Continuous Distribution Arising from the Three
Gap Theorem
Geremı́as Polanco Encarnación
Elementary Analytic and Algorithmic Number Theory
Athens, Georgia
June 8, 2015
Hampshire college, MA
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Happy Birthday Prof. Carl Pomerance
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Continuous Distribution
CONTINUOUS DISTRIBUTION ARISING FROM THE THREE
GAP THEOREM
(Joint work with D. Schultz, and A. Zaharescu)
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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 SQ (α) and consider the gaps between consecutive
elements, then there are at most three gap sizes in SQ (α).
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Example of Steinhaus Theorem
√
√
α = 2 and Q = 10. The points S10 ( 2) along with the three
gaps labeled A, B, and C are shown here.
A
80Α< 85Α<
C
A
B
A
83Α< 88Α<
81Α< 86Α<
C
√
Figure : S10 ( 2)
A
B
A
B
84Α< 89Α<
82Α< 87Α<
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The Main Problem
Denote the nth element of SQ (α) by {σn α}
{σn+1 α} − {σn α} ≥
1
.
Q
(1)
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The Main Problem
Denote the nth element of SQ (α) by {σn α}
{σn+1 α} − {σn α} ≥
λ
.
Q
(1)
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The Main Problem
Denote the nth element of SQ (α) by {σn α}
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
λ
}.
Q
(1)
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The Main Problem
Denote the nth element of SQ (α) by {σn α}
Z1
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
λ
}dα.
Q
(1)
0
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The Main Problem
Denote the nth element of SQ (α) by {σn α}
Z1
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
Q
λ
Q}
dα.
(1)
0
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The Main Problem
Denote the nth element of SQ (α) by {σn α} and consider the
function:
Z1
g (λ; Q) = lim
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
Q
Q→∞
λ
Q}
dα.
0
(1)
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The Main Problem
Denote the nth element of SQ (α) by {σn α} and consider the
function:
g1γ,η (λ; Q)
1
=
η
γ+η
Z
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
Q
λ
Q}
dα.
γ
(1)
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The Main Problem
Denote the nth element of SQ (α) by {σn α}
g1γ,η (λ; Q)
1
=
η
γ+η
Z
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
Q
λ
Q}
dα.
γ
(1)
Does this limit exist?
Is it a continuous function of λ?
Is it differentiable?
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The Main Problem
More generally, we consider the join distribution
gk (λ1 , λ2 , . . . , λk ), which describes the average distribution of
k-tuples of consecutive gaps over the interval [a, b], defined as
gk (λ1 , λ2 , . . . , λk ) = lim gkα,η (λ1 , . . . , λk ; Q),
Q→∞
(2)
where
gkγ,η (λ1 , . . . , λk ; Q)
1
=
η
γ+η
Z
#{ω1 · · · ωk ∈ GQ,k (α) : ∀i≤k ωi ≥
Q
λi
Q}
dα.
γ
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A Probability question
For k = 1
For a given λ1 , what is the probability that a randomly selected
gap will be greater than λ1 times the average gap?
And for the the k dimensional case (say k = 2)
for given λ1 and λ2 what is the probability that a gap and its
neighbor to the right are both greater than the average times λ1
and λ2 respectively?
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A Differentiable distribution
1
Figure : g13
1
, 10
(λ; 1000)
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A Differentiable distribution
1
Figure : g13
1
, 10
(λ; 1000)
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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
g1 (λ) :=
γ,η
lim g
(λ; Q)
Q→∞ 1










6 
=
2

π 








π2
6
− λ,
0<λ<1
2
2 log 2−λ
log2 (2) − 2π
−1+ λ
− λ
2 λ−1
3
4
1
λ
− log λ log(λ) + 4 Li2 λ + 2 Li2 2
−1 +
λ
2
2
− λ
log
λ−2
λ−1
+ 3λ
log
2
λ
λ−1
+ 3λ
log
2
+ 4 Li2
λ
λ−1
1
λ
,
− 2 Li2
1<λ<2
2
λ
,
,
2<λ
where the Dilogarithm is defined for |z| ≤ 1 by
Li2 (z) =
∞ n
X
z
n=1
n2
.
More precisely, we have the error estimate:
α,η
Q −1/2
g
(λ; Q) − g1 (λ) (1 + λ)
.
1
η
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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 (λ1 ) 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.
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Gaps are related to Farey Sequence
A
80Α< 85Α<
C
A
B
A
83Α< 88Α<
81Α< 86Α<
C
Figure : SQ (α)
A
B
A
B
84Α< 89Α<
82Α< 87Α<
Recall the Farey sequence of order Q is given by
FQ = {0 ≤ qa ≤ 1|(a, q) = 1, q ≤ Q}
Example: F4 = { 10 , 41 , 13 , 21 , 32 , 43 , 11 }
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Gaps are related to Farey Sequence
A
80Α< 85Α<
C
A
B
A
83Α< 88Α<
81Α< 86Α<
C
Figure : SQ (α)
A
B
A
B
84Α< 89Α<
82Α< 87Α<
Recall the Farey sequence of order Q is given by
FQ = {0 ≤ qa ≤ 1|(a, q) = 1, q ≤ Q}
Example: F4 = { 10 , 41 , 13 , 21 , 32 , 43 , 11 }
Choose consecutive fractions
a1
q1
and
a2
q2
in FQ−1 , so that
a1
a2
<α< .
q1
q2
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Gaps are related to Farey Sequence
Lemma
The three gaps that can appear have lengths
A = q1 α − a1 = {q1 α} ,
B = a2 − q2 α = 1 − {q2 α},
C = A + B,
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Gaps are related to Farey Sequence
Lemma
The three gaps that can appear have lengths
A = q1 α − a1 = {q1 α} ,
B = a2 − q2 α = 1 − {q2 α},
C = A + B,
The permutation σ of the set {0, 1, . . . , Q − 1} that increasingly
orders the sequence of fractional parts is
{σ0 α}, {σ1 α}, . . . , {σQ−1 α}
and satisfies:
σ0 = 0,

if σi ∈ [0, Q − q1 )
 q1 ,
q1 − q2 , if σi ∈ [Q − q1 , q2 )
σi+1 − σi =

−q2 ,
if σi ∈ [q2 , Q)
(A Gap)
(C Gap) .
(B Gap)
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Gaps are related to Farey Sequence
The numbers of A gaps, B gaps and C gaps are, respectively,
Q − q1 ,
Q − q2 ,
q1 + q2 − Q.
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Gaps are related to Farey Sequence
The numbers of A gaps, B gaps and C gaps are, respectively,
Q − q1 ,
Q − q2 ,
q1 + q2 − Q.
Note that this agrees with
(Q − q1 ) + (Q − q2 ) + (q1 + q2 − Q) = Q,
(Q − q1 )A + (Q − q2 )B + (q1 + q2 − Q)C = 1.
This last equality is equivalent to the determinant property of
Farey fractions:
a2 q1 − a1 q2 = 1.
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A Lemma Via Kloosterman Sums (Boca, Cobeli, and
Zaharescu)
Notation
Let
X
X
f :=
Ω
γ≤
( qQ1 , qQ2 )
a1
a2
q1 < q2
f.
∈Ω
≤γ+η
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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
ZZ
1 X q q 1
6
1
2
f (x, y )dxdy f
,
− 2
2
η
Q Q Q
π
Ω
Ω
∂f ∂f Area(Ω) log Q
(1 + Len(∂Ω)) log Q
+ kf k∞
+
∂x + ∂y ηQ
ηQ
∞
∞
mf kf k∞
,
ηQ 1/2−
where mf is an upper bound for the number of intervals of
monotonicity of each of the maps y → f (x, y ).
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Proof of the Main Theorem for K = 1
Contribution of the A gaps
Recall the integral in the main theorem
g1γ,η (λ; Q)
1
=
η
γ+η
Z
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
Q
λ
Q}
dα.
γ
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Proof of the Main Theorem for K = 1
Contribution of the A gaps
Recall the integral in the main theorem
g1γ,η (λ; Q)
1
=
η
γ+η
Z
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
Q
λ
Q}
dα.
γ
Partition the interval [γ, γ + η] into Farey arcs with denominators
strictly less than Q.
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Proof of the Main Theorem for K = 1
Contribution of the A gaps
Recall the integral in the main theorem
g1γ,η (λ; Q)
1
=
η
γ+η
Z
#{0 ≤ n < Q : {σn+1 α} − {σn α} ≥
Q
λ
Q}
dα.
γ
Partition the interval [γ, γ + η] into Farey arcs with denominators
strictly less than Q.
In the interval [ qa11 , qa22 ], the A gaps contribute the amount
a2
Zq2
a1
q1
1
Q
Q − q1 , q1 α − a1 ≥ λ/Q
dα,
0,
q1 α − a1 < λ/Q
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Proof of the Main Theorem for K = 1
If a new integration variable t defined by
α=
t
a1
+
q1 q1 q2
(3)
is introduced, using a2 q1 − a1 q2 = 1, the integral becomes
Z1
0
dt
Qq1 q2
(
Q − q1 , t ≥
0,
t<
λq2
Q
λq2
Q
(
=
Q−q1
Qq1 q2
0,
1−
λq2
Q
, 0≤
1<
λq2
Q
λq2
Q
≤1
.
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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
λq2
X
1
1
−
, 0 ≤ qQ2 ≤ λ1
Qq1 q2
Q
.
q2
1
η a1 a2
0,
<
λ
Q
γ≤ < ≤γ+η
q1
q2
By Lemma 4, this sum converges to the integrals

R1 R1 (1−x)(1−λy )



dxdy , 0 ≤ λ ≤ 1

xy
6  0 1−y
,
1
π2 
Rλ R1 (1−x)(1−λy )


dxdy , 1 < λ


xy
0 1−y
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Proof of the Main Theorem for K = 1
This further equals
2
6
−1 − λ2 + π6 ,
1
π2
−1 − 2λ
+ (1 − λ) log 1 − λ1 + Li2
0≤λ≤1
,
, 1<λ
(4)
The B and C gaps are handled similarly, and we add their
respective contributions.
1
λ
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Proof of the Main Theorem for K = 1
The result is the expression:

2

0≤λ≤1
1 − π12 ,






1

log2 (2)

1
π2
1
λ
2
2−λ

−
+
+
+
log
−
1
−
log
−

2
3
2λ
λ

, 1<λ<2
21
4
λ
λ−1


λ−1
λ
1
λ

log λ + 2 log 4 log(λ) + Li2 λ + Li2 2

6
π2 
log2 (2)

3
π2

+ log(2)
λ=2

4 − 12 −
2
2 ,





1


1
1
λ
2
1
λ−1

+
+
log
1
−
−
log
1
−
+
log

2
2λ
4
λ
λ
λ
λ−2 + , 2 < λ


 Li 1 − Li 2 2 λ
2 λ
This function is differentiable. The error term is handled with the
previous lemma.
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Thank You
Thanks For Your Attention!
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