Linear Recurrence Relations with Non-constant Coefficients ......Linear Recurrence Relations of...

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

mengxiw@umich.eduls12@williams.edu

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

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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Examples with First Digit Bias

Fibonacci numbers

Most common iPhone passcodes

Twitter users by # followers

Distance of stars from Earth

4

Introduction General Theory Linear Appendix (Multi) Open Questions

Examples with First Digit Bias

Fibonacci numbers

Most common iPhone passcodes

Twitter users by # followers

Distance of stars from Earth

5

Introduction General Theory Linear Appendix (Multi) Open Questions

Examples with First Digit Bias

Fibonacci numbers

Most common iPhone passcodes

Twitter users by # followers

Distance of stars from Earth

6

Introduction General Theory Linear Appendix (Multi) Open Questions

Examples with First Digit Bias

Fibonacci numbers

Most common iPhone passcodes

Twitter users by # followers

Distance of stars from Earth

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Introduction General Theory Linear Appendix (Multi) Open Questions

Applications

Analyzing round-off errors.

Determining the optimal way to storenumbers.

Detecting tax and image fraud, and dataintegrity.

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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Outline

Benford’s Law.

Linear Recurrence relations (degree 2 and higherdegree).

Multiplicative Recurrence relations.

Open problems and references.

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Introduction General Theory Linear Appendix (Multi) Open Questions

Equidistributionand

Benford’s Law

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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Logarithms and Benford’s Law

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations

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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations of Degree 2

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations of Degree 2

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations of Degree 2

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations of Degree 2

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations of Degree 2

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations of Degree 2

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Linear Recurrence Relations of Degree 2

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

AppendixMultiplicative Recurrence Relations

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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Generalization to Multiplicative Recurrence Relations

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Generalization to Multiplicative Recurrence Relations

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Generalization to Multiplicative Recurrence Relations

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Generalization to Multiplicative Recurrence Relations

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

Open Questionsand

References

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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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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Introduction General Theory Linear Appendix (Multi) Open Questions

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