From Cookie Monster to the IRS: Some Fruitful Interactions between Probability, Combinatorics and Number Theory

Transcription

1 From Cookie Monster to the IRS: Some Fruitful Interactions between Probability, Combinatorics and Number Theory Steven J Miller Williams College Steven.J.Miller@williams.edu UNC Charlotte, February 1, 2011

2 Outline Summarize some of my interests, highlighting interplays across fields. Joint with many faculty, grad students and undergrads. Gaussian behavior in Zeckendorf decompositions: Gene Kopp, Murat Koloğlu, Yinghui Wang. More Sums Than Differences Sets: Peter Hegarty, Brooke Orosz, Dan Scheinerman. Classical Random Matrix Theory: Eduardo Dueñez, Chris Hughes, Jon Keating, Nina Snaith, Duc Khiem Huynh, Tim Novikoff, Chris Hammond, Steven Jackson, Gene Kopp, Murat Koloğlu, Adam Massey, Thuy Pham, Anthony Sabelli, John Sinsheimer. Benford s Law: Chaouki Abdallah, Gregory Heileman, Mark Nigrini, Fernando Perez-Gonzalez, Tu-Thach Quach, Alex Kontorovich, Dennis Jang, Jung Uk Kang, Alex Kruckman, Jun Kudo.

3 Fibonacci Numbers

4 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F 2 = 2, F 3 = 3, F 4 = 5,...

5 5 Fibonacci RMT Circulant Ensemble Benford History / Apps Benford Good 3x + 1 Problem Products F MSTD Sets Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F 2 = 2, F 3 = 3, F 4 = 5,... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers.

6 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F 2 = 2, F 3 = 3, F 4 = 5,... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 2011 = = F 16 + F 13 + F 8 + F 3.

7 7 Fibonacci RMT Circulant Ensemble Benford History / Apps Benford Good 3x + 1 Problem Products F MSTD Sets Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F 2 = 2, F 3 = 3, F 4 = 5,... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 2011 = = F 16 + F 13 + F 8 + F 3. Lekkerkerker s Theorem (1952) The average number of summands in the Zeckendorf decomposition for integers in [F n, F n+1 ) tends to n.276n, where φ = 1+ 5 is the golden mean. φ

8 Central Limit Type Theorem Central Limit Type Theorem As n, the distribution of the number of summands, i.e., a 1 + a 2 + +a m in the generalized Zeckendorf decomposition m i=1 a ih i for integers in [H n, H n+1 ) is Gaussian

9 Far-difference Representation Theorem (Alpert, 2009) (Analogue to Zeckendorf) Every integer can be written uniquely as a sum of the ±F n s, such that every two terms of the same (opposite) sign differ in index by at least 4 (3). Example: 1900 = F 17 F 14 F 10 + F 6 + F 2. K : # of positive terms, L: # of negative terms. Generalized Lekkerkerker s Theorem As n, E[K] and E[L] n/10, E[K] E[L] = φ/ Central Limit Type Theorem As n, K and L converges to a bivariate Gaussian. corr(k, L) = (21 2φ)/(29+2φ).551. K + L and K L are independent.

10 New Approach: Case of Fibonacci Numbers p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: N [F n+1, F n+2 ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n 2,k +

11 New Approach: Case of Fibonacci Numbers p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: N [F n+1, F n+2 ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n 2,k + p n,k+1 = p n 2,k + p n 3,k +

12 New Approach: Case of Fibonacci Numbers p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: N [F n+1, F n+2 ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n 2,k + p n,k+1 = p n 2,k + p n 3,k + p n+1,k+1 = p n,k+1 + p n 1,k.

13 New Approach: Case of Fibonacci Numbers p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: N [F n+1, F n+2 ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n 2,k + p n,k+1 = p n 2,k + p n 3,k + p n+1,k+1 = p n,k+1 + p n 1,k. Generating function: p n,kx k y n y = n,k>0 1 y xy 2. Partial fraction expansion: ( ) y 1 y xy = y 1 2 y 1 (x) y 2 (x) y y 1 (x) 1 y y 2 (x),where y 1 (x) and y 2 (x) are the roots of 1 y xy 2 = 0. Coefficient of y n : g(x) = n,k>0 p n,kx k.

14 New Approach: Case of Fibonacci Numbers (Continued) K n : random variable associated with k. g(x) = n,k>0 p n,kx k. Differentiating identities: g(1) = n,k>0 p n,k = F n+1 F n, g (x) = n,k>0 kp n,kx k 1, g (1) = g(1)e[k n ], (xg (x)) = n,k>0 k 2 p n,k x k 1, (xg (x)) x=1 = g(1)e[kn 2 ],... Method of moments (for normalized K n ): E[(K n )2m ]/(SD(K n ))2m (2m 1)!!, E[(K n) 2m 1 ]/(SD(K n)) 2m 1 0.

15 New Approach: General Case Let p n,k = # {N [H n, H n+1 ): the generalized Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: Fibonacci: p n+1,k+1 = p n,k+1 + p n,k. sm+1 1 j=s m p n m,k j. General: p n+1,k = L 1 m=0 where s 0 = 0, s m = c 1 + c 2 + +c m. Generating function: Fibonacci: y 1 y xy 2. General: n L p n,kx k y n L 1 sm+1 1 m=0 1 L 1 m=0 j=s m x j y m+1 n<l m p n,kx k y n sm+1. 1 j=s m x j y m+1

16 New Approach: General Case (Continued) Partial fraction expansion: ( y Fibonacci: y 1 (x) y 2 (x) General: 1 sl 1 j=s L 1 x j L i=1 B(x, y) = L 1 p n,k x k y n n L 1 1 y y 1 (x) y y 2 (x) ). B(x, y) (y y i (x)) j =i (y j(x) y i (x)). m=0 s m+1 1 j=s m x j y m+1 n<l m y i (x): root of 1 L 1 sm+1 1 m=0 j=s m x j y m+1 = 0. Coefficient of y n : g(x) = n,k>0 p n,kx k. Differentiating identities Method of moments: K n Gaussian. p n,k x k y n,

17 Random Matrix Theory

18 Fundamental Problem: Spacing Between Events General Formulation: Studying system, observe values at t 1, t 2, t 3,... Question: What rules govern the spacings between the t i? Examples: Spacings b/w Energy Levels of Nuclei. Spacings b/w Eigenvalues of Matrices. Spacings b/w Primes. Spacings b/w n k α mod 1. Spacings b/w Zeros of L-functions.

19 Origins of Random Matrix Theory Classical Mechanics: 3 Body Problem Intractable. Heavy nuclei (Uranium: 200+ protons / neutrons) worse! Get some info by shooting high-energy neutrons into nucleus, see what comes out. Fundamental Equation: Hψ n = E n ψ n H : matrix, entries depend on system E n : energy levels ψ n : energy eigenfunctions

20 Origins of Random Matrix Theory Statistical Mechanics: for each configuration, calculate quantity (say pressure). Average over all configurations most configurations close to system average. Nuclear physics: choose matrix at random, calculate eigenvalues, average over matrices (real Symmetric A = A T, complex Hermitian A T = A).

21 Random Matrix Ensembles A = a 11 a 12 a 13 a 1N a 12 a 22 a 23 a 2N a 1N a 2N a 3N a NN = AT, a ij = a ji Fix p, define Prob(A) = p(a ij ). 1 i j N This means Prob(A : a ij [α ij,β ij ]) = 1 i j N βij Want to understand eigenvalues of A. x ij =α ij p(x ij )dx ij.

22 Eigenvalue Distribution δ(x x 0 ) is a unit point mass at x 0 : f(x)δ(x x0 )dx = f(x 0 ). To each A, attach a probability measure: b a μ A,N (x) = 1 N ( δ x λ ) i(a) N 2 N i=1 { } # λ i : λ i(a) 2 [a, b] N μ A,N (x)dx = N N k th i=1 moment = λ i(a) k. 2 k N k 2 +1

23 SKETCH OF PROOF: Eigenvalue Trace Lemma Want to understand A s eigenvalues, but it s A s elements that are chosen randomly and independently. Eigenvalue Trace Lemma Let A be an N N matrix with eigenvalues λ i (A). Then Trace(A k ) = N λ i (A) k, n=1 where Trace(A k ) = N N a i1 i 2 a i2 i 3 a in i 1. i 1 =1 i k =1

24 SKETCH OF PROOF: Correct Scale Trace(A 2 ) = By the Central Limit Theorem: N λ i (A) 2. i=1 N N Trace(A 2 ) = a ij a ji = N λ i (A) 2 N 2 i=1 i=1 j=1 N N a 2 ij N 2 i=1 j=1 Gives NAve(λ i (A) 2 ) N 2 or Ave(λ i (A)) N.

25 SKETCH OF PROOF: Averaging Formula Recall k-th moment of μ A,N (x) is Trace(A k )/2 k N k/2+1. Average k-th moment is Trace(A k ) p(a 2 k N k/2+1 ij )da ij. i j Proof by method of moments: Two steps Show average of k-th moments converge to moments of semi-circle as N ; Control variance (show it tends to zero as N ).

26 SKETCH OF PROOF: Averaging Formula for Second Moment Substituting into expansion gives N 2 N i=1 N a 2 ji p(a 11 )da 11 p(a NN )da NN j=1 Integration factors as a ij = a 2 ijp(a ij )da ij (k,l) =(i,j) k<l a kl = p(a kl )da kl = 1. Higher moments involve more advanced combinatorics (Catalan numbers).

27 SKETCH OF PROOF: Averaging Formula for Higher Moments Higher moments involve more advanced combinatorics (Catalan numbers). 1 2 k N k/2+1 N i 1 =1 N i k =1 a i1 i 2 a ik i 1 i j p(a ij )da ij. Main contribution when the a il i l+1 s matched in pairs, not all matchings contribute equally (if did would get a Gaussian and not a semi-circle; this is seen in Real Symmetric Palindromic Toeplitz matrices).

28 Real Symmetric m-circulant Ensemble

29 Circulant Matrices Study circulant matrices, period m on diagonals. 6-by-6 real symmetric period 2-circulant matrix: c 0 c 1 c 2 c 3 c 2 d 1 c 1 d 0 d 1 d 2 c 3 d 2 c 2 d 1 c 0 c 1 c 2 c 3. c 3 d 2 c 1 d 0 d 1 d 2 c 2 c 3 c 2 d 1 c 0 c 1 d 1 d 2 c 3 d 2 c 1 d 0 Look at the expected value for the moments: M n (N) := E(M n (A, N)) 1 = N n i 1,...,i n N E(a i1 i 2 a i2 i 3 a ini 1 ).

30 Matchings Rewrite: M n (N) = 1 η( )m N n 2 +1 d1 ( ) m dl ( ). where the sum is over equivalence relations on {(1, 2),(2, 3),...,(n, 1)}. The d j ( ) denote the sizes of the equivalence classes, and the m d the moments of p. Finally, the coefficient η( ) is the number of solutions to the system of Diophantine equations: Whenever (s, s + 1) (t, t + 1), i s+1 i s i t+1 i t (mod N) and i s i t (mod m), or i s+1 i s (i t+1 i t )(mod N) and i s i t+1 (mod m).

31 Matchings i s+1 i s i t+1 i t (mod N) and i s i t (mod m), or i s+1 i s (i t+1 i t )(mod N) and i s i t+1 (mod m). Figure: Red edges same orientation and blue, green opposite.

32 Contributing Terms As N, the only terms that contribute to this sum are those in which the entries are matched in pairs and with opposite orientation.

33 Contributing Terms: Algebraic Topology Think of pairings as topological identifications, the contributing ones give rise to orientable surfaces. Contribution from such a pairing is m 2g, where g is the genus (number of holes) of the surface. Proof: combinatorial argument involving Euler characteristic.

34 Computing the Even Moments Theorem: Even Moment Formula M 2k = k/2 g=0 ε g (k)m 2g + O k ( 1 N ), with ε g (k) the number of pairings of the edges of a (2k)-gon giving rise to a genus g surface. J. Harer and D. Zagier (1986) gave generating functions for the ε g (k).

35 Computing the Even Moments Harer and Zagier where k/2 g=0 1+2 ε g (k)r k+1 2g = (2k 1)!! c(k, r) c(k, r)x k+1 = k=0 ( ) r 1+x. 1 x Thus, we write M 2k = m (k+1) (2k 1)!! c(k, m).

36 Computing the Even Moments A multiplicative convolution and Cauchy s residue formula yields the characteristic function of the distribution (inverse Fourier transform of the density). (it) 2k M 2k φ(t) = (2k)! k=0 ( (1+z ) 1 m 1 1 = 1) 2πim z =2 2z 1 1 z 1 = 1 m ( ) ( ) m 1 t 2 l 1 m e t2 2m. l (l 1)! m l=1 e t2 z/2m dz z

37 Results Fourier transform and algebra yields Theorem: Kopp, Koloğlu and M The limiting spectral density function f m (x) of the real symmetric m-circulant ensemble is given by the formula f m (x) = mx 2 e 2 m 2πm r=0 1 m r (2r)! (2r + 2s)! (r + s)!s! s=0 ( m r + s + 1 ) ( 1 2) s (mx 2 ) r. As m, the limiting spectral densities approach the semicircle distribution.

38 Results (continued) Figure: Plot for f 1 and histogram of eigenvalues of 100 circulant matrices of size

39 Results (continued) Figure: Plot for f 2 and histogram of eigenvalues of circulant matrices of size

40 Results (continued) Figure: Plot for f 3 and histogram of eigenvalues of circulant matrices of size

41 Results (continued) Figure: Plot for f 4 and histogram of eigenvalues of circulant matrices of size

42 Results (continued) Figure: Plot for f 8 and histogram of eigenvalues of circulant matrices of size

43 Results (continued) Figure: Plot for f 20 and histogram of eigenvalues of circulant matrices of size

44 Results (continued) Figure: Plot of convergence to the semi-circle.

45 Benford History and Applications

46 Benford s Law: Newcomb (1881), Benford (1938) Statement For many data sets, ( probability of observing a first digit of d base B is log d+1 ) B d. First 60 values of 2 n (only displaying 30) digit # Obs Prob Benf Prob

47 Examples recurrence relations special functions (such as n!) iterates of power, exponential, rational maps products of random variables L-functions, characteristic polynomials iterates of the 3x + 1 map differences of order statistics hydrology and financial data

48 Applications analyzing round-off errors determining the optimal way to store numbers detecting tax fraud and data integrity

49 Caveats! A math test indicating fraud is not proof of fraud: unlikely events, alternate reasons.

50 Caveats! A math test indicating fraud is not proof of fraud: unlikely events, alternate reasons.

51 Detecting Fraud Bank Fraud Bank audit: huge spike of numbers starting with 48 and 49, most due to one person.

52 Detecting Fraud Bank Fraud Bank audit: huge spike of numbers starting with 48 and 49, most due to one person. Write-off limit of $5,000. Officer had friends applying, run up balances just under $5,000...

53 Data Integrity: Stream Flow Statistics: 130 years, 457,440 records

54 Election Fraud: Iran 2009 Numerous protests/complaints over Iran s 2009 elections. Lot of analysis; data moderately suspicious: First and second leading digits; Last two digits (should almost be uniform); Last two digits differing by at least 2. Warning: enough tests, even if nothing wrong will find a suspicious result (but when all tests are on the boundary...).

55 55 Fibonacci RMT Circulant Ensemble Benford History / Apps Benford Good 3x + 1 Problem Products F MSTD Sets New Test for Fraud

56 New Test for Fraud Victoria Cuff, Allie Lewis, M (2010)

57 Image Analysis Pictures aren t Benford s law, but coefficients of Discrete Cosine Transform (DCT) very close (slightly modified law). Analysis of coefficients, from Generalized Gaussian Distributions: f X (x) = A exp( βx c ). 57 Application: detect compression, steganography (hidden message in picture by modifying least significant bit in pixels).

58 Image Analysis (continued) Figure Man used in the experiments.

59 Image Analysis (continued) (a) Histogram of the luminance values of Man in Benford (log 10 mod 1) domain; (b) Distribution of first digits from Man.

60 Image Analysis (continued) Histogram of the DCT values of Man in Benford (log 10 mod 1) domain; (b) Distribution of first digits from Man.

61 Logarithms and Benford s Law Fundamental Equivalence Data set {x i } is Benford base B if {y i } is equidistributed mod 1, where y i = log B x i.

62 Logarithms and Benford s Law Fundamental Equivalence Data set {x i } is Benford base B if {y i } is equidistributed mod 1, where y i = log B x i. 0 log 2 log 10 1

63 Logarithms and Benford s Law Fundamental Equivalence Data set {x i } is Benford base B if {y i } is equidistributed mod 1, where y i = log B x i. 0 log 2 log

64 Logarithms and Benford s Law Fundamental Equivalence Data set {x i } is Benford base B if {y i } is equidistributed mod 1, where y i = log B x i. Kronecker-Weyl Theorem If β Q then nβ mod 1 is equidistributed. (Thus if log B α Q, then α n is Benford.)

65 Example of Equidistribution: n π mod n π mod 1 for n 10

66 Example of Equidistribution: n π mod n π mod 1 for n 100

67 Example of Equidistribution: n π mod n π mod 1 for n 1000

68 Example of Equidistribution: n π mod n π mod 1 for n 10, 000

69 Logarithms and Benford s Law χ 2 values for α n, 1 n N (5% 15.5). N χ 2 (γ) χ 2 (e) χ 2 (π)

70 Logarithms and Benford s Law: Base 10 log(χ 2 ) vs N for π n (red) and e n (blue), n {1,..., N}. Note π , (5%, log(χ 2 ) 2.74)

71 Benford Good Processes

72 Poisson Summation and Benford s Law: Definitions Feller, Pinkham (often exact processes)

73 Poisson Summation and Benford s Law: Definitions Feller, Pinkham (often exact processes) data Y T,B = log B X T (discrete/continuous): P(A) = lim T #{n A : n T} T

74 Poisson Summation and Benford s Law: Definitions Feller, Pinkham (often exact processes) data Y T,B = log B X T (discrete/continuous): P(A) = lim T #{n A : n T} T Poisson Summation Formula: f nice: f(l) = ˆf(l), l= Fourier transformˆf(ξ) = l= f(x)e 2πixξ dx.

75 75 Fibonacci RMT Circulant Ensemble Benford History / Apps Benford Good 3x + 1 Problem Products F MSTD Sets Benford Good Process X T is Benford Good if there is a nice f st y ( ) 1 t CDF Y (y) = T,B T f dt+e T (y) := G T (y) T and monotonically increasing h (h( T ) ):

76 Benford Good Process X T is Benford Good if there is a nice f st y ( ) 1 t CDF Y (y) = T,B T f dt+e T (y) := G T (y) T and monotonically increasing h (h( T ) ): Small tails: G T ( ) G T (Th(T)) = o(1), G T ( Th(T)) G T ( ) = 0(1).

77 77 Fibonacci RMT Circulant Ensemble Benford History / Apps Benford Good 3x + 1 Problem Products F MSTD Sets Benford Good Process X T is Benford Good if there is a nice f st y ( ) 1 t CDF Y (y) = T,B T f dt+e T (y) := G T (y) T and monotonically increasing h (h( T ) ): Small tails: G T ( ) G T (Th(T)) = o(1), G T ( Th(T)) G T ( ) = 0(1). Decay of the Fourier Transform: ˆf(Tl) = o(1). l =0 l

78 Benford Good Process X T is Benford Good if there is a nice f st y ( ) 1 t CDF Y (y) = T,B T f dt+e T (y) := G T (y) T and monotonically increasing h (h( T ) ): Small tails: G T ( ) G T (Th(T)) = o(1), G T ( Th(T)) G T ( ) = 0(1). Decay of the Fourier Transform: ˆf(Tl) = o(1). l =0 l Small translated error: E(a, b, T)) = l Th(T) [E T(b+l) E T (a+l)] = o(1).

79 Main Theorem Theorem (Kontorovich and M, 2005) X T converging to X as T (think spreading Gaussian). If X T is Benford good, then X is Benford.

80 Main Theorem Theorem (Kontorovich and M, 2005) X T converging to X as T (think spreading Gaussian). If X T is Benford good, then X is Benford. Examples L-functions characteristic polynomials (RMT) 3x + 1 problem geometric Brownian motion.

81 Sketch of the proof Structure Theorem: main term is something nice spreading out apply Poisson summation

82 Sketch of the proof Structure Theorem: main term is something nice spreading out apply Poisson summation Control translated errors: hardest step techniques problem specific

83 Sketch of the proof (continued) l= ( P a+l ) Y T,B b +l

84 Sketch of the proof (continued) = l= l Th(T) ( P a+l ) Y T,B b +l [G T (b+l) G T (a+l)]+o(1)

85 Sketch of the proof (continued) = = l= l Th(T) b a ( P a+l ) Y T,B b +l l Th(T) [G T (b+l) G T (a+l)]+o(1) ( ) 1 t T f dt +E(a, b, T)+o(1) T

86 Sketch of the proof (continued) = = l= l Th(T) b a ( P a+l ) Y T,B b +l l Th(T) [G T (b+l) G T (a+l)]+o(1) = ˆf(0) (b a)+ l =0 ( ) 1 t T f dt +E(a, b, T)+o(1) T ˆf(Tl) e 2πibl e 2πial 2πil + o(1).

87 Some Results Theorem (Kontorovich and M, 2005) L(s, f) a good L-function, as T, L(σ T + it, f) is Benford. Theorem (Kontorovich and M, 2005) As N, the distribution of digits of the absolute values of the characteristic polynomials of N N unitary matrices (with respect to Haar measure) converges to the Benford probabilities.

88 The 3x + 1 Problem and Benford s Law

89 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1.

90 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) = 1.

91 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) = 1. 7

92 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) =

93 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) =

94 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) =

95 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) =

96 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) =

97 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) = ,

98 3x + 1 Problem Kakutani (conspiracy), Erdös (not ready). x odd, T(x) = 3x+1 2 k, 2 k 3x + 1. Conjecture: for some n = n(x), T n (x) = , 2-path (1, 1), 5-path (1, 1, 2, 3, 4). m-path: (k 1,..., k m ).

99 Heuristic Proof of 3x + 1 Conjecture a n+1 = T(a n )

100 Heuristic Proof of 3x + 1 Conjecture a n+1 = T(a n ) ( ) 1 E[log a n+1 ] 2 log 3an k 2 k k=1

101 Heuristic Proof of 3x + 1 Conjecture a n+1 = T(a n ) ( ) 1 E[log a n+1 ] 2 log 3an k 2 k k=1 ( ) 3 = log a n + log. 4 Geometric Brownian Motion, drift log(3/4) < 1.

102 Structure Theorem: Sinai, Kontorovich-Sinai P(A) = lim N #{n N:n 1,5 mod 6,n A} #{n N:n 1,5 mod 6}.

103 Structure Theorem: Sinai, Kontorovich-Sinai P(A) = lim N #{n N:n 1,5 mod 6,n A} #{n N:n 1,5 mod 6}. (k 1,..., k m ): two full arithm progressions: 6 2 k 1+ +k m p + q.

104 Structure Theorem: Sinai, Kontorovich-Sinai P(A) = lim N #{n N:n 1,5 mod 6,n A} #{n N:n 1,5 mod 6}. (k 1,..., k m ): two full arithm progressions: 6 2 k 1+ +k m p + q. Theorem (Sinai, Kontorovich-Sinai) k i -values are i.i.d.r.v. (geometric, 1/2): [ ] x log m 2 ( P ( 3 4) m x 0 a 2m = P Sm 2m 2m ) a

105 Structure Theorem: Sinai, Kontorovich-Sinai P(A) = lim N #{n N:n 1,5 mod 6,n A} #{n N:n 1,5 mod 6}. (k 1,..., k m ): two full arithm progressions: 6 2 k 1+ +k m p + q. Theorem (Sinai, Kontorovich-Sinai) k i -values are i.i.d.r.v. (geometric, 1/2): [ ] x log m ( ) 2 P ( 4) 3 m x 0 (log 2 B) 2m a = P S m 2m (log 2 B) 2m a

106 Structure Theorem: Sinai, Kontorovich-Sinai P(A) = lim N #{n N:n 1,5 mod 6,n A} #{n N:n 1,5 mod 6}. (k 1,..., k m ): two full arithm progressions: 6 2 k 1+ +k m p + q. Theorem (Sinai, Kontorovich-Sinai) k i -values are i.i.d.r.v. (geometric, 1/2): [ ] x log m B P ( 3 4) m (S x 0 m 2m) a 2m = P log 2 B a 2m

107 3x + 1 and Benford Theorem (Kontorovich and M, 2005) As m, x m /(3/4) m x 0 is Benford. Theorem (Lagarias-Soundararajan 2006) X 2 N, for all but at most c(b)n 1/36 X initial seeds the distribution of the first N iterates of the 3x + 1 map are within 2N 1/36 of the Benford probabilities.

108 Sketch of the proof Failed Proof: lattices, bad errors.

109 Sketch of the proof Failed Proof: lattices, bad errors. CLT: (S m 2m)/ 2m N(0, 1): P(S m 2m = k) = η(k/ m) m ( + O 1 g(m) m ).

110 Sketch of the proof Failed Proof: lattices, bad errors. CLT: (S m 2m)/ 2m N(0, 1): P(S m 2m = k) = η(k/ m) m ( + O 1 g(m) m ). Quantified Equidistribution: I l = {lm,...,(l+1)m 1}, M = m c, c < 1/2

111 Sketch of the proof Failed Proof: lattices, bad errors. CLT: (S m 2m)/ 2m N(0, 1): P(S m 2m = k) = η(k/ m) m ( + O 1 g(m) m ). Quantified Equidistribution: I l = {lm,...,(l+1)m 1}, M = m c, c < 1/2 ) η( ) k 1, k 2 I l : k1 η( m m k2 small

112 Sketch of the proof Failed Proof: lattices, bad errors. CLT: (S m 2m)/ 2m N(0, 1): P(S m 2m = k) = η(k/ m) m ( + O 1 g(m) m ). Quantified Equidistribution: I l = {lm,...,(l+1)m 1}, M = m c, c < 1/2 ) η( ) k 1, k 2 I l : k1 η( m m k2 small C = log B 2 of irrationality type κ < : #{k I l : kc [a, b]} = M(b a)+o(m 1+ε 1/κ ).

113 Irrationality Type Irrationality type α has irrationality type κ if κ is the supremum of all γ with lim q q γ+1 min p α p q = 0. Algebraic irrationals: type 1 (Roth s Thm). Theory of Linear Forms: log B 2 of finite type.

114 Linear Forms Theorem (Baker) α 1,...,α n algebraic numbers height A j 4, β 1,...,β n Q with height at most B 4, Λ = β 1 logα 1 + +β n logα n. If Λ = 0 then Λ > B CΩ logω, with d = [Q(α i,β j ) : Q], C = (16nd) 200n, Ω = j log A j, Ω = Ω/ log A n. Gives log 10 2 of finite type, with κ < : log 10 2 p/q = q log 2 p log 10 /q log 10.

115 Quantified Equidistribution Theorem (Erdös-Turan) D N = sup [a,b] N(b a) #{n N : x n [a, b]} N There is a C such that for all m: ( m 1 D N C m N h N h=1 n=1 e 2πihx n )

116 Proof of Erdös-Turan Consider special case x n = nα, α Q. Exponential sum 1 sin(πhα) 1 2 hα. Must control m 1 h=1 h hα, see irrationality type enter. type κ, m 1 h=1 h hα = O( m κ 1+ε), take m = N 1/κ.

117 3x + 1 Data: random 10,000 digit number, 2 k 3x ,514 iterations ((4/3) n = a 0 predicts 80,319); χ 2 = 13.5 (5% 15.5). Digit Number Observed Benford

118 3x + 1 Data: random 10,000 digit number, 2 3x ,344 iterations, χ 2 = 11.4 (5% 15.5). Digit Number Observed Benford

119 5x + 1 Data: random 10,000 digit number, 2 k 5x ,004 iterations, χ 2 = 1.8 (5% 15.5). Digit Number Observed Benford

120 5x + 1 Data: random 10,000 digit number, 2 5x ,344 iterations, χ 2 = (5% 15.5). Digit Number Observed Benford

121 Products and Chains of Random Variables

122 Preliminaries X 1 X n Y 1 + +Y n mod 1, Y i = log B X i Density Y i is g i, density Y i + Y j is (g i g j )(y) = 1 0 g i (t)g j (y t)dt. h n = g 1 g n, ĝ(ξ) = ĝ 1 (ξ) ĝ n (ξ). Dirac delta functional: δ α (y)g(y)dy = g(α).

123 Fourier input Fejér kernel: F N (x) = N n= N Fejér series: T N f(x) = N (f F N )(x) = n= N ( 1 n ) e 2πinx. N ( 1 n N ) ˆf(n)e 2πinx. Lebesgue s Theorem: f L 1 ([0, 1]). As N, T N f converges to f in L 1 ([0, 1]). T N (f g) = (T N f) g: convolution assoc.

124 Modulo 1 Central Limit Theorem Theorem (M and Nigrini 2007) {Y m } independent continuous random variables on [0, 1) (not necc. i.i.d.), densities {g m }. Y 1 + +Y M mod 1 converges to the uniform distribution as M in L 1 ([0, 1]) iff n = 0, lim M ĝ 1 (n) ĝ M (n) = 0.

125 Generalizations Levy proved for i.i.d.r.v. just one year after Benford s paper. Generalized to other compact groups, with estimates on the rate of convergence. Stromberg: n-fold convolution of a regular probability measure on a compact Hausdorff group G converges to normalized Haar measure in weak-star topology iff support of the distribution not contained in a coset of a proper normal closed subgroup of G.

126 Non-Benford Product Distribution of digits (base 10) of 1000 products X 1 X 1000, where g 10,m = φ 11 m. φ m (x) = m if x 1/8 1/2m (0 otherwise)

127 Proof of Modulo 1 CLT Density of sum is h l = g 1 g l. Suffices show ε: lim M 1 0 h M(x) 1 dx < ε. Lebesgue s Theorem: N large, h 1 T N h 1 1 = 1 0 h 1 (x) T N h 1 (x) dx < ε 2. Claim: above holds for h M for all M.

128 Proof of Modulo 1 CLT Show lim M h M 1 1 = 0. Triangle inequality: h M 1 1 h M T N h M 1 + T N h M 1 1. Choices of N and ε: h M T N h M 1 < ε/2. Show T N h M 1 1 < ε/2.

129 Proof of Modulo 1 CLT T N h M 1 1 = 1 0 N n= N n =0 N n= N n =0 ( 1 n ) N ( 1 n ) ĥm(n) N ĥ M (n)e 2πinx dx ĥ M (n) = ĝ 1 (n) ĝ M (n) M 0. For fixed N and ε, choose M large so that ĥm(n) < ε/4n whenever n = 0 and n N.

130 Conditions for Chains of Random Variables Conditions {D i (θ)} i I : one-parameter distributions, densities f Di (θ) on [0, ). p : N I, X 1 D p(1) (1), X m D p(m) (X m 1 ). m 2, f m (x m ) = ( ) xm f Dp(m) (1) f m 1 (x m 1 ) dx m 1 x m 1 x m 1 0 lim n l= l =0 n m=1 (Mf Dp(m) (1)) ( 1 2πil ) log B = 0

131 Behavior of Chains of Random Variables Theorem (JKKKM) If conditions hold, as n the distribution of leading digits of X n tends to Benford s law. The error is a nice function of the Mellin transforms: if Y n = log B X n, then Prob(Y n mod 1 [a, b]) (b+a) n (b a) (Mf Dp(m) (1)) m=1 l= l =0 ( 1 2πil log B )

132 Example: All X i Unif(0, k) X i Unif(0, k): without loss of generality k [1, 10). P n (s) = Prob(M 10 (Ξ n ) s). P n (s) log 10 (s) k (log k) n 1 s Γ(n) + ( ζ(n) 1 ) 2 log n 2.7 n 10 s.

133 Example: All X i Exp(1) X i Exp(1), Y n = log B Ξ n. Needed ingredients: 0 exp( x)x s 1 dx = Γ(s). Γ(1+ix) = πx/ sinh(πx), x R. P n (s) log 10 (s) log B s l=1 ( 2π 2 ) n/2 l/ log B. sinh(2π 2 l/ log B)

134 Example: All X i Exp(1) (continued) Bounds on the error P n (s) log 10 s log B s if n = 2, log B s if n = 3, log B s if n = 5, and log B s if n = 10. Error at most ( ) n/ l log 10 s.057 n log exp(8.5726l) s 10 l=1

135 More Sums Than Difference Sets

136 Statement A finite set of integers, A its size. Form Sumset: A+A = {a i + a j : a j, a j A}. Difference set: A A = {a i a j : a j, a j A}. Definition We say A is difference dominated if A A > A+A, balanced if A A = A+A and sum dominated (or an MSTD set) if A+A > A A.

137 Questions Expect generic set to be difference dominated: addition is commutative, subtraction isn t: Generic pair (x, y) gives 1 sum, 2 differences. Questions Do there exist sum-dominated sets? If yes, how many?

138 Examples Conway: {0, 2, 3, 4, 7, 11, 12, 14}. Marica (1969): {0, 1, 2, 4, 7, 8, 12, 14, 15}. Freiman and Pigarev (1973): {0, 1, 2, 4, 5, 9, 12, 13, 14, 16, 17, 21, 24, 25, 26, 28, 29}. Computer search: random subsets of {1,..., 100}: {2, 6, 7, 9, 13, 14, 16, 18, 19, 22, 23, 25, 30, 31, 33, 37, 39, 41, 42, 45, 46, 47, 48, 49, 51, 52, 54, 57, 58, 59, 61, 64, 65, 66, 67, 68, 72, 73, 74, 75, 81, 83, 84, 87, 88, 91, 93, 94, 95, 98, 100}.

139 Binomial model Binomial model, parameter p(n) Each k {0,..., n} is in A with probability p(n). Consider uniform model (p(n) = 1/2): Let A {0,..., n}. Most elements in {0,..., 2n} in A+A and in { n,..., n} in A A. E[ A+A ] = 2n 11, E[ A A ] = 2n 7.

140 Martin and O Bryant 06 Theorem A from {0,..., N} by binomial model with constant parameter p (so k A with probability p). At least k SD;p 2 N+1 subsets are sum dominated. k SD;1/2 10 7, expect about Proof (p = 1/2): Generically A = N 2 + O( N). about N 4 N k 4 ways write k A+A. about N 4 k 4 ways write k A A.

141 Notation X f(n) means ε 1,ε 2 > 0, N ε1,ε 2 st N N ε1,ε 2 Prob(X [(1 ε 1 )f(n),(1 +ε 1 )f(n)]) < ε 2. S = A+A, D = A A, S c = 2N + 1 S, D c = 2N + 1 D. New model: Binomial with parameter p(n): 1/N = o(p(n)) and p(n) = o(1); Prob(k A) = p(n). Conjecture (Martin-O Bryant) As N, A is a.s. difference dominated.

142 Main Result Theorem (Hegarty-Miller) p(n) as above, g(x) = 2 e x (1 x) x. p(n) = o(n 1/2 ): D 2S (Np(N)) 2 ; ( p(n) = cn 1/2 : D g(c 2 )N, S g (c 0, D/S 2; c, D/S 1); N 1/2 = o(p(n)): S c 2D c 4/p(N) 2. c 2 2 ) N Can generalize to binary linear forms, still have critical threshold.

143 Inputs Key input: recent strong concentration results of Kim and Vu (Applications: combinatorial number theory, random graphs,...). Example (Chernoff): t i iid binary random variables, Y = n i=1 t i, then ( λ > 0 : Prob Y E[Y] ) λn 2e λ/2. Need to allow dependent random variables. Sketch of proofs: X {S,D,S c,d c }. 1 Prove E[X] behaves asymptotically as claimed; 2 Prove X is strongly concentrated about mean.

144 Setup for Proofs Note: only need strong concentration for N 1/2 = o(p(n)).

145 Setup for Proofs Note: only need strong concentration for N 1/2 = o(p(n)). Will assume p(n) = o(n 1/2 ) as proofs are elementary (i.e., Chebyshev: Prob( Y E[Y] kσ Y ) 1/k 2 )).

146 Setup for Proofs Note: only need strong concentration for N 1/2 = o(p(n)). Will assume p(n) = o(n 1/2 ) as proofs are elementary (i.e., Chebyshev: Prob( Y E[Y] kσ Y ) 1/k 2 )). For convenience let p(n) = N δ, δ (1/2, 1). IID binary indicator variables: { 1 with probability N δ X n;n = 0 with probability 1 N δ. X = N i=1 X n;n, E[X] = N 1 δ.

147 Proof Lemma P 1 (N) = 4N (1 δ), O = #{(m, n) : m < n {1,...,N} A}. With probability at least 1 P 1 (N) have 1 X [ 1 2 N1 δ, 3 2 N1 δ] N1 δ ( 1 2 N1 δ 1) 2 O 3 2 N1 δ ( 3 2 N1 δ 1) 2.

148 Proof Lemma P 1 (N) = 4N (1 δ), O = #{(m, n) : m < n {1,...,N} A}. With probability at least 1 P 1 (N) have 1 X [ 1 2 N1 δ, 3 2 N1 δ] N1 δ ( 1 2 N1 δ 1) 2 O 3 2 N1 δ ( 3 2 N1 δ 1) 2. Proof: (1) is Chebyshev: Var(X) = NVar(X n;n ) N 1 δ. (2) follows from (1) and ( r 2) ways to choose 2 from r.

149 Concentration Lemma f(δ) = min ( 1 2, ) 3δ 1 2, g(δ) any function st 0 < g(δ) < f(δ). p(n) = N δ, δ (1/2, 1), P 1 (N) = 4N (1 δ), P 2 (N) = CN (f(δ) g(δ)). With probability at least 1 P 1 (N) P 2 (N) have D/S = 2+O(N g(δ) ).

150 Concentration Lemma f(δ) = min ( 1 2, ) 3δ 1 2, g(δ) any function st 0 < g(δ) < f(δ). p(n) = N δ, δ (1/2, 1), P 1 (N) = 4N (1 δ), P 2 (N) = CN (f(δ) g(δ)). With probability at least 1 P 1 (N) P 2 (N) have D/S = 2+O(N g(δ) ). Proof: Show D 2O + O(N 3 4δ ), S O + O(N 3 4δ ). As O is of size N 2 2δ with high probability, need 2 2δ > 3 4δ or δ > 1/2.

151 Analysis of D Contribution from diagonal terms lower order, ignore.

152 Analysis of D Contribution from diagonal terms lower order, ignore. Difficulty: (m, n) and (m, n ) could yield same differences.

153 Analysis of D Contribution from diagonal terms lower order, ignore. Difficulty: (m, n) and (m, n ) could yield same differences. Notation: m < n, m < n, m m, Y m,n,m,n = { 1 if n m = n m 0 otherwise.

154 Analysis of D Contribution from diagonal terms lower order, ignore. Difficulty: (m, n) and (m, n ) could yield same differences. Notation: m < n, m < n, m m, Y m,n,m,n = { 1 if n m = n m 0 otherwise. E[Y] N 3 N 4δ + N 2 N 3δ 2N 3 4δ. As δ > 1/2, Expected number bad pairs O.

155 Analysis of D Contribution from diagonal terms lower order, ignore. Difficulty: (m, n) and (m, n ) could yield same differences. Notation: m < n, m < n, m m, Y m,n,m,n = { 1 if n m = n m 0 otherwise. E[Y] N 3 N 4δ + N 2 N 3δ 2N 3 4δ. As δ > 1/2, Expected number bad pairs O. Claim: σ Y N r(δ) with r(δ) = 1 2 max(3 4δ, 5 7δ). This and Chebyshev conclude proof of theorem.

156 Proof of claim Cannot use CLT as Y m,n,m,n are not independent.

157 Proof of claim Cannot use CLT as Y m,n,m,n are not independent. Use Var(U + V) 2Var(U)+2Var(V).

158 Proof of claim Cannot use CLT as Y m,n,m,n are not independent. Use Var(U + V) 2Var(U)+2Var(V). Write Ym,n,m,n = U m,n,m,n + V m,n,n with all indices distinct (at most one in common, if so must be n = m ). Var(U) = Var(U m,n,m,n )+2 CoVar(U m,n,m,n, U m,ñ, m,ñ ). (m,n,m,n ) = ( m,ñ, m,ñ )

159 Analyzing Var(U m,n,m,n ) At most N 3 tuples. Each has variance N 4δ N 8δ N 4δ. Thus Var(U m,n,m,n ) N3 4δ.

160 Analyzing CoVar(U m,n,m,n, U m,ñ, m,ñ ) All 8 indices distinct: independent, covariance of 0. 7 indices distinct: At most N 3 choices for first tuple, at most N 2 for second, get E[U (1) U (2) ] E[U (1) ]E[U (2) ] = N 7δ N 4δ N 4δ N 7δ. Argue similarly for rest, get N 5 7δ + N 3 4δ.