Continuous Distribution Arising from the Three Gap Theorem

Transcript

1 Continuous Distribution Arising from the Three Gap Theorem Geremías Polanco Encarnación Elementary Analytic and Algorithmic Number Theory Athens, Georgia June 8, 2015 Hampshire college, MA 1 / 24

2 Happy Birthday Prof. Carl Pomerance 2 / 24

3 Continuous Distribution CONTINUOUS DISTRIBUTION ARISING FROM THE THREE GAP THEOREM (Joint work with D. Schultz, and A. Zaharescu) 3 / 24

4 The Three Gap Theorem Theorem (Steinhaus Theorem or Steinhaus Conjecture) Let α be an irrational number. Consider the fractional parts {nα} 0 n<q arranged in increasing order and placed on the interval [0, 1] with 0 and 1 identified so that Q gaps appear. If we denote this sequence by S Q (α) and consider the gaps between consecutive elements, then there are at most three gap sizes in S Q (α). 4 / 24

5 Example of Steinhaus Theorem α = 2 and Q = 10. The points S 10 ( 2) along with the three gaps labeled A, B, and C are shown here. A C A B A C A B A B 0Α 5Α 3Α 8Α 1Α 6Α 4Α 9Α 2Α 7Α Figure : S 10 ( 2) 5 / 24

6 The Main Problem Denote the n th element of S Q (α) by {σ n α} {σ n+1 α} {σ n α} 1 Q. (1) 6 / 24

7 The Main Problem Denote the n th element of S Q (α) by {σ n α} {σ n+1 α} {σ n α} λ Q. (1) 6 / 24

8 The Main Problem Denote the n th element of S Q (α) by {σ n α} #{0 n < Q : {σ n+1 α} {σ n α} λ }. (1) Q 6 / 24

9 The Main Problem Denote the n th element of S Q (α) by {σ n α} 1 0 #{0 n < Q : {σ n+1 α} {σ n α} λ }dα. (1) Q 6 / 24

10 The Main Problem Denote the n th element of S Q (α) by {σ n α} 1 0 #{0 n < Q : {σ n+1 α} {σ n α} λ Q } dα. (1) Q 6 / 24

11 The Main Problem Denote the n th element of S Q (α) by {σ n α} and consider the function: 1 g(λ; Q) = lim Q 0 #{0 n < Q : {σ n+1 α} {σ n α} λ Q } dα. Q (1) 6 / 24

12 The Main Problem Denote the n th element of S Q (α) by {σ n α} and consider the function: g γ,η 1 (λ; Q) = 1 η γ+η γ #{0 n < Q : {σ n+1 α} {σ n α} λ Q } dα. Q (1) 6 / 24

13 The Main Problem Denote the n th element of S Q (α) by {σ n α} g γ,η 1 (λ; Q) = 1 η γ+η Does this limit exist? Is it a continuous function of λ? Is it differentiable? γ #{0 n < Q : {σ n+1 α} {σ n α} λ Q } dα. Q (1) 6 / 24

14 The Main Problem More generally, we consider the join distribution g k (λ 1, λ 2,..., λ k ), which describes the average distribution of k-tuples of consecutive gaps over the interval [a, b], defined as where g k (λ 1, λ 2,..., λ k ) = lim Q g α,η k (λ 1,..., λ k ; Q), (2) g γ,η k (λ 1,..., λ k ; Q) = 1 η γ+η γ #{ω 1 ω k G Q,k (α) : i k ω i λ i Q Q } dα. 7 / 24

15 A Probability question For k = 1 For a given λ 1, what is the probability that a randomly selected gap will be greater than λ 1 times the average gap? And for the the k dimensional case (say k = 2) for given λ 1 and λ 2 what is the probability that a gap and its neighbor to the right are both greater than the average times λ 1 and λ 2 respectively? 8 / 24

16 A Differentiable distribution Figure : g 1 3, (λ; 1000) 9 / 24

17 A Differentiable distribution Figure : g 1 3, (λ; 1000) 10 / 24

18 Solution: A Differentialble Distribution Theorem (G.P.;D.S. and A.Z.) Assume η > Q (t 1/2), where t > 0. As Q, the nearest neighbor distribution in (19) is independent of γ and η, and we have g 1 (λ) := lim g γ,η Q 1 (λ; Q) = 6 π 2 π 2 6 λ, 0 < λ < 1 log 2 (2) 2π2 3 log ( ) 1 + λ2 λ 2 log ( ) ( ) 2 λ + λ 1 3λ 2 log λ λ 1 ( 4λ ) log(λ) + 4 Li 2 ( 1λ ) + 2 Li 2 ( λ2 ), 1 < λ < 2 ( ) ( ) ( ) ( ) ( ) 1 + λ2 λ 2 log λ 2 + λ 1 3λ 2 log λ λ Li 1λ 2 2 Li 2λ 2, 2 < λ, where the Dilogarithm is defined for z 1 by z n Li 2 (z) = n=1 n 2. More precisely, we have the error estimate: α,η g 1 (λ; Q) g 1 (λ) ɛ (1 + λ) Qɛ 1/2. η 11 / 24

19 Key steps in the proof The proof of this theorem have three key stages 1 We first establish a key connection of the elements of the sequence of fractional parts with Farey fractions. 2 Use the help of some lemmas that allow us to write a sum over Farey fractions in a given region as an integral that is easier to handle. The main ingredients of these lemmas are Kloosterman sums. 3 In the last part we write g(λ 1 ) as a suitable sum for which we can apply the lemmas we mentioned above. After some other miscellaneous work we complete the proof of the Theorem. 12 / 24

20 Gaps are related to Farey Sequence A C A B A C A B A B 0Α 5Α 3Α 8Α 1Α 6Α 4Α 9Α 2Α 7Α Figure : S Q (α) Recall the Farey sequence of order Q is given by F Q = {0 a q 1 (a, q) = 1, q Q} Example: F 4 = { 0 1, 1 4, 1 3, 1 2, 2 3, 3 4, 1 1 } 13 / 24

21 Gaps are related to Farey Sequence A C A B A C A B A B 0Α 5Α 3Α 8Α 1Α 6Α 4Α 9Α 2Α 7Α Figure : S Q (α) Recall the Farey sequence of order Q is given by F Q = {0 a q 1 (a, q) = 1, q Q} Example: F 4 = { 0 1, 1 4, 1 3, 1 2, 2 3, 3 4, 1 1 } Choose consecutive fractions a 1 q 1 and a 2 q 2 a 1 q 1 < α < a 2 q 2. in F Q 1, so that 13 / 24

22 Gaps are related to Farey Sequence Lemma The three gaps that can appear have lengths A = q 1 α a 1 = {q 1 α}, C = A + B, B = a 2 q 2 α = 1 {q 2 α}, 14 / 24

23 Gaps are related to Farey Sequence Lemma The three gaps that can appear have lengths A = q 1 α a 1 = {q 1 α}, C = A + B, B = a 2 q 2 α = 1 {q 2 α}, The permutation σ of the set {0, 1,..., Q 1} that increasingly orders the sequence of fractional parts is {σ 0 α}, {σ 1 α},..., {σ Q 1 α} and satisfies: σ 0 = 0, q 1, if σ i [0, Q q 1 ) (A Gap) σ i+1 σ i = q 1 q 2, if σ i [Q q 1, q 2 ) (C Gap) q 2, if σ i [q 2, Q) (B Gap). 14 / 24

24 Gaps are related to Farey Sequence The numbers of A gaps, B gaps and C gaps are, respectively, Q q 1, Q q 2, q 1 + q 2 Q. 15 / 24

25 Gaps are related to Farey Sequence The numbers of A gaps, B gaps and C gaps are, respectively, Note that this agrees with Q q 1, Q q 2, q 1 + q 2 Q. (Q q 1 ) + (Q q 2 ) + (q 1 + q 2 Q) = Q, (Q q 1 )A + (Q q 2 )B + (q 1 + q 2 Q)C = 1. This last equality is equivalent to the determinant property of Farey fractions: a 2 q 1 a 1 q 2 = / 24

26 A Lemma Via Kloosterman Sums (Boca, Cobeli, and Zaharescu) Notation Let f := Ω ( q 1 Q, q 2 γ a 1 q 1 < a 2 q 2 Q ) Ω γ + η f. 16 / 24

27 A Lemma Via Kloosterman Sums (Boca, Cobeli, and Zaharescu) Lemma Then if Ω is a convex subregion of [0, 1] [0, 1] with rectifiable boundary, and f is a C 1 function on Ω, we have 1 ( q1 f η Q, q ) 2 1 Q Q 2 6 π 2 f (x, y)dxdy Ω Ω ( ) f ɛ x + f Area(Ω) log Q (1 + Len( Ω)) log Q y + f ηq + ηq m f f ηq 1/2 ɛ, where m f is an upper bound for the number of intervals of monotonicity of each of the maps y f (x, y). 17 / 24

28 Proof of the Main Theorem for K = 1 Contribution of the A gaps Recall the integral in the main theorem g γ,η 1 (λ; Q) = 1 η γ+η γ #{0 n < Q : {σ n+1 α} {σ n α} λ Q } dα. Q 18 / 24

29 Proof of the Main Theorem for K = 1 Contribution of the A gaps Recall the integral in the main theorem g γ,η 1 (λ; Q) = 1 η γ+η γ #{0 n < Q : {σ n+1 α} {σ n α} λ Q } dα. Q Partition the interval [γ, γ + η] into Farey arcs with denominators strictly less than Q. 18 / 24

30 Proof of the Main Theorem for K = 1 Contribution of the A gaps Recall the integral in the main theorem g γ,η 1 (λ; Q) = 1 η γ+η γ #{0 n < Q : {σ n+1 α} {σ n α} λ Q } dα. Q Partition the interval [γ, γ + η] into Farey arcs with denominators strictly less than Q. In the interval [ a 1 q 1, a 2 q 2 ], the A gaps contribute the amount a 2 q 2 a 1 q 1 { 1 Q q1, q 1 α a 1 λ/q Q 0, q 1 α a 1 < λ/q dα, 18 / 24

31 Proof of the Main Theorem for K = 1 If a new integration variable t defined by α = a 1 + t (3) q 1 q 1 q 2 is introduced, using a 2 q 1 a 1 q 2 = 1, the integral becomes 1 0 dt Qq 1 q 2 { Q q 1, t λq 2 Q 0, t < λq 2 Q = { ( Q q1 Qq 1 q 2 1 λq 2 Q ), 0 λq 2 Q 1 0, 1 < λq 2 Q. 19 / 24

32 Proof of the Main Theorem for K = 1 Sum this expression over consecutive pairs of Farey fractions in the interval [γ, γ + η] as { ( ) Q q1 Qq 1 q 2 1 λq 2 Q, 0 q 2 Q 1 λ 1 η γ a 1 q1 < a 2 q2 γ+η 1 0, λ < q 2 Q. By Lemma 4, this sum converges to the integrals 6 π y 1 λ y (1 x)(1 λy) xy dxdy, 0 λ 1 (1 x)(1 λy) xy dxdy, 1 < λ, 20 / 24

33 Proof of the Main Theorem for K = 1 This further equals { 6 1 λ π π2 6, 0 λ λ + (1 λ) log ( 1 λ) 1 ( + 1 ), Li2 λ, 1 < λ The B and C gaps are handled similarly, and we add their respective contributions. (4) 21 / 24

34 Proof of the Main Theorem for K = 1 The result is the expression: 1 π2 12, 0 λ π2 3 + log2 (2) λ + λ 4 log ( ( ) 2 λ 1) 1 λ log 2 λ λ 1 log ( ) λ 1 λ log ( ) ( λ 4 log(λ) + 1 ( Li2 λ) + λ ), 1 < λ < 2 Li2 2 6 π π2 12 log2 (2) 2 + log(2) 2, λ = λ + λ 4 log ( ) ( ) ( ) 1 2 λ log 1 1 λ + 1 λ log λ 1 λ 2 + ( Li 1 ( 2 λ) 2 ), 2 < λ Li2 λ This function is differentiable. The error term is handled with the previous lemma. 22 / 24

35 Thank You Thanks For Your Attention! 23 / 24