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module
public import FormalConjecturesUtilEquidistributed Sequences
Corollary 4.2 of Chapter 1 states that the sequence $(x^n), n = 1, 2, ... ,$ is equidistributed modulo 1 for almost all x > 1. And a little bit further down: "one does not know whether sequences such as $(e^n)$, $(π^n)$, or even $((\frac 3 2)^n)$" are equidistributed modulo 1 or not.
References:
Uniform Distribution of Sequences by L. Kuipers and H. Niederreiter, 1974
[Mat80] de Mathan, Bernard. "Numbers contravening a condition in density modulo 1." Acta Mathematica Hungarica 36.3-4 (1980): 237-241.
[Pol79] Pollington, Andrew Douglas. "On the density of sequence ${n_ {k}\xi} $." Illinois Journal of Mathematics 23.4 (1979): 511-515.
@[expose] public sectionnamespace Equidistributionopen scoped Topology
A point x is an accumulation point of a sequence s_0, s_1, ...
if any neighbourhood of x contains a point of the sequence distinct
from x.
def IsAccumulationPoint (x : ℝ) (s : ℕ → ℝ) : Prop :=
x ∈ closure (Set.range s \ {x})
If a point x is an accumulation point of a sequence s_0, s_1, ... then
there is a subsequence of s that tends to x
x:ℝs:ℕ → ℝhx:IsAccumulationPoint x sU:Set ℝhU:U ∈ 𝓝 xN:ℕhcon:∀ (b : ℕ), N ≤ b → s b ∉ UhV:U ∩ ⋂ n ∈ Finset.range N, {y | s n ≠ x → y ≠ s n} ∈ 𝓝 xn:ℕhyU:s n ∈ UhyI:∀ i ∈ Finset.range N, s n ∈ {y | s i ≠ x → y ≠ s i}hyx:s n ∉ {x}⊢ False
x:ℝs:ℕ → ℝhx:IsAccumulationPoint x sU:Set ℝhU:U ∈ 𝓝 xN:ℕhcon:∀ (b : ℕ), N ≤ b → s b ∉ UhV:U ∩ ⋂ n ∈ Finset.range N, {y | s n ≠ x → y ≠ s n} ∈ 𝓝 xn:ℕhyU:s n ∈ UhyI:∀ i ∈ Finset.range N, s n ∈ {y | s i ≠ x → y ≠ s i}hyx:s n ∉ {x}hn:n < N⊢ Falsex:ℝs:ℕ → ℝhx:IsAccumulationPoint x sU:Set ℝhU:U ∈ 𝓝 xN:ℕhcon:∀ (b : ℕ), N ≤ b → s b ∉ UhV:U ∩ ⋂ n ∈ Finset.range N, {y | s n ≠ x → y ≠ s n} ∈ 𝓝 xn:ℕhyU:s n ∈ UhyI:∀ i ∈ Finset.range N, s n ∈ {y | s i ≠ x → y ≠ s i}hyx:s n ∉ {x}hn:n ≥ N⊢ False
x:ℝs:ℕ → ℝhx:IsAccumulationPoint x sU:Set ℝhU:U ∈ 𝓝 xN:ℕhcon:∀ (b : ℕ), N ≤ b → s b ∉ UhV:U ∩ ⋂ n ∈ Finset.range N, {y | s n ≠ x → y ≠ s n} ∈ 𝓝 xn:ℕhyU:s n ∈ UhyI:∀ i ∈ Finset.range N, s n ∈ {y | s i ≠ x → y ≠ s i}hyx:s n ∉ {x}hn:n < N⊢ False All goals completed! 🐙
x:ℝs:ℕ → ℝhx:IsAccumulationPoint x sU:Set ℝhU:U ∈ 𝓝 xN:ℕhcon:∀ (b : ℕ), N ≤ b → s b ∉ UhV:U ∩ ⋂ n ∈ Finset.range N, {y | s n ≠ x → y ≠ s n} ∈ 𝓝 xn:ℕhyU:s n ∈ UhyI:∀ i ∈ Finset.range N, s n ∈ {y | s i ≠ x → y ≠ s i}hyx:s n ∉ {x}hn:n ≥ N⊢ False All goals completed! 🐙
The sequence (3/2)^n is equidistributed modulo 1.
@[category research open, AMS 11]
theorem isEquidistributedModuloOne_three_halves_pow :
IsEquidistributedModuloOne (fun n => (3 / 2 : ℝ)^n) := ⊢ IsEquidistributedModuloOne fun n ↦ (3 / 2) ^ n
All goals completed! 🐙
It is not true that for every transcendental number x the sequence x * (3 / 2) ^ n is
equidistributed modulo 1. The sequence (3 / 2) ^ n is lacunary, so by the theorem of
Pollington [Pol79] and de Mathan [Mat80] the set of real numbers x for which x * (3 / 2) ^ n
is not even dense modulo 1 has Hausdorff dimension 1. This set is uncountable, so it contains
transcendental numbers. Alternatively, an elementary nested-interval construction (see issue
#5003) gives a Cantor set of x with Int.fract (x * (3 / 2) ^ (8 * j)) ∈ [0, 1 / 10] for all
j, which already rules out equidistribution for uncountably many, hence some transcendental,
x. By Koksma's metric theorem (Kuipers–Niederreiter, Chapter 1, Section 4), the sequence
x * (3 / 2) ^ n is equidistributed modulo 1 for almost all x.
@[category research solved, AMS 11]
theorem isEquidistributedModuloOne_transcendental_three_halves_pow :
¬ ∀ x : ℝ, Transcendental ℚ x →
IsEquidistributedModuloOne (fun n ↦ x * (3 / 2 : ℝ) ^ n) := ⊢ ¬∀ (x : ℝ), Transcendental ℚ x → IsEquidistributedModuloOne fun n ↦ x * (3 / 2) ^ n
All goals completed! 🐙
The sequence (3/2)^n has infinitely many accumulation points modulo 1.
@[category research solved, AMS 11]
theorem isAccumulationPoint_three_halves_pow_infinite :
{x | IsAccumulationPoint x (fun n => Int.fract <| (3 / 2 : ℝ)^n)}.Infinite := ⊢ {x | IsAccumulationPoint x fun n ↦ Int.fract ((3 / 2) ^ n)}.Infinite
All goals completed! 🐙
Find an accumulation point of the sequence (3/2)^n modulo 1.
@[category research open, AMS 11]
theorem isAccumulationPoint_three_halves_pow :
IsAccumulationPoint answer(sorry) (fun n => Int.fract <| (3 / 2 : ℝ)^n) := ⊢ IsAccumulationPoint sorry fun n ↦ Int.fract ((3 / 2) ^ n)
All goals completed! 🐙
The values of (3/2)^n modulo 1 are pairwise distinct: if (3/2)^n - (3/2)^m were an
integer for m < n, then 3^n - 3^m * 2^(n - m) would be even.
n:ℕm:ℕhnm:(fun n ↦ Int.fract ((3 / 2) ^ n)) n = (fun n ↦ Int.fract ((3 / 2) ^ n)) mhne:¬n = mhlt:m < nz:ℤhz:(3 / 2) ^ n - (3 / 2) ^ m = ↑zh2:2 ^ n ≠ 0key:3 ^ n - 3 ^ m * 2 ^ (n - m) = z * 2 ^ nhmod:1 ^ n - 1 ^ m * 0 = ↑z * 0⊢ False
All goals completed! 🐙
There is an accumulation point of the sequence (3/2)^n modulo 1: its values are pairwise
distinct and lie in [0, 1], so they have an accumulation point by compactness.
hinf:(Set.range fun n ↦ Int.fract ((3 / 2) ^ n)).Infinitehsub:(Set.range fun n ↦ Int.fract ((3 / 2) ^ n)) ⊆ Set.Icc 0 1⊢ ∃ p, IsAccumulationPoint p fun n ↦ Int.fract ((3 / 2) ^ n)
hinf:(Set.range fun n ↦ Int.fract ((3 / 2) ^ n)).Infinitehsub:(Set.range fun n ↦ Int.fract ((3 / 2) ^ n)) ⊆ Set.Icc 0 1x:ℝhx:AccPt x (Filter.principal (Set.range fun n ↦ Int.fract ((3 / 2) ^ n)))⊢ ∃ p, IsAccumulationPoint p fun n ↦ Int.fract ((3 / 2) ^ n)
All goals completed! 🐙end Equidistribution