Linear Recurrence Relations with Non-constant Coefficients ......Linear Recurrence Relations of Degree 2 a n+1 = f(n)a n +g(n)a n 1 with non-constant coefﬁcients f(n) and g(n). Explore

Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations withNon-constant Coefficients and Benford’s

Law

Mengxi Wang, Lily ShaoUniversity of Michigan, Williams College

[email protected]@williams.edu

Young Mathematics Conference, Ohio State University,August 10, 2018

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Motivating Question: For a nice data set, such as theFibonacci numbers, stock prices, street addresses ofcollege employees and students, ..., what percent of theleading digits are 1?

Answer: Benford’s law!

2


Motivating Question: For a nice data set, such as theFibonacci numbers, stock prices, street addresses ofcollege employees and students, ..., what percent of theleading digits are 1?

Answer: Benford’s law!

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Examples with First Digit Bias

Fibonacci numbers

Most common iPhone passcodes

Twitter users by # followers

Distance of stars from Earth

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Fibonacci numbers




5



Fibonacci numbers




6



Fibonacci numbers




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Applications

Analyzing round-off errors.

Determining the optimal way to storenumbers.

Detecting tax and image fraud, and dataintegrity.

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An Interesting Question

From previous works: sequences generated by linearrecurrence relations with constant coefficients obeyBenford’s Law.

Example: The Fibonacci Sequence 1, 2, 3, 5, 8, 13,21, 34, . . . .

Question: Non-constant coefficients?

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Outline

Benford’s Law.

Linear Recurrence relations (degree 2 and higherdegree).

Multiplicative Recurrence relations.

Open problems and references.

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Equidistributionand

Benford’s Law

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Benford’s Law

Benford’s LawA set of numbers is Benford (base b) if the probability of

observing a first digit of d is logb

(1 + 1

d

).

Strong BenfordA set of numbers is Strong Benford (base b) if theprobability of observing a significand in [1, s) is logb(s).Then the probability of observing a significand in [s, s + 1)

is logb

(1 + 1

s

).

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Benford’s Law

Benford’s LawA set of numbers is Benford (base b) if the probability of

observing a first digit of d is logb

(1 + 1

d

).

Strong BenfordA set of numbers is Strong Benford (base b) if theprobability of observing a significand in [1, s) is logb(s).Then the probability of observing a significand in [s, s + 1)

is logb

(1 + 1

s

).

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Equidistribution and Benford’s Law

Equidistribution{yn}∞n=1 is equidistributed modulo 1 if probabilityyn mod 1 ∈ [a,b] tends to b − a:

#{n ≤ N : yn mod 1 ∈ [a,b]}N

→ b − a.

Theoremβ 6∈ Q, nβ is equidistributed mod 1.

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Logarithms and Benford’s Law

Fundamental EquivalenceData set {xi} is Benford base B if {yi} is equidistributedmod 1, where yi = logB xi .

x = S10(x) · 10k then

log10 x = log10 S10(x) + k = log10 S10x mod 1.

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x = S10(x) · 10k then


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x = S10(x) · 10k then


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Prob(leading digit d)= log10(d + 1)− log10(d)= log10

(d+1

d

)= log10

(1 + 1

d

).

Have Benford’s law↔mantissa (fractional part) oflogarithms of data areuniformly distributed

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Examples

Remember: β 6∈ Q, nβ is equidistributed mod 1.

2n is Benford base 10 as log10 2 6∈ Q.

Fibonacci numbers are Benford base 10.Binet: an = 1√

5

(1+√

52

)n− 1√

5

(1−√

52

)n.

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Examples




5

(1+√

52

)n− 1√

5

(1−√

52

)n.

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Examples




5

(1+√

52

)n− 1√

5

(1−√

52

)n.

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Useful Theorem

It suffices to analyze the main term of the sequence:

TheoremIf a sequence {an} is Benford and lim

n→∞bn = an then {bn}

is Benford as well.

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Linear Recurrence Relations

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Linear Recurrence Relations of Degree 2

an+1 = f (n)an + g(n)an−1 with non-constantcoefficients f (n) and g(n).

Explore conditions on f and g such that the sequencegenerated obeys Benford’s Law for all initial values.

First solve the closed form of the sequence (an), thenanalyze its main term.

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To solve for the closed form of the sequence:

Main idea: reduce the degree of recurrence.

Define an auxiliary sequence {bn}∞n=1 bybn = an+1 − λ(n)an for n ≥ 1.((an) recurrent of degree 2, so (bn) of degree 1).

Set an+1 − λ(n)an = µ(n)(an − λ(n − 1)an−1) forn ≥ 2.

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To solve for the closed form of the sequence:

Main idea: reduce the degree of recurrence.

Define an auxiliary sequence {bn}∞n=1 bybn = an+1 − λ(n)an for n ≥ 1.((an) recurrent of degree 2, so (bn) of degree 1).

Set an+1 − λ(n)an = µ(n)(an − λ(n − 1)an−1) forn ≥ 2.

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an+1 = (λ(n) + µ(n))an − µ(n)λ(n − 1)an−1, andcompare the coefficients:

f (n) = λ(n) + µ(n)g(n) = −λ(n − 1)µ(n).

We show that for any given pair of f and g, such λand µ always exist.

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an+1 = (λ(n) + µ(n))an − µ(n)λ(n − 1)an−1, andcompare the coefficients:

f (n) = λ(n) + µ(n)g(n) = −λ(n − 1)µ(n).

We show that for any given pair of f and g, such λand µ always exist.

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Recurrence relations of degree 1:

an+1 = λ(n)an + bn

bn = µ(n)bn−1.

an+1 = r(n)(

1 +n∑

k=3

n∏i=k

λ(i)µ(i) +

a2

b1

n∏i=2

λ(i)µ(i)

), where

r(n) := b1

n∏i=2

µ(i).

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Recurrence relations of degree 1:

an+1 = λ(n)an + bn

bn = µ(n)bn−1.

an+1 = r(n)(

1 +n∑

k=3

n∏i=k

λ(i)µ(i) +

a2

b1

n∏i=2

λ(i)µ(i)

), where

r(n) := b1

n∏i=2

µ(i).

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Asymptotic analysis: show the main term dominatesfor suitable choices of µ and λ.

Let λ(n)µ(n) be a non-increasing function and

limn→∞

λ(n)µ(n) = 0, then lim

n→∞(an+1 − r(n)) = 0.

limn→∞

λ(n)µ(n) = 0 implies lim

n→∞g(n)f (n)2 = 0.

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limn→∞


n→∞(an+1 − r(n)) = 0.

limn→∞


n→∞g(n)f (n)2 = 0.

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limn→∞


n→∞(an+1 − r(n)) = 0.

limn→∞


n→∞g(n)f (n)2 = 0.

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Benford-ness of the Main Term

Main term of an+1 is r(n) = b1

n∏i=2

µ(i).

Since (strong) Benford-ness is preserved under

translation and dilation we let r(n) =n∏

i=1µ(i) for

simplicity.

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Benford-ness of the Main Term

Main term of an+1 is r(n) = b1

n∏i=2

µ(i).

Since (strong) Benford-ness is preserved under

translation and dilation we let r(n) =n∏

i=1µ(i) for

simplicity.

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Examples when f and g are functions

If µ(k) = k , then r(n) = n!.

If µ(k) = kα where α ∈ R, then r(n) = (n!)α.

If µ(k) = exp(αh(k)) where α is irrational and h(k) is a

monic polynomial, then log r(n) = αn∑

k=1h(k).

LemmaThe sequence {αp(n)} is equidistributed mod 1 if α /∈ Qand p(n) a monic polynomial.

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k=1h(k).


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k=1h(k).


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Examples when f and g are random variables

Take µ(n) ∼ h(n)Un where the Un’s are independentuniform distributions on [0,1], and h(n) is a

deterministic function in n such thatn∏

i=1h(i) is Benford.

Then r(n) =n∏

i=1h(i)

n∏i=1

Ui is Benford.

Take µ(n) ∼ exp(Un) where the Un’s are i.i.d. randomvariables. Then take logarithm and sum up log(µ(n)).Apply Central Limit Theorem and get a Gaussiandistribution with increasing variance.

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Examples when f and g are random variables

Take µ(n) ∼ h(n)Un where the Un’s are independentuniform distributions on [0,1], and h(n) is a

deterministic function in n such thatn∏

i=1h(i) is Benford.

Then r(n) =n∏

i=1h(i)

n∏i=1

Ui is Benford.

Take µ(n) ∼ exp(Un) where the Un’s are i.i.d. randomvariables. Then take logarithm and sum up log(µ(n)).Apply Central Limit Theorem and get a Gaussiandistribution with increasing variance.

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Linear Recurrences of Higher Degree

Use recurrence relation of degree 3 as an example.Similar main idea: reduce the degree.

Define the sequence {an}∞n=1 byan+1 = f1(n)an + f2(n)an−1 + f3(n)an−2.

Define an auxiliary sequence (bn)∞n=1 by

bn = an+1 − λ(n)an. Then (bn) is degree 2.

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AppendixMultiplicative Recurrence Relations

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Generalization to Multiplicative Recurrence Relations

Define sequence (An)∞n=1 by the recurrence relation

An+1 = Af (n)n Ag(n)

n−1 with initial values A1, A2.

Then the closed form of An is An = Axn2 Ayn

1 , where theexponents (xn) and (yn) satisfy the linear recurrencerelations

xn+1 = f (n)xn + g(n)xn−1

yn+1 = f (n)yn + g(n)yn−1

with initial values

x1 = 0, x2 = 1,y1 = 1, y2 = 0.

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Define sequence (An)∞n=1 by the recurrence relation

An+1 = Af (n)n Ag(n)

n−1 with initial values A1, A2.

Then the closed form of An is An = Axn2 Ayn

1 , where theexponents (xn) and (yn) satisfy the linear recurrencerelations

xn+1 = f (n)xn + g(n)xn−1

yn+1 = f (n)yn + g(n)yn−1

with initial values

x1 = 0, x2 = 1,y1 = 1, y2 = 0.

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Again, solve for (xn) and (yn) with auxiliary functionsλ(n) and µ(n).

Let λ(n) and µ(n) satisfy limn→∞

λ(n)µ(n)

= 0.

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Again, solve for (xn) and (yn) with auxiliary functionsλ(n) and µ(n).

Let λ(n) and µ(n) satisfy limn→∞

λ(n)µ(n)

= 0.

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Take the main terms:xn+1 → (x2 − λ(1)x1)

n∏i=2

µ(i) =n∏

i=2µ(i),

yn+1 → (y2 − λ(1)y1)n∏

i=2µ(i) = −λ(1)

n∏i=2

µ(i).

By the closed form An = Axn2 Ayn

1 ,

log(An+1) = xn log(A2) + yn log(A1)

→ (n∏

i=2

µ(i))(log(A2)− λ(1) log(A1))

as n→∞.

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Take the main terms:xn+1 → (x2 − λ(1)x1)

n∏i=2

µ(i) =n∏

i=2µ(i),

yn+1 → (y2 − λ(1)y1)n∏

i=2µ(i) = −λ(1)

n∏i=2

µ(i).

By the closed form An = Axn2 Ayn

1 ,

log(An+1) = xn log(A2) + yn log(A1)

→ (n∏

i=2

µ(i))(log(A2)− λ(1) log(A1))

as n→∞.

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Benford-ness of the main term

(An) is a Benford sequence if the main term of log(An)is equidistributed mod 1.

We choose µ and the initial values such that

(log(A2)− λ(1) log(A1))n∏

i=2µ(i) is equidistributed mod

1.

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Benford-ness of the main term

(An) is a Benford sequence if the main term of log(An)is equidistributed mod 1.

We choose µ and the initial values such that

(log(A2)− λ(1) log(A1))n∏

i=2µ(i) is equidistributed mod

1.

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Examples

Remeber: The sequence {αp(n)} is equidistributed mod1 if α /∈ Q and p(n) a monic polynomial.

Our construction:log(A2)−λ(1) log(A1)

µ(1) =: α /∈ Q,

µ(i) = p(i)p(i−1) where p(n) is a non-vanishing monic

polynomial.

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Examples

Remeber: The sequence {αp(n)} is equidistributed mod1 if α /∈ Q and p(n) a monic polynomial.

Our construction:log(A2)−λ(1) log(A1)

µ(1) =: α /∈ Q,

µ(i) = p(i)p(i−1) where p(n) is a non-vanishing monic

polynomial.

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Open Questionsand

References

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Open Questions

1 We mainly consider the case when λ(n)µ(n) → 0 as

n→∞. What about when λ(n)µ(n) →∞? In this case,

there is no simple dominating term.

2 Applications of recurrence relations when f (n) andg(n) are random variables?

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Open Questions

1 We mainly consider the case when λ(n)µ(n) → 0 as

n→∞. What about when λ(n)µ(n) →∞? In this case,

there is no simple dominating term.

2 Applications of recurrence relations when f (n) andg(n) are random variables?

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Acknowledgement

The presenters are partially supported by NSF grants, theUniversity of Michigan, and Williams College. We aregrateful to our excellent advisor Professor Steven J. Millerfor valuable guidance, and thanks the other contributorsMadeleine Farris and Noah Luntzlara for help.

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References

P. Diaconis, The Distribution of Leading Digits and UniformDistribution Mod 1, The Annals of Probability 5 (1977), No. 1,72–81.

D. Jang, J. U. Kang, A. Kruckman, J. Kudo, and S. J. Miller,Chains of Distributions, Hierarchical Bayesian Models andBenford’s Law, preprint, 2008.https://arxiv.org/pdf/0805.4226.pdf.

A. Kontorovich and S. J. Miller, Benford’s Law, Values ofL-functions and the 3x + 1 Problem, Acta Arithmetica, 120(2005), 269–297.

P. K. Romano and H. Mclaughlin, On Non-Linear RecursiveSequences and Benford’s Law, Fibonacci Quarterly, 49 (2011),134–138.

T. Tao, 254B, Notes 1: Equidistribution of polynomial sequencesin tori 254A, 2010,https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/.63

https://arxiv.org/pdf/0805.4226.pdf

https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/

https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/

Documents

Linear Recurrence Relations with Non-constant Coefficients ......Linear Recurrence Relations of Degree 2 a n+1 = f(n)a n +g(n)a n 1 with non-constant coefﬁcients f(n) and g(n). Explore