#rieman hyp_prime numbers & physics

Introduction Prime number theorem Proof of prime number theorem Riemann’s idea Prime symphony

Counting prime numbers, or listening to them

Byungchul Cha

Muhlenberg College

Mount St. Mary’s UniversityACM/MAA Lecture Series

February 9, 2011

1 / 29


Warm-up

Let’s try to factor some integers:

28

= 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)

2011 is a prime number.Must check all primes 2, 3, 5, . . . , 41, 43 ≤

√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.

21727 is a prime number. (What a prime location!)Must check 34 primes: 2, 3, 5, . . . , 137, 139 ≤

√21727.

2 / 29


Warm-up


28 = 4 · 7

= 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300

= 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163

= 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163

. (163 is a prime number.)2011 is a prime number.Must check all primes 2, 3, 5, . . . , 41, 43 ≤

√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)

2011

is a prime number.Must check all primes 2, 3, 5, . . . , 41, 43 ≤

√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021

= 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4

= 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22

= (45 − 2)(45 + 2) = 43 · 47.21727 is a prime number. (What a prime location!)Must check 34 primes: 2, 3, 5, . . . , 137, 139 ≤

√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.

21727

is a prime number. (What a prime location!)Must check 34 primes: 2, 3, 5, . . . , 137, 139 ≤

√21727.

2 / 29


Warm-up


28 = 4 · 7 = 22· 7.

16300 = 100 · 163 = 22· 52· 163. (163 is a prime number.)


√2011 = 44.8 . . . .

2021 = 2025 − 4 = 452− 22 = (45 − 2)(45 + 2) = 43 · 47.


√21727.

2 / 29


Bit more challenging

188198812920607963838697239461650439807163563379

417382700763356422988859715234665485319060606504

743045317388011303396716199692321205734031879550

656996221305168759307650257059 =

398075086424064937397125500550386491199064362342

526708406385189575946388957261768583317

*

472772146107435302536223071973048224632914695302

097116459852171130520711256363590397527

This number (188...), whose nickname is RSA-576, wasfactored in 2003 by a team led by J. Franke and T. Kleinjung.The point of these exercises is

Primes numbers seem to behave randomly.

3 / 29



188198812920607963838697239461650439807163563379

417382700763356422988859715234665485319060606504

743045317388011303396716199692321205734031879550

656996221305168759307650257059 =

398075086424064937397125500550386491199064362342

526708406385189575946388957261768583317

*

472772146107435302536223071973048224632914695302

097116459852171130520711256363590397527

This number (188...), whose nickname is RSA-576, wasfactored in 2003 by a team led by J. Franke and T. Kleinjung.

The point of these exercises is


3 / 29



188198812920607963838697239461650439807163563379

417382700763356422988859715234665485319060606504

743045317388011303396716199692321205734031879550

656996221305168759307650257059 =

398075086424064937397125500550386491199064362342

526708406385189575946388957261768583317

*

472772146107435302536223071973048224632914695302

097116459852171130520711256363590397527

This number (188...), whose nickname is RSA-576, wasfactored in 2003 by a team led by J. Franke and T. Kleinjung.The point of these exercises is


3 / 29


Moral of this talk

Each prime number behaves randomly.

Yet, they behave regularly in aggregate.

4 / 29


Moral of this talk

Each prime number behaves randomly.Yet, they behave regularly in aggregate.

4 / 29


Randomness vs Regularity, according to Don Zagier

D. Zagier (1951–): There are two factsabout the distribution of prime numberswhich I hope to convince you sooverwhelmingly that they will bepermanently engraved in your hearts.

(continued on the next slide..)

5 / 29


Zagier’s quote: continued

The first is that . . . the prime numbers belong to the most arbitraryand ornery objects studied by mathematicians: they grow like weedsamong the natural numbers, seeming to obey no other law than that ofchance, and nobody can predict where the next one will sprout.

The second fact is even more astonishing, for it states just theopposite: that the prime numbers exhibit stunning regularity, thatthere are laws governing their behavior, and that they obey these lawswith almost military precision.

In this talk, we focus on regularity, which becomes especiallyprominent when we try to count them.

6 / 29


Zagier’s quote: continued

The first is that . . . the prime numbers belong to the most arbitraryand ornery objects studied by mathematicians: they grow like weedsamong the natural numbers, seeming to obey no other law than that ofchance, and nobody can predict where the next one will sprout.The second fact is even more astonishing, for it states just theopposite: that the prime numbers exhibit stunning regularity, thatthere are laws governing their behavior, and that they obey these lawswith almost military precision.

In this talk, we focus on regularity, which becomes especiallyprominent when we try to count them.

6 / 29


There are enough supply of primes

Theorem (Euclid)

There are infinitely many primes.

Proof.Start with any finite list of primes {p1, p2, . . . , pn}. Then, thenumber

p1 · p2 · · · pn + 1

is not divisible by any prime in the list, therefore, contains anew prime number as its factor. So, it is always possible to addone more to the list! By iterating this process, you can create alist of primes as long as you wish. �

7 / 29


Prime counting function

Define π(x) by

π(x) = the number of primes p ≤ x.

We just proved that

π(x)→∞, as x→∞.

Question: How fast does π(x) go to infinity?

8 / 29


The function π(x)


9 / 29


The function π(x)


Computing π(x) exactly for large x is a very hard computationalchallenge. Current record of π(x) for the largest x is

π(1024) = 18435599767349200867866,

[Found by Buethe, Franke, Jost, Kleinjung in July 2010]

Asymptotic expression for π(x)

It would be nice to have a function, say f (x), with a closed-formformula such that

π(x) ∼ f (x)

(meaning: limx→1

π(x)/f (x) = 1.)

9 / 29


The function π(x)


Computing π(x) exactly for large x is a very hard computationalchallenge. Current record of π(x) for the largest x is

π(1024) = 18435599767349200867866,

[Found by Buethe, Franke, Jost, Kleinjung in July 2010]

Asymptotic expression for π(x)

It would be nice to have a function, say f (x), with a closed-formformula such that

π(x) ∼ f (x) (meaning: limx→1

π(x)/f (x) = 1.)

9 / 29


Guessing asymptotic formula with some data

x π(x) x/π(x)

10 4 2.5100 25 4.0

1,000 168 6.010,000 1,229 8.1100,000 9,592 10.4

1,000,000 78,498 12.710,000,000 664,579 15.0100,000,000 5,761,455 17.4

1,000,000,000 50,847,534 19.710,000,000,000 455,052,511 22.0100,000,000,000 4,118,054,813 24.3

10 / 29



x π(x) x/π(x)

10 4 2.5100 25 4.0

1,000 168 6.010,000 1,229 8.1100,000 9,592 10.4

1,000,000 78,498 12.710,000,000 664,579 15.0100,000,000 5,761,455 17.4

1,000,000,000 50,847,534 19.710,000,000,000 455,052,511 22.0100,000,000,000 4,118,054,813 24.3

x/π(x) ≈ (the # of digits in x) · 2.3

10 / 29



x π(x) x/π(x)

10 4 2.5100 25 4.0

1,000 168 6.010,000 1,229 8.1100,000 9,592 10.4

1,000,000 78,498 12.710,000,000 664,579 15.0100,000,000 5,761,455 17.4

1,000,000,000 50,847,534 19.710,000,000,000 455,052,511 22.0100,000,000,000 4,118,054,813 24.3

x/π(x) ≈ log10(x) · ln(10)

10 / 29



x π(x) x/π(x)

10 4 2.5100 25 4.0

1,000 168 6.010,000 1,229 8.1100,000 9,592 10.4

1,000,000 78,498 12.710,000,000 664,579 15.0100,000,000 5,761,455 17.4

1,000,000,000 50,847,534 19.710,000,000,000 455,052,511 22.0100,000,000,000 4,118,054,813 24.3

π(x) ∼x

ln(x)10 / 29


Carl Friedrich Gauss

C. F. Gauss (1777–1855), when he was14 years old, conjectured that

π(x) ∼∫ x

2

dtln t

.

After he read a book that containedlogarithms of numbers up to 7 digitsand a table of primes up to 10,009.From the table of primes, he observedthat “the density (of prime numbers nearx) is, on the average, inverselyproportional to the logarithm of x”.

11 / 29


Gauss’s guess on π(x)

Challenge for Calculus students

Prove thatlimx→∞

x/ ln x∫ x2 (1/ ln t) dt

= 1.

Hint: L’Hopital’s rule and Fundamental Theorem of Calculus!

Because of the above, Gauss’s formula

π(x) ∼∫ x

2

dtln t

is compatible with our formula

π(x) ∼x

ln x.

12 / 29


Prime number theorem

Almost 100 years after Gauss stated his conjecture on π(x), theso-called Prime number theorem was finally proved:

Theorem (Hadamard and de la Vallee Poussin in 1896)As x→∞,

π(x) ∼x

ln x.

We will sketch some ideas behind the proof of this theorem.

13 / 29


Proof of prime number theorem

Theorem (Prime number theorem)As x→∞,

π(x) ∼x

ln x.

Sketch of proof (adapted from presentation of T. Tao):

Artistic

1 Create a noise, ateach prime p.

2 Listen to the primesymphony.

3 (hardest) Writedown the notesyou’re listening.

Technical

1 Defineψ(x) :=

∑p≤x ln p.

2 Take (sort of) Fouriertransform of ψ(x).

3 (hardest) Locate thefrequencies.

14 / 29




π(x) ∼x

ln x.


Artistic1 Create a noise, at

each prime p.



Technical

1 Defineψ(x) :=

∑p≤x ln p.



14 / 29




π(x) ∼x

ln x.



each prime p.



Technical1 Defineψ(x) :=

∑p≤x ln p.



14 / 29




π(x) ∼x

ln x.



each prime p.2 Listen to the prime

symphony.



∑p≤x ln p.



14 / 29




π(x) ∼x

ln x.



each prime p.2 Listen to the prime

symphony.3 (hardest) Write

down the notesyou’re listening.


∑p≤x ln p.



14 / 29


What is Fourier transform?

A (sample) soundwave:

(a periodic function)

(2) sin(x)

(−1) sin(2x)

(2/3) sin(3x)

2 sin(x) − sin(2x) +(2/3) sin(3x)

Fourier transform isa process of

decomposing aperiodic function

into several simplewaves of different

frequencies.

15 / 29


Your ears do Fourier transform all the time.

When you hearsound like this..

Hairs in your earresonate...

Amplitude: 2

))

Amplitude: −1)Amplitude: 2/3

)

And, your braincollects the signals

from hairs!

16 / 29


Visualizing Fourier transformation

Java applet demonstration:

http://www.falstad.com/fourier/

written by Paul Falstad.

17 / 29


Georg Friedrich Bernhard Riemann

G. F. B. Riemann (1826–1866) madefundamental contributions in manybranches of mathematics. Especially,he was a pioneer of the theory offunctions, which is nowadays coveredin an undergraduate course oncomplex analysis.

18 / 29


The Riemann’s zeta function ζ(s)

In 1864, Riemann wrote a short memoir, titled Uber die Anzahlder Primzahlen unter einer gegebenen Grosse.

(Translation: On theNumber of Primes Less than a Given Magnitude)

Riemann’s revolutionary idea

Study prime numbers by the zeta function

ζ(s) := 1 +12s +

13s +

14s + · · · ,

where s is a complex variable. (Simply put, a complex numbers = σ + it is just a pair (σ, t) of real numbers.)

Complex analysis techniques can show that the domain of ζ(s)can be extended to the entire complex plane, except s = 1.

19 / 29


The Riemann’s zeta function ζ(s)

In 1864, Riemann wrote a short memoir, titled Uber die Anzahlder Primzahlen unter einer gegebenen Grosse. (Translation: On theNumber of Primes Less than a Given Magnitude)

Riemann’s revolutionary idea

Study prime numbers by the zeta function

ζ(s) := 1 +12s +

13s +

14s + · · · ,

where s is a complex variable. (Simply put, a complex numbers = σ + it is just a pair (σ, t) of real numbers.)

Complex analysis techniques can show that the domain of ζ(s)can be extended to the entire complex plane, except s = 1.

19 / 29


The zeros σ + it of ζ(s)

σ

t

11/2

10

20

30

40

50 If ζ(σ + it) = 0, then we say that σ + it is azero of ζ(s). Riemann discovered:

ζ(s) has infinitely many zeros.

All (at least the interesting ones) areon the strip 0 ≤ σ ≤ 1.Riemann himself found a few zerosexplicitly (pictured left), all of whichhappen to be on the line σ = 1/2.Finally, Riemann conjectured that allthe zeros are on the line σ = 1/2.

The last item deserves some attention.

20 / 29



σ

t

11/2

10

20

30

40


ζ(s) has infinitely many zeros.All (at least the interesting ones) areon the strip 0 ≤ σ ≤ 1.

Riemann himself found a few zerosexplicitly (pictured left), all of whichhappen to be on the line σ = 1/2.Finally, Riemann conjectured that allthe zeros are on the line σ = 1/2.


20 / 29



σ

t

11/2

10

20

30

40


ζ(s) has infinitely many zeros.All (at least the interesting ones) areon the strip 0 ≤ σ ≤ 1.Riemann himself found a few zerosexplicitly (pictured left), all of whichhappen to be on the line σ = 1/2.

Finally, Riemann conjectured that allthe zeros are on the line σ = 1/2.


20 / 29



σ

t

11/2

10

20

30

40


ζ(s) has infinitely many zeros.All (at least the interesting ones) areon the strip 0 ≤ σ ≤ 1.Riemann himself found a few zerosexplicitly (pictured left), all of whichhappen to be on the line σ = 1/2.Finally, Riemann conjectured that allthe zeros are on the line σ = 1/2.


20 / 29


Riemann Hypothesis

The last item (the conjecture thatall the zeta zeros are on σ = 1/2) iscalled Riemann Hypothesis. Itis now literally a million-dollarquestion, thanks to the bountyoffered by the ClayMathematics Institute.

So, what do the zeta zeroes have to do with prime-countingand Fourier analysis?

21 / 29


What Riemann heard

σ

t

11/2

HHJL

Each zero of ζ(s) corresponds to oneparticular frequency, or one note.Riemann had good enough ears torecord a few notes in the primesymphony.To finish the proof of Prime numbertheorem, one needs to prove thatthere is no note on the line σ = 1,which was technically the hardestpart, and took almost 30 years afterRiemann.

22 / 29


Prime symphony

Since Riemann, number theorists have understood that, themore we know about the locations of zeros of the Riemann zetafunction ζ(s), the better we understand the distribution ofprime numbers.

Artistically speaking, this means that, to best understand thedistribution of prime numbers, we need to write down as manynotes as possible, while listening to the prime symphony. (Recall:there are infinitely many notes.)

23 / 29


A maestro, who has (computationally) perfect pitch

In the 80’s, A. Odlyzko, who is amaster in various number theoreticalcomputations, succeeded in locatingbillions of zeta zeros. His data gave avery strong statistical support to anearlier stunning observation made byMontgomery in early 70’s, which was. . .

24 / 29


Prime symphony sounds a lot like... heavy nuclei?

Montgomery-Odlyzko Law

The distribution of zeta zeros statistically follows that of theenergy levels of heavy nuclei.

In order to describe the energy levels of heavy nuclei, whoseexact determination is virtually impossible, physicists use astatistical model, called GUE (Gaussian Unitary Ensemble),derived from the distribution of eigenvalues of large unitarymatrices.

25 / 29


Zeta zeros vs GUE, I

100 zeta zeros starting att = 1200, wrapped around a

unit circle once

100 eigenvalues of a random100 × 100 unitary matrix (the

GUE model)

26 / 29


Zeta zeros vs GUE, II

Nearest neighborspacings among1000 zeta zeroes,versus the GUE

model

27 / 29



Nearest neighborspacings among 70

million zetazeroes, versus the

GUE model

27 / 29





GUE model

P. Diaconis (statistician): “I’ve been a statistician all my life, andI’ve never seen such a good fit of data.”

27 / 29





GUE model

Katz and Sarnak (number theorists): “At the phenomenologicallevel this is perhaps the most striking discovery about zeta sinceRiemann.”

27 / 29


Number theory and physics

This unexpected connection between number theory andphysics has recently generated a lot of interesting research andcontinued to be a prime topic in modern number theory.

28 / 29


Closing audio clip

(From NPR, Fresh Air) Interview with Brian May, the leadguitarist in the British rock band Queen, about the sound effectin the song We will rock you.

29 / 29

#rieman hyp_prime numbers & physics

Documents

prime location