn

b2 −1

b3 − . . .1bn

b2 −1

b3 − . . .1bn

The hyperbolic geometry of continued fractions The Farey graph

The Farey graph

We work in the hyperbolic upper half-plane H. Let L denote the linesegment joining 0 to∞.

The Farey graph, F , is the orbit of L under Γ: Edges are images of L,and vertices are images of∞, under elements of Γ.

The Farey graphF is an infinite graph: The skeleton of a tessellation of H by idealhyperbolic triangles.

Vertices lie on R ∪ {∞}. They are precisely the rational numbers, plus∞. Each vertex has infinite valency. The vertices neighbouring∞ arethe integers.

Paths in the Farey graph

[b1,b2, . . . ,bn] = Sb1TSb2T . . .SbnT (∞)

so finite continued fractions are vertices of F . In particular, eachconvergent Ci of an infinite continued fraction is a vertex of F .

TheoremA sequence of vertices∞ = C0,C1,C2 . . . forms an infinite path in F ifand only if they are the consecutive convergents of an integercontinued fraction expansion.

We can reformulate questions about continued fractions into questionsabout paths. In particular, an infinite continued fraction converges ifand only if its corresponding path converges.

[b1,b2, . . . ,bn] = Sb1TSb2T . . .SbnT (∞)

Take, for example, [0,−2,1,3, . . . ]

C0 =∞, C1 = 0, C2 =12, C3 =

13, C4 =

27, . . .

Reformulating the problemThe coefficient bi corresponds to the number of triangle segmentsbetween the edges 〈Ci−2,Ci−1〉 and 〈Ci−1,Ci〉. We call bi the turningnumber at Ci−1.

Can we characterise convergent continued fractions in terms of thecoefficients bi?

↔Can we characterise convergent paths in terms of the turning numbers

Reformulating the problemThe coefficient bi corresponds to the number of triangle segmentsbetween the edges 〈Ci−2,Ci−1〉 and 〈Ci−1,Ci〉. We call bi the turningnumber at Ci−1.

Can we characterise convergent continued fractions in terms of thecoefficients bi?

↔Can we characterise convergent paths in terms of the turning numbers

The hyperbolic geometry of continued fractions A geometric approach to convergence

Convergence of integer continued fractionsTheoremAn infinite path in F with vertices∞ = C0,C1,C2, . . . converges to anirrational number x if and only if the sequence C0,C1,C2, . . . containsno constant subsequence.

Proof.=⇒ Clear. ⇐= Assume that {Ci} diverges, so has two limit points, v1and v2. There is some edge of F , with endpoints u and w ‘separating’v1 and v2.

Thus the path must pass through one of u or v infinitely many times,and has a convergent subsequence.

Proof.=⇒ Clear.

⇐= Assume that {Ci} diverges, so has two limit points, v1and v2. There is some edge of F , with endpoints u and w ‘separating’v1 and v2.

Convergence of paths in F

The integer continued fraction [b1,b2, . . . ] converges↔

The path with vertices C0,C1,C2, . . . converges↔

The sequence C0,C1,C2, . . . contains no constant subsequence

So to determine when an integer continued fractions converges, weneed to determine when there is some vertex that a path returns toinfinitely many times. Such a path can be split into a sequence ofconsecutive closed paths.

Can we characterise closed paths in F in terms of their turningnumbers?

So to determine when an integer continued fractions converges, weneed to determine when there is some vertex that a path returns toinfinitely many times.

Such a path can be split into a sequence ofconsecutive closed paths.

Closed paths in the Farey Graph Closed paths in the Farey graph

A Euclidean represenataion

A closed path surrounds a collection of faces forming a finite subgraphof F , the dual graph to which is a tree. The turning numbers determinethe number of triangles meeting at a vertex.

We view the faces as Euclidean triangles for simplicity - this does notaffect the structure of the graph.

Sleszynski-PringsheimWe can easily obtain the Sleszynski-Pringsheim theorem.

Theorem (Sleszynski-Pringsheim)If bi ∈ Z with |bi | > 2 for all i , then [b1,b2,b3, . . . ] converges.

The condition that bi 6= 0,±1, prevents there from being any closedsubpaths in the path∞,C1,C2, . . . .

This condition is far from being necessary as convergent paths canhave infinitely many closed subpaths.

Sleszynski-PringsheimWe can easily obtain the Sleszynski-Pringsheim theorem.

Theorem (Sleszynski-Pringsheim)If bi ∈ Z with |bi | > 2 for all i , then [b1,b2,b3, . . . ] converges.

The condition that bi 6= 0,±1, prevents there from being any closedsubpaths in the path∞,C1,C2, . . . .

This condition is far from being necessary as convergent paths canhave infinitely many closed subpaths.

Some graph theoryWe can obtain a condition using basic graph theory. Let L denote thelength of a closed path, and N the number of triangles it surrounds.

Notice that L = N + 2.

V = L = N + 2

E = 2N + 1 = 2L− 3

F = N + 1 = L− 1

Some graph theoryWe can obtain a condition using basic graph theory. Let L denote thelength of a closed path, and N the number of triangles it surrounds.Notice that L = N + 2.

V = L = N + 2

E = 2N + 1 = 2L− 3

F = N + 1 = L− 1

Some graph theoryWe can obtain a condition using basic graph theory. Let L denote thelength of a closed path, and N the number of triangles it surrounds.Notice that L = N + 2.

V = L = N + 2

E = 2N + 1 = 2L− 3

F = N + 1 = L− 1

Some graph theory

Now ∑bi = 3N = 3(L− 2).

∑bi = 27 = 3(11− 2) = 3(L− 2)

Some graph theoryThis condition is necessary but not sufficient.

Both∑

bi and L are invariant under permutation of the bi , but theproperty that a path is closed is not.

1,3,1,3,1,3 1,3,3,1,1,3

We need a different approach.

Both∑

1,3,1,3,1,3 1,3,3,1,1,3

Both∑

1,3,1,3,1,3 1,3,3,1,1,3

Both∑

1,3,1,3,1,3 1,3,3,1,1,3

Closed paths in the Farey Graph An algorithmic approach

Reducing paths

Can we find an algorithm to place a path in a simplified form thatallows us to easily tell whether or not it converges?

The idea is to ‘shrink’ closed paths to the simplest possible closedpath.

Then we can easily identify closed paths.

Reducing paths

Reducing closed pathsWe use three operations.

1 Replace . . . k ,1, k ′, . . . with . . . k − 1, k ′ − 1, . . . .

2 Replace . . . k ,−1, k ′, . . . with . . . k + 1, k ′ + 1, . . . .

3 Replace . . . k ,0, k ′, . . . with . . . k + k ′, . . . .

Reducing a closed path

We can apply these operations to reduce a closed path.

k ,2,4,1,3,3,1,2,5,1,4, k ′

k ,2,3,2,3,1,2,5,1,4, k ′

k ,2,3,2,2,1,5,1,4, k ′

k ,2,3,2,1,4,1,4, k ′

k ,2,3,1,3,1,4, k ′

k ,2,2,2,1,4, k ′

k ,2,2,1,3, k ′

k ,2,1,2, k ′

k ,1,1, k ′

A reduction algorithmWe have a procedure that reduces closed paths to closed pathstraversing one triangle (subsequences k ,±1,±1 with k 6= ±1). We callthese loops.

We can apply the procedure systematically to place a path intoreduced form.

A path returns to some vertex infinitely many times if and only if itsreduced form contains infintiely many loops starting at the same point.

A reduction algorithmWe have a procedure that reduces closed paths to closed pathstraversing one triangle (subsequences k ,±1,±1 with k 6= ±1). We callthese loops.

We can apply the procedure systematically to place a path intoreduced form.

A path returns to some vertex infinitely many times if and only if itsreduced form contains infintiely many loops starting at the same point.

Some consequences

If a path in reduced form has a sequence of turning numbersb1,b2,b3, . . . eventually equal to

k1,±1,±1, k2,±1,±1, k3,±1,±1 . . .

with ki 6= ±1 then the continued fraction diverges.

If after some point there are no occurances of a subsequencek ,±1,±1 with k 6= ±1 then the path converges.

Some consequences

If a path in reduced form has a sequence of turning numbersb1,b2,b3, . . . eventually equal to

k1,±1,±1, k2,±1,±1, k3,±1,±1 . . .

with ki 6= ±1 then the continued fraction diverges.

If after some point there are no occurances of a subsequencek ,±1,±1 with k 6= ±1 then the path converges.

The problems

In all other cases, we have infinitely many loops, but only finitely manyare ever consecutive.

Such paths can either converge or diverge.

How can we tell?

Thanks for listening :)

convergence of integer continued fractions: a geometric approach · 2019-10-27 · convergence of...

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