Nine Chapters of Analytic Number Theory [2mm] in …In this work: only multiplicative number theory (primes, divisors, etc.) Much of the formalised material is not particularly analytic

Nine Chapters of Analytic Number Theoryin Isabelle/HOL

Manuel EberlTechnische Universität München12 September 2019

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Manuel Eberl

Rodrigo Raya

library

unformalised173 18

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What is Analytic Number Theory?Studying the multiplicative and additive structure of the integers

using analytic methods

In this work: only multiplicative number theory (primes, divisors, etc.)Much of the formalised material is not particularly analytic.

Some of these results have already been formalised by other people (Avigad,Harrison, Carneiro, . . . ) – but not in the context of a large library.

What is Analytic Number Theory?Studying the multiplicative and additive structure of the integersusing analytic methods




In this work: only multiplicative number theory (primes, divisors, etc.)

Much of the formalised material is not particularly analytic.








Material• Various number-theoretic functions (executable and with many properties

proven):

− Euler’s totient ϕ and Carmichael’s λ− Divisor function σx− Möbius’s µ− Liouville’s λ− Prime-counting functions: π(n), ϑ(n), ψ(n)• Dirichlet series (both formal and analytic)• Multiplicative characters• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions


proven):− Euler’s totient ϕ and Carmichael’s λ

− Divisor function σx− Möbius’s µ− Liouville’s λ− Prime-counting functions: π(n), ϑ(n), ψ(n)• Dirichlet series (both formal and analytic)• Multiplicative characters• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions


proven):− Euler’s totient ϕ and Carmichael’s λ− Divisor function σx

− Möbius’s µ− Liouville’s λ− Prime-counting functions: π(n), ϑ(n), ψ(n)• Dirichlet series (both formal and analytic)• Multiplicative characters• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions


proven):− Euler’s totient ϕ and Carmichael’s λ− Divisor function σx− Möbius’s µ

− Liouville’s λ− Prime-counting functions: π(n), ϑ(n), ψ(n)• Dirichlet series (both formal and analytic)• Multiplicative characters• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions


proven):− Euler’s totient ϕ and Carmichael’s λ− Divisor function σx− Möbius’s µ− Liouville’s λ

− Prime-counting functions: π(n), ϑ(n), ψ(n)• Dirichlet series (both formal and analytic)• Multiplicative characters• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions


proven):− Euler’s totient ϕ and Carmichael’s λ− Divisor function σx− Möbius’s µ− Liouville’s λ− Prime-counting functions: π(n), ϑ(n), ψ(n)

• Dirichlet series (both formal and analytic)• Multiplicative characters• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions


proven):− Euler’s totient ϕ and Carmichael’s λ− Divisor function σx− Möbius’s µ− Liouville’s λ− Prime-counting functions: π(n), ϑ(n), ψ(n)• Dirichlet series (both formal and analytic)

• Multiplicative characters• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions


proven):− Euler’s totient ϕ and Carmichael’s λ− Divisor function σx− Möbius’s µ− Liouville’s λ− Prime-counting functions: π(n), ϑ(n), ψ(n)• Dirichlet series (both formal and analytic)• Multiplicative characters

• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions


proven):− Euler’s totient ϕ and Carmichael’s λ− Divisor function σx− Möbius’s µ− Liouville’s λ− Prime-counting functions: π(n), ϑ(n), ψ(n)• Dirichlet series (both formal and analytic)• Multiplicative characters• Riemann’s ζ, Hurwitz’s ζ, Dirichlet’s L functions

Interesting resultsPrime Number Theorem

|{p | p prime ∧ p ≤ x}| ∼ x log x

Dirichlet’s Theorem

gcd(k ,m) = 1 =⇒ {p | p prime ∧ p ≡ k (mod m)} is infinite

Asymptotics• lcm(1, . . . ,n) = eψ(n) where ψ(x) ∼ x• On average, an integer n has log n + 2γ− 1 divisors.• The set of square-free integers has density 6/π2 ≈ 60.8%.• The set of fractions p/q for p,q prime is dense in R.• Prime harmonic series: ∑p≤x

1p = log log x + M + O(1/log x)











Asymptotics• lcm(1, . . . ,n) = eψ(n) where ψ(x) ∼ x

• On average, an integer n has log n + 2γ− 1 divisors.• The set of square-free integers has density 6/π2 ≈ 60.8%.• The set of fractions p/q for p,q prime is dense in R.• Prime harmonic series: ∑p≤x






Asymptotics• lcm(1, . . . ,n) = eψ(n) where ψ(x) ∼ x• On average, an integer n has log n + 2γ− 1 divisors.

• The set of square-free integers has density 6/π2 ≈ 60.8%.• The set of fractions p/q for p,q prime is dense in R.• Prime harmonic series: ∑p≤x






Asymptotics• lcm(1, . . . ,n) = eψ(n) where ψ(x) ∼ x• On average, an integer n has log n + 2γ− 1 divisors.• The set of square-free integers has density 6/π2 ≈ 60.8%.

• The set of fractions p/q for p,q prime is dense in R.• Prime harmonic series: ∑p≤x






Asymptotics• lcm(1, . . . ,n) = eψ(n) where ψ(x) ∼ x• On average, an integer n has log n + 2γ− 1 divisors.• The set of square-free integers has density 6/π2 ≈ 60.8%.• The set of fractions p/q for p,q prime is dense in R.

• Prime harmonic series: ∑p≤x1p = log log x + M + O(1/log x)







Dirichlet seriesA Dirichlet series is a formal series of the form:

F (s) =∞

∑n=1

an

ns

They are the ‘number-theoretic analogue’ of formal power series:

• They form a commutative ring (plus many other operations: division,derivative, exponential, logarithm, etc.)

• They have a deep connection to number-theoretic functions.• When they converge, the corresponding complex functions often contain

useful information.• Transfer between the analytic world and the algebraic world is often possible

(in both directions).• Even when they do not converge, they can be very useful.


F (s) =∞

∑n=1

an

ns







F (s) =∞

∑n=1

an

ns



• They have a deep connection to number-theoretic functions.

• When they converge, the corresponding complex functions often containuseful information.

• Transfer between the analytic world and the algebraic world is often possible(in both directions).

• Even when they do not converge, they can be very useful.


F (s) =∞

∑n=1

an

ns




useful information.

• Transfer between the analytic world and the algebraic world is often possible(in both directions).



F (s) =∞

∑n=1

an

ns





(in both directions).



F (s) =∞

∑n=1

an

ns






The ζ functionsRiemann ζ function:

ζ(s) =∞

∑n=1

1ns

Dirichlet L functions:

L(s,χ) =∞

∑n=0

χ(n)ns

Hurwitz ζ function:

ζ(s,a) =∞

∑n=0

1(n + a)s

Definition only valid for Re(s) > 1; elsewhere by analytic continuation.

Luckily, analytic continuation for ζ(s,a) yields continuations for ζ(s) and L(s,χ)for free.


ζ(s) =∞

∑n=1

1ns


L(s,χ) =∞

∑n=0

χ(n)ns


ζ(s,a) =∞

∑n=0

1(n + a)s




ζ(s) =∞

∑n=1

1ns


L(s,χ) =∞

∑n=0

χ(n)ns


ζ(s,a) =∞

∑n=0

1(n + a)s




ζ(s) =∞

∑n=1

1ns


L(s,χ) =∞

∑n=0

χ(n)ns


ζ(s,a) =∞

∑n=0

1(n + a)s




ζ(s) =∞

∑n=1

1ns


L(s,χ) =∞

∑n=0

χ(n)ns


ζ(s,a) =∞

∑n=0

1(n + a)s



The ζ functionsMany basic properties were proven:

• ζ(2) = π2/6, ζ(0) = −12, ζ(−1) = − 1

12

• Reflection formula:

ζ(1− s) = 2(2π)−s cos(πs/2)Γ(s)ζ(s)

• Euler product:

ζ(s) = ∏p

11− p−s for Re(s) > 1

• Connection between Γ and ζ:

Γ(s)ζ(a,s) =∫ ∞

0

xs−1e−ax

1− e−x dx


• ζ(2) = π2/6, ζ(0) = −12, ζ(−1) = − 1

12


ζ(1− s) = 2(2π)−s cos(πs/2)Γ(s)ζ(s)

• Euler product:

ζ(s) = ∏p

11− p−s for Re(s) > 1


Γ(s)ζ(a,s) =∫ ∞

0

xs−1e−ax

1− e−x dx


• ζ(2) = π2/6, ζ(0) = −12, ζ(−1) = − 1

12


ζ(1− s) = 2(2π)−s cos(πs/2)Γ(s)ζ(s)

• Euler product:

ζ(s) = ∏p

11− p−s for Re(s) > 1


Γ(s)ζ(a,s) =∫ ∞

0

xs−1e−ax

1− e−x dx


• ζ(2) = π2/6, ζ(0) = −12, ζ(−1) = − 1

12


ζ(1− s) = 2(2π)−s cos(πs/2)Γ(s)ζ(s)

• Euler product:

ζ(s) = ∏p

11− p−s for Re(s) > 1


Γ(s)ζ(a,s) =∫ ∞

0

xs−1e−ax

1− e−x dx


• ζ(2) = π2/6, ζ(0) = −12, ζ(−1) = − 1

12


ζ(1− s) = 2(2π)−s cos(πs/2)Γ(s)ζ(s)

• Euler product:

ζ(s) = ∏p

11− p−s for Re(s) > 1


Γ(s)ζ(a,s) =∫ ∞

0

xs−1e−ax

1− e−x dx

The ζ functionsRiemann hypothesis can be stated:

theorem RH:assumes Re s ∈ {1/2 < ..< 1} and zeta s = 0shows Re s = 1/2

sorry

Generalised Riemann hypothesis:

theorem GRH:assumes dcharacter m χassumes s /∈R≤0 and Dirichlet_L m χ s = 0shows Re s = 1/2sorry


theorem RH:assumes Re s ∈ {1/2 < ..< 1} and zeta s = 0shows Re s = 1/2sorry




theorem RH:assumes Re s ∈ {1/2 < ..< 1} and zeta s = 0shows Re s = 1/2sorry



Non-vanishing of ζTheorem: ζ(s) 6= 0 for Re(s) ≥ 1.

Proof:For Re(s) > 1: almost for free from Dirichlet series libraryFor Re(s) = 1:

• Suppose ζ(1 + it) = 0 for t ∈R>0.• By symmetry, ζ(1− it) = 0 as well.• Let h(s) := ζ(s)2ζ(s + it)ζ(s− it) and note that:1. h is analytic everywhere

(since the pole of each factor is cancelled by the zero of another factor).2. The Dirichlet series of h(s) is non-negative

(since the coefficients of log h(s) are non-negative by inspection)• By the Pringsheim–Landau Theorem, it then converges everywhere.• This is clearly wrong, since even the subseries for 2−s diverges for s→ 0+.

Easily generalises to Dirichlet L-functions.

Non-vanishing of ζTheorem: ζ(s) 6= 0 for Re(s) ≥ 1.Proof:For Re(s) > 1: almost for free from Dirichlet series library

For Re(s) = 1:• Suppose ζ(1 + it) = 0 for t ∈R>0.• By symmetry, ζ(1− it) = 0 as well.• Let h(s) := ζ(s)2ζ(s + it)ζ(s− it) and note that:1. h is analytic everywhere




Non-vanishing of ζTheorem: ζ(s) 6= 0 for Re(s) ≥ 1.Proof:For Re(s) > 1: almost for free from Dirichlet series libraryFor Re(s) = 1:






• Suppose ζ(1 + it) = 0 for t ∈R>0.

• By symmetry, ζ(1− it) = 0 as well.• Let h(s) := ζ(s)2ζ(s + it)ζ(s− it) and note that:1. h is analytic everywhere





• Suppose ζ(1 + it) = 0 for t ∈R>0.• By symmetry, ζ(1− it) = 0 as well.

• Let h(s) := ζ(s)2ζ(s + it)ζ(s− it) and note that:1. h is analytic everywhere





• Suppose ζ(1 + it) = 0 for t ∈R>0.• By symmetry, ζ(1− it) = 0 as well.• Let h(s) := ζ(s)2ζ(s + it)ζ(s− it) and note that:

1. h is analytic everywhere(since the pole of each factor is cancelled by the zero of another factor).

2. The Dirichlet series of h(s) is non-negative(since the coefficients of log h(s) are non-negative by inspection)

• By the Pringsheim–Landau Theorem, it then converges everywhere.• This is clearly wrong, since even the subseries for 2−s diverges for s→ 0+.









(since the pole of each factor is cancelled by the zero of another factor).

2. The Dirichlet series of h(s) is non-negative(since the coefficients of log h(s) are non-negative by inspection)











(since the coefficients of log h(s) are non-negative by inspection)






(since the coefficients of log h(s) are non-negative by inspection)• By the Pringsheim–Landau Theorem, it then converges everywhere.

• This is clearly wrong, since even the subseries for 2−s diverges for s→ 0+.












Problems

Problem: The Residue Theorem

ε

2iπ

4iπ

2Niπ

2(N+1)iπ

−2iπ

−4iπ

−2Niπ

−2(N+1)iπ

Showing Hurwitz’s formula was tricky.One problem in particular:

Residue Theorem:∮C

f (z)dz = ∑z0 is pole

indC(z0) ·Res(f ,z0)

Problem: The horizontal line segmentsof C lie on the branch cut!

And we take a different branch of f oneach side.


ε

2iπ

4iπ

2Niπ

2(N+1)iπ

−2iπ

−4iπ

−2Niπ

−2(N+1)iπ








ε

2iπ

4iπ

2Niπ

2(N+1)iπ

−2iπ

−4iπ

−2Niπ

−2(N+1)iπ








ε

2iπ

4iπ

2Niπ

2(N+1)iπ

−2iπ

−4iπ

−2Niπ

−2(N+1)iπ








Leonid 2 [CC BY-SA 3.0]

Mathematical justification:f is a multi-valued function.

We are not really integratingover a function C→ C,but over a Riemann surface.

But we don’t have Riemannsurfaces in Isabelle (yet).

What can we do instead?













Problem: Residue Theorem

ε

2iπ

4iπ

2Niπ

2(N+1)iπ


• Only look at the upper half• Lower half follows by symmetry• Move the branch cut out of the way

Now the Residue Theorem can easilybe applied.

Other problems:• Geometry of the integration contour• Establishing homotopy of paths• Evaluating winding numbers

The last one was made much easierby Wenda Li’s winding-number tactic.


ε

2iπ

4iπ

2Niπ

2(N+1)iπ

What can we do instead?• Only look at the upper half

• Lower half follows by symmetry• Move the branch cut out of the way





ε

2iπ

4iπ

2Niπ

2(N+1)iπ

What can we do instead?• Only look at the upper half• Lower half follows by symmetry

• Move the branch cut out of the wayNow the Residue Theorem can easilybe applied.




ε

2iπ

4iπ

2Niπ

2(N+1)iπ

What can we do instead?• Only look at the upper half• Lower half follows by symmetry• Move the branch cut out of the way





ε

2iπ

4iπ

2Niπ

2(N+1)iπ






ε

2iπ

4iπ

2Niπ

2(N+1)iπ



Other problems:

• Geometry of the integration contour• Establishing homotopy of paths• Evaluating winding numbers



ε

2iπ

4iπ

2Niπ

2(N+1)iπ



Other problems:• Geometry of the integration contour

• Establishing homotopy of paths• Evaluating winding numbers



ε

2iπ

4iπ

2Niπ

2(N+1)iπ



Other problems:• Geometry of the integration contour• Establishing homotopy of paths

• Evaluating winding numbersThe last one was made much easierby Wenda Li’s winding-number tactic.


ε

2iπ

4iπ

2Niπ

2(N+1)iπ






ε

2iπ

4iπ

2Niπ

2(N+1)iπ





Problem: Ugly Limits

2C/R + C/N + 3MπR log N

+3RMδπNδ

< ε for suff. large N

n1−s

1− s− (n + 1)1−s

1− s∈ O(1) for n→∞

∫ ∞

ε

xse−ax

1− e−x dx exists becausexse−ax

1− e−x ∈ O(e−ax/2)

Just kidding: Our ‘real_asymp’ tactic eats problems like that for breakfast.,(see ISSAC’19)



+3RMδπNδ


n1−s

1− s− (n + 1)1−s

1− s∈ O(1) for n→∞

∫ ∞

ε

xse−ax


1− e−x ∈ O(e−ax/2)




+3RMδπNδ


n1−s

1− s− (n + 1)1−s

1− s∈ O(1) for n→∞

∫ ∞

ε

xse−ax

1− e−x dx exists

becausexse−ax

1− e−x ∈ O(e−ax/2)




+3RMδπNδ


n1−s

1− s− (n + 1)1−s

1− s∈ O(1) for n→∞

∫ ∞

ε

xse−ax


1− e−x ∈ O(e−ax/2)




+3RMδπNδ


n1−s

1− s− (n + 1)1−s

1− s∈ O(1) for n→∞

∫ ∞

ε

xse−ax


1− e−x ∈ O(e−ax/2)


Problem: Pole CancellationPart of Newman’s proof of PNT:

ζ ′(s)/ζ(s) has a simple pole at s = 1 with residue −1

, i.e.

ζ ′(s)/ζ(s) ∼ −1s− 1

+ O(1) (for s→ 1)

So there exists some constant C for such thatζ ′(s)

(s− 1)ζ(s)− ζ ′(s)− cζ(s)

is analytic at s = 1.

Such arguments are still a bit tedious in Isabelle.


ζ ′(s)/ζ(s) has a simple pole at s = 1 with residue −1, i.e.

ζ ′(s)/ζ(s) ∼ −1s− 1

+ O(1) (for s→ 1)


(s− 1)ζ(s)− ζ ′(s)− cζ(s)





ζ ′(s)/ζ(s) ∼ −1s− 1

+ O(1) (for s→ 1)


(s− 1)ζ(s)− ζ ′(s)− cζ(s)





ζ ′(s)/ζ(s) ∼ −1s− 1

+ O(1) (for s→ 1)


(s− 1)ζ(s)− ζ ′(s)− cζ(s)



Conclusion• Formalised most of the content of an entire undergraduate mathematics

textbook on an advanced subject

• Many new results that had not been formalised elsewhere• Some nicer, more ‘high-level’ proofs for things already formalised• Formalisation was smooth and pleasant; no major problems• About 25,000 lines of Isabelle, spread over 5 AFP entries• Not only can it be done, but it can be done by a single person as a side

project,• Future Work: Maybe the second book?


textbook on an advanced subject• Many new results that had not been formalised elsewhere

• Some nicer, more ‘high-level’ proofs for things already formalised• Formalisation was smooth and pleasant; no major problems• About 25,000 lines of Isabelle, spread over 5 AFP entries• Not only can it be done, but it can be done by a single person as a side



textbook on an advanced subject• Many new results that had not been formalised elsewhere• Some nicer, more ‘high-level’ proofs for things already formalised

• Formalisation was smooth and pleasant; no major problems• About 25,000 lines of Isabelle, spread over 5 AFP entries• Not only can it be done, but it can be done by a single person as a side



textbook on an advanced subject• Many new results that had not been formalised elsewhere• Some nicer, more ‘high-level’ proofs for things already formalised• Formalisation was smooth and pleasant; no major problems

• About 25,000 lines of Isabelle, spread over 5 AFP entries• Not only can it be done, but it can be done by a single person as a side



textbook on an advanced subject• Many new results that had not been formalised elsewhere• Some nicer, more ‘high-level’ proofs for things already formalised• Formalisation was smooth and pleasant; no major problems• About 25,000 lines of Isabelle, spread over 5 AFP entries

• Not only can it be done, but it can be done by a single person as a sideproject,

• Future Work: Maybe the second book?


textbook on an advanced subject• Many new results that had not been formalised elsewhere• Some nicer, more ‘high-level’ proofs for things already formalised• Formalisation was smooth and pleasant; no major problems• About 25,000 lines of Isabelle, spread over 5 AFP entries• Not only can it be done, but it can be done by a single person as a side

project,

• Future Work: Maybe the second book?


textbook on an advanced subject• Many new results that had not been formalised elsewhere• Some nicer, more ‘high-level’ proofs for things already formalised• Formalisation was smooth and pleasant; no major problems• About 25,000 lines of Isabelle, spread over 5 AFP entries• Not only can it be done, but it can be done by a single person as a side


Documents

Nine Chapters of Analytic Number Theory [2mm] in …In this work: only multiplicative number theory (primes, divisors, etc.) Much of the formalised material is not particularly analytic