Convergence rates in generalized Zeckendorf decomposition problems

Transcription

1 Convergence rates in generalized Zeckendorf decomposition problems Ray Li 1 Steven J Miller (sjm1@williams.edu) 2 Zhao Pan (zhaop@andrew.cmu.edu) 3 Huanzhong Xu (huanzhox@andrew.cmu.edu) 4 1 Carnegie Mellon 2 Williams College 3 Carnegie Mellon 4 Carnegie Mellon Workshop on Combinatorial and Additive Number Theory New York, NY, May 26, 2016

2 Summary Review Zeckendorf-type decompositions

3 Summary Review Zeckendorf-type decompositions Discuss new approaches to asymptotic behavior of variance

4 Summary Review Zeckendorf-type decompositions Discuss new approaches to asymptotic behavior of variance Discuss new results on Gaussian behavior of gaps between summands

5 Previous Results

6 Definitions: Zeckendorf Decomposition Theorem (Zeckendorf) Let {F n } n N denote the Fibonacci numbers with F 1 = 1 and F 2 = 2. Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers.

7 Definitions: Zeckendorf Decomposition Theorem (Zeckendorf) Let {F n } n N denote the Fibonacci numbers with F 1 = 1 and F 2 = 2. Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example 101 = = F 10 + F 5 + F 3 + F 1

8 Definitions: Positive Linear Recurrence Sequence Definition A Positive Linear Recurrence Sequence (PLRS) is a sequence {G n } satisfying G n = c 1 G n c L G n L with nonegative integer coefficients c i with c 1, c L, L 1 and initial conditions G 1 = 1 and G n = c 1 G n 1 + c 2 G n c n 1 G for 1 n L.

9 Examples of PLRS 1 Fibonacci numbers: L = 2, c 1 = c 2 = 1. G 1 = 1, G 2 = 2, G 3 = 3, G 4 = 5,....

10 Examples of PLRS 1 Fibonacci numbers: L = 2, c 1 = c 2 = 1. G 1 = 1, G 2 = 2, G 3 = 3, G 4 = 5, Powers of b: L = 2, c 1 = b 1, c 2 = b. G 1 = 1, G 2 = b, G 3 = b 2, G 4 = b 3,....

11 Examples of PLRS 1 Fibonacci numbers: L = 2, c 1 = c 2 = 1. G 1 = 1, G 2 = 2, G 3 = 3, G 4 = 5, Powers of b: L = 2, c 1 = b 1, c 2 = b. G 1 = 1, G 2 = b, G 3 = b 2, G 4 = b 3, d-bonacci numbers: L = d, c 1 = c 2 = = c d = 1, G n = 2 n 1 for n d.

12 Definition: Generalized Zeckendorf Decomposition Definition (Generalized Zeckendorf Decomposition) Let {G n } be a PLRS and m be a positive integer. Then m = N a i G N+1 i i=1 is a legal decomposition if a 1 > 0 and the other a i 0, and one of the following conditions holds. 1 We have N < L and a i = c i for 1 i N. 2 There exists an s {1,..., L} such that a 1 = c 1, a 2 = c 2,..., a s 1 = c s 1, a s < c s, and {b i } N s i=1 (with b i = a s+i ) is either legal or empty.

13 Example Consider the PLRS: G n = 3G n 1 + 2G n 2 + 2G n 4. Examples of legal decompositions: m = 3G 9 + 2G 8 + G 6 + 3G 5 + G 4 + 2G 1. m = 3G 9 + 2G 8 + G 6 + 3G 5 + G 4 + 3G 1. Examples of NOT legal decompositions: m = 4G 9. m = 3G 9 + 2G 8 + G 7. m = 3G 9 + 2G 8 + 2G 6.

14 Theorem: Generalized Zeckendorf Decomposition Theorem Let {G n } be a PLRS. Then there is a unique legal decomposition for every positive integer m.

15 Definitions and Notations Definition Probability Space Ω n : The set of legal decompositions of integers in [G n, G n+1 ). Probability Measure: Let each of the G n+1 G n legal decompositions be weighted equally. Random Variables K n : Set K n (ω) equal to the number of summands of ω Ω n.

16 Asymptotic Behavior of Variance

17 Old Result Theorem (Miller and Wang, 2012) When {G n } is a PLRS, there exists constants A, B, C, D, γ 1 (0, 1), γ 2 (0, 1) such that E[K n ] = An + B + o(γ n 1) (1) Var[K n ] = Cn + D + o(γ n 2) (2) Remark Proof of (1) is easy. Proof of (2) is hard. Key difficulty: bound of C.

18 Main New Result Theorem When {G n } is a PLRS with length L, there exists C min > 0 such that Var[K n ] > C min n for all n > L.

19 Definition of Blocks Definition We define a Type 1 block as an integer sequence corresponding to Condition 1. We define a Type 2 block as an integer sequence corresponding to Condition 2. Example G n = 2G n 1 + 2G n 2 + 2G n 4 Suppose m = G 5 + 2G 3 + G 2 + 2G 1, we write the decomposition into an integer sequence: [1, 0, 2, 1, 2]. The Type 2 Blocks: [1], [0], [2, 1], [1]. The Type 1 Block: [2].

20 Definition of Blocks Definition We define the size of a block as the total number of summands in it. We define the length of a block as the total number of indices in it. Example G n = 2G n 1 + 3G n 3 + 2G n 4 A possible Type 2 Block is [2, 0, 3, 1]. It has size 6 and length 4.

21 Key Observations of Type 1 Blocks Properties: It appears at most once in any legal decomposition. So, It has to be the last block if it does exist. Type 1 block matters little when n is large. Second to last block has to be a Type 2 block.

22 Key Observations of Type 2 Blocks The length of a Type 2 block is fully determined by its size. So we can define a function l(t) such that a Type 2 block with size t has length l(t).

23 Key Observations of Type 2 Blocks The length of a Type 2 block is fully determined by its size. So we can define a function l(t) such that a Type 2 block with size t has length l(t). A Type 2 block always has nonnegative size and strictly positive length.

24 Key Observations of Type 2 Blocks The length of a Type 2 block is fully determined by its size. So we can define a function l(t) such that a Type 2 block with size t has length l(t). A Type 2 block always has nonnegative size and strictly positive length. A legal decomposition will stay legal if we add a Type 2 block to it or remove a Type 2 block from it.

25 Key Idea Lemma For a fixed t, there is a bijection h t between: all legal decompositions with total length n and the second to last block with size t, and all legal decompositions with total length n l(t). Example G n = 2G n 1 + 2G n G n 4. m = G 6 + G 5 + 2G 4 + G 1. Its block representation: [1], [1], [2, 0], [0], [1]. After removing [0]: [1], [1], [2, 0], [1]. The resulting legal decomposition: G 5 + G 4 + 2G 3 + G 1.

26 Key Idea Set Z n (ω) equal to the size of the second to last block for the legal decomposition ω Ω n. Then if Z n (ω) = t, we have K n (ω) = K n l(t) (h t (ω)) + t.

27 Key Idea Set Z n (ω) equal to the size of the second to last block for the legal decomposition ω Ω n. Then if Z n (ω) = t, we have K n (ω) = K n l(t) (h t (ω)) + t. In other words, we get the conditional expectations: E[K n Z n = t] = E[K n l(t) ] + t E[Kn 2 Z n = t] = E[(K n l(t) + t) 2 ] = E[Kn l(t)] 2 + 2t E[K n l(t) ] + t 2.

28 Proof of Main Theorem We first explicitly choose C min. Then, we use strong induction on n to prove Var[K n ] > C min n.

29 Proof of Main Theorem We first explicitly choose C min. Then, we use strong induction on n to prove Var[K n ] > C min n. From previous results, we know E[K n ] = An + B + f (n) for some f (n) = o(γ n 1 ).

30 Proof of Main Theorem We first explicitly choose C min. Then, we use strong induction on n to prove Var[K n ] > C min n. From previous results, we know E[K n ] = An + B + f (n) for some f (n) = o(γ n 1 ). From induction hypothesis, we can bound E[K 2 n l(t) ] by estimating Var[K n l(t) ] + (E[K n l(t) ]) 2.

31 Proof of Main Theorem We first explicitly choose C min. Then, we use strong induction on n to prove Var[K n ] > C min n. From previous results, we know E[K n ] = An + B + f (n) for some f (n) = o(γ n 1 ). From induction hypothesis, we can bound E[K 2 n l(t) ] by estimating Var[K n l(t) ] + (E[K n l(t) ]) 2. In inductive step, we will be able to bound Var[K n ] by estimating E[K 2 n ] (E[K n ]) 2.

32 Further Works Can we lift the constraint that the recurrence relationship is positive linear?

33 Further Works Can we lift the constraint that the recurrence relationship is positive linear? Can we generalize to more general sequences? Even without a recurrence relation?

34 Further Works Can we lift the constraint that the recurrence relationship is positive linear? Can we generalize to more general sequences? Even without a recurrence relation? Can we define legal decompositions and blocks for other sequences?

35 Gaussian Behavior of Gaps

36 Technical Note For the rest of the talk, assume PLRS refers to recurrences with c i > 0 for 1 i L. G n = c 1 G n c L G n L (Previoiusly, we only needed c 1, c L > 0 and c i 0 for 1 < i < L.)

37 Gaps in Decompositions Definition Let m be a positive integer with decomposition m = N a i G N+1 i = G i1 + G i2 + + G ik i=1 and i 1 i 2 i k. Then the gaps in the decomposition of m are the numbers i 1 i 2, i 2 i 3,..., i k 1 i k.

38 Gaps in Decompositions Definition Let m be a positive integer with decomposition m = N a i G N+1 i = G i1 + G i2 + + G ik i=1 and i 1 i 2 i k. Then the gaps in the decomposition of m are the numbers i 1 i 2, i 2 i 3,..., i k 1 i k. Example The gaps for the decomposition of m = 101 are 5, 2, = F 10 + F 5 + F 3 + F 1.

39 Gaps in Decompositions Recall: Let k(m) be the number of summands in the decomposition of m. Let K n be k(m) for an m chosen uniformly in [G n, G n+1 ).

40 Gaps in Decompositions Let k(m) be the number of summands in the decomposition of m. Let K n be k(m) for an m chosen uniformly in [G n, G n+1 ). Let k g (m) be the number of size-g gaps in the decomposition of m. Let K g,n be k g (m) for an m chosen uniformly in [G n, G n+1 ).

41 Gaps in Decompositions k(m) : number of summands in m k g (m) : number of size-g gaps in m Example 101 = F 10 + F 5 + F 3 + F 1 k(101) = 4, k 2 (101) = 2, k 3 (101) = k 4 (101) = 0, k 5 (101) = 1

42 Gaps in Decompositions k(m) : number of summands in m k g (m) : number of size-g gaps in m Example 101 = F 10 + F 5 + F 3 + F 1 k(101) = 4, k 2 (101) = 2, k 3 (101) = k 4 (101) = 0, k 5 (101) = 1 Fact k(m) = 1 + k g (m) g=0

43 Results Understand: Distribution of number of summands: K n

44 Results Understand: Distribution of number of summands: K n Don t Understand: Distribution of number of size-g gaps: K g,n

45 Results: Mean and Variance K n : number of summands in random m [G n, G n+1 ) Theorem (Lekkerkerker, 1951) When {G n } is Fibonacci, E[K n ] = C Lek n + O(1) for C Lek = 1 ϕ (C lek 0.276) Theorem (Miller and Wang, 2012) When {G n } is a PLRS, there exists A > 0, B, and γ 1 (0, 1) such that E[K n ] = An + B + O(γ n 1 ). Theorem (Miller and Wang, 2012) When {G n } is a PLRS, there exists C > 0, D, and γ 2 (0, 1) such that Var[K n ] = Cn + D + O(γ n 2 ).

46 Results: Mean and Variance K g,n : number of size-g gaps in random m [G n, G n+1 ) Theorem (Main Result 1: Lekkerkerker for gaps) When {G n } is a PLRS, for every integer g 0, there exists A g, B g and γ g,1 (0, 1) such that E[K g,n ] = A g n + B g + O(γ n g,1 ). By Bower et al. 2013, A g is known for all PLRS {G n } and g. Theorem (Main Result 2: Variance is linear for gaps) When {G n } is a PLRS, for every integer g 0, there exists C g, D g and γ g,2 (0, 1) such that Var[K g,n ] = C g n + D g + O(γ n g,2 ).

47 Results: Asymptotic Gaussianity K n : number of summands in random m [G n, G n+1 ) K g,n : number of size-g gaps in random m [G n, G n+1 ) Theorem (Kopp, Kolo glu, Miller, and Wang, 2011) When {G n } is Fibonacci, as n, K n approaches Gaussian. Theorem (Miller and Wang, 2012) When {G n } is PLRS, as n, K n approaches Gaussian. Theorem (Main Result 3: Gaussian Behavior for Gaps) When {G n } is PLRS, as n, K g,n approaches Gaussian.

48 Results: Summary Theorem (Main Results, Summary) The mean µ g,n and variance σ 2 g,n of K g,n grow linearly in n, and (K g,n µ g,n )/σ g,n converges to the standard normal N(0, 1) as n.

49 Proof Sketch Theorem (Main Result 3: Fibonacci Numbers, g = 2) When {G n } is Fibonacci, as n, (K 2,n µ 2,n )/σ 2,n converges to standard normal.

50 Proof Sketch Theorem (Main Result 3: Fibonacci Numbers, g = 2) When {G n } is Fibonacci, as n, (K 2,n µ 2,n )/σ 2,n converges to standard normal. Let p 2,n,k = #{m [F n, F n+1 ) with k size-2 gaps}. Pr[K 2,n = k] p 2,n,k.

51 Proof Sketch p 2,n,k = #{m [F n, F n+1 ) with k size-2 gaps}.

52 Proof Sketch p 2,n,k = #{m [F n, F n+1 ) with k size-2 gaps}. Lemma p 2,n,k = p 2,n 1,k + p 2,n 2,k 1 + p 2,n 3,k 1 p 2,n 3,k

53 Proof Sketch Lemma (Key Lemma) If a n,k is a nice two dimensional homogenous recurrence a n,k = i 0 i=1 j 0 j=0 t i,j a n i,k j for constants t i,j, then the random variable X n given by Pr[X n = k] a n,k approaches Gaussian as n.

54 Proof Sketch Lemma (Key Lemma) If a n,k is a nice two dimensional homogenous recurrence then the random variable X n given by Pr[X n = k] a n,k approaches Gaussian as n. Famous example: a n,k = a n 1,k + a n 1,k 1 gives...

55 Proof Sketch Lemma (Key Lemma) If a n,k is a nice two dimensional homogenous recurrence then the random variable X n given by Pr[X n = k] a n,k approaches Gaussian as n. Famous example: a n,k = a n 1,k + a n 1,k 1 gives... the binomials: a n,k = ( n k)!

56 Proof Sketch Lemma (Key Lemma) If a n,k is a nice two dimensional homogenous recurrence then the random variable X n given by Pr[X n = k] a n,k approaches Gaussian as n. Applications a n,k = a n 1,k + a n 1,k 1 X n = # heads after n coin flips (Pr[X n = k] = k) (n ) 2 n a n,k = a n 1,k + a n 2,k 1 + a n 3,k 1 a n 3,k X n = # size-2 gaps of random m [F n, F n+1 ) (a n,k = p 2,n,k, X n = K 2,n ) a n,k = a n 1,k + a n 2,k 1 X n = # summands of random m [F n, F n+1 ) (X n = K n )

57 Proof Sketch Lemma (Key Lemma) If a n,k is a nice two dimensional homogenous recurrence then the random variable X n given by Pr[X n = k] a n,k approaches Gaussian as n. Let µ n (m) = E[(X n µ n ) m ]. Lemma (Method of Moments) Suffices to prove lim n µ n (2m) µ n (2m + 1) = (2m 1)!! lim µ n (2) m n µ n (2) m+ 1 2 = 0.

58 Proof Sketch Let µ n (m) = E[(X n µ n ) m ]. We have µ n (m) = m l=0 ( m l ) t i,j 0 F n i t i,j F n (j + µ n i µ n ) l µ n i (m l).

59 Proof Sketch Let µ n (m) = E[(X n µ n ) m ]. We have µ n (m) = m l=0 ( m l ) t i,j 0 F n i t i,j F n (j + µ n i µ n ) l µ n i (m l). Lemma For each integer m 0, there exist degree m polynomials Q 2m, Q 2m+1 and γ 2m, γ 2m+1 (0, 1) such that µ n (2m) = Q 2m (n) + O(γ n 2m) µ n (2m + 1) = Q 2m+1 (n) + O(γ n 2m+1). Furthermore, if C 2m is the leading coefficient of Q 2m, then for all m, C 2m = (2m 1)!! C m 2.

60 Future Work Are there other meaningful two-dimensional recurrences to which we can apply our asymptotic Gaussianity result?

61 Future Work Are there other meaningful two-dimensional recurrences to which we can apply our asymptotic Gaussianity result? Can you lift the constraint that every coefficient c i must be positive?

62 Future Work Are there other meaningful two-dimensional recurrences to which we can apply our asymptotic Gaussianity result? Can you lift the constraint that every coefficient c i must be positive? What is the rate at which our K g,n converges to a normal distribution?