Uniform Distribution Modulo One

Theory and ApplicationAram TangboonduangjitMahidol University International CollegeMathematics SeminarMay 29, 2019

Abstract

We discuss the distribution of fractional parts of real numbers in theunit interval. This concept was originated in 1916 by Hermann Weylin his celebrated paper titled “Über die Gleichverteilung von Zahlenmod. Eins.” A surprising application of the theory is to show that theset of the Fibonacci numbers is extendable in any base.

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Uniform Distribution Modulo One

Definition

The sequence ω = (xn), n = 1, 2, . . . , of real numbers is said to beuniformly distributed modulo 1 (u.d. mod 1) if for every pair a, b ofreal numbers with 0 ≤ a < b ≤ 1 we have

limN→∞

A([a, b);N;ω)

N= b− a,

where A(E;N;ω) = Card{xn : 1 ≤ n ≤ N, {xn} ∈ E}.

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Basic Properties

• The sequence (xn) is u.d. mod 1 if and only if for everycomplex-valued continuous function f on R with period 1 wehave

limN→∞

1

N

N∑n=1

f(xn) =∫ 1

0f(x) dx.

• If the sequence (xn) is u.d. mod 1, and if (yn) is a sequence withthe property limn→∞(xn − yn) = α, a real constant, then (yn)is u.d. mod 1.

• If (xn) is u.d. mod 1, then the sequence ({xn}) of fractionalparts is every dense in I = [0, 1]. The converse is not true. Forexample, the sequence ({sin n}) is dense in I but not u.d.mod 1. (For proof, see Appendix 1.)

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Basic Properties

• The sequence (xn) is u.d. mod 1 if and only if for everycomplex-valued continuous function f on R with period 1 wehave

limN→∞

1

N

N∑n=1

f(xn) =∫ 1

0f(x) dx.

• If the sequence (xn) is u.d. mod 1, and if (yn) is a sequence withthe property limn→∞(xn − yn) = α, a real constant, then (yn)is u.d. mod 1.

• If (xn) is u.d. mod 1, then the sequence ({xn}) of fractionalparts is every dense in I = [0, 1]. The converse is not true. Forexample, the sequence ({sin n}) is dense in I but not u.d.mod 1. (For proof, see Appendix 1.)

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Basic Properties

• The sequence (xn) is u.d. mod 1 if and only if for everycomplex-valued continuous function f on R with period 1 wehave

limN→∞

1

N

N∑n=1

f(xn) =∫ 1

0f(x) dx.

• If the sequence (xn) is u.d. mod 1, and if (yn) is a sequence withthe property limn→∞(xn − yn) = α, a real constant, then (yn)is u.d. mod 1.

• If (xn) is u.d. mod 1, then the sequence ({xn}) of fractionalparts is every dense in I = [0, 1]. The converse is not true. Forexample, the sequence ({sin n}) is dense in I but not u.d.mod 1. (For proof, see Appendix 1.) 3/22

Uniform Distribution Modulo One

Example

The sequence

0

1,0

2,1

2,0

3,1

3,2

3, . . . ,

0

k,1

k, . . . ,

k− 1

k, . . .

is u.d. mod 1.

Proof.

See Appendix 2.

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Uniform Distribution Modulo One

Example

The sequence

0

1,0

2,1

2,0

3,1

3,2

3, . . . ,

0

k,1

k, . . . ,

k− 1

k, . . .

is u.d. mod 1.

Proof.

See Appendix 2.

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Uniform Distribution Modulo One

Theorem (Weyl Criterion)

The sequence (xn) is u.d. mod 1 if and only if

limN→∞

1

N

N∑n=1

e2πihxn = 0 for all integers h 6= 0.

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Uniform Distribution Modulo One

Theorem (Weyl Criterion)

The sequence (xn) is u.d. mod 1 if and only if

limN→∞

1

N

N∑n=1

e2πihxn = 0 for all integers h 6= 0.

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Sequence of Multiples of an Irrational Number

Theorem

Let θ be an irrational number. Then the sequence (nθ) is u.d. mod 1.

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Sequence of Multiples of an Irrational Number

Theorem

Let θ be an irrational number. Then the sequence (nθ) is u.d. mod 1.

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Sequence of Multiples of an Irrational Number

Theorem

Let θ be an irrational number. Then the sequence (nθ) is u.d. mod 1.

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Sequence of Multiples of an Irrational Number

Let θ be an irrational number.

• Kronecker’s Theorem states that ({nθ}) is everywhere densein [0, 1].

• Not every subsequence of (nθ) is u.d. mod 1. For example, thesequence (n!e) is not u.d. mod 1.

• The sequence (pnθ) is u.d. mod 1, where (pn) is the sequenceof primes arranged in increasing order. (Vinogradov, Hua)

• The sequence (ω(n)θ) is u.d. mod 1, where ω(n) is the numberof prime divisors of n. (Erdös, Delange)

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Sequence of Multiples of an Irrational Number

Let θ be an irrational number.

• Kronecker’s Theorem states that ({nθ}) is everywhere densein [0, 1].

• Not every subsequence of (nθ) is u.d. mod 1. For example, thesequence (n!e) is not u.d. mod 1.

• The sequence (pnθ) is u.d. mod 1, where (pn) is the sequenceof primes arranged in increasing order. (Vinogradov, Hua)

• The sequence (ω(n)θ) is u.d. mod 1, where ω(n) is the numberof prime divisors of n. (Erdös, Delange)

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Sequence of Multiples of an Irrational Number

Let θ be an irrational number.

• Kronecker’s Theorem states that ({nθ}) is everywhere densein [0, 1].

• Not every subsequence of (nθ) is u.d. mod 1. For example, thesequence (n!e) is not u.d. mod 1.

• The sequence (pnθ) is u.d. mod 1, where (pn) is the sequenceof primes arranged in increasing order. (Vinogradov, Hua)

• The sequence (ω(n)θ) is u.d. mod 1, where ω(n) is the numberof prime divisors of n. (Erdös, Delange)

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Sequence of Multiples of an Irrational Number

Let θ be an irrational number.

• Kronecker’s Theorem states that ({nθ}) is everywhere densein [0, 1].

• Not every subsequence of (nθ) is u.d. mod 1. For example, thesequence (n!e) is not u.d. mod 1.

• The sequence (pnθ) is u.d. mod 1, where (pn) is the sequenceof primes arranged in increasing order. (Vinogradov, Hua)

• The sequence (ω(n)θ) is u.d. mod 1, where ω(n) is the numberof prime divisors of n. (Erdös, Delange)

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An Open Problem Concerning (nθ) by Erdös

For each positive integer h, let

Ah = limspN→∞

N∑n=1

e2πihnθ

.

Then Ah is finite for every positive integer h. However, is it true thatlimsph→∞ Ah is infinite?

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A Generalization

Theorem

Let p(x) = αmxm + αm−1xm−1 + · · · + α0, m ≥ 1, be a polynomialwith real coefficients and let at least one of the coefficients αj withj > 0 be irrational. Then the sequence (p(n)), n = 1, 2, . . . , is u.d.mod 1.

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Application

Definition

Suppose that S is an infinite set of positive integers. We say that S isextendable in base b if for each positive integer x, there are positiveintegers y and n, with y < bn, such that xbn + y is in S.

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ApplicationTheorem

The set {Fn : n = 1, 2, . . .} of Fibonacci numbers is extendable inany base.

Proof.

1. The set S = {s1, s2, . . .} is extendable in base b if and only ifthe sequence ({logb sn}) of fractional parts is dense in I. (See[1].)

2. If a sequence (xn) has the property thatΔxn = xn+1 − xn→ θ (irrational) as n→∞, then the sequence(xn) is u.d. mod 1. (See Appendix 3.)

3. The sequence (logb Fn) is u.d. mod 1 for every integer b > 1.

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ApplicationTheorem

The set {Fn : n = 1, 2, . . .} of Fibonacci numbers is extendable inany base.

Proof.

1. The set S = {s1, s2, . . .} is extendable in base b if and only ifthe sequence ({logb sn}) of fractional parts is dense in I. (See[1].)

2. If a sequence (xn) has the property thatΔxn = xn+1 − xn→ θ (irrational) as n→∞, then the sequence(xn) is u.d. mod 1. (See Appendix 3.)

3. The sequence (logb Fn) is u.d. mod 1 for every integer b > 1.

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ApplicationTheorem

The set {Fn : n = 1, 2, . . .} of Fibonacci numbers is extendable inany base.

Proof.

1. The set S = {s1, s2, . . .} is extendable in base b if and only ifthe sequence ({logb sn}) of fractional parts is dense in I. (See[1].)

2. If a sequence (xn) has the property thatΔxn = xn+1 − xn→ θ (irrational) as n→∞, then the sequence(xn) is u.d. mod 1. (See Appendix 3.)

3. The sequence (logb Fn) is u.d. mod 1 for every integer b > 1.

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ApplicationTheorem

The set {Fn : n = 1, 2, . . .} of Fibonacci numbers is extendable inany base.

Proof.

1. The set S = {s1, s2, . . .} is extendable in base b if and only ifthe sequence ({logb sn}) of fractional parts is dense in I. (See[1].)

2. If a sequence (xn) has the property thatΔxn = xn+1 − xn→ θ (irrational) as n→∞, then the sequence(xn) is u.d. mod 1. (See Appendix 3.)

3. The sequence (logb Fn) is u.d. mod 1 for every integer b > 1.11/22

References

[1] R. S. Bird, Integers with given initial digits, Amer. Math. Monthly,79 (1972), 367–370.

[2] Y. Bugeaud, Distribution Modulo One and DiophantineApproximation Cambridge University Press, Cambridge, 2012.

[3] P. Erdös, Problems and results on diophantine approximations,Compos. Math., 16 (1964), 52–65.

[4] L. Kuipers and H. Niederreiter, Uniform Distribution of SequencesJohn Wiley & Sons, New York-London-Sydney, 1974.

[5] H. Weyl, Über die Gleichverteilung von Zahlen mod. Eins, Math.Ann., 77 (1916), 313–352.

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Appendix 1Let x be a real number in (0, 1). For each ϵ > 0, we want to showthat there exists a k such that |x− {xk}| < ϵ.Without loss ofgenerality, we may assume that x− ϵ/2 ≥ 0 and x+ ϵ/2 ≤ 1. Since(xn) is u.d. mod 1, there exists an N such that

A([x− ϵ/2, x+ ϵ/2);N)

N− ϵ

< ϵ.

This implies A([x− ϵ/2, x+ ϵ/2);N) > 0. Hence there exists a k(1 ≤ k ≤ N) such that x− ϵ/2 ≤ {xk} < x+ ϵ/2, so that

|x− {xk}| < ϵ,

as desired. For the cases x = 0 and x = 1, we consider the intervals[0, ϵ) and [1− ϵ, 1), respectively, instead. 13/22

Appendix 2

Let N be a positive integer and 0 ≤ a < b ≤ 1. We try to computeA([a, b);N). Let m be the largest positive integer such that12m(m+ 1) ≤ N. Then the first N terms of the sequence can be listedas follows:

0

1,0

2,1

2,0

3,1

3,2

3, . . . ,

0

m,1

m,2

m, . . . ,

m− 1

m,

0

m+ 1,

1

m+ 1, . . . ,

ℓ

m+ 1,

where 0 ≤ ℓ < m if 12m(m+ 1) < N. Now we partition this finite

sequence intom+ 1 subpartitions by letting the jth subpartition(1 ≤ j ≤ m) consist of 0

j ,1j , . . . ,

j−1j and the last subpartition (could

be empty if 12m(m+ 1) = N) consist of 0

m+1 ,1

m+1 , . . . ,ℓ

m+1 .

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Appendix 2 (cont.)

Now we count the number of elements in the jth subpartition(1 ≤ j ≤ m) that are in [a, b), that is, we count the integers k in[0, j− 1) such that a ≤ k

j < b or ja ≤ k < jb. This number isj(b− a) + θj where |θj| < 1. The number of elements in the lastsubpartition is at mostm. Thus,

m∑j=1

j(b− a) + θj ≤ A([a, b);N) ≤m∑j=1

j(b− a) + θj + m.

Since −1 < θj < 1 and∑m

j=1 j =12m(m+ 1), we have further that

1

2(b−a)m(m+ 1)−m < A([a, b);N) <

1

2(b−a)m(m+ 1)+m+m.

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Appendix 2 (cont.)

From the way the integerm is defined, we have12m(m+ 1) ≤ N <

12(m+ 1)(m+ 2), so that,

(b− a)(N− (m+ 1)) − m < A([a, b);N) < (b− a)m+ 2m,

or, equivalently,

(b− a) −b− a

N−

m

N(1+ b− a) <

A([a, b);N)

N< (b− a) +

2m

N.

The integerm can be approximated by 12m

2 < N <12(m+ 2)2 or

p2N− 2 < m <

p2N, so that

p2N−2N <

mN <

p2NN . Consequently,

mN → 0 as N→∞.

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Appendix 2 (cont.)

Combining with the last inequality that approximates A([a,b);N)N , we

have

limN→∞

A([a, b);N)

N= b− a,

as desired.

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Appendix 3

Let q be a positive integer, then since Δxn→ θ as n→∞, thereexists an integer g0 = g0(q) such that for any integers n > g ≥ g0,|Δxj − θ| ≤ 1

q2 for j = g, g+ 1, . . . , n− 1. Since∑b−1

j=a Δxj = xb − xafor 1 ≤ a < b, we have

|xn−xg−(n−g)θ| =

n−1∑j=g

(Δxj − θ)

≤ n−1∑

j=g

|Δxj−θ| ≤n−1∑j=g

1

q2=

n− g

q2.

For arbitrary real numbers u and v, we have

|e2πiu − e2πiv| =eπi(u+v)(eπi(u−v) − e−πi(u−v))

=

2i sin π(u− v)

≤ 2π|u− v|.

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Appendix 3 (cont.)

Hence, if h 6= 0 is an integer, then

e2πihxn − e2πih(xg+(n−g)θ) ≤2π|h|(n− g)

q2.

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Appendix 3 (cont.)

By the triangle inequality, it follows thatg+q−1∑

n=ge2πihxn

≤

g+q−1∑

n=ge2πih(xg+(n−g)θ)

+ 2π|h|

q2

g+q−1∑n=g(n− g) ≤ K,

where K = | sin πhθ|−1 + π|h|. Thus, by induction, for every positiveinteger H,

g−1+Hq∑n=g

e2πihxn

≤ HK.

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Appendix 3 (cont.)

Let N ≥ g be an integer. Choose the largest integer H such thatg− 1+ Hq < N. It follows that

g− 1+ Hq ≤ N− 1 =⇒ H ≤N− g

q

andg− 1+ (H+ 1)q ≥ N =⇒ N− g− Hq+ 1 ≤ q.

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Appendix 3 (cont.)

We have N∑n=1

e2πihxn

=

g−1∑n=1

e2πihxn +g−1+Hq∑n=g

e2πihxn +N∑

n=g+Hqe2πihxn

≤ (g− 1) + HK + (N− g− Hq+ 1)

≤ g− 1+N− g

qK + q.

Keeping q fixed, we obtain

limspN→∞

1N

N∑n=1

e2πihxn

≤ K

q.

Since q can be as large as we please, the theorem follows. 22/22