Satyadev Nandakumar Santosh Kumar Vangepalli Indian ...PDF/FSDNLN.pdf · 1 / 21 Normality and...

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Normality and Finite State Dimension of Liouville Numbers

Satyadev Nandakumar Santosh Kumar VangepalliIndian Institute of Technology Kanpur

September 2013

Introduction

⊲ Introduction

Definition

Examples

Properties

Transcendence

Normality andDiophantineApproximation

Normality ofLiouville numbers

Generalization:Finite StateDimension

Open Questions

2 / 21

Normal Numbers

Introduction

⊲ Definition

Examples

Properties

Transcendence

Open Questions

3 / 21

Definition. [Normal Binary Sequence]

Normal Numbers

Introduction

⊲ Definition

Examples

Properties

Transcendence

Open Questions

3 / 21

An infinite binary sequence ω is said to be normal in base 2 if forevery natural number k and every k-long string w,

limn→∞

|{i ∈ N | ω[i . . . i+ n− 1] = w}|n

Normal Numbers

Introduction

⊲ Definition

Examples

Properties

Transcendence

Open Questions

3 / 21

limn→∞

|{i ∈ N | ω[i . . . i+ n− 1] = w}|n

For example, the asymptotic density of 0s is 1/2,

Normal Numbers

Introduction

⊲ Definition

Examples

Properties

Transcendence

Open Questions

3 / 21

limn→∞

|{i ∈ N | ω[i . . . i+ n− 1] = w}|n

For example, the asymptotic density of 0s is 1/2, the asymptoticdensity of 01 is 1

4 , and so on.

Normal Numbers

Introduction

⊲ Definition

Examples

Properties

Transcendence

Open Questions

3 / 21

limn→∞

|{i ∈ N | ω[i . . . i+ n− 1] = w}|n

For example, the asymptotic density of 0s is 1/2, the asymptoticdensity of 01 is 1

4 , and so on.

A real number in [0,1] will be called normal if its binaryexpansion is a normal sequence.

Examples of Normal Numbers

Introduction

Definition

⊲ Examples

Properties

Transcendence

Open Questions

4 / 21

Are there normal numbers in [0,1]?

Introduction

Definition

⊲ Examples

Properties

Transcendence

Open Questions

4 / 21

Theorem (Borel, 1909). The set of normal numbers in [0,1] hasLebesgue measure 1.

Introduction

Definition

⊲ Examples

Properties

Transcendence

Open Questions

4 / 21

But are there any constructions known?

Introduction

Definition

⊲ Examples

Properties

Transcendence

Open Questions

4 / 21

A simple example of a normal binary sequence:

Introduction

Definition

⊲ Examples

Properties

Transcendence

Open Questions

4 / 21

Introduction

Definition

⊲ Examples

Properties

Transcendence

Open Questions

4 / 21

0 1 00 01 10 11

Introduction

Definition

⊲ Examples

Properties

Transcendence

Open Questions

4 / 21

0 1 00 01 10 11 000 . . .

Introduction

Definition

⊲ Examples

Properties

Transcendence

Open Questions

4 / 21

0 1 00 01 10 11 000 . . .

This is known as the Champernowne sequence (1933).

Properties of Normal Numbers

Introduction

Definition

Examples

⊲ Properties

Transcendence

Open Questions

5 / 21

No normal number can be rational.

Introduction

Definition

Examples

⊲ Properties

Transcendence

Open Questions

5 / 21

Definition. A real number is algebraic if it is the root of apolynomial with integer coefficients. e.g.

√2 is the root of the

polynomial x2 − 2.

Introduction

Definition

Examples

⊲ Properties

Transcendence

Open Questions

5 / 21

polynomial x2 − 2.Non-algebraic irrational numbers are called transcendental. (e.g.π, e etc.)

Introduction

Definition

Examples

⊲ Properties

Transcendence

Open Questions

5 / 21

Are normal numbers always transcendental? (We do not know)

Introduction

Definition

Examples

⊲ Properties

Transcendence

Open Questions

5 / 21

Are normal numbers always transcendental? (We do not know)

Can we construct normal numbers which are provablytranscendental? What can we say about normal numbers inspecial classes of transcendental numbers?

Transcendence and Normality

Introduction

Definition

Examples

Properties

⊲ Transcendence

Open Questions

6 / 21

Champernowne number is a transcendental number. (Mahler,1937)

Introduction

Definition

Examples

Properties

⊲ Transcendence

Open Questions

6 / 21

Some transcendental numbers are far from normal.

Example. Liouville constant1 defined as

α =∞∑

is a transcendental number.

Introduction

Definition

Examples

Properties

⊲ Transcendence

Open Questions

6 / 21

Some transcendental numbers are far from normal.

Example. Liouville constant1 defined as

α =∞∑

is a transcendental number.

α cannot be normal - e.g. 111 never appears in α.

1in base 2

Normality and Diophantine Approximation

Introduction

Liouville Numbers

Open Questions

7 / 21

Liouville Numbers

8 / 21

Theorem (Liouville, 1854). A real number γ is an algebraic number of degreed. Then there is a constant C such that for every pair of natural numbers aand b,

∣γ − a

Liouville Numbers

8 / 21

∣γ − a

We can define the following class of transcendentals.

Liouville Numbers

8 / 21

∣γ − a

Definition. A real number β is called a Liouville Number if there is a

sequence of rationals(

i=1such that

β − piqi

Liouville Numbers

8 / 21

∣γ − a

Definition. A real number β is called a Liouville Number if there is a

sequence of rationals(

i=1such that

β − piqi

Liouville’s constant is a Liouville number. (first number provedtranscendental).

Normality of Liouville numbers

Introduction

⊲Normality ofLiouville numbers

Question

de Bruijn strings

Construction

Liouvilleness

Normality

Open Questions

9 / 21

Normality of Liouville Numbers

Introduction

⊲ Question

de Bruijn strings

Construction

Liouvilleness

Normality

Open Questions

10 / 21

Liouville numbers have Lebesgue measure 0. (e.g. Oxtoby, 1980)

Introduction

⊲ Question

de Bruijn strings

Construction

Liouvilleness

Normality

Open Questions

10 / 21

Liouville numbers have Lebesgue measure 0. (e.g. Oxtoby, 1980)They even have Hausdorff dimension 0. (e.g. Oxtoby, 1980).Further, they have constructive Hausdorff Dimension 0. (Staiger,2002)

Introduction

⊲ Question

de Bruijn strings

Construction

Liouvilleness

Normality

Open Questions

10 / 21

No easy measure-theoretic existence proof. (A measure 0 setmay or may not meet a measure 1 set.)

Introduction

⊲ Question

de Bruijn strings

Construction

Liouvilleness

Normality

Open Questions

10 / 21

Liouville’s constant is non-normal.

Introduction

⊲ Question

de Bruijn strings

Construction

Liouvilleness

Normality

Open Questions

10 / 21

Liouville’s constant is non-normal.

Surprisingly, there are normal Liouville numbers.

Idea: de Bruijn Strings

Introduction

Question

⊲ de Bruijn strings

Construction

Liouvilleness

Normality

Open Questions

11 / 21

A binary string w of length 2n is called a de-Bruijn string oforder n if every n length string occurs exactly once in w, whenlooked in a cyclic manner.

Introduction

Question

Construction

Liouvilleness

Normality

Open Questions

11 / 21

e.g. 0110 is a de-Bruin sequence of order 2:

Introduction

Question

Construction

Liouvilleness

Normality

Open Questions

11 / 21

0110 0110

Introduction

Question

Construction

Liouvilleness

Normality

Open Questions

11 / 21

0110 0110 0110

Introduction

Question

Construction

Liouvilleness

Normality

Open Questions

11 / 21

0110 0110 0110 0110

Introduction

Question

Construction

Liouvilleness

Normality

Open Questions

11 / 21

0110 0110 0110 0110

Theorem (N. G. de Bruijn 1946, I. J. Good 1946). Let Σ be afinite alphabet, and k ∈ N. Then there exists a de Bruijn stringfrom the alphabet Σ of order k. This string has length |Σ|k.

Introduction

Question

Construction

Liouvilleness

Normality

Open Questions

11 / 21

0110 0110 0110 0110

Theorem (N. G. de Bruijn 1946, I. J. Good 1946). Let Σ be afinite alphabet, and k ∈ N. Then there exists a de Bruijn stringfrom the alphabet Σ of order k. This string has length |Σ|k.

de Bruijn sequences are a standard tool in the study of normality.

Construction

Introduction

Question

de Bruijn strings

⊲ Construction

Liouvilleness

Normality

Open Questions

12 / 21

Denote the lexicographically least binary de Bruijn string of orderk as B(k). Consider the number defined as follows.

Construction

Introduction

Question

de Bruijn strings

⊲ Construction

Liouvilleness

Normality

Open Questions

12 / 21

Construction

Introduction

Question

de Bruijn strings

⊲ Construction

Liouvilleness

Normality

Open Questions

12 / 21

β = B(1)11

Construction

Introduction

Question

de Bruijn strings

⊲ Construction

Liouvilleness

Normality

Open Questions

12 / 21

β = B(1)11

B(2)22

Construction

Introduction

Question

de Bruijn strings

⊲ Construction

Liouvilleness

Normality

Open Questions

12 / 21

β = B(1)11

B(2)22

. . . B(k)kk

Construction

Introduction

Question

de Bruijn strings

⊲ Construction

Liouvilleness

Normality

Open Questions

12 / 21

β = B(1)11

B(2)22

. . . B(k)kk

β is a normal Liouville number.

Construction

Introduction

Question

de Bruijn strings

⊲ Construction

Liouvilleness

Normality

Open Questions

12 / 21

β = B(1)11

B(2)22

. . . B(k)kk

Intuition (Liouville Number): The stages have lengthsapproximately k! as in Liouville’s constant. Hence we can try toform a diophantine approximation as in Liouville’s constant.

Construction

Introduction

Question

de Bruijn strings

⊲ Construction

Liouvilleness

Normality

Open Questions

12 / 21

β = B(1)11

B(2)22

. . . B(k)kk

Intuition (Liouville Number): The stages have lengthsapproximately k! as in Liouville’s constant. Hence we can try toform a diophantine approximation as in Liouville’s constant.

Intution (Normality): The basic building block of each stage k isa balanced string of all orders ≤ k.

β is a Liouville number

Introduction

Question

de Bruijn strings

Construction

⊲ Liouvilleness

Normality

Open Questions

13 / 21

The length of stage k is 2k kk.

Introduction

Question

de Bruijn strings

Construction

⊲ Liouvilleness

Normality

Open Questions

13 / 21

The length of stage k is 2k kk. The length of the prefix up toand including stage k is Lk =

∑ki=0 2

iii = O((2k)k).

Introduction

Question

de Bruijn strings

Construction

⊲ Liouvilleness

Normality

Open Questions

13 / 21

∑ki=0 2

iii = O((2k)k).

Consider the rational number bk at level k defined to have thesame binary expansion as β up to stage k, followed by a periodicpattern of B(k + 1).

Introduction

Question

de Bruijn strings

Construction

⊲ Liouvilleness

Normality

Open Questions

13 / 21

∑ki=0 2

iii = O((2k)k).

Consider the rational number bk at level k defined to have thesame binary expansion as β up to stage k, followed by a periodicpattern of B(k + 1). The length of its denominator is O(Lk).

Introduction

Question

de Bruijn strings

Construction

⊲ Liouvilleness

Normality

Open Questions

13 / 21

∑ki=0 2

iii = O((2k)k).

|bk − β| ≤ 1

2(2(k+1))k+1≤ 1

(2O((2k)k))k

Introduction

Question

de Bruijn strings

Construction

⊲ Liouvilleness

Normality

Open Questions

13 / 21

∑ki=0 2

iii = O((2k)k).

|bk − β| ≤ 1

2(2(k+1))k+1≤ 1

(2O((2k)k))k

Hence β is a Liouville number.

β is a normal number

Introduction

Question

de Bruijn strings

Construction

Liouvilleness

⊲ Normality

Open Questions

14 / 21

In B(k), every string of length k appears once (wrappingaround).

Introduction

Question

de Bruijn strings

Construction

Liouvilleness

⊲ Normality

Open Questions

14 / 21

In B(k), every string of length k appears once (wrappingaround). Hence every string w of length n ≤ k appears with theright frequency 1

2k−n .

Introduction

Question

de Bruijn strings

Construction

Liouvilleness

⊲ Normality

Open Questions

14 / 21

2k−n .

Hence for all k ≥ |w|, the frequency of any string w is 12w at the

end of every block B(k).

Introduction

Question

de Bruijn strings

Construction

Liouvilleness

⊲ Normality

Open Questions

14 / 21

2k−n .

end of every block B(k). What about within B(k)?

Introduction

Question

de Bruijn strings

Construction

Liouvilleness

⊲ Normality

Open Questions

14 / 21

2k−n .

For every large enough k, length(B(k)) = o(Lk−1).

Introduction

Question

de Bruijn strings

Construction

Liouvilleness

⊲ Normality

Open Questions

14 / 21

2k−n .

For every large enough k, length(B(k)) = o(Lk−1). Hence thedeviations from balance within B(k) is insignificant for lengthsfrom Lk−1 to Lk−1 + length(Bk).

Generalization: Finite State Dimension

Introduction

Definition

Normality and FSD

Construction

Open Questions

15 / 21

Finite-State Dimension

16 / 21

Consider the n-long prefix of an infinite binary sequence ω. Then for k < n,the frequency of a k-long word w in ω[0 . . . n− 1] is

π(ω[0 . . . n− 1], w) =|{i | 0 ≤ i ≤ n− k − 1 ω[i . . . i+ k − 1] = w}|

n− k − 1.

16 / 21

π(ω[0 . . . n− 1], w) =|{i | 0 ≤ i ≤ n− k − 1 ω[i . . . i+ k − 1] = w}|

n− k − 1.

The entropy of this probability distribution is

Hk,n(ω) =∑

w∈Σk

−π(ω[0 . . . n− 1], w) log π(ω[0 . . . n− 1], w).

16 / 21

π(ω[0 . . . n− 1], w) =|{i | 0 ≤ i ≤ n− k − 1 ω[i . . . i+ k − 1] = w}|

n− k − 1.

Hk,n(ω) =∑

w∈Σk

−π(ω[0 . . . n− 1], w) log π(ω[0 . . . n− 1], w).

The k-entropy rate of ω is defined as

Hk(ω) =1

klim infn→∞

Hk,n(ω).

16 / 21

π(ω[0 . . . n− 1], w) =|{i | 0 ≤ i ≤ n− k − 1 ω[i . . . i+ k − 1] = w}|

n− k − 1.

Hk,n(ω) =∑

w∈Σk

−π(ω[0 . . . n− 1], w) log π(ω[0 . . . n− 1], w).

The k-entropy rate of ω is defined as

Hk(ω) =1

klim infn→∞

Hk,n(ω).

The finite-dimensional entropy rate of ω is defined as H(ω) = infk Hk(ω).

Normality and Finite-state dimension

Introduction

Definition

⊲Normality andFSD

Construction

Open Questions

17 / 21

A binary sequence is normal if and only if it has finite statedimension 1 (Schnorr and Stimm 1971,

Introduction

Definition

⊲Normality andFSD

Construction

Open Questions

17 / 21

A binary sequence is normal if and only if it has finite statedimension 1 (Schnorr and Stimm 1971, Becher and Heiber 2012.)

Introduction

Definition

⊲Normality andFSD

Construction

Open Questions

17 / 21

Every binary sequence has a finite-state dimension between 0and 1.

Introduction

Definition

⊲Normality andFSD

Construction

Open Questions

17 / 21

Finite-state dimension thus provides a quantification of the“non-normality” of binary expansions of reals.

Introduction

Definition

⊲Normality andFSD

Construction

Open Questions

17 / 21

Finite-state dimension thus provides a quantification of the“non-normality” of binary expansions of reals.

Can we produce Liouville numbers of any given finite-statedimension in [0, 1]?

Liouville numbers of any FSD between 0 and 1

18 / 21

Let m,n be positive integers, m < n. We construct a Liouville number withfinite-state dimension m/n.

αm/n = 0·(

)n−mB(1)m)11 (

)n−mB(2)m)22

. . .(

)n−mB(k)m)kk

Liouville numbers of any FSD between 0 and 1

18 / 21

Let m,n be positive integers, m < n. We construct a Liouville number withfinite-state dimension m/n.

αm/n = 0·(

)n−mB(1)m)11 (

)n−mB(2)m)22

. . .(

)n−mB(k)m)kk

αm/n is a Liouville number with finite-state dimension m/n.

Open Questions

Introduction

⊲ Open Questions

Open Questions

19 / 21

Open Questions

Introduction

Open Questions

⊲ Open Questions

20 / 21

� Can we construct absolutely normal Liouville numbers? Theyexist. (Bugeaud, 2002)

Open Questions

Introduction

Open Questions

⊲ Open Questions

20 / 21

� Can we construct absolutely normal Liouville numbers? Theyexist. (Bugeaud, 2002)

� Can we construct algebraic normal numbers?

Introduction

Open Questions

21 / 21

Thank You!

Satyadev Nandakumar Santosh Kumar Vangepalli Indian ...PDF/FSDNLN.pdf · 1 / 21 Normality and...

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