“On the Number of Primes Less Than a Given Magnitude” Asilomar - December 2009 Bruce Cohen Lowell High School, SFUSD [email protected] .

“On the Number of Primes Less Than a Given Magnitude”

Asilomar - December 2009

Bruce CohenLowell High School, SFUSD

[email protected]://www.cgl.ucsf.edu/home/bic

David SklarSan Francisco State University

[email protected]

Ver. 20.00

Eric [email protected]

Worksheet

x

0 if 0 is a , . Sketch a graph of 9 .

1 if 0

xu x unit step function u x u x

x

(d)

x

J x

u x

Worksheet

-1 -1 -1 -1 1 110 0 01

2

1Li log

lim lim

log log

x

x x

d

dx

d

dx

dtx t

x xx x

Worksheet

and 0,a

1

1 1limbs s

a abx dx x dx

lim lim 0

bs s s s s

b ba

x b a a a

s s s s s

log as limlogx

xx x

x

1

1lim lim

1x xx

x

2

1Li

log

xx dt

t as x

2

1log

lim1 1 1

log log

x

x

xx xx

1lim

11

log

x

x

Lilog

xx

xHence

b

c

2

1xdt

t

The methods he introduced became the foundation of modern analytic number theory.

Georg Friedrich Bernhard Riemann (1826-1866) was one of the greatest mathematicians of the 19th century. Most of his work concerns analysis and its developments, but in 1859 he published his only paper on number theory. It was entitled “Über die Anzahl der Primzahlen unter einer gegebenen Grösse” (On the Number of Primes less than a given Magnitude). In this eight page paper, he obtained a formula for the function (x) (the number of primes less than or equal to x).

This presentation is an attempt to provide a comprehensible overview of the essential content of Riemann’s paper.

Riemann and Prime Numbers

What Are Prime Numbers and Why Do We Care About Them?

Answer: A Prime number is an integer greater than 1 whose only divisors are 1 and itself. (Note: 1 is not prime.)

Answer: Primes are important because they may be viewed as the multiplicative “building blocks” of the natural numbers. This idea is embodied in

The first few primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ... , 91, 97, 101

Every integer greater than one can be written as a product of prime numbers in one and only one way (aside from a reordering of the factors).

260 3 20 3 4 5 3 2 2 5 2 3 5

260 4 15 2 2 3 5 2 3 5 Example:

The Fundamental Theorem of Arithmetic:

Note: this is a step function whose value is zero for real x less than 2 and jumps up by one at each prime.

1 0, 2 1, 2.5 1, 3 2, 4 2, 5 3, 6 3,

An important function describing the distribution of primes among the natural numbers is

the number of primes less than or equal to x x

Analytic Number Theory

Analytic Number Theory: The study of the integers using the methods of analysis

The Prime Number Theorem:

2

, where Li log

x dtx

t

(Note: The theorem was conjectured by Gauss in 1796 and proved in 1896 using methods that were introduced by Riemann 1859.)

A central result of analytic number theory concerns the distribution of primes. Specifically it concerns smooth approximations to (x) for large x.

Analysis: The theory of functions of a real or complex variable, especially using methods of the calculus and its generalizations

If we say that f (x) is asymptotically equal to g(x) or simply that f (x) is asymptotic to g(x).

( ) ( )f x g x

( )lim 1

( )x

f x

g xNotation: means . ( ) ( )f x g x

log

x

x Li( )x x

( )x

Li( )x

log

x

x

Prime Numbers and the Zeta Function

One of the most important objects in analytic number theory is the zeta function, usually denoted by (s). This notation is due to Riemann, but the function was first studied by Euler in the 1730’s. The most basic definition for this function, and the one first considered by Euler, is in terms of an infinite series:

1

1( )

sn

sn

2 2 21

1 1 1(2) 1

2 3n n

1 1 1 11 1 1 1

1 1 1 12 3 5 7s s s s

The importance of the zeta function in the theory of prime numbers is due to a remarkable alternative representation of (s) discovered by Euler in 1737. This new version, known today as Euler’s product, takes the form of an infinite product over the prime numbers p:

This is in contrast to the situation in Taylor series, in which the independent variable is the base in each term .n

na s

1s For , Example:

1

1 1( )

11

sn p

s

sn

p

Note that here the independent variable s appears as the exponent of each term.

A Brief Outline of Riemann’s Paper

He begins with Euler's product representation for (s) (the zeta function). He points out that the sum and product define the same function of a complex variable s where they converge, but “They converge only when the real part of s is greater than 1”.

I.

Then, with the phrase “however, it is easy to find an expression of the function which is always valid” he sketches a theory of the zeta function in which he:

II.

Extends the domain of to include all complex numbers s except

1s A.Shows that the extended (s) has zeros ( all with ) Re 1s B.Proves, states and conjectures certain results about these zeros(including the Riemann Hypothesis)

C.

Obtains a product representation of in terms of its zerosD.

He obtains log (s) as an integral involving a step function J(x) that jumps at prime powers

III.

Uses Fourier inversion to express J(x) as an integral involving log (s)IV.

Evaluates the integral to obtain a formula for J(x)V.

Uses Möbius inversion to obtain a formula for (x) in terms of J(x)VI.

Proposes an improved approximation to (x) in terms of Li(x)VII.

A Ramble Through of Riemann’s Paper

2( ) Li( ) log 2 Li

( 1) logx

dtJ x x x

t t t

2( ) Li( ) log 2 Li

( 1) logx

dtJ x x x

t t t

Riemann’s main formula is an expression for J (x) a step function that jumps only at prime powers. The height of the jump at pn is 1/n .

V VI III IV I II V VI VII ?( )

Li x

, a sum over the “mysterious” zeros of zeta is a sum of “oscillating”

functions (“the music of the primes”), provides the terms that converge to the jumps.

( first movie )

sin( log )2

log

x x

x

Li x

Assuming the Riemann Hypothesis this sum can expressed in terms of the imaginary parts of the zeros of zeta.

Euler’s Product for (s)

1

1 1( ) , where ranges over all primes.

11

sn p

s

s pn

p

Euler noted that this result implies that there are infinitely many primes.

Now, the sum on the left is just the harmonic series, which diverges. Hence,the product on the right must have infinitely many factors (or it would be finite).

In 1737 Euler discovered a remarkable product representation for the zeta function.Specifically, he found (stated in modern terms) for 1s

Consider what happens as s approaches 1 (from the right). This would give

But there is exactly one prime for each factor in the product, so there must beinfinitely many primes. This was the first new proof of this fact since Euclid’sproof, 2000 years earlier.

1 1 1 11 1 1 1

1 1 1 12 3 5 7

1

11p

p

1

1

n n

2

and existence implies that every term (except 1) is eventually eliminated.

A Derivation of Euler’s Product for (s)

At the k th step in this process all of the terms with denominators ns, where the smallest prime factor of n is the k th prime, are subtracted from the sum. The Fundamental Theorem of Arithmetic implies that every n has a unique smallest prime factor. Uniqueness implies that no term is subtracted twice,

1 1 1 1( ) 1

2 3 4 5s s s ss

1 1 1 1 1 1( )

2 2 4 6 8 10s s s s s ss

1 1 1 1 1 11 ( ) 1

2 3 5 7 9 11s s s s s ss

1

1 1 1 1 1 1 11 ( )

3 2 3 9 15 21 27s s s s s s ss

1 1 1 1 1 1 11 1 ( ) 1

3 2 5 7 11 13 17s s s s s s ss

1 prime

1 11

1s

n ps

sn

p

A Derivation of Euler’s Product for (s)

So

and

31 1 1 1 1 1 11 1 1 ( ) 1

5 3 2 7 11 13 17s s s s s s ss

1 1 1 11 1 1 1 ( ) 1

7 5 3 2s s s ss

1( )

1 1 1 11 1 1 1

2 3 5 7s s s s

s

Riemann’s formula for log (s) in terms of J(x)

Riemann begins his 1859 paper by deriving a formula for log ( ) in terms of the

prime-power step-function ( ). He starts by taking the log of Euler's product:

s

J x

1log 1

sp p

He substitutes this in the RHS of his result above for log ( ) and getss

1

1log ( ) .

sn

p n

s pn

1

1log 1 n

n

x xn

11Riemann uses this with to obtain a series for log 1 in prime powers:ss p

xp

1log ( ) log

11p

s

s

p

Now, recall the Taylor series for , log 1 x

1

1 1 1log 1

n

s snp n p

1

1 ns

n

pn

1

1s

n

n

pn

Riemann’s formula for log (s) in terms of J(x)

1

1log ( ) .

sn

p n

s pn

At this stage, Riemann has

These facts allow him to rewrite his double sum as a sum over prime powers,

1with terms of the form , in order of increasing numerical value of :

sn n

single

p pn

prime powers

1log ( )

n

sn

p

s pn

1 1 2 1 1 3 21 1 1 1 1 1 12 3 2 5 7 2 3

1 1 2 1 1 3 2

s s s s s s s

1 1 2 1

Next, he notes that each prime power occurs in this double sum,

and also that the prime powers may be placed in a natural order, ie., in order of

increasing numerical value: 2 3 2 5

np exactly once

1 3 2 17 2 3 11 .

Riemann’s formula for log (s) in terms of J(x)

1 .n

sn s

pp s x dx

and therefore

0, 0( )

1, 0

xu x

x

Now, if we define u(x), the unit step function, by

1

1

sn n sp s u x p x dx then we may write

x

u(x)

1 0s s s

s

aa

x a ax dx

s s s

1s s

aa s x dx

so

Riemann now converts his single sum to an integral essentially as follows.(Actually, we have made the argument slightly more modern, but equivalent,by use of the unit step function). First, note that for 1s

Riemann’s formula for log (s) in terms of J(x) IV

1

1

sn n sp s u x p x dx Using

We may substitute in Riemann's single sum for log ( ) which we obtained

earlier, and then exchange order of summation and integration, to get

s

1log ( )

n

sn

p

s pn

1

1

1

n

n s

p

s u x p x dxn

1

1

1

n

n s

p

s u x p x dxn

?? J x

21 2 2u x

2u x

1n

n

p x

u x pn

( )J x

3u x

the number of primes x 1the number of prime squares

2x

1the number of prime cubes

3x

112

1u x 11

31

u x 212

2u x

1

n

n

p

u x pn

1

1log ( ) ( ) ss s J x x dx

Riemann’s formula for log (s) in terms of J(x) IV

Since the sum in the integrand is the step function J(x) we have arrived at

Riemann’s formula for log (s) in terms of J(x)

1

1

sn n sp s u x p x dx Using

We may substitute in Riemann's single sum for log ( ) which we obtained

earlier, and then exchange order of summation and integration, to get

s

1log ( )

n

sn

p

s pn

1

1

1

n

n s

p

s u x p x dxn

1

1

1

n

n s

p

s u x p x dxn

1 log ( )( ) , Real and 1.

2

a i s

a i

sJ x x ds a

i s

Riemann’s Formula for J(x) in terms of log (s)

1

1log ( ) ( ) ss s J x x dx

Riemann had now obtained a formula for log (s) in terms of the prime-power step function J(x):

He was able to do this by the method of Fourier inversion, yielding

The integral on the right is a contour integral along a path in the complex plane. We can’t develop the theories of contour integration and Fourier inversion in this talk, but the details of these theories will not be needed to understand the essentials of what follows. What matters is that Riemann was able to evaluate this integral by using certain special properties of (s) that he discovered.

Now, if he could just solve this equation to get J(x) in terms of log (s) he would have obtained complete information about prime powers (and thus about primes) from an expression involving only the smooth function (s) and some elementary functions.

Integration Contour for Riemann’s Formula

1 log ( )( ) , and 1

2

a i s

a i

sJ x x ds a a

i s

Re

Im

1

a

a i

a i

Riemann’s Plan to Evaluate His Integral

1 log ( )( ) , Real and 1.

2

a i s

a i

sJ x x ds a

i s

Riemann now had his integral formula for J(x):

First, note that the path of integration is not the real line. Thus the integrand must be evaluated for complex values of s. The original definition of (s) as an infinite series would allow a direct evaluation, but is not sufficient to support Riemann’s attack on the integral. He was able to construct an analytic continuation of s) which has meaning throughout the complex plane, except at s = 1.

To actually evaluate the integral, his idea was to write (s) as a product of simpler functions. Then log (s), and the integral, would break up into a sum of terms simple enough that the individual integrals could be evaluated more easily.

Before he could write (s) as a product, he needed to clean it up a bit. Recall that (s) blows up at s = 1. This behavior is a problem if one is to factor a function in the way Riemann had in mind. In short, he ended up defining a new function (s) that is well behaved throughout the complex plane:

2( ) 1 12

s ss s s

where refers to the Gamma function. The details of this choice were driven by a certain symmetry (the “functional equation”) that Riemann had discovered, as well as by the need for a function well behaved throughout the complex plane.

This function has the further property that its only zeros are the zeros of (s) that don’t lie on the real line. These are known as the non-trivial zeros of (s).

Riemann’s Plan to Evaluate His Integral

The Zeros of the Zeta Function

The two specific forms for the zeta function (infinite series and Euler product) are intrinsically incomplete due to the fact that they are only defined for s > 1. In fact, both the series and product blow up for .1s

He found that this extended function does have zeros on the negative real axis (the “trivial” zeros) and in the strip of the complex plane with real part between 0 and 1. It is these latter “non-trivial” zeros which play a major part in the theory.

This is a problem, because a central property of the zeta function is its set of roots or “zeros”, ie. values of s where the function value is zero. However, a glance at either of the two representations reveals that the zeta function is never zero for s > 1 (every term of the sum and every factor of the product is > 0). Thus the zeta function has no zeros for .1s

Now, Riemann was able to find an analytic continuation of the zeta function to the entire real line and, in fact, to the entire complex plane, except for the point .1s

1 prime

1 11

1s

n ps

sn

p

1s For ,

The Zeta Function in the Complex Plane and The Riemann Hypothesis

The Riemann HypothesisAll of the non-trivial zeros of the zeta function have real part 1/2 (ie., they lie on the critical line).

Re

Im

12-1-2

Trivial Zeros

14.1

21.025.030.432.9

Non-TrivialZeros

Critical Line1

Re( )2

s

1

34

2

1

34

2

5

5

10

Region Representedby the

Infinite SeriesExtended

Region

Riemann’s Plan to Evaluate His Integral

1

1( ) 1

2 k k

ss

This is directly analogous to factorization of a polynomial ps) in terms of its roots 1 , 2 , … , n in the form

01

1n

k k

sa

1 11 01 1 0( ) n n n nn

n n nn n

a ap s a s a s a s a a s s

a a

11

1 1n

n

n nk k

sa

1 2n na s s s

2( ) 1 12

s ss s s

Since is defined and well behaved

throughout the complex plane, Riemann could factor it in terms of its zeros 1 , 2 … (the nontrivial zeros of zeta):

Riemann’s Plan to Evaluate His Integral

2( ) 1 1 .2

s ss s s

1( ) 1

2

ss

At this point Riemann had written his new function s) in two ways:

He then used these results to eliminate s) and to solve for s):

2

11

2( )

1 12

s

s

ss

s

This is the desired factorization of s). It yields immediately

log ( )s log 1s log2

s log 12

s

log 2 log 1s

1 log ( )( )

2

a i s

a i

sJ x x ds

i s

Evaluation of Riemann’s Integral

log 1log1 log( 1) 1 1 22( )2 2 2

a i a i a is s s

a i a i a i

sss

J x x ds x ds x dsi s i s i s

1

log ( ) log 1 log log 1 log 2 log 12 2 k k

s s ss s

1

1 log 2 1 1log 1 .

2 2

a i a is s

a i a ik k

sx ds x ds

i s i s

So, to evaluate his formula

Riemann could substitute

and evaluate term by term to get

Evaluation of Riemann’s Integral

Riemann was a master of complex analysis. He was able to evaluate each of these complex contour integrals with the following results:

1 log( 1)

2

a i s

a i

sx ds

i s

Li( )x

log1 22

a i s

a i

s

x dsi s

log

4

a i s

a ix ds

i

0

log 121

2

a i s

a i

s

x dsi s

2( 1) logx

dt

t t t

1 log 2

2

a i s

a ix ds

i s

log 2

2

sa i

a i

xds

i s

log 2

1 1log 1

2

a i s

a i

sx ds

i s

Li( )x

2( ) Li( ) 0 log 2 Li

( 1) logx

dtJ x x x

t t t

2( ) Li( ) log 2 Li

( 1) logx

dtJ x x x

t t t

Riemann’s formula for J(x)

Putting all this together he had

Or, rearranging slightly

2( ) Li( ) log 2

( 1) logs x

dtJ x x

t t t

This is Riemann's formula for ( ). Note that all the "wiggles" are in the

term Li . Thus the smooth part of ( ), which we'll call ( ), iss

J x

x J x J x

Riemann’s Formula’s for J and

2( ) Li( ) log 2 Li

( 1) logx

dtJ x x x

t t t

2( ) Li( ) log 2

( 1) logs x

dtJ x x

t t t

x J x 1 212 J x 1 31

3 J x 1 515 J x 1 61

6 J x 1 717 J x

1 2 1 3 1 4 1 51 1 1 12 3 4 5 J x x x x x x

Riemann pointed out that

and used Möbius inversion to obtain

1 2 1 3 1 4 1 51 1 1 12 3 4 5 2 3 4 5 x J x J x J x J x J x

1 2 1 3 1 5 1 6 1 71 1 1 1 15 72 3 6Li( ) Li( ) Li( ) Li( ) Li( ) Li( )x x x x x x x

Riemann then suggested an improved version of the Prime Number Theorem

Bibliography

[2] T. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976

[8] http://en.wikipedia.org/wiki/Prime_number_theorem

[1] M. Abramowitz, I. Stegun, Handbook of Mathematical Functions, Dover, New York, 1965

[3] Brian Conrey, The Riemann Hypothesis, Notices of the AMS, March 2003

[4] H. M. Edwards, Riemann’s Zeta Function, Dover, New York, 2001 (Republication of the 1974 edition from Academic Press)

[5] Andrew Granville, Greg Martin, Prime Number Races, American Mathematical Monthly, Volume 113, January 2006

[9] http://en.wikipedia.org/wiki/Riemann_zeta_function

[6] Bernhard Riemann, Gesammelte Werke, Teubner, Leipzig, 1892. (Reprinted by Dover Books, New York, 1953.)

[7] http://www.claymath.org/millennium/

END OF MAIN TALK

21

sin( log )( ) Li( ) 2 log 2

log ( 1) logk

xk k

xx dtJ x x

x t t t

1

log2

Li1log log log loglog2

i i i xx x x x x xex

x x i x i xi x

cos( log ) sin( log ) sin( log ) cos( log )

log log

x x i x x x i x

x i x

sin( log )2Re Li 2

log

x xx

x

1 1

sin( log )2Re Li 2

logk k

k k k

xxx

x

Oscillations and Li(x)

(assuming RH)

225 15primes less than or equal to are

The same reasoning leads us to the general result,

1 2 x x More generally, the number of prime squares

2 2 2 2 2 2 22 3 5 7 11 13 17prime squares less than or equal to 225 areExample

Using (x) to Count Prime Powers

225 225 15 6 Hence the number of prime squares

1 nx x The number of prime nth powers

J (x) the number of primes x 1/2 the number of prime squares x

1/3 the number of prime cubes x

We can use this result to write J(x) in terms of (x).

J x x 1 212 x 1 31

3 x 1 4 1 51 14 5 x x

2 3 5 7 11 13 17

1 2 1 3 1 4 1 51 1 1 12 3 4 5 2 3 4 5 x J x J x J x J x J x

Möbius inversion, a technique from elementary number theory, can be used to write (x) in terms of J(x).

1 2 1 3 1 4 1 51 1 1 12 3 4 5 J x x x x x x We have

Relating (x) and J(x)

x J x 1 212 J x 1 31

3 J x 1 515 J x 1 61

6 J x 1 717 J x

2log x

Note: These sums contain only finitely many terms, they terminate when 1/ 2 nx

since for . Thus the number of terms is .

2 ( ) ( ) 0x J x x

1 if 1

0 if is divisible by the square of a prime

if is the product of distinct primes1k

n

n n

n k

where the Möbius function is defined as

Zeta Between 0 and 1

1 1 1 1 1

1 1 1 1 1 1( )

22 1 2 2 1s s ss s s

n k k k k

sn kk k k

1

( ) ( )2s

s s

1

1 1 1

1 1 1 1( ) ( ) ( )

22 1 2

n

s ss sn k k

s s sn k k

2( ) ( ) ( )

2ss s s

11 1

1 11

11 1( ) 1 ( ) 1

2 2

n

s s sn

s sn

The sum on the right converges for and provides an analytic continuation of into the interval between 0 and 1.

0 1s

For 1s

Start with a simple staircase function,

Speculations About Riemann’s Motivation“Music of the Primes”

An example from classical Fourier (harmonic) analysisfind a nice smooth approximation,

subtract to get a periodic “sawtooth function”, expand in a Fourier series

12x x

x

2x

2x x

1

sin 22

2n

nx

n

sin 2 sin 4 sin 6

2 3

x x x

Speculations About Riemann’s Motivation“Music of the Primes”

In our example from classical Fourier (harmonic) analysis

We started with a simple staircase function, found a nice smooth approximation, subtracted to get a periodic “sawtooth function”, expanded in a Fourier series

sin 2 sin 4 sin 62

2 4 6

x x x

12x x

1

sin 22

2n

nx

n

By analogy

Start with a step function that describes the distribution of primes, find a nice smooth approximation,subtract to get an oscillatory “sawtooth function”, find appropriate oscillatory functions, and a function whose zeros provide the appropriate frequencies,

Note: the sine function plays two roles in this example; it provides the periodic oscillations, and its zeros provide the frequencies.

expand in a Fourier-like series

x s x

( ) sx x

1

2Re Li k

k

x

12

1

2 Re Li ki

k

x

1

sin( log )2

logk

k k

xx

x

zz n

z z z n

1 1

, for all z except 0, -1, -2, -3, …

z n z n z n z n z n z z z 1 1 1 2 1

1 1

, for Re(z) > 0 z t e dtz t

1

0

1z z z ( by integration by parts )

( the functional equation for the gamma function )

n n 1 ! ! 1z z ( gamma interpolates the factorial function )

The functional equation provides an analytic continuation for the gamma function

The Gamma Function

1 11 1

0, 0, 1, 2,z z z n

zz z n z

21

( ) Li( ) Li log 2( 1) log

k

xk

dtJ x x x

t t t

1 2 1 3 1 4 1 51 1 1 12 3 4 5 2 3 4 5 x J x J x J x J x J x

Definitions

21

_ ( ) Li( ) Li log 2( 1) log

k

n

xk

dtr J x x x

t t t

1

1

( ) 1 1_ Li arctan

log logn

Sn

npi smooth x x

n x x

1

1 1

log( )Li 1

1 !

k

n

n n

xnx

n k k k

1

log 1 11 arctan

1 ! log log

k

n

x

k k k x x

1

1

( ) 1 1Riemann (1859): ( ) Li arctan

log logn

n

nx x

n x x

1 2 1 3 11 1 12 3_ _ 2 _ 3 _ 3 _ m

mr x r J x r x r J x r J x

0 2Li( ) 1.045...

log log

x xdt dtx cpv

t t ,