Harvey Mudd Senior Thesis - Zeta Function.pdf

8/9/2019 Harvey Mudd Senior Thesis - Zeta Function.pdf

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An Exploration of the Riemann Zeta Function

and its Application to the Theory of PrimeNumber Distribution

Elan Segarra

Professor Darryl Yong, Advisor

Professor Lesley Ward, Reader

May, 2006

Department of Mathematics


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Copyright c 2006 Elan Segarra.

The author grants Harvey Mudd College the nonexclusive right to make this workavailable for noncommercial, educational purposes, provided that this copyrightstatement appears on the reproduced materials and notice is given that the copy-ing is by permission of the author. To disseminate otherwise or to republish re-quires written permission from the author.


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Contents

Abstract v

Acknowledgments vii

1 History 1

2 Fundamentals and Eulers Zeta Series 5

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Infinite Series and Convergence . . . . . . . . . . . . . . . . . 52.3 Bernoulli Numbers . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Bernoulli Polynomials . . . . . . . . . . . . . . . . . . . . . . 112.5 Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Evaluation of Eulers Zeta Series 15

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Integration of Infinite Series . . . . . . . . . . . . . . . . . . . 163.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.4 Differentiation of Infinite Series . . . . . . . . . . . . . . . . . 233.5 Values of Eulers Zeta Series at the Even Integers . . . . . . . 253.6 Euler-Maclaurin Summation Formula . . . . . . . . . . . . . 303.7 Approximation of Eulers Zeta Series Elsewhere . . . . . . . 33

4 Extension of the Zeta Function to the Complex Plane 35

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.2 Jacobis Theta Function . . . . . . . . . . . . . . . . . . . . . . 36

4.3 The Functional Equation for(s) . . . . . . . . . . . . . . . . 434.4 Values of Riemanns Zeta Function at the Negative Integers 46


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iv Contents

5 The Riemann Zeta Hypothesis and the Consequences of its Truth 49

5.1 The Riemann Zeta Hypothesis . . . . . . . . . . . . . . . . . . 495.2 The Prime Number Theorem . . . . . . . . . . . . . . . . . . 505.3 Additional Consequences . . . . . . . . . . . . . . . . . . . . 515.4 Lindelof s Hypothesis . . . . . . . . . . . . . . . . . . . . . . 53

Bibliography 57


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Abstract

Identified as one of the 7 Millennium Problems, the Riemann zeta hypoth-esis has successfully evaded mathematicians for over 100 years. Simplystated, Riemann conjectured that all of the nontrivial zeroes of his zeta func-tion have real part equal to 1/2. This thesis attempts to explore the theorybehind Riemanns zeta function by first starting with Eulers zeta seriesand building up to Riemanns function. Along the way we will developthe math required to handle this theory in hopes that by the end the readerwill have immersed themselves enough to pursue their own explorationand research into this fascinating subject.


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Acknowledgments

I would like to thank Professor Yong, Professor Ward, and especially Pro-fessor Raugh for helping this thesis come into being. I could not have ac-

complished this without their input, their faith and their nearly unendingpatience. I would also like to thank Libby Beckman for being my late nightthesis buddy on many occasions.


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Chapter 1

History

Every great story has a beginning, and like many adventures in mathemat-ics, this story begins with Leonard Euler. In 1737 Euler proved that theinfinite sum of the inverses of the primes

1

2+

1

3+

1

5+

1

7+

1

11+

diverges [Edw]. Euler no doubt realized that this observation not only sup-ported the already known fact that the primes are infinite, but also hintedthat the primes are fairly dense among the integers. In addition to this,Euler also proposed what is now known as the Euler product formula,

n=1

1

ns =

primep

1

1+ 1ps,

although it was not rigorously proven until 1876 by Leopold Kronecker[Dun]. Amazingly, this formula links the worlds of number theory andanalysis and as a result allows tools from each field to combine and minglelike never before. These two facts represent the gateway and the key to oneof the most intriguing and mathematically diverse theories, the theory ofthe Riemann zeta function.

Sometime later, Gauss found himself working on a related problem, al-though he probably didnt know it. It was a habit of Gauss to keep a table

of prime number counts which he continually added to. At some pointGauss noticed an interesting pattern. Gauss states in a letter written in1849 that he has observed as early as 1792 or 1793 that the density of primenumbers appears on average to be 1/ log x.[Edw] This observation wouldlater become known as the prime number theorem.


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2 History

Theorem 1.0.1(Prime Number Theorem).

(x) xln x

as x ,

where(x)is the number of primes less than or equal to x (where 1 is not consid-ered to be prime).

Gauss later refined this to the following statement.

Theorem 1.0.2(Prime Numer Theorem II).

(x) Li(x) = x

2

dt

ln t as x.

Over the years, many notable mathematicians worked to prove thisconjecture. In 1800 Legendre published hisTheorie des Nombres in whichhe proposed an explicit formula for (x) that basically amounted to theprimes having a density of 1/ ln x. Unfortunately, his proof was not rigor-ous, leaving the problem just as open as it was before.

In 1850 Chebyshev proved some major results, notably that

(0.89) x

2

dt

ln t< (x) < (1.11)

x2

dt

ln t,

(0.89)Li(x) < (x) < (1.11)Li(x),

for all sufficiently largex. He was also able to show thatifthe ratio of(x)to Li(x) approached a limit as x goes to infinity, then the limit must be 1.Then Bernhard Riemann attempted to work on the problem, and he

was the first to see the potential in Eulers product formula. He used it as ajump off point to create his zeta function by extending the product functionnot only to the entire real line, but to the entire complex plane with theexception of a singularity at s = 1. This of course was only natural giventhat he was right on the frontier of complex variable analysis. In 1859 hepublished his monumental paperOn the Number of Primes Less Than a GivenMagnitude.

In this paper Riemann attempted to show that (x)(x)where

(x) = Li(x)

Li() ln 2+

x

dtt(t2 1) ln t ,

and runs over the non trivial zeros of his zeta function. Unfortunatelyhe was unable to show (or at least he did not show in the paper) whether


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3

Li(x) was the dominating term or even if the series converged. The real

importance of his paper lay in the novel tools he developed and utilized.For the first time a mathematician was using tools like Fourier transformsand complex analysis to say something about the nature of prime numberdistribution. It truly was revolutionary, and it is not surprising that therewas no more substantial work in the field for 30 years after his publica-tion, almost as if it took the mathematical world that much time to digestRiemanns ideas.[Edw]

Then in 1896 Jacques Hadamard and C. J. de la Vallee Poussin simulta-neously and independently proved the prime number theorem using manyof the tools that Riemann had developed and used. Their proofs showedthat Riemanns formula for (x) was in fact correct and that the largest

term was Li(x). Even though the main ideas that Riemann proposed werenow rigorously proved, there was still one more conjecture that remainedunverified.

In his 1859 paper, Riemann mentioned that he considers it very likelythat the nontrivial zeros of(s), which will be defined during the courseof this thesis, all lie on the line(s) = 1/2. This statement would laterbecome known as the Riemann zeta hypothesis and has managed to stumpmathematicians for over a century. The purpose of this thesis is to exploresome of the different facets of Riemanns zeta function and introduce someof the math that will assist us. Chapter 2 will start by introducing Eulerszeta series, which is essentially the predecessor of Riemanns zeta function.The chapter will also introduce concepts such as convergence, Barenoulli

numbers, and the gamma function in order to construct a solid base tolaunch from. Chapter 3 is concerned with the values of Eulers zeta series.It will present 2 proofs for the values of the function at the even integersas well as provide a method to approximate the values elsewhere on thepositive real line. In order to accomplish the proofs presented, the chapteralso introduces Fourier series and the Euler Maclaurin summation formula.Chapter 4 uses a transformational property of Jacobis theta function tohelp derive the functional equation in order to extend the Euler zeta seriesto the complex plane. This extension of the series then becomes Riemannszeta function. Finally, Chapter 5 examines Riemanns famous hypothesisand the many consequences of its truth including Lindelofs hypothesis.


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6 Fundamentals and Eulers Zeta Series

Clearly if we sum an infinite number of 1s we will end up with infinity:

k=1

1= .

What about the following infinite sum (known as the harmonic series)

k=1

1

k =1+

1

2+

1

3+

1

4+

1

5+

1

6+

1

7+ ?

Even though there are an infinite number of terms, we can see that theterms get smaller and smaller, and toward the end of the sequence theterms are practically zero. It seems like this would imply that the sum is

finite, but consider the following grouping

k=1

1

k = (1) +

1

2

+

1

3+

1

4

+

1

5+

1

6+

1

7+

1

8

+

1

2

+

1

2

+

1

4+

1

4

+

1

8+

1

8+

1

8+

1

8

+

=

1

2

+

1

2

+

1

2

+

1

2

+

With this grouping, we can see that the harmonic series sums to somethingthat is greater than an infinite number of halves whose sum is clearly infi-

nite. Therefore the harmonic series shows that even if the terms of a seriesapproach zero, this is not sufficient for a series to sum to a finite num-ber. The question then becomes, when will an infinite series sum to a finitenumber? To answer that we introduce the notion of convergence.

Given a sequence{an} of real numbers. We say that the seriesan isconvergentandconvergestoSasn if for all > 0 there exists anNsuchthat

n

k=1

ak S < for all n >N

If no suchSexists, then we say that the series diverges [Kno].In other words, our series converges to S if the partial sums become

arbitrarily close to S as n goes to infinity. Note that thepartial sum of aseries is simply the sum of the first n numbers, wheren is finite. Anotherway to think about convergence is that the partial sums of the series are allbounded.


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Infinite Series and Convergence 7

Theorem 2.2.1. The infinite series anconverges if and only if there exists a real

M such that

N

n=1

an

M for all N 1.From the divergent harmonic series, we can see that convergence is not

always as simple as we would hope. Therefore mathematicians have devel-oped several methods for determining convergence without directly show-ing that a Nexists for every as the definition demands. We already sur-reptitiously used the following method when we showed that the harmonicseries was divergent.

Theorem 2.2.2. Given infinite series

k=1

ak and

k=1

bk

of non-negative terms (meaning ak 0and bk 0), with akbk. Then we have Ifakconverges, then bkconverges. Ifbkdiverges, then akdiverges.Another useful convergence test is known as the integral test, which

will help us to show that Eulers zeta function does in fact make sense.

Theorem 2.2.3. Given the an infinite series

n=1

an

of non-negative terms, if f(x)is decreasing and continuous for x 1and f(n) =anfor all n 1, then

1

f(x)dx and

n=1

an

either both diverge or both converge. In other words, if the integral is finite thenthe series converges. Otherwise the series diverges.

Using the integral test we can now finally prove that(s)converges fors > 1 and thus makes sense.


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Proof. If we let

f(x) = 1xs

wheresis taken to be a constant greater than 1, then clearly

(s) =

n=1

f(n).

Note also thatf(x)is continuous, positive and decreasing (given thats > 1)and that

1f(x)dx =

1s

1

limx

1

xs1 1

1s1=

1s

1

(0 1) = 1s

1

Thus the integral test tells us that since

1 f(x)dxconverges then 1 f(n)

converges as well fors > 1.

Note that the requirement thats be greater than 1 makes sense because(1)is the harmonic series which we showed diverges. Whats remakableis that as soon asscrosses that threshold, than the series instantly becomesfinite.

Although they will not be useful until the next chapter, the Bernoullinumbers and Bernoulli polynomials play a very important role in the the-ory behind both Eulers zeta series and Riemanns zeta function and thusmerit mention in the chapter on Fundamentals. The next two sections will

be devoted to their descriptions and properties.


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Bernoulli Numbers 9

2.3 Bernoulli Numbers

Bernoulli was well aware of the formulas for the partial sums of the powersof the integers, the first few being:

n

k=1

k = n(n+ 1)

2

n

k=1

k2 = n(n+ 1)(2n+ 1)

6

n

k=1

k3 = n2(n+ 1)2

4 .

Given the parallel nature between increasing exponents in the summandsand the exponents in the explicit formulas, it is natural to search for somegeneralization. The result gave rise to an interesting sequence of num-bers, called the Bernoulli Numbers; named by Euler in honor of his teacher.Viewed as a formal expression (meaning thatBk is actuallyBk) we have thegeneral formula

n1k=0

km = 1

m+ 1

(n+B)m+1 Bm+1

, (2.1)

whereBkis thekth Bernoulli number. Using the binomial formula on (2.1)yields

n1k=0

km = 1m+ 1

m

j=0

m+ 1

j

nm+1jBj

. (2.2)

From here, we can produce the recursive relationship between the Bernoullinumbers by pluggingn = 1 into equation (2.2), resulting in

0=m

j=0

m+ 1

j

Bj. (2.3)

Using the recursive formula and B0 = 1 we have a fairly straight forward,although not very speedy, method for computing the Bernoulli numbers.For reference, the first couple Bernoulli numbers are:

B0= 1, B1 =12

, B2=1

6, B3= 0, B4 =1

30, B5 = 0.

Notice that both the third and fifth Bernoulli numbers are zero, and in factBk=0 for all oddk> 1. This fact will be proved rigorously momentarily.


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Another important aspect of the Bernoulli numbers is that they are gen-

erated [IK] by the Taylor series approximation of

x

ex 1 =

k=0

Bkxk

k! . (2.4)

This relation is very important since the frequent appearance of this gener-ating function amid formulas involving Riemanns zeta function will be anindication of the involvement of the Bernoulli numbers.

Theorem 2.3.1. For odd k> 1, Bk= 0.

Proof. After moving thek= 1 term of the sum over to the other side, equa-

tion (2.4) becomes x

ex 1B1x

1! =

k=0,k=1

Bkxk

k! (2.5)

Focusing on the left hand side of the equation we have

x

ex 1B1x

1! =

x

ex 1+x

2

= x

2

2

ex 1+ex 1ex 1

= x

2 ex + 1

ex

1

= x

2

ex/2

ex/2

ex/2 +ex/2

ex/2 ex/2

.

Now notice that substituting xfor x in the last line leaves the expressionunchanged. Therefore the right hand side of equation (2.5) must also re-main unchanged by the same substitution. This means that

k=0,k=1

Bkxk

k! =

k=0,k=1

Bk(x)kk!

(2.6)

and therefore thatBk= (1)k

Bkfor allk=1. While this is trivially true forevenk, this forcesBk=0 for all oddk=1.


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Bernoulli Polynomials 11

2.4 Bernoulli Polynomials

Next after the Bernoulli numbers we have the Bernoulli polynomials. Inshort, these are the polynomials (for n 0) uniquely defined by the fol-lowing three characteristics:

(i) B0(x) =1.

(ii) Bk(x) =kBk1(x), k 1.

(iii) 1

0Bk(x)dx = 0, k 1.

Using these facts, we can easily compute the first couple Bernoulli polyno-

mials:

B0(x) = 1

B1(x) = x 12

B2(x) = x2 x+ 1

6

B3(x) = x3 3

2x2 +

1

2x

B4(x) = x4 2x3 +x2 1

30.

Euler found a generating function [IK] for these functions as well:

xezx

ex 1 =

k=0

Bk(z)xk

k! . (2.7)

You may have noticed that the Bernoulli polynomials that we listed take ontheir corresponding Bernoulli numbers at x = 0. This is in fact true for allBernoulli polynomials, as evidenced by the result of substitutingz = 1 intothe generating function,

k=0

Bk(0)xk

k! =

xe(0)x

ex 1 = x

ex 1 =

k=0

Bkxk

k! .

Here are some other useful properties of the Bernoulli polynomials.

Theorem 2.4.1. Bn(1 x) = (1)nBn(x)


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Proof. Using the (2.7) we have

k=0

Bk(1 z)xkk!

= xe(1z)x

ex 1 =(ex)

(ex)

xezx

1 ex =(x)ez(x)

ex 1 =

k=0

Bk(z)(x)kk!

Equating the coefficients then yieldsBn(1 x) = (1)nBn(x).Corollary 2.4.2. Bn(1) = (1)nBn(0). Furthermore, since Bn(0) = Bn, andBn= 0 for all odd n > 1, then Bn(1) =Bn(0) =Bnfor all n=1.Theorem 2.4.3. Bn(x+ 1) Bn(x) =nxn1.Proof.

k=0

(Bk(x+ 1) Bk(x))tkk!

= te(x+1)t

et 1 text

et 1

= tetext text

et 1=

text (et 1)et 1

= text

=

k=0

t(xt)k

k!

=

k=0

(k+ 1)xktk+1

(k+ 1)!

=

k=1

(kxk1)tk

k!

The index difference between the beginning and the end is not a problemsince the first term of the left hand side is zero because

B0(x+ 1) B0(x) =1 1= 0.ThusBk(x+ 1) Bk(x) =kxk1.


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Gamma Function 13

2.5 Gamma Function

Another useful function when dealing with the theory surrounding Eulerszeta series is the gamma function. Essentially, the gamma function is anextension of the factorial function to all real numbers greater than 0 [Sto].

(s) =

0exxs1dx fors > 0. (2.8)

While Euler was the first to introduce this function, he defined it slightlydifferently and Legendre was the first to denote it as (s).

We should note that this function can actually be extended to all com-plex numbers so that it only has simple poles at the negative integers (and

zero). And it is for this reason that we can continue to use

(s)even whenwe begin to work with Riemanns zeta function in the realm of complexnumbers.

Be careful when researching this subject using different sources becauseGauss [Edw] also defined this function, except that he denotes it by (s)where

(s) = (s+ 1).

This thesis will be using the more popular notation of(s) as defined in(2.8).

The rest of this section will derive (or sketch the idea of the derivation)of several identities which will prove useful as we manipulate (s)and (s)

over the course of this thesis.What is known as the functional equation of(s),

(s+ 1) =s(s), (2.9)

can easily be obtained by applying integration by parts ons(s)and usingthe definition given in (2.8). Using this as a recursive relation and seeingthat (1) =1 shows us that for any positive integer nwe have

(n) = (n 1)!,confirming that (s)is in fact an extension of the factorial function.

We mentioned earlier that (S)can be extended over the entire complexplane with the exception of the negative integers. This is accomplished byredefining the function as

(s+ 1) = limN

1 2 3 N(s+ 1)(s+ 2)(s+ 3) (s+N) (N+ 1)

s. (2.10)


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Since

(N+ 1)s = (1)s(1 + 1)s(2)s(2 + 1)s (N 1)s(N 1 + 1)s(N)s(N+ 1)s

Then we can write (2.10) as an infinite product instead of a limit, yielding

(s+ 1) =

n=1

nns(n+ 1)s

(s+n) .

A little manipulation of this then gives us

(s+ 1) =

n=1

1+ s

n1

1+1

n

s.

Now, ifsis not an integer we have

(s+ 1)(s+ 1) =

n=1

1+

s

n

1 1+

1

n

s n=1

1+

sn

1 1+

1

n

s

=

n=1

1+

s

n

1 1+

1

n

s 1+

sn

1 1+

1

n

s

=

n=1

1 s

2

n2

1. (2.11)

Although it wont be proved until section 3.3, let us take for granted thatwe have the product formula for sine,

sinx= x

n=1

1 x

2

n2

.

Plugging (2.11) into the sine formula then yields the identity

sin(s) = s

(s+ 1)(1 s) . (2.12)

Finally, a special case of the Legendre relation gives us the additionalidentity

(s+ 1) =2s

s2

s 1

2

1/2. (2.13)


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Chapter 3

Evaluation of Eulers ZetaSeries

3.1 Introduction

Now that we have defined and verified that (s)converges for reals > 1 itwould be nice to figure out what the series converges to. This goal dividesitself nicely into two parts: the values at the even integers and the valueseverywhere else.

Two proofs will be presented of the formula for the values of Eulerszeta series at the even integral points, but before we proceed we need to

introduce Fourier series and prove that we can integrate an infinite seriesbecause both of which are required for the proofs that will be presented.Therefore the next section will introduce infinite series of functions andintegration while the following section will introduce the theory of Fourierseries and help establish a useful identity which will serve as the startingpoint for the first proof.


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16 Evaluation of Eulers Zeta Series

3.2 Integration of Infinite Series

Several proofs in this chapter require that we swap an infinite sum andan integral. In other words, we want to know whether we can integratean infinite sum term by term and still arrive at the integral of the entiresum. Before we present that theorem we should reintroduce convergencesince we are now talking about infinite sums of functions, which are subtlydifferent then infinite sums of numbers. All of the definitions and theoremsin this section are discussed in more detail in [Kno].

Given a sequence of functions fn(x), we say that the infinite sum

n=1

fn(x)

ispoint-wise convergentto f(x)in a region D if given any >0 andx Dthere existsn0(which is dependent on and x) such that

|fn(x) f(x)| < for all n > n0We say that the infinite sum

n=1

fn(x)

isuniformly convergentto f(x)in a regionD if given any >0 there existsn0(which isonlydependent on) such that

|fn(x) f(x)| < for all n > n0for allxD .

As a side note, clearly uniform convergence is stronger than point-wiseconvergence since any function which converges uniformly obviously con-verges at each point.

Just like in the case of convergence, strictly using the definition to checkfor uniform convergence can be a daunting task, therefore mathematicianshave developed several tests to help spot and show uniform convergence.Here are two of these tests.

Theorem 3.2.1(Abels Test for Uniform Convergence). Given a series

f(x) =

n=1

un(x)

with un(x) =anfn(x). If


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Integration of Infinite Series 17

n=1anis convergent. fn+1(x) fn(x)for all n. fn(x)is bounded for x[a, b], meaning0 fn(x)M.

then n=1un(x)uniformly converges to f(x)in[a, b].

Theorem 3.2.2(Weierstrass Test for Uniform Convergence). Given a series

f(x) =

n=1

fn(x),

if there exists a positive n for each n such that fn(x)

n in the interval[a, b]

(same interval for all n) and n is convergent, then fn(x)is uniformly con-vergent in[a, b]. Essentially, this theorem says that if you can bound the sequenceof functions by a convergent sequence then the sequence of functions convergesuniformly.

Now that we have established convergence in terms of infinite sums offunctions and introduced some tests for convergence we can present a veryuseful theorem regarding the integration of these series.

Theorem 3.2.3. If a given infinite sum

f(x) =

n=1

fn(x)

is uniformly convergent in the region D and fn(x) is integrable in [a, b] (where[a, b] D) for all n N, then the integral of f(x)on [a, b]may be obtained bythe term by term integration of the series. In other words

ba

f(x)dx= b

a

n=1

fn(x)

dx=

n=1

ba

fn(x)dx

.


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3.3 Fourier Series

The theory and implementation of Fourier series is a very useful tool inmathematics. For the purposes of this thesis, this section will provide thebackground needed.

The whole idea behind Fourier series is to write a periodic functionf(x)as an infinite sum of sines and cosines:

f(x) = a0

2 +

n=1

(ancos(nx) +bnsin(nx)) , (3.1)

where the coefficients of the sines and cosines are given by Eulers formu-las:

an = 1

f(x) cos(nx) dx (3.2)

bn = 1

f(x) sin(nx) dx. (3.3)

The previous formulas can be derived by multiplying equation (3.1) byeither cos(mx)or sin(mx), integrating x over(,), and then using theorthogonality relations:

sin(nx) sin(mx) dx =

cos(nx) cos(mx) dx =

0 ifn=m ifn = m=0

and

sin(nx) cos(mx) dx = 0 for integralm, n.

Note that if f(x)is an odd function (meaning thatf(x) =f(x)) thenthe integrand of equation (3.2) is an odd function (becausef(x) cos(nx) =f(x) cos(nx)) which means that integrating over a symmetric intervalwill yield 0. Therefore an = 0 and thus the Fourier series of f(x) is asine series. By similar logic, if f(x) is an even function (meaning thatf(x) = f(x)) then its Fourier series consists only of cosines.

Now that we have defined what a Fourier series is, we are naturallydriven to ask which functions can be represented as Fourier series? Giventhat we would be representing the function as a sum of functions that all

have periods that divide 2, it is clear that f(x)must not only be periodic,but must also have a period that divides 2. Furthermore, in order forthe series to have any chance of converging, the sequence{ancos(nx) +bnsin(nx)} must tend to 0 as n goes to infinity. This last property is guar-anteed by the following lemma.


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Fourier Series 19

Lemma 3.3.1. If a function k(x), and its first derivative k(x) are sectionally

continuous in the interval[a, b], then the integral

K = b

ak(x) sin(x) dx

tends to zero as.A proof of this Lemma, as given by [CJ], is easily obtained through

integration by parts ofK. To clarify, by sectionally continuous we meanthat f(x) can be broken up into a finite number of subintervals such thatf(x)is continuous on the open subintervals and approaches definite limitsat the endpoints. This essentially means that f(x) can only have a finite

number of discontinuities on the interval [a, b]. If both f(x)and f(x)aresectionally continuous we say that f(x)issectionally smooth. In addition tothis we require that the value of f(x)at any discontinuity be the average ofthe limiting values from each side.

To summarize, a function f(x) can be represented as a Fourier seriesprovided the following properties hold:

f(x) = f(x+ 2) f(x)is sectionally smooth. At every pointx we have

f(x) = 12

limax+

f(a) + limax

f(a)

As an exercise we will now derive the product formula for sine, whichwill be useful later on in another proof. This derivation comes from [CJ].

To start off our derivation we will find the Fourier series of cos(x)onthe interval(,)where is a real number that is not an integer. Notethat all three properties are satisfied, so our quest for a Fourier series isfeasible. Since cos(x)is an even function we immediately know that ourFourier series will be a cosine series, thus bn= 0. Then using equation (3.2)


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we have

an = 1

cos(x) cos(nx)dx

= 1

0

cos(( +n)x) +cos(( n)x)dx

= 1

sin(( +n))

+n +

sin(( n)) n

= 1

( n) sin(( +n)) + ( +n) sin(( n))

2 n2

= 1

2 sin(x) cos(nx)

2 n2

= 2(1)n

(2 n2)sin()

Just to make the arithmetic clear, we used the trigonometric identities

sin(x y) = sin(x) cos(y) cos(x) sin(y) andcos(x y) = cos(x) cos(y) sin(x) sin(y)

in the previous series of steps. Now that we have found our coefficients,we have the Fourier series representation

cos(x) = sin()

+2 sin()

n=1

(

1)n

cos(nx)

2

n2.

Since our function satisfied the property that the value at every point (specif-ically the endpoints and ) was the average of the limits from bothsides, we can plugx = into our Fourier expression without any trouble.After doing this and dividing by sin(x) (which we can do since is notan integer) we get

cot() = 1

+

2

n=1

1

2 n2 .

Moving terms around yields

cot() 1

= 2

n=1

n2 2 . (3.4)

From here we can use Abels test to verify that the series is uniformly con-vergent, so that we can integrate the series term by term. We can write the


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Fourier Series 21

previous series as anfn()where

an= 1

n2 and fn() =

1 2n2

.

We know that an converges because an = (2) and in section 2.2 weshowed that(s)converges for all reals > 1. Inspection shows us that

fn+1() =

1 2(n+1)2

k2

42k2

u +u > k2

142k2

u +u 0 to be

(t) =

k=

ek2t. (4.1)

One of the most important theorems related to Jacobis theta function isthat the functiont1/4(t)is unchanged by the substitution of 1/tfor t. Inother words we have the following theorem.

Theorem 4.2.1.

1

t= t1/2(t) (4.2)

Before we begin to prove this theorem, we will prove a couple of smallfacts which will help us along the way. To prevent myself from reinventingthe wheel, I will take for granted that

limn

1+

x

n

n=ex.

Since ex is clearly continuous this tells us that the left hand side is alsocontinuous, and therefore given any sequence{xn} whose limit is x as ngoes to infinity we have

limn 1+

xn

n

n=e x. (4.3)

Make sure to remember this since we will be using it several times over thecourse of our proof of theorem 4.2.1.

Another small fact, which will provide the starting point for our proof,begins by lettingz C and lettingm be a positive integer. Using the bino-mial expansion formula and then substitutingk= m+j yields the follow-ing series of steps:

z1/2 +z1/2

2m=

2m

k=0

2m

k

zk/2z(2mk)/2

=m

j=m

2mm+j

z(m+j)/2z(2mmj)/2

=m

j=m

2mm+j

zj. (4.4)


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Jacobis Theta Function 37

We are now prepared to prove Theorem 4.2.1. The proof will be broken up

into three lemmas, the first of which proves an equality while the secondand third prove that the left hand side and right hand side of this equalityapproach the left and right hand sides of(4.2)respectively. All of the proofsof these lemmas were provided by [Sto].

Lemma 4.2.2. Let m and l be positive integers and let = e2i/l. Then

l2 n< l2

n/2 + n/2

2

2m=

m/l

k=m/l

l

22m

2mm+kl

.

Proof. Letm andl be positive integers and let = e2i/l. Note that

l =e2i =cos(2) +i sin(2) =1.

Now letz = n and sum equation (4.4) over l/2n


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38 Extension of the Zeta Function to the Complex Plane

Finally, since 0 n < l 1 implies that 1 n+1 < land forn = l 1 wehave

(n+1)j = l j =1 = (0)j

then(n+1)j for 0n


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Furthermore, the Maclaurin series for cosh gives us the approximation

cosh(x) =1+x2

2 + O(x4).

Utilizing this approximation yields

n/2 + n/2

2

2m=

1+

(in)2

2l2 + O

in

l

42m.

At this point we can simplify, sum both sides overl/2 n < l/2 andtake the limit asmgoes to infinity. Note that sincel =mt,mgoing toinfinity forcesl to go to infinity as well. Thus

limm l2 n< l2

n/2 + n/2

2

2m= lim

m l2 n< l2

1+

2n22l2

+ O

(in)4

l4

2m

= limm

n=

1+

2n22l2

+ O

(in)4

l4

2m.

Sincem l2/twe can rewrite the exponent and rearrange the terms inthe parentheses to obtain

limm l2 n< l2n/2 + n/2

2

2m

= limm

1

+

n2/t

2l2/t

2l2/t

.

We were able to drop the big-O term since it becomes insignificant aslgoesto infinity. Then, since 2l2/t asl , we can use equation (4.3) toconclude that the right hand side goes to en2/t asl andm go to infinity.Thus we have our intended result of

limm l2 n< l2

n/2 + n/2

2

2m=en

2/t =

1

t

. (4.8)

Lemma 4.2.4.

limm

m/l

k=m/l

mt22m

2mm+kl

=

t

n=

en2t =

t(t).


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Proof. Just as before,t is still fixed andl =mt. Therefore

mt tm.Thus

limm

m/l

k=m/l

mt22m

2mm+kl

=

t lim

m

m/l

k=m/l

m

22m

2mm+kl

.

If we focus on the summand of the right hand side we have

m

22m 2mm+kl =

m

22m(2m)!

(m+kl)!(m

kl)!

.

Using Stirlings Formula,

n!n

e

n 2n,

on all the factorial parts yields

m

22m

2m

e

2m 22m

e

m+kl

m+kl 12(m+kl)

e

m klmkl 1

2(m kl) .

After canceling out thees ands and 2s we are left with

m2m

(m+kl)m+kl (m kl)mkl m

m+kl

m kl . (4.9)

For the right term of (4.9) we have

mm+kl

m kl =

mm2 k2l2 =

11 k2l2/m2 1 as m.

For the left term of (4.9) we have

m2m

(m+kl)m+kl (m

kl)mkl

= m2m

(m2

k2l2)m

(m kl)kl

(m+kl)kl. (4.10)

Now, when we dissect the right hand side of (4.10), the left term yields

m2m

(m2 k2l2)m = 1

(1 k2l2/m2)m 1

(1 k2t/m)m


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because l2 mt as l . Furthermore we can use equation (4.3) toobtain

m2m

(m2 k2l2)m 1

(1+ (k2t)/m)m 1

ek2t as m.

Now we can focus on the right term of (4.10). Recalling that the Maclaurinseries of(1 x)/(1+x)yields the approximation

1 x1+x

=1 2x+ O(x2),

we can then apply this to the right term of (4.10) to get

(m kl)kl(m+kl)kl

=

1 lk/m1+lk/m

lk

=

1 2lk

m + O

lk

m

2

lk

.

Using the fact thatl/mt/lgives us

(m kl)kl(m+kl)kl

=

1 2tk

2

lk + O

tk

l

2lk

1 2tk2

lk

lk.

We can drop the big-O because it is insignificant as l . Finally usingequation (4.3) yet again will yield

1+

2tk2lk

lk e2k2t as m

becauselk asl . To summarize everything we have done in thisproof:

limm

m/l

k=m/l

mt22m

2mm+kl


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=

t limm

m/l

k=m/l

m22m

2mm+kl

=

t

m=

m2m

(m+kl)m+kl (m kl)mkl

mm+kl

m kl

=

t

m=

m2m

(m2 k2l2)m(m kl)kl(m+kl)kl

[1]

=

t

m=

1

ek2t e2k2t

[1]

= t

m= ek

2

t

=

t(t).

Thus we have proven lemma 4.2.4.

Finally, utilizing lemmas 4.2.2, 4.2.3 and 4.2.4 we have that

1

t

= lim

m l2 n< l2

n/2 + n/2

2

2m

= limm

m/l

k=m/l mt

22m 2mm+kl =

t(t)

which proves theorem 4.2.1.


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The Functional Equation for(s) 43

4.3 The Functional Equation for(s)

Theorem 4.3.1(The Functional Equation for(s)).

(1s)/2((1 s)/2)(1 s) =s/2(s/2)(s)In other words if we let (s) =s/2(s/2)(s)then

(1 s) = (s).Proof. Consider the following integral

0

en2tts/2

dt

t

.

Using the change of variablesx = n2t(and thus dx/x= dt/t), wherenisan integer, we have

0

en2tts/2

dt

t =

0

exs/2nsxs/21dx

x

= 1

nss/2

0

ex xs/21dx

= 1

nss/2(s/2).

Let

(s) > 0, and then sum both sides fromn = 1 to to get

(s) =(s)s/2(s/2) =

n=1

1

nss/2(s/2) =

n=1

0

en2tts/2

dt

t .

To keep from getting sidetracked we will simply switch the order of thesum and the integral while noting that to prove this is legal is not a trivialprocedure. Thus we have

(s) =

0f(t)ts/2

dt

t where f(t) =

n=1

en2t. (4.11)

Recall Jacobis theta function (section 4.2) and the its corresponding trans-


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formation invariant, theorem 4.2.1. Using those facts, we have

1+ 2f(t) = 1+ 2

n=1

en2t

= e(0)2t +

n=1

en2t +

1

n=

en2t

=

n=

en2t

= (t)

= t1/2(t1)= t1/2(1+ 2f(t1)).

Then solving for f(t)in the first and last lines yields

f(t) =1

2(t1/2 1) +t1/2f(t1). (4.12)

Returning to (s)and using equation (4.12) we have

(s) =

0f(t)ts/2

dt

t

= 1

0f(t)ts/2

dt

t +

1

f(t)ts/2dt

t

=

1

0

12

(t1/2 1) +t1/2f(t1)

ts/2 dtt

+

1f(t)ts/2 dt

t

= 1

0

1

2(ts/23/2 ts/21)dt+

10

t1/2f(t1)ts/2dt

t +

1

f(t)ts/2dt

t .

Integrating the first term yields

10

1

2(ts/23/2 ts/21)dt= 1

2

ts/21/2

s/2 1/2ts/2

s/2

t=1

t=0

= 1

s 11

s.

(4.13)Using the change of variables x = 1/t (and thus dx/x =

dt/t) on the

second term results in 10

t(s1)/2f(t1)dt

t =

1

x(s1)/2f(x)dx

x =

1

x(1s)/2f(x)dx

x .

(4.14)


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The Functional Equation for(s) 45

Then after substitutingt for x in this, we can recombine and rearrange ev-

erything to obtain

(s) = 1

s 11

s+

1

t(1s)/2f(t)dt

t +

1

f(t)ts/2dt

t

= 1

s(s 1)+

1(t(1s)/2 +ts/2)f(t)

dt

t .

Inspection shows us that in fact (1 s) = (s), which finishes the proof.


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4.4 Values of Riemanns Zeta Function at the Negative

Integers

Now that we have established the functional equation, we can use it toevaluate the Riemann zeta function in areas that Eulers zeta series was notdefined, for example the negative integers.

4.4.1 Negative Even Integers

The functional equation we proved has the form

(1s)/2((1 s)/2)(1 s) =s/2(s/2)(s),

however, we can use identities of the -function to rewrite this in anotherform. After lots of algebraic manipulation using properties (2.9), (2.12) and(2.13) we eventually come to the equivalent functional form

(s) = (s)(2)s12sin(s/2)(1 s).In this form we can clearly see that Riemanns zeta function is zero at thenegative even integers because of the sine term. These zeros at

s=2, 4, 6 , . . .are known as the trivial zeros of the zeta function.

4.4.2 Negative Odd Integers

We can arrive at a formula for the values of Riemanns zeta function atthe negative odd integers by simply combining the formula we derived insection 3.5 and the functional equation. To review, for k 1

(2k) =

n=1

1

n2k =

(1)k1B2k(2)2k2(2k)!

and the functional equation tells us that

(1s)/2((1

s)/2)(1

s) =s/2(s/2)(s).

Therefore, fork 1(1(2k))/2((1 (2k))/2)(1 (2k)) = (2k)/2((2k)/2)(2k)

1/2+k(1/2 k)(1 2k) = k(k)(2k).


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Values of Riemanns Zeta Function at the Negative Integers 47

After substituting (k) = k! (since kis a positive integer) and collecting

terms on one side we have

(1 2k) = k!2k1/2(1/2 k) (2k)

= k!

2k1/2(1/2 k)(1)k1B2k(2)2k

2(2k)!

= (1)k1B2k22k1k!1/2(2k)!(1/2 k) .

Which finally gives us the formula for the values of the Riemann zeta func-tion at the odd integers:

(1 2k) = (1)k1B2k22k1k!

1/2(2k)!(1/2 k) .


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Chapter 5

The Riemann Zeta Hypothesisand the Consequences of itsTruth

5.1 The Riemann Zeta Hypothesis

First we define

Z={s C | (s) =0 and 0 (s) 1}.

In words,Z is the set of nontrivial zeros of the Riemann zeta function. An-other helpful definition is that whenever we mentionor t we are talkingabout the real and complex components of the variable plugged into (s).Meaing that ifs C thens = +it. The classic Riemann zeta hypothesisthen becomes:

Z {s C | (s) =1/2},which is sometimes also seen as Z {s C | = 1/2}. As a side note,another useful convention is that usually represents a nontrivial zero of(s).

There are also other, though more unnatural, ways to express this hy-pothesis. For example, the following two hypotheses, that were supplied

by S. J. Patterson, are equivalent to Riemanns zeta hypothesis.

(i) (s)/(s) + (s 1)1 isholomorphicin {s C | (s) > 1/2}.(ii) log{(s 1)(s)} isholomorphicin {s C | (s) > 1/2}.


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50 The Riemann Zeta Hypothesis and the Consequences of its Truth

Just to be clear, holomorphic is just another way to say that the function is

analytic, which simply means that the function is complex differentiable atevery point in the specified region.

5.2 The Prime Number Theorem

Now that we have clarified what the Riemann zeta hypothesis is, we canstart looking at some of the numerous results that pop up if the hypothesisis assumed to be true. The following two theorems, that are proved inPatterson, create a direct link between the Riemann zeta hypothesis andthe distribution of the primes.

Theorem 5.2.1. Suppose there exists < 1such that

Z {s C | 1 (s)}.Then as X

nX

(n) =X+ O(X(log X)2).

Theorem 5.2.2. Suppose there exists an < 1such that as X

nX

(n) =X+ O(X)

then Z {s C | (s)}.Just to clarify, (n)is Mangoldts function and is defined as

(n) =

logp ifn = pk for some prime p and some integralk> 00 otherwise

.

Currently, these theorems have not been terribly useful because no one hasfound either aorthat satisfies them. However when combined with thevalidity of the Riemann hypothesis they say a lot regarding the theory sur-rounding the prime number theorem. Although it is obvious that a proof of

the prime number theorem does not require the Riemann zeta hypothesisto be true (because if it did we wouldnt have a proof of the prime num-ber theorem) it does rely on a small part of the hypothesis, mainly that if is a nontrivial zero of(s) , then Re()= 1 However, this does not atall say that a proof of Riemanns zeta hypothesis would not improve the


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Additional Consequences 51

prime number theorem. The prime number theorem is usually stated in

two equivalent forms, the first is more intuitive while the second is moreuseful for the purposes of this section. The following two theorems comefrom [Pat].

Theorem 5.2.3 (Intuitive Prime Number Theorem). There exists a constantc > 0such that

(X) = X

2(log u)1 du + O(Xec(log X)

1/2).

Theorem 5.2.4(Useful Prime Number Theorem). There exists a constant c >0so that as X

n 0is not the ordinate (where theordinateof a pointz is the imaginary compo-nent ofz) of a zero, let

S(T) =1arg(1/2+iT)


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52 The Riemann Zeta Hypothesis and the Consequences of its Truth

obtained by continuous variation along the straight lines joining 2, 2+iT,

and 1/2+iTstarting at 0. IfT> 0 is an ordinate of a zero then let

S(T) =S(T+ 0).

The Riemann hypothesis provides many characteristics about S(T) and avariant S1(T) such as big-O and little-o approximations and mean valuetheorems for them.

It turns out that proving the convergence of

n=1

(n)

ns (s > 1/2)

is not only a necessary condition for the Riemann hypothesis, but also asufficient one. Recall that ifn= p11 p

22 pkk is the prime factorization ofn

where the pis are distinct primes and alli >0, then the Mobius functionis defined as

(n) =

1 ifn = 1(1)k if1 = 2= = k= 10 otherwise

.

Be careful not to confuse this function with ()which I will discuss in thenext section.

There are also some interesting results surrounding the function

M(x) = nx

(n).

Without any help from the Riemann hypothesis we have

M(x) =O

x1/2

A

log x

loglog x

.

However, showing that

M(x) =O(x12+)

for some > 0 is a necessary and sufficient condition for the Riemann

hypothesis.In addition to this we have the Mertens hypothesis, which is that

|M(x)| < n (n > 1).


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Lindelofs Hypothesis 53

It turns out that although this is not implied by the Riemann hypothesis, it

does itself imply the Riemann hypothesis.Titchmarsh notes that the weaker hypothesis,M(x) =O(x1/2), is eerily

similar to the function

(x) x= nx

(n) x,

because the Riemann hypothesis does imply that(x) x= O(x 12+).Now consider the function

F(x) =

k=1

(1)k+1xk(k 1)!(2k) .

On its own we can show that F(x) = O(x12+). However showing that

F(x) = O(x14+) is another necessary and sufficient condition on the Rie-

mann hypothesis.You can also derive yet another necessary and sufficient condition on

the Riemann zeta hypothesis through the clever use of Farey series [Tit].

5.4 Lindel ofs Hypothesis

Lindelofs hypothesis is a consequence of the Riemann zeta hypothesishowever, the effects of its veracity are much more accessible and far reach-

ing and thus it merits it own section.Before we get to the hypothesis we must define a new function. Let

() =inf{aR | (+it) =O(|t|a) ast }.In other words,|t|() is the closest upper bound we can get to the zetafunction restricted to the line{c C | (c) = }. To get a feeling for thiswe can figure out() for select areas right off the bat. For example, weknow that(+it)is bounded for all > 1. Since it is bounded we knowthat(+it) =O(|t|)for all > 0, therefore() =0 for> 1. We canthen use this result in conjunction with the functional equation to arrive at

() =

1

2 for 0.

The fact that the Riemann zeta hypothesis implies the Lindelof hypoth-esis is probably not easily recognizable in this form. However the followinghypothesis is equivalent to that of Lindelof. First we will define for T>0and> 1/2

N(, T) =Card{Z | () > and 0 < () < T}.ThusN(, T)is the number of non-trivial zeroes in the rectangle boundedby and 1 on the real axis, and 0 and T on the complex axis. Then the

Lindelof hypothesis is the same as saying for all> 1/2

N(, T+ 1) N(, T) =o(log T) as T .With this form, we can see that the Riemann hypothesis forces N(, T) =0for allTand > 1/2 and thus would imply the Lindelof hypothesis.

Hardy and Littlewood continue the work by showing even more state-ments that are equivalent to this hypothesis.

(i) For allk 1 and > 0

T1 T

0 1

2+it

k

dt =O(T).

(ii) For all> 1/2,k 1 and > 0

T1 T

0|(+it)|kdt =O(T) asT .


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Lindelofs Hypothesis 55

There are 3 more statements by Hardy and Littlewood that are equivalent,

however since they deal with Plitzs generalization of Dirichlets divisorproblem (which extends far beyond the scope of this thesis) they will notbe discussed [Tit].


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Bibliography

[Apo] Tom M. Apostol. Introduction to Analytic Number Theory. Springer-Verlag, 1976.

[BBC] Jonathan M. Borwein, David M. Bradley, and Richard E. Crandall.Computational strategies for the riemann zeta function. J. Comput.Appl. Math., 121(1-2):247296, 2000. Numerical analysis in the 20thcentury, Vol. I, Approximation theory.

[CJ] Richard Courant and Fritz John.Introduction to Calculus and Analysis,volume 1. Interscience Publishers, 1965.

[Dun] William Dunham. Euler: The Master of Us All. The MathematicalAssociation of America, 1999.

[Edw] H. M. Edwards. Riemanns Zeta Function. Dover Publications Inc.,

1974.[IK] Henryk Iwaniec and Emmanuel Kowalski.Analytic Number Theory.

The American Mathematical Society, 2004.

[Ivi] Aleksandar Ivic. The Riemann Zeta-Function: The Theory of the Rie-mann Zeta-Function wih Applications. John Wiley & Sons, 1985.

[Kno] Konrad Knopp.Theory and Application of Inifinite Series. Dover Pub-lications Inc., second edition, 1990.

[LR] Norman Levinson and Raymond M. Redheffer. Complex Variables.Holden-Day Inc., 1970.

[Pat] S. J. Patterson. An Introduction to the Theory of the Riemann Zeta-Function. Cambridge University Press, 1995.

[Sto] Jeffrey Stopple.A Primer of Analytic Number Theory : From Pythagorasto Riemann. Cambridge University Press, 2003.


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58 Bibliography

[Tit] E. C. Titchmarsh.The Theory of the Riemann Zeta Function. Clarendon

Press, second edition, 1986.

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