prime numbers and the riemann...

Prime Numbers From Euler to Riemann A ‘Wonderful Formula’ The End

Prime Numbers and the Riemann Hypothesis

Benjamin Klopsch

Mathematisches InstitutHeinrich-Heine-Universität zu Düsseldorf

Tag der Forschung ◦ November 2005

“Investigation of the Distribution of Prime Numbers”

What will we encounter in the next 45 minutes?

The ‘Building Blocks of Numbers’What are prime numbers?How many prime numbers are there?

Riemann – a Pioneer in Complex AnalysisInfinite Series and Complex FunctionsRiemann and the Zeta Function ζ(s)

A ‘Wonderful Formula’Riemann’s FormulaThe Riemann Hypothesis

What are prime numbers?DefinitionA prime number is a natural number p admitting precisely twodistinct divisors, namely 1 and p.Thus the first seven prime numbers are 2, 3, 5, 7, 11, 13, 17.

Theorem (Fundamental Theorem of Arithmetic)Every natural number can written uniquely as a product ofprime numbers, up to re-ordering the factors.For instance, the fairy tale number satisfies

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

and similarly

112005 = 32 · 5 · 19 · 131.

What are prime numbers?

DefinitionA prime number is a natural number p admitting precisely twodistinct divisors, namely 1 and p.Thus the first seven prime numbers are 2, 3, 5, 7, 11, 13, 17.

1001 = 7 · 143.

and similarly112005 = 32 · 5 · 19 · 131.

1001 = 7 · 11 · 13

and similarly112005 = 32 · 5 · 19 · 131.

1001 = 7 · 11 · 13

and similarly112005 = 32 · 5 · 19 · 131.

1001 = 7 · 11 · 13

and similarly112005 = 32 · 5 · 19 · 131.

The Sieve of Erathostenes

Among the first 100 numbers there are 25 prime numbers.

1 2 3 4 5 6 7 8 9 10 11 1213 14 15 16 17 18 19 20 21 22 23 2425 26 27 28 29 30 31 32 33 34 35 3637 38 39 40 41 42 43 44 45 46 47 4849 50 51 52 53 54 55 56 57 58 59 6061 62 63 64 65 66 67 68 69 70 71 7273 74 75 76 77 78 79 80 81 82 83 8485 86 87 88 89 90 91 92 93 94 95 9697 98 99

2 3 4 5 6 7 8 9 10 11 1213 14 15 16 17 18 19 20 21 22 23 2425 26 27 28 29 30 31 32 33 34 35 3637 38 39 40 41 42 43 44 45 46 47 4849 50 51 52 53 54 55 56 57 58 59 6061 62 63 64 65 66 67 68 69 70 71 7273 74 75 76 77 78 79 80 81 82 83 8485 86 87 88 89 90 91 92 93 94 95 9697 98 99

2 3 5 7 9 1113 15 17 19 21 2325 27 29 31 33 3537 39 41 43 45 4749 51 53 55 57 5961 63 65 67 69 7173 75 77 79 81 8385 87 89 91 93 9597 99

2 3 5 7 1113 17 19 2325 29 31 3537 41 43 4749 53 55 5961 65 67 7173 77 79 8385 89 91 9597

2 3 5 7 1113 17 19 23

29 3137 41 43 4749 53 5961 67 7173 77 79 83

89 9197

2 3 5 7 1113 17 19 23

29 3137 41 43 4749 53 5961 67 7173 77 79 83

89 9197

2 3 5 7 1113 17 19 23

29 3137 41 43 47

53 5961 67 7173 79 83

2 3 5 7 1113 17 19 23

29 3137 41 43 47

53 5961 67 7173 79 83

There are infinitely many prime numbers

Theorem (Euclid, around 300 BC)There are infinitely many prime numbers.

Proof.

• Assume, p1 = 2,p2 = 3, . . . ,pr are all the prime numbers.• Set N := p1 · p2 · · · pr + 1.• Then N ≥ 2, but N is not divisible by any prime number.• Contradiction!

Question: How many is “infinitely many”?

Proof.

Number of Primes up to a Given Size

DefinitionFor every real number x let π(x) denote thenumber of primes between 1 and x .

x π(x) x/π(x)

100 25 4.01000 168 ≈ 6.0

10,000 1,229 ≈ 8.1100,000 9,592 ≈ 10.4

1,000,000 78,498 ≈ 12.710,000,000 664,579 ≈ 15.0

100,000,000 5,761,455 ≈ 17.4

x π(x) x/π(x)

100 25 4.01000 168 ≈ 6.0

10,000 1,229 ≈ 8.1100,000 9,592 ≈ 10.4

1,000,000 78,498 ≈ 12.710,000,000 664,579 ≈ 15.0

100,000,000 5,761,455 ≈ 17.4

x π(x) x/π(x)

100 25 4.01000 168 ≈ 6.0

10,000 1,229 ≈ 8.1100,000 9,592 ≈ 10.4

1,000,000 78,498 ≈ 12.710,000,000 664,579 ≈ 15.0

100,000,000 5,761,455 ≈ 17.4

The Logarithmic Integral Function

Observation: Viewed from a distance, the function π(x) issurprisingly smooth. The ‘density’ of primes around a largenumber x is about 1/log(x).Remark: log(x) ≈ 2.3 · (# digits before the decimal point of x).Thus one has approximately π(x) ≈ x/ log(x).Gauß suggested the following logarithmic integral Li(x) as amore precise approximation:

Li(x) :=

1log(t)

dt ≈ 1log(2)

log(3)+ . . .+

1logbxc

Li(x) :=

1log(t)

dt ≈ 1log(2)

log(3)+ . . .+

1logbxc

Li(x) :=

1log(t)

dt ≈ 1log(2)

log(3)+ . . .+

1logbxc

Li(x) :=

1log(t)

dt ≈ 1log(2)

log(3)+ . . .+

1logbxc

Li(x) :=

1log(t)

dt ≈ 1log(2)

log(3)+ . . .+

1logbxc

Li(x) :=

1log(t)

dt ≈ 1log(2)

log(3)+ . . .+

1logbxc

Li(x) :=

1log(t)

dt ≈ 1log(2)

log(3)+ . . .+

1logbxc

Li(x) :=

1log(t)

dt ≈ 1log(2)

log(3)+ . . .+

1logbxc

We build infinite series

• The harmonic series diverges to infinity:

∞∑n=1

= 1 +12

+ . . . = ∞.

to the proof

For∞∑

= 1 +12

)+ . . .

For∞∑

= 1 +12

)+ . . .

≥ 1 +12

)+ . . .

For∞∑

= 1 +12

)+ . . .

≥ 1 +12

)+ . . .

= 1 +12

+ . . .

For∞∑

= 1 +12

)+ . . .

≥ 1 +12

)+ . . .

= 1 +12

+ . . .

= 1 +12

+ . . . = ∞.

∞∑n=1

= 1 +12

+ . . . = ∞.

to the proof

∞∑n=1

= 1 +12

+ . . . = ∞.

• In contrast, the sum of quadratic reciprocals

∞∑n=1

1n2 = 1 +

132 + . . . = 1 +

+ . . .

is bounded above and converges. to the proof

For∞∑

1n2 = 1 +

∞∑n=2

For∞∑

1n2 = 1 +

∞∑n=2

1n2 ≤ 1 +

∞∑n=2

1(n − 1)n

For∞∑

1n2 = 1 +

∞∑n=2

1n2 ≤ 1 +

∞∑n=2

1(n − 1)n

= 1 +∞∑

n − (n − 1)

(n − 1)n= 1 +

∞∑n=2

n − 1− 1

For∞∑

1n2 = 1 +

∞∑n=2

1n2 ≤ 1 +

∞∑n=2

1(n − 1)n

= 1 +∞∑

n − (n − 1)

(n − 1)n= 1 +

∞∑n=2

n − 1− 1

)= 1 +

(1− 1

(12− 1

(13− 1

)+ . . .

For∞∑

1n2 = 1 +

∞∑n=2

1n2 ≤ 1 +

∞∑n=2

1(n − 1)n

= 1 +∞∑

n − (n − 1)

(n − 1)n= 1 +

∞∑n=2

n − 1− 1

)= 1 +

(1− 1

(12− 1

(13− 1

)+ . . .

= 1 + 1 +

)+ . . .

For∞∑

1n2 = 1 +

∞∑n=2

1n2 ≤ 1 +

∞∑n=2

1(n − 1)n

= 1 +∞∑

n − (n − 1)

(n − 1)n= 1 +

∞∑n=2

n − 1− 1

)= 1 +

(1− 1

(12− 1

(13− 1

)+ . . .

= 1 + 1 +

)+ . . .

= 1 + 1 + 0 + 0 + 0 + . . . = 2.

∞∑n=1

= 1 +12

+ . . . = ∞.

∞∑n=1

1n2 = 1 +

132 + . . . = 1 +

+ . . .

is bounded above and converges. to the proof

∞∑n=1

= 1 +12

+ . . . = ∞.

∞∑n=1

1n2 = 1 +

132 + . . . = 1 +

+ . . .

is bounded above and converges.

Question: What is the sum of all the quadratic reciprocals?

Euler and the ‘Real Zeta Function’

One has∞∑

6≈ 1.6449,

where π = 3.1415....

Euler considered more generally the real function

ζ(s) =∞∑

1ns for s > 1

and he computed ζ(2m) for all even numbers 2, 4,6, 8, . . .

One has∞∑

6≈ 1.6449,

where π = 3.1415....

ζ(s) =∞∑

1ns for s > 1

One has∞∑

6≈ 1.6449,

where π = 3.1415....

ζ(s) =∞∑

1ns for s > 1

The Euler Product FormulaThe Euler Product Formula yields a interesting connectionbetween the function ζ(s) and prime numbers:

ζ(s) = 1 +12s +

16s + . . .

122s +

123s + . . .

1 +13s +

132s +

133s + . . .

1 +15s +

152s +

153s + . . .

)· · ·

p prime

1p2s +

1p3s + . . .

∏p prime

11− p−s .

The Complex Number Plane

The familiar real number line Rextends to the so-calledcomplex number planeC = R + iR.

reelle Achse

1 2 3−1

imaginäre Achse

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

The complex nmber i , called the imaginary unit, has theproperty that its square equals −1.

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

3.5 + 1.5 i

z + w =

w = 1 − 0.5 i

The basic arithmetic operationsaddition, subtraction,multiplication and divisionextend from the real numbersto the complex numbers.

reelle Achse

1 2 3−1

imaginäre Achse

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

3.5 + 1.5 i

z + w =

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

3.5 + 1.5 i

z + w =

w = 1 − 0.5 i

Complex Functions

Simple transformations of the complex number plane to itself,such as the polynomial function

f (z) = f (x + iy) := z3 − 64z,

can be visualised as landscapes.

Complex Functions

f (z) = f (x + iy) := z3 − 64z,

Georg Friedrich Bernhard Riemann

• born 1826 in Breselenz, Lüchow-Dannenberg• student in Göttingen (initially in Theology) and Berlin• 1851 PhD with the seminal Inauguraldissertation

“Grundlagen für eine allgemeine Theorie der Functioneneiner veränderlichen complexen Grösse”

• 1854 Habilitation lecture “Über die Hypothesen, welche derGeometrie zu Grunde liegen” - a cornerstone of modernDifferential Geometry

• 1859 successor of Dirichlet to the Gauß-Chair, publicationof the article “Ueber die Anzahl der Primzahlen unter einergegebenen Grösse”

• deceased 1866 from tuberculosis in Selasca, LagoMaggiore, Italy

The Riemann Zeta Function

The infinite series

ζ(s) =∞∑

1ns = 1 +

14s + . . .

converges for all complex numbers s = x + yi with real partx > 1.Moreover, the function ζ(s), extends uniquely to the entirecomplex number plane, apart from a pole at s = 1.

The infinite series

ζ(s) =∞∑

1ns = 1 +

14s + . . .

The infinite series

ζ(s) =∞∑

1ns = 1 +

14s + . . .

The Critical Strip

The ‘real’ function ζ(s) can be extended to theentire complex number plane by means of theFunctional Equation

Λ(s) := π−s/2Γ(s/2)ζ(s) = Λ(1− s).

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

triviale Nullstellen für s = −2, −4, −6, ...

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

0.5 + i (14.13...)

0.5 + i (21.02...)

0.5 − i (21.02...)

0.5 − i (14.13...)

Nullstellen

nicht−triviale

Of particular importance are the zeros of Λ(s).They correspond to the non-trivial zeros of ζ(s)and lie in the critical stripS := {x + yi | 0 ≤ x ≤ 1}.

All known zeros of Λ(s) lie on the lineG := {x + yi | x = 1/2}.

The Critical Strip

Λ(s) := π−s/2Γ(s/2)ζ(s) = Λ(1− s).

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

0.5 + i (14.13...)

0.5 + i (21.02...)

0.5 − i (21.02...)

0.5 − i (14.13...)

Nullstellen

nicht−triviale

The Critical Strip

Λ(s) := π−s/2Γ(s/2)ζ(s) = Λ(1− s).

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

0.5 + i (14.13...)

0.5 + i (21.02...)

0.5 − i (21.02...)

0.5 − i (14.13...)

Nullstellen

nicht−triviale

The Chebyshev ψ-Function

Recall: We are to show that, for large x , the number π(x) ofprime numbers less than or equal to x is closely approximatedby the logarithmic integral Li(x). Grafik

More handy than the prime countingfunction π(x) =

∑p≤x 1 is a variant,

introduced by Chebyshev,

ψ(x) :=∑pk≤x

log(p),

which counts prime numbers with alogarithmic weight.

Loosely speaking, the approximation π(x) ≈ Li(x)is equivalent to ψ(x) ≈ x .

ψ(x) :=∑pk≤x

log(p),

ψ(x) :=∑pk≤x

log(p),

ψ(x) :=∑pk≤x

log(p),

Riemann’s Formula for the Prime Counting Function

The step function ψ(x) can be expressed precisely in terms ofthe complex zeros of the Riemann zeta function:

ψ·(x) = x − log(2π) −∑ρ

ρ− 1

1− 1x2

The following graphics visualise how the prime countingfunction ψ(x) is stepwise approximated using the first 300 pairsof zeros of the Riemann zeta function. Graphics

ψ·(x) = x − log(2π) −∑ρ

ρ− 1

1− 1x2

ψ·(x) = x − log(2π) −∑ρ

ρ− 1

1− 1x2

Stepwise Approximation of ψ(x)

The Riemann Hypothesis

Riemann HypothesisAll non-trivial zeros of the Riemann zeta function lie on the lineG = {x + yi | x = 1/2}.To this day, Riemann’s conjecture has not been proved ordisproved.The Riemann Hypothesis is equivalent to the followingassertion.

Equivalent Formulation of the Riemann HypothesisThere is a constant C such that π(x) does not deviate fromLi(x) by more than C ·

√x log(x).

"‘. . . Man findet nun in der That etwa so viel reelleWurzeln innerhalb dieser Grenzen, und es ist sehrwahrscheinlich, dass alle Wurzeln reell sind. Hiervonwäre allerdings ein strenger Beweis zu wünschen; ichhabe indess die Aufsuchung desselben nach einigenflüchtigen vergeblichen Versuchen vorläufig bei Seitegelassen, da er für den nächsten Zweck meinerUntersuchung entbehrlich schien. . . . "’

Riemann

Following Riemann’s Pioneering Work . . .

. . . von Mangoldt . . . Hadamard . . . de la Vallée Poussin. . . Hardy . . . Littlewood . . . Selberg . . . Montgomery . . .

The End

Many Thanksfor your kind attention

and credits toTobias Ebel

for the computer-technical assistance

Appendix with Pictures and Graphics

Proof.

x π(x) x/π(x)

100 25 4.01000 168 ≈ 6.0

10,000 1,229 ≈ 8.1100,000 9,592 ≈ 10.4

1,000,000 78,498 ≈ 12.710,000,000 664,579 ≈ 15.0

100,000,000 5,761,455 ≈ 17.4

x π(x) x/π(x)

100 25 4.01000 168 ≈ 6.0

10,000 1,229 ≈ 8.1100,000 9,592 ≈ 10.4

1,000,000 78,498 ≈ 12.710,000,000 664,579 ≈ 15.0

100,000,000 5,761,455 ≈ 17.4

Euler and the ‘Real Zeta Function’One has∞∑

6≈ 1.6449,

where π = 3.1415....

ζ(s) =∞∑

1ns for s > 1

One has∞∑

6≈ 1.6449,

where π = 3.1415....

ζ(s) =∞∑

1ns for s > 1

reelle Achse

1 2 3−1

imaginäre Achse

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

3.5 + 1.5 i

z + w =

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

3.5 + 1.5 i

z + w =

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

3.5 + 1.5 i

z + w =

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

w = 1 − 0.5 i

reelle Achse

1 2 3−1

imaginäre Achse

z = 2.5 + 2 i

3.5 + 1.5 i

z + w =

w = 1 − 0.5 i

Complex Functions

f (z) = f (x + iy) := z3 − 64z,

The infinite series

ζ(s) =∞∑

1ns = 1 +

14s + . . .

The infinite series

ζ(s) =∞∑

1ns = 1 +

14s + . . .

The Critical Strip

Λ(s) := π−s/2Γ(s/2)ζ(s) = Λ(1− s).

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

0.5 + i (14.13...)

0.5 + i (21.02...)

0.5 − i (21.02...)

0.5 − i (14.13...)

Nullstellen

nicht−triviale

The Critical Strip

Λ(s) := π−s/2Γ(s/2)ζ(s) = Λ(1− s).

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

0.5 + i (14.13...)

0.5 + i (21.02...)

0.5 − i (21.02...)

0.5 − i (14.13...)

Nullstellen

nicht−triviale

The Critical Strip

Λ(s) := π−s/2Γ(s/2)ζ(s) = Λ(1− s).

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

0.5 + i (14.13...)

0.5 + i (21.02...)

0.5 − i (21.02...)

0.5 − i (14.13...)

Nullstellen

nicht−triviale

The Critical Strip

Λ(s) := π−s/2Γ(s/2)ζ(s) = Λ(1− s).

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

−10 i

−20 i

1−1−2−3−4

reelle Achse

imaginäre Achse

kritischer Streifen

0.5 + i (14.13...)

0.5 + i (21.02...)

0.5 − i (21.02...)

0.5 − i (14.13...)

Nullstellen

nicht−triviale

ψ(x) :=∑pk≤x

log(p),

prime numbers and the riemann...

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