The Riemann Hypothesis 9780883859896

The Riemann hypothesis concerns the prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 ...Ubiquitous a

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The Riemann Hypothesis
 9780883859896

Table of contents :
Contents......Page 8
Preface......Page 10
1.1 Primes as elementary building blocks......Page 14
1.2 Counting primes......Page 16
1.3 Using the logarithm to count powers......Page 20
1.4 Approximations for π(x)......Page 22
1.6 Counting prime powers logarithmically......Page 24
1.7 The Riemann hypothesis—a look ahead......Page 27
1.8 Additional exercises......Page 29
2.1 Infinite sums......Page 34
2.2 Series for well-known functions......Page 39
2.3 Computation of ζ(2)......Page 42
2.4 Euler’s product formula......Page 45
2.6 Additional exercises......Page 47
3.1 Euler’s discovery of the product formula......Page 54
3.2 Extending the domain of the zeta function......Page 56
3.3 A crash course on complex numbers......Page 58
3.4 Complex functions and powers......Page 60
3.5 The complex zeta function......Page 63
3.6 The zeroes of the zeta function......Page 64
3.7 The hunt for zeta zeroes......Page 67
3.8 Additional exercises......Page 68
4 Primes and the Riemann hypothesis......Page 72
4.1 Riemann’s functional equation......Page 73
4.2 The zeroes of the zeta function......Page 76
4.3 The explicit formula for ψ(x)......Page 79
4.4 Pairing up the non-trivial zeroes......Page 82
4.5 The prime number theorem......Page 85
4.6 A proof of the prime number theorem......Page 86
4.7 The music of the primes......Page 89
4.8 Looking back......Page 91
4.9 Additional exercises......Page 94
Appendix A. Why big primes are useful......Page 100
Appendix B. Computer support......Page 104
Appendix C. Further reading and internet surfing......Page 112
Appendix D. Solutions to the exercises......Page 114
Index......Page 156

Citation preview

The Riemann Hypothesis A Million Dollar Problem

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The half title page figure shows the graph of the function 1 f (y) = |ζ ( + i y)| 2 for 0 ≤ y ≤ 32 against a background in which the primes less than 100 are marked (see also figure 3.7 on page 54).

Original title: De Riemann-hypothese – Een miljoenenprobleem Epsilon Uitgaven, Utrecht, 2011 Translated by the authors. c 2015 by  The Mathematical Association of America (Incorporated) Library of Congress Catalog Card Number 2015959833 Print edition ISBN 978-0-88385-650-5 Electronic edition ISBN 978-0-88385-989-6 Printed in the United States of America Current Printing (last digit): 10 9 8 7 6 5 4 3 2 1

The Riemann Hypothesis A Million Dollar Problem

Roland van der Veen Leiden University and

Jan van de Craats University of Amsterdam

Published and Distributed by The Mathematical Association of America

Council on Publications and Communications Jennifer J. Quinn, Chair Committee on Books Fernando Gouvˆea, Chair Anneli Lax New Mathematical Library Editorial Board Karen Saxe, Editor Timothy G. Feeman John H. McCleary Katharine Ott Katherine S. Socha James S. Tanton Jennifer M Wilson

ANNELI LAX NEW MATHEMATICAL LIBRARY 1. 2. 3. 4. 5.

Numbers: Rational and Irrational by Ivan Niven What is Calculus About? by W. W. Sawyer An Introduction to Inequalities by E. F. Beckenbach and R. Bellman Geometric Inequalities by N. D. Kazarinoff The Contest Problem Book I Annual High School Mathematics Examinations 1950–1960. Compiled and with solutions by Charles T. Salkind 6. The Lore of Large Numbers by P. J. Davis 7. Uses of Infinity by Leo Zippin 8. Geometric Transformations I by I. M. Yaglom, translated by A. Shields 9. Continued Fractions by Carl D. Olds 10.  Replaced by NML-34 11. Hungarian Problem Books I and II, Based on the E¨otv¨os Competitions 12. 1894–1905 and 1906–1928, translated by E. Rapaport 13. Episodes from the Early History of Mathematics by A. Aaboe 14. Groups and Their Graphs by E. Grossman and W. Magnus 15. The Mathematics of Choice by Ivan Niven 16. From Pythagoras to Einstein by K. O. Friedrichs 17. The Contest Problem Book II Annual High School Mathematics Examinations 1961–1965. Compiled and with solutions by Charles T. Salkind 18. First Concepts of Topology by W. G. Chinn and N. E. Steenrod 19. Geometry Revisited by H. S. M. Coxeter and S. L. Greitzer 20. Invitation to Number Theory by Oystein Ore 21. Geometric Transformations II by I. M. Yaglom, translated by A. Shields 22. Elementary Cryptanalysis by Abraham Sinkov, revised and updated by Todd Feil 23. Ingenuity in Mathematics by Ross Honsberger 24. Geometric Transformations III by I. M. Yaglom, translated by A. Shenitzer 25. The Contest Problem Book III Annual High School Mathematics Examinations 1966–1972. Compiled and with solutions by C. T. Salkind and J. M. Earl 26. Mathematical Methods in Science by George P´olya 27. International Mathematical Olympiads—1959–1977. Compiled and with solutions by S. L. Greitzer 28. The Mathematics of Games and Gambling, Second Edition by Edward W. Packel 29. The Contest Problem Book IV Annual High School Mathematics Examinations 1973–1982. Compiled and with solutions by R. A. Artino, A. M. Gaglione, and N. Shell 30. The Role of Mathematics in Science by M. M. Schiffer and L. Bowden 31. International Mathematical Olympiads 1978–1985 and forty supplementary problems. Compiled and with solutions by Murray S. Klamkin 32. Riddles of the Sphinx by Martin Gardner 33. U.S.A. Mathematical Olympiads 1972–1986. Compiled and with solutions by Murray S. Klamkin 34. Graphs and Their Uses by Oystein Ore. Revised and updated by Robin J. Wilson

35. Exploring Mathematics with Your Computer by Arthur Engel 36. Game Theory and Strategy by Philip D. Straffin, Jr. 37. Episodes in Nineteenth and Twentieth Century Euclidean Geometry by Ross Honsberger 38. The Contest Problem Book V American High School Mathematics Examinations and American Invitational Mathematics Examinations 1983–1988. Compiled and augmented by George Berzsenyi and Stephen B. Maurer 39. Over and Over Again by Gengzhe Chang and Thomas W. Sederberg 40. The Contest Problem Book VI American High School Mathematics Examinations 1989–1994. Compiled and augmented by Leo J. Schneider 41. The Geometry of Numbers by C. D. Olds, Anneli Lax, and Giuliana P. Davidoff 42. Hungarian Problem Book III, Based on the E¨otv¨os Competitions 1929–1943, translated by Andy Liu 43. Mathematical Miniatures by Svetoslav Savchev and Titu Andreescu 44. Geometric Transformations IV by I. M. Yaglom, translated by A. Shenitzer 45. When Life is Linear: from computer graphics to bracketology by Tim Chartier 46. The Riemann Hypothesis: A Million Dollar Problem by Roland van der Veen and Jan van de Craats Other titles in preparation.

MAA Service Center P.O. Box 91112 Washington, DC 20090-1112 1-800-331-1MAA FAX: 1-240-396-5647

Contents Preface 1 Prime numbers 1.1 Primes as elementary building blocks . . . . . . . . . . . . . . . . . . . . . 1.2 Counting primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Using the logarithm to count powers . . . . . . . . . . . . . . . . . . . . . 1.4 Approximations for π(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 The prime number theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Counting prime powers logarithmically . . . . . . . . . . . . . . . . . . . 1.7 The Riemann hypothesis—a look ahead . . . . . . . . . . . . . . . . . . 1.8 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 The zeta function 2.1 Infinite sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Series for well-known functions . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Computation of ζ (2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Euler’s product formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Looking back and a glimpse of what is to come . . . . . . . . . . . 2.6 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix 1 1 3 7 9 11 11 14 16 21 21 26 29 32 34 34

3 The Riemann hypothesis 3.1 Euler’s discovery of the product formula . . . . . . . . . . . . . . . . . . 3.2 Extending the domain of the zeta function . . . . . . . . . . . . . . . . 3.3 A crash course on complex numbers . . . . . . . . . . . . . . . . . . . . . 3.4 Complex functions and powers . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 The complex zeta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 The zeroes of the zeta function. . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 The hunt for zeta zeroes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41 41 43 45 47 50 51 54 55

4 Primes and the Riemann hypothesis 4.1 Riemann’s functional equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The zeroes of the zeta function. . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 The explicit formula for ψ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Pairing up the non-trivial zeroes . . . . . . . . . . . . . . . . . . . . . . . . .

59 60 63 66 69 vii

viii

Contents

4.5 The prime number theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 A proof of the prime number theorem . . . . . . . . . . . . . . . . . . . . 4.7 The music of the primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Looking back . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Additional exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix A. Why big primes are useful

72 73 76 78 81 87

Appendix B. Computer support Appendix C. Further reading and internet surfing

91 99

Appendix D. Solutions to the exercises Index

101 143

Preface Mathematics is full of unsolved problems and other mysteries, but none more important and intriguing than the Riemann hypothesis. Baffling the greatest minds for more than a hundred and fifty years, the Riemann hypothesis is at the very core of mathematics. A proof of it would mean an enormous advance. In addition, the Riemann hypothesis was chosen as one of the seven Millennium Problems1 by the Clay Mathematics Institute. This means that proving the Riemann hypothesis will not only make you world famous, but also earns you a one million dollar prize. The Riemann hypothesis concerns the prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, . . . i.e., the integer numbers greater than 1 that are not divisible by any smaller number (except 1). Ubiquitous and fundamental in mathematics as they are, it is important and interesting to know as much as possible about these numbers. Simple questions would be: how are the prime numbers distributed among the positive integers? How many prime numbers are there? What is the number of prime numbers of one hundred digits? Of one thousand digits? These questions were the starting point of a groundbreaking paper by Bernhard Riemann written in 1859. As an aside in his article Riemann formulated his now famous hypothesis, that so far nobody has come close to proving: The Riemann Hypothesis. All nontrivial zeroes of the zeta function lie on the critical line. Hidden behind this at first mysterious statement, lies a whole mathematical universe of prime numbers, infinite sequences, infinite products and complex functions. The present book is a first exploration of this fascinating world.

1

see http://www.claymath.org/millennium/Riemann Hypothesis/

ix

x

Preface

The Riemann hypothesis as an online course The Riemann hypothesis was the subject of a web class, a four-week online course organized four times in the years 2006 to 2010 by the KortewegDe Vries Institute for Mathematics at the University of Amsterdam. It was aimed at mathematically talented secondary school students. This book originated from the course material. The goal of these courses was not to solve the Riemann hypothesis, but to introduce interested and talented students to challenging university level mathematics. At the end of the course the participants had a good idea of what the Riemann hypothesis is and why it is such an important and interesting problem in mathematics. After attending the web class, many of them decided upon university study in mathematics. A web class at the University of Amsterdam is a four-week course in which secondary school students from all over the country work at home and communicate with each other and with university staff via the internet. They study the course material, solve the exercises, send their solutions to the staff and get feedback within a few days. The four chapters of this book correspond to the material of the four weeks of the web class, one for every week. For some of the exercises the participants could use computer algebra worksheets made available by the university. The readers of this book may use free mathematical software on the internet instead. In the appendix Computer support on page 91 we give suggestions and instructions. During the web class there was much interest in the material by mathematics teachers, secondary school students and undergraduate university students. Therefore we decided to make it available as a book, including full solutions to all exercises, and extended with quite a few challenging additional exercises at the end of each chapter. The latter are somewhat more difficult and may be skipped on first reading. To maximize accessibility, we have limited the prerequisites for the text to a minimal amount of elementary calculus, excluding integration.

How to read a mathematics text? Mathematics texts are usually written in a very concise manner: each and every detail counts. This means that they can never be read both fast and thoroughly at the same time. Fast reading is possible, but only to get a first impression. Thereafter you should reread, focussing on more details. And

Preface

xi

then again and again, if necessary. If you get stuck, our advice is simple: read on and come back later. The exercises are the ultimate test: if you can solve them, you understand the matter. Do not look at the solution before trying to solve an exercise on your own, but afterwards always compare your solution with the one given at the end of the book.

1 Prime numbers How are the prime numbers distributed among the other numbers? How many primes are there? What is the number of primes of one hundred digits? Of one thousand digits? These questions were the starting point of a seminal paper by Bernhard Riemann (1826–1866) written in 1859. As an aside in his article Riemann formulated his now famous hypothesis that so far nobody has come close to proving. In this first chapter you will get to know the primes, two distinct functions that count primes, and various approximations of these functions. The natural logarithm plays an important role. At the end of the chapter the Riemann hypothesis itself will make its first appearance.

1.1 Primes as elementary building blocks Counting is more than twenty thousand years old. Long before written language was invented, people already were tallying with notches on pieces of bone1 . Counting and arithmetic are arguably the oldest concepts in mathematics and yet numbers still exhibit mysterious patterns that we do not fully understand. For example, if you ask how numbers can be constructed by multiplication alone, you quickly hit upon one of the biggest riddles of mathematics. Using a single number as a building block, you can never construct all numbers by multiplication only. Take 2 for example. All you can get is 2 itself, 2 × 2 = 4, 2 × 2 × 2 = 8, 16, 32, . . . and the other powers of 2. If you add 3 as a building block, then you can also construct 3, 2 × 3 = 6, 9, 12, . . . . However, the number 5 is still out of reach. By adding 5 to the list it is possible to construct all marked numbers in table 1.1. 1

The so-called Ishango bone, see, e.g., nrich.maths.org/6013

1

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1. Prime numbers

10 20 30 40 50 60 70 80 90

1 11 21 31 41 51 61 71 81 91

2 12 22 32 42 52 62 72 82 92

3 13 23 33 43 53 63 73 83 93

4 14 24 34 44 54 64 74 84 94

5 15 25 35 45 55 65 75 85 95

6 16 26 36 46 56 66 76 86 96

7 17 27 37 47 57 67 77 87 97

8 18 28 38 48 58 68 78 88 98

9 19 29 39 49 59 69 79 89 99

Table 1.1. The numbers 2, 3 and 5 as building blocks for multiplication. Adding 7, the first number not yet constructed, as a building block, more numbers come within reach, but 11 and 13 still elude us. Will this process finally come to an end, or are we forced to add ever more building blocks to construct all numbers by multiplication? The ancient Greeks already knew the answer to this question: there is no end to the list of building blocks. The reason is that at any given moment there are special numbers that clearly cannot be constructed with the building blocks you have at that time. For example, suppose your building blocks are just 2, 3, 5 and 7. The special number that will certainly be out of reach with these blocks is 211 = 2 × 3 × 5 × 7 + 1. Why? Well, first of all it is not divisible by any of the building blocks since division by 2, 3, 5 or 7 leaves a remainder of 1. However, all constructible numbers must be divisible by at least one of the four building blocks. The numbers 2, 3, 5, 7, 11, . . . are called the prime numbers (or primes, for short). They are the elementary building blocks from which all numbers can be constructed by multiplication. A prime cannot be divisible by any smaller number except 1, since otherwise it would not be necessary as a building block. The argument in the previous paragraph shows that the list of prime numbers never ends. In other words: there are infinitely many prime numbers. The number 1 is a bit different. It is of little use as a building block since 1 × n = n for all numbers n. This, among other things, causes mathematicians not to include 1 in the list of primes. Exercise 1.1. a. Using Table 1.1 find all primes smaller than 100. b. What numbers smaller than 100 can you construct by multiplication, using only the numbers 3 and 8 as building blocks?

1.2. Counting primes

3

c. Is it possible to construct 103 = 2 × 3 × 17 + 1 using only 2, 3 and 17? What about 104? d. Is the number 2 × 3 × 5 × 7 × 11 + 1 prime? And what about the number 2 × 3 × 5 × 7 × 11 × 13 + 1?

1.2 Counting primes How are the primes distributed among the other numbers? That is what this book is all about. We already know that there are infinitely many primes, but that does not tell much about their distribution. For some other infinite sequences of numbers it is easy to see how they are distributed. The odd numbers for example. Although there are infinitely many of them, we know exactly how they are distributed among the other numbers: every other number is odd. With the primes things are more subtle. The sequence of prime numbers is not caught as easily in a regular pattern or in a formula. It is easy to say what the thousandth odd number is (as an aside: what is the thousandth odd number?), but it is another matter if you want to know the thousandth prime. All you can do is run through the sequence of primes one by one. Nobody can predict what the next prime will be. And yet we will see that there exists a mysterious sense of order in the seemingly erratic sequence of primes. This combination of unpredictability and regularity is especially striking when you count the number of primes below a certain bound. From exercise 1.1 we know that there are exactly 25 primes below 100, but how many are there below 200? And how many primes are less than 1000? To study this question systematically we denote the number of primes less than or equal to x by π (x). For example, π(100) = 25 because there are exactly 25 primes less than or equal to 100; π(2) = 1 since there is only a single prime less than or equal to 2, the number 2 itself. π(10) = 4, because there are four prime numbers below 10; these are 2, 3, 5 and 7. Finally π (17.351) = 7, since there are seven prime numbers less than or equal to 17.351, namely 2, 3, 5, 7, 11, 13 and 17. The function π(x) is called the prime counting function because it counts the number of primes in a certain sense. It is a function that is defined on all real numbers (not just on the natural numbers). The graph of π (x) is a sort of staircase that makes a jump of 1 above every prime number. In this chapter we will also meet other counting functions. Be careful: the notation π(x) has nothing to do with the famous number π = 3.1415926 . . . . In mathematics the same symbol sometimes has several different meanings. The intended meaning is usually clear from the context.

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1. Prime numbers 25 20 15 10 5 0

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Figure 1.1. The prime counting function π (x). Although there is no easy formula for the function π(x), it is still possible to graph it and gain valuable information on the prime numbers. Figure 1.1 shows a graph of π(x) for 0 ≤ x ≤ 100. Using the computer it is easy to further examine the graph of π(x). This is the subject of the next exercise. For suggestions on how to use the computer to complete this and future exercises, see the appendix Computer support on page 91. Exercise 1.2. a. Using the computer, examine graphs of π(x) on various domains. First take 0 ≤ x ≤ 100, as in Figure 1.1. This graph should also be consistent with the outcome of exercise 1.1.a. Check this! b. Also try the domains 0 ≤ x ≤ 1 000, 0 ≤ x ≤ 10 000, . . . , 0 ≤ x ≤ 1 000 000. c. Find the smallest prime number greater than 1 000 000. d. What is the ten-thousandth prime number? From nearby, the graph of π(x) may look like a chaotic staircase, but by choosing larger and larger domains you will have noticed a beautiful regularity. The smooth, almost straight graph contains many of the secrets of the primes. To unlock these secrets we will try to find formulas to describe the graph. Is there a simple function whose graph looks like that of π (x), at least when viewed from a distance? Here are some experiments using the computer to find approximations of π (x). The approximating functions we tried are linear functions of the

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Figure 1.2. The prime counting function π (x) and the approximations 0.25 x and x 0.7 on the interval [0, 100] (top) and on the interval [0, 400] (bottom).

form cx and power functions of the form x a . In Figure 1.2 you see the graph of π(x) together with the approximations 0.25 x and x 0.7 . On the interval [0, 100] both approximations look pretty good but on a larger interval they fail completely. The first graph increases way too fast and the second approximation does not grow quickly enough. Figure 1.3 shows our second attempt at approximation. On the large interval [0, 10000] the functions 0.125 x and x 0.773 seem good approximations but again on the larger interval [0, 40000] they fail miserably. What is striking in Figure 1.3 is that π(x) now seems just as smooth as the other two graphs. The graph of π(x) increases but the growth is slower and slower. This is not strange, since the bigger a number the more possible

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1. Prime numbers

1200 1000 800 600 400 200 0

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10000

5000 4000 3000 2000 1000 0

10000

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Figure 1.3. The prime counting function π (x) and the approximations 0.125 x and x 0.773 on the interval [0, 10000] (top) and on the interval [0, 40000] (bottom).

divisors there are. In that sense it becomes harder and harder for a number to be prime. This suggests that no linear function will be a good approximation of π (x). The power functions at first sight seem to be doing a better job, but they also fail when we pass to larger and larger domains. Exercise 1.3. a. Using the computer, you can check that π(40 000) = 4203. Find functions f (x) = cx and g(x) = x a that satisfy f (40 000) = g(40 000) = 4203. Find the constants c and a up to three decimals accuracy. b. On the interval [0, 40 000] the functions f (x) and g(x) approximate π (x) reasonably well, but they will fail on a larger interval. Check this by comparing f (100 000) and g(100 000) with π (100 000) = 9592.

7

1.3. Using the logarithm to count powers

1.3 Using the logarithm to count powers We are looking for functions that approximate π(x) reasonably well on every domain, because only such functions will genuinely tell us something about the distribution of the primes. Before being more explicit about what we mean by a reasonable approximation, let us first consider a simpler example. We will look at a different counting function and will see that it is approximated very well by a logarithm. Instead of counting the number of primes less than or equal to x we can also count how many powers of 2 there are below this bound x. This provides us with a new counting function that we call T2 (x). By powers of 2 we mean the numbers 21 = 2, 22 = 4, 23 = 8, . . . and so on. For example, we have T2 (1) = 0, T2 (3) = 1 and T2 (99) = 6. Figure 1.4 shows the graph of the power counting function T2 (x) together with a graph of the function f (x) = log2 x (logarithm to the base 2). The graph of the logarithm intersects the corners of the stairs exactly (do you see why?). Since every step of the staircase has height 1, we see that 0 ≤ f (x) − T2 (x) < 1 for all x ≥ 1. 6 5 4 3 2 1 0

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x

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100

Figure 1.4. The power counting function T2 (x) and its approximating function f (x) = log2 x.

Exercise 1.4. a. Using a computer draw the graphs of T2 (x) and f (x) = log2 x on various domains. b. Use f (x) to give an estimate for the number T2 (1050 ). How good is your estimate?

8

1. Prime numbers

The difference between f (x) = log2 x and T2 (x) fluctuates between 0 and 1. So the error in the approximation does not tend to zero in any way. And yet we say that f (x) is a good approximation because the relative error f (x) − T2 (x) f (x) does go to zero as x goes to infinity. This is because 0 ≤ f (x) − T2 (x) < 1 implies 0≤

1 f (x) − T2 (x) < . f (x) f (x)

Since f (x) = log2 x goes to infinity as x → ∞, the relative error will go to 0 as x → ∞. Viewed from a distance the two graphs look exactly alike. Exercise 1.5. a. Is it possible to approximate T2 (x) well by using a power function of the form f (x) = x a ? Do some experiments on the computer as follows. Choose a domain, find an approximation that looks good on this domain and enlarge the domain to see if the same approximation is still good. 2 (x) b. Also experiment with the relative error x −T and see if it seems to go to xa 0 as x → ∞. If you do it right, you will see that this is not true, whatever value of a you choose. a

The previous exercise suggests that T2 (x) may not be approximated well by a power of x. The relative error does not look like it goes to 0, irrespective of the a you choose. The larger the domain, the smaller the exponent a has to be. This has to do with the fact that the logarithm grows slower than any power of x. On the other hand, the power function with minimal exponent x 0 cannot be a good approximation either. We will now prove that the relative error cannot go to 0. Indeed, it goes to 1 instead. The proof is not hard but it takes some time to carefully check all steps. Since these steps are not essential for the rest of the text, the reader may wish to skip over the details upon first reading. Proof. As a preparation we will first prove that the function f (x) = log2 x (which is a good approximation for T2 (x)) grows slower than any positive power of x. More precisely: for all a > 0 we have that the quotient logx a2 x tends to zero as x → ∞. This is expressed by log2 x = 0. xa In this notation, ‘lim’ is an abbreviation for ‘limit’. lim

x→∞

9

1.4. Approximations for π(x)

To prove this limit, we set y = ln x a so x a = e y . According to the rules for the logarithm y = ln x a = a ln x

and

log2 x =

ln x , ln 2

and so ln x y/a 1 y log2 x = = = . a a y x (ln 2) x (ln 2) e a ln 2 e y If x goes to infinity, y = ln x a also goes to infinity because a > 0. It is wellknown that limy→∞ eyy = 0 (the exponential function e y grows much faster than y as y goes to infinity; for example, e100 is already a 44 digit number!). Hence we have log2 x = 0. x→∞ x a lim

Now we are almost done. Since 0 ≤ log2 x − T2 (x) < 1 for all x, we have x a − T2 (x) T2 (x) T2 (x) − log2 x + log2 x = 1 − lim = 1 − lim a a x→∞ x→∞ x→∞ x x xa T2 (x) − log2 x log2 x = 1 − lim − lim a x→∞ x→∞ x xa = 1 − 0 − 0 = 1. lim

So indeed the relative error does not go to 0; it goes to 1 instead. End of proof! What we did with T2 (x), the counting function for the powers of 2, can also be done with counting functions for the powers of any other prime number p. For any given prime p we denote by Tp (x) the number of powers of p less than or equal to x. For example, T5 (100) = 2 because only 5 and 52 = 25 are less than or equal to 100. Just like we did for T2 (x) you can show that Tp (x) ln x in the sense that the relative error goes is approximated well by logp x = ln p to 0. We will make use of this fact later in this chapter.

1.4 Approximations for π (x) We are now ready to look for good approximations of the prime counting function π (x). A function f (x) is called a good approximation of π (x) if the relative error goes to zero, i.e., if f (x) − π (x) = 0. x→∞ f (x) lim

10

1. Prime numbers

We will abbreviate this as π(x) ∼ f (x). An equivalent, and often more convenient way to write the above limit, is lim

x→∞

π(x) = 1, f (x)

which intuitively makes it clear that π (x) and f (x) go to infinity “in a similar way”. It also shows that π(x) ∼ f (x) implies f (x) ∼ π (x) and vice versa. We already tried approximating π(x) with powers x a , but that did not work. The function x 1 = x grows too fast, but all functions x a for 0 < a < 1 did not increase fast enough. We are looking for a function that is something in between these two extremes. In the beginning of the nineteenth century, it became clear that a good candidate might be L(x) = lnxx . Indeed the natural logarithm ln x grows slower than any power x b with b > 0, so our candidate L(x) = lnxx will grow faster than xxb = x 1−b for any b > 0 but slower than x, just like the prime counting function π (x). 1200

π(x)

1000 800

L(x)

600 400 200 0

2000

4000

x

6000

8000

10000

Figure 1.5. The graph of the prime counting function π (x) and the graph of the approximation L(x) = lnxx on the domain 2 ≤ x ≤ 10000. Using the computer, you can verify experimentally how good this approximation is. It is also useful to plot the relative error. In these plots you will see that the absolute error |π(x) − L(x)| does not go to zero but that relative error might do so. Exercise 1.6. a. Plot the functions π(x) and L(x) and the relative error on various domains.

1.6. Counting prime powers logarithmically

11

b. What grows faster eventually? Sort the functions below, first the function that grows slowest as x → ∞, and last the function that grows fastest as x → ∞: √ x ln x, x/(ln x), x/(ln x), x 1.001 , x 0.4 .

1.5 The prime number theorem The function L(x) = lnxx looks like a decent approximation of the prime counting function π (x). Although the approximation is not as good as we might have hoped, it looks like it does not get worse as the domain gets larger. This is indeed the case. At the end of the nineteenth century it was proven that the relative error does indeed go to zero, so in that sense L(x) is indeed a good approximation for π(x). This was a major breakthrough and is know as the prime number theorem. The proof is not easy, but nowadays it is accessible to undergraduate mathematics students. Most proofs make use of complex numbers (we will meet them in chapter 3). Chapter 4 contains the gist of the proof of the prime number theorem. Prime number theorem (Hadamard, De la Vall´ee Poussin (1896)) The relative error between lnxx and π(x) goes to zero as x goes to infinity. Written as a formula, this reads: π(x) ∼ lnxx . As a direct consequence of the prime number theorem, the number of primes below x is about lnxx , at least for large x. Exercise 1.7. a. Use the function L(x) from the prime number theorem to find approximate answers to the following questions: How many 5 digit primes are there? How many 100 digit primes? Is the number of primes with 100 digits greater or less than the number of elementary particles in the universe? (The latter is estimated to be an 80 digit number) b. For the first question in part (a) compare your answer with the exact answer that can be found by computer using the π (x) function. (Warning: do not try to get an exact value for π(10100 ) by computer!)

1.6 Counting prime powers logarithmically In the second half of the nineteenth century people came to realize that to unlock the secrets of the prime numbers it was important to not only count the prime numbers themselves, but also the prime powers. Prime powers are numbers of the form p k where p is prime and k = 1, 2, 3, . . . .

12

1. Prime numbers

Around 1850 the Russian mathematician Pafnuti Chebyshev introduced a new ‘logarithmic’ prime counting function ψ(x) (ψ is the Greek letter psi). Chebyshev defined his counting function ψ(x) to be the number of prime powers less than or equal to x, except that not all primes are counted equally. The powers of the prime p are counted with weight ln p. This is useful for putting the primes on a more equal footing. For example, the powers of 2 are much more common than the powers of the larger prime 101. This is accounted for because the powers of 2 are only counted as ln 2 = 0.631 . . . while the powers of 101 each count as ln 101 = 4.615. . . . Here is an example of how to calculate ψ(x). Take x = 20. The prime powers less than or equal to 20 are 2, 4, 8, 16 (the powers of 2), 3, 9 (the powers of 3) and the prime numbers 5, 7, 11, 13, 17, 19. Indeed, the primes themselves are also prime powers (with exponent 1). To calculate ψ(20) we need to count the four powers of 2 with weight ln 2, the two powers of 3 with weight ln 3, and the six other primes all with a weight that is their own logarithm. This means that ψ(20) = 4 × ln 2 + 2 × ln 3 + ln 5 + ln 7 + ln 11 + ln 13 + ln 17 + ln 19 ≈ 19.266. This is almost 20 and that is no coincidence. The function ψ(x) turns out to be approximated well by the function x. In chapter 4 we will show why the relative error should indeed go to zero so that ψ(x) ∼ x, which means that lim

x→∞

ψ(x) − x = 0. x

Figure 1.6 shows a graph of ψ(x) drawn on two domains. The graph looks very close to the diagonal y = x. There is an intimate connection between what we said about ψ(x) (i.e., that ψ(x) ∼ x) and the prime number theorem that says π (x) ∼ lnxx . This is true since ψ(x) ∼ π(x) ln x, as we will now show. To better understand the connection between π (x) and ψ(x), let us first consider ψ(20) again. We already know that ψ(20) = 4 × ln 2 + 2 × ln 3 + ln 5 + · · · + ln 19. The factors 4 and 2 in front of ln 2 and ln 3 indicate that there are exactly 4 powers of 2 less than or equal to 20 and 2 powers of 3 less than or equal to 20. Going back to section 1.3, we can express these factors in terms of the counting functions Tp (x). Recall that Tp (x) is the number of powers of p less

1.6. Counting prime powers logarithmically

13

Figure 1.6. The graph of the function ψ(x) with the line y = x on two domains.

14

1. Prime numbers

than or equal to x. Using the functions Tp (x), we can rewrite the computation for ψ(20) as ψ(20) = T2 (20) ln 2 + T3 (20) ln 3 + T5 (20) ln 5 + · · · + T19 (20) ln 19 (notice that T5 (20) = · · · = T19 (20) = 1). In the general case we can do exactly the same thing: ψ(x) = T2 (x) ln 2 + T3 (x) ln 3 + T5 (x) ln 5 + · · · , where the sum runs over all primes less than or equal to x. Therefore this sum has exactly π(x) terms. Here comes the big surprise. From section 1.3 we know that we can approximate Tp (x) well using logp x. So every term Tp (x) ln p is approximately equal to (logp x) × (ln p) = ln x (see page 9). So every term is approximately equal to ln x and there are π (x) terms in total. It follows that ψ(x) is approximated well by π (x) ln x. With some additional mathematical technique one may show that ψ(x) ∼ x implies π(x) ln(x) ∼ x, and that the prime number theorem π (x) ∼ lnxx then also follows. Figure 1.7 shows a graph of the function π (x) ln x together with the line y = x. The relative error is also plotted in the figure next to it. Perhaps you are not convinced by this plot that the relative error goes to 0, but in exercise 1.8 you will be able to confirm this experimentally for yourself. Exercise 1.8. a. Compute ψ(10) and ψ(15) to three decimals accuracy. b. Use the computer to experiment with the graph of π (x) ln x and the relative error in the approximation x.

1.7 The Riemann hypothesis—a look ahead In this book we will explain an explicit formula for the logarithmic prime counting function ψ(x). The amazing thing is that such a formula exists at all: a formula in which all irregularities of the prime numbers have been accounted for. Here is this formula; in chapter 4 we will say much more about it:  xρ ψ(x) = x − ln(2π ) − . ρ ρ The right-hand side starts with the term x that should be no surprise, since we already know on account of the prime number theorem that ψ(x) ∼ x. The second term ln(2π) includes the famous number π ≈ 3.1415926. . . . This term is a constant that does not depend on x. It is a modest number:

15

1.7. The Riemann hypothesis—a look ahead

π(x) ln(x)

y=x

Figure 1.7. The plot on the top shows the function π (x) ln x together with the diagonal y = x while on the bottom plot the relative error π (x) lnx x−x is shown.

ln(2π ) ≈ 1.837877067. . . . The real mystery is in the last term. It is a sum  ρ (a of infinitely many terms that all have the same form: xρ . The letter Greek capital sigma) is called the summation sign. It is perhaps familiar from statistics. The letter ρ is the Greek letter rho. It denotes the zeroes of the zeta function. This is the function that is the main subject of Riemann’s article from 1859 in which he also formulated his famous hypothesis. The meaning  ρ of the term ρ xρ is that for every zero ρ of the zeta function one has to ρ compute the expression xρ and add it to the rest. It turns out that there are infinitely many such zeroes so we need to deal with infinite sums! Such sums will be explained in the next chapter.

16

1. Prime numbers

Looking ahead a bit further, the zeta function will make its first real appearance in the next chapter. In chapter 3 you will learn much more about it, including a crash course on complex numbers. Finally in chapter 4 the above explicit formula for ψ(x) will be explained further. Then we will make clear what the zeta function has to do with the prime number theorem and the Riemann hypothesis. The Riemann hypothesis is the following statement: all non-trivial zeroes of the zeta function are on the critical line. At the end of this first chapter we do not know what this mysterious zeta function is, what non-trivial zeroes are and what the critical line is. However we do know that these zeroes, whatever they are, play the main role in the explicit formula for the prime counting function ψ(x). They contain the secrets of the prime numbers.

1.8 Additional exercises Exercise 1.9. Two families of prime numbers. The odd primes can be organized into two families: the primes of the form 4n + 1, for example 5 and 13, and those of the form 4n + 3 such as 3 and 7. We know there are infinitely many primes, but how are they distributed among these two families? In this exercise we will modify the proof that there are infinitely many primes to construct a proof that there are also infinitely many primes of the form 4n + 3. a. Suppose M = 4n + 3. Show that M must have at least one prime divisor of the form 4n + 3. b. Suppose p1 , . . . , pN are prime numbers of the form 4n + 3. Show that the numbers M1 = p1 × p2 × · · · × pN + 2 and M2 = p1 × p2 × · · · × pN + 4 are not divisible by the primes p1 , . . . , pN . c. Prove: if N is even, then M1 = p1 × p2 × · · · × pN + 2 is of the form 4n + 3 for some n. Conclude that in that case M1 is divisible by a prime of the form 4n + 3 that is not included in the original list p1 , . . . , pN . d. Prove: if N is odd, then M2 = p1 × p2 × · · · × pN + 4 is of the form 4n + 3. Deduce that in this case M2 is divisible by a prime of the form 4n + 3 that is not included in the original list p1 , . . . , pN . e. Conclude that there are infinitely many prime numbers of the form 4n + 3. What does this say about the number of primes in the other family, i.e., those of the form 4n + 1?

17

1.8. Additional exercises

Exercise 1.10. The nth prime number. a. Write a computer program that converts the input of a number n into an output that is the nth prime number. b. Estimate how the size of the nth prime number depends on n. Exercise 1.11. Counting functions. a. The square counting function K(x) is defined as follows. K(x) is the √ number of squares less than or equal to x. Show that f (x) = x is a good approximation of K(x) in the sense that the relative error goes to zero for x → ∞. b. For any increasing sequence of numbers {a} = a1 < a2 < · · · we can define a counting function Ta (x). To do so, define T{a} (x) to be the number of a1 , a2 , . . . that are less than or equal to x. How are the graph of T{a} (x) and the graph of {a} related? By the graph of {a} we mean the graph of the function that is equal to an for all n ≤ x < n + 1. Exercise 1.12. Rate of growth. a. Come up with a function f (x) that grows faster than x but slower than all functions cx a for a > 1 and c > 0 as x → ∞. In other words, find a function f (x) such that lim

x→∞

x =0 f (x)

and

lim

x→∞

f (x) = 0. cx a

b. Come up with a function g(x) that goes to infinity faster than x but slower than the function f (x) you found in part (a). c. Find a function h(x) that goes to infinity faster than x but slower than the function g(x) you found in part (b). Is it possible to keep going on like this? Exercise 1.13. The function ψ(x). a. Show that T2 (x) ≤ log2 x. b. Show that ψ(x) ≤ π (x) ln x. c. Is it also true that for every x > 2 we have ψ(x) < π (x) ln x?

18

1. Prime numbers

d. Prove that ψ(x) is equal to the natural logarithm of the least common multiple of the numbers 1, 2, 3, . . . , x . Here x means that we need to round down x to the nearest integer. Remark. To solve the next exercise, you have to know some elementary techniques of integration. It is the only place in this book where integrals are used, so you may safely skip this exercise if you do not yet have this knowledge. Exercise 1.14. A better approximation for the prime counting function. According to the prime number theorem, the prime counting function can be approximated well by the function L(x) = lnxx . However, comparing the graphs shows a significant mismatch. We can try to improve the approximation L(x) by looking at how fast the graph grows. a. To do this, first show that L (x) =

1 1 . − ln(x) (ln x)2

L(x) grows a bit too slowly. If we remove the term − (ln1x)2 from L (x) we will get a function that grows a bit faster. The resulting function is called the logarithmic integral Li(x) and satisfies Li (x) = ln1x . To fix matters, we set Li(e) = 0 and write  x dt . Li(x) = ln t e b. Using the computer, test whether Li(x) actually approximates the prime counting function. Make sure you check a couple of domains. See also Figure 1.8 on page 19. For the following items you need to know the technique of partial integration. Part (b) suggests that Li(x) and π (x) agree to some extent, but what about the relative error? We will see that the relative error, as expected, converges to zero. c. First show by partial integration that Li(x) =

x −e + ln x

 e

x

dt . (ln t)2

d. Next prove, again by partial integration, that  x dt x x + 2 − 2e . Li(x) = + 2 3 ln x (ln x) e (ln x)

19

1.8. Additional exercises

Below we will show that the integral from part (d) is less than √ Cx x+ (ln x)2 for certain C > 0. e. First prove that this implies that indeed the relative error between Li(x) and L(x) goes to zero. f. Does this automatically imply that the relative error between Li(x) and π (x) also goes to zero? x We will show that the integral e (lndtx)3 from part (d) is less than √ x + (lnCxx)2 . √ g. Divide the integral into two pieces: the first part from e to x and the √ √ second part from x to x. Prove that the integral from e to x is less √ than x. h. For the second integral we will argue as follows. Show that  x  x dt dt < x . √ (ln t)3 √ t(ln t)3 x x Compute the last integral using the substitution u = ln t and show in this way that the integral is less than (lnCx)2 .

Figure 1.8. The functions Li(x), π (x) and L(x). Figure 1.8 shows the graphs of the functions Li(x), π (x) and L(x) = It looks like the function Li(x) is always a bit greater than π (x).

x . ln(x)

20

1. Prime numbers

This seems to be confirmed by plotting the graphs on other domains, but things are not what they seem. In 1914, John E. Littlewood showed that the graphs will intersect infinitely many times. How far do you need to go to see the first intersection? Littlewood’s student Stanley Skewes computed that 1034 one will encounter an intersection before x = 1010 . More recently, this unreasonably large bound has been reduced quite a bit. In 2000 Carter Bays and Richard Hudson proved that the first point of intersection occurs before the value x = 1.39822 × 10316 . Even this improved upper bound, however, is still incredibly far from being of any practical use for actually finding the first intersection point.

2 The zeta function In the first chapter, while investigating the number of primes below a certain bound, we hit upon a striking pattern. The number of primes less than x is about lnxx . This statement is known as the prime number theorem. In this chapter we will delve deeper into the mathematical machinery that allowed Riemann and others to establish a remarkable connection between differentiable functions and the prime numbers. This resulted in explicit formulas for the prime counting functions π (x) and ψ(x). An intriguing formula for ψ(x) was already shown in the previous chapter. Riemann’s primary tool was the zeta function (ζ is the Greek letter zeta): ζ (x) = 1 +

1 1 1 1 + x + x + x + ··· x 2 3 4 5

This formula means that for every number x we can compute the value of ζ (x) by adding infinitely many terms of which we have written out only the first few. But why would infinite sums have any sensible outcome at all? Does the zeta function qualify as an honest function with a formula like that? Can we at least compute some of its values like ζ (1) or ζ (2)? What would the graph look like? And what exactly does the zeta function have to do with the prime numbers? This and more you will discover in the present chapter.

2.1 Infinite sums A good starting point in our discussion is the summation formula for the geometric series 1 + x + x2 + x3 + x4 + · · · that you may have seen before. Just like the zeta function the dots at the end of the formula indicate that we are dealing with an infinite sum. Such infinite 21

22

2. The zeta function

sums are often called series. In this case we are dealing with the geometric series with terms 1, x, x 2 , x 3 , . . . The x that appears here is traditionally called the ratio. Whether or not anything sensible comes out depends on the value 1 provided −1 < x < 1. for x you plug in. We will see that the answer is 1−x The usual terminology for this is to say that the series converges to the 1 . limit 1−x Before proving this fact, let us first look at some examples. 1 = 1−1 1 = 2 so we should get If x = 12 we have 1−x 2

1+ If x =

1 3

we have

1 1−x

=

1 1 1 1 + + + + · · · = 2. 2 4 8 16 1 1− 13

1+ And if x = − 12 , then

1 1−x

=

3 2

so we expect

1 3 1 1 + + + ··· = . 3 9 27 2

=

1 1+ 12

=

2 3

so

1 2 1 1 1 + − + − ··· = . 2 4 8 16 3 To see if this makes sense, you could, for example, compute the sum of the first 100 terms (see the appendix Computer support). You will see that this approximates the answer very well, but a mathematician would rather see a proof that the series converges. Here is such a proof: Look at the beginning of the geometric series, say the sum of the first n terms for some n. Let us call this sum Mn (x), so 1−

Mn (x) = 1 + x + x 2 + · · · + x n−1 . Notice that multiplying by x has a very simple effect: xMn (x) = x + x 2 + x 3 + · · · + x n . Subtracting the two equations yields (1 − x)Mn (x) = 1 − x n , since on the right-hand side all intermediate terms cancel out. Provided that x = 1, we can divide both sides by 1 − x to obtain 1 xn 1 − xn = − . 1−x 1−x 1−x If −1 < x < 1 and n gets bigger and bigger, the last term vanishes, since then limn→∞ x n = 0. We conclude that Mn (x) does indeed tend to the limit 1 , as we set out to prove. 1−x Mn (x) =

23

2.1. Infinite sums

Just like in the case of the zeta function, we would like to view the value of the geometric series in x as the value of a function of x. Let us call this function M(x): M(x) = 1 + x + x 2 + x 3 + · · · We learned that the series converges for −1 < x < 1, so for such values the function M(x) makes sense and is defined properly. However, in the cases x > 1 or x < −1 the series has no sum since the terms themselves grow ever larger and larger. We cannot expect that anything finite will come out once we sum all those terms. In such cases we say that the series diverges. Exercise 2.1. a. Try to draw a graph of M(x) by hand on the domain (−1, 1). b. What happens to the geometric series at the points x = 1 and x = −1? c. Using the function M(x), try to decide what would be a reasonable value at x = −1. We now have seen how one can use a series to define functions like M(x) and ζ (x), but one has to be very careful: this may not work for all x you wish to plug into the series, since the series needs to converge in order for the definition to make sense. The values of x for which the series does converge are called the domain of the function. For example in the case of the zeta function, the sum ζ (−2) = 1 + 22 + 32 + 42 + 52 + · · · obviously diverges (it tends to infinity). The same holds for all negative values of x since then the zeta series will be a sum of terms that grow bigger and bigger. From this we conclude that the domain of the zeta function can only consist of positive numbers x. But are all positive values of x in the domain of the zeta function? Does the sum ζ (2) = 1 + 14 + 19 + · · · converge? And what about ζ (1) = 1 + 12 + 13 · · · ? or ζ ( 21 )? To answer such questions we will make use of a trick that goes back to the 14th century French bishop Nicholas Oresme. Let us first look at ζ (1) = 1 + 12 + 13 + 14 + · · · . This series is known as the harmonic series. Oresme discovered that this series is divergent. In other words by adding enough terms in the sum your answer will exceed every

24

2. The zeta function 1

1_ 2

1

1_ 3

1_ 4

_1 2 _1 4

_1 4

_1 8

1_ 5

_1 8

1_ 6 _1 8

1_ 7

1_ 8

_1 8

1_ 9

1_ 1_ 10 11 1_ 1_ _1 _1 12 13 14 15 1_ 16

Figure 2.1. Oresme’s trick to prove that ζ (1) = ∞. bound. Figure 2.1 shows an illustration of the fact that 1 1 1 1 1 1 1 + + + + + + + ··· 2 3 4 5 6 7 8 1 1 1 1 1 1 1 > 1 + + + + + + + +··· 2 4  4 8 8  8 8

ζ (1) = 1 +

= 1+

1 + 2

1 2

+

1 2

+ ···

=∞ because in the third line we are adding infinitely many copies of 12 . From ζ (1) = ∞, we can conclude that certainly also ζ (x) = ∞ for all x < 1 since then all terms are even bigger: if x < 1, then k x < k 1 = k so 1 > k1 . kx However, we will see that the series does converge for x > 1. Surprisingly, we can employ the same trick in reverse to do this. Figure 2.2 on page 25 shows how. Suppose x > 1 (in the picture we have taken x = 1.1). We estimate 1 1 1 1 1 1 1 1 1 + x + x + x + x + x + x + x + x + ··· 2x 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 < 1 + x + x + x + x + x + x + x + x + x + ··· 2 2 4 4 4 4 8 8 8

ζ (x) = 1 +

25

2.1. Infinite sums 1

1 _ 2x

1

1 _ 2x

1 _ 2x

1 _ 4x 1 _ 3x

1 _ 4x

1 _ 5x

1 _ 4x

1 _ 4x

1 _ 4x 1 _ 8x

1 _ 6x

1 _ 7x

1 _ 8x

1 _ 8x

1 _ 8x

1_ 8x

1_ 8x

1_ 8x

1_ 8x

1_ 8x

1 _ 9x

1_ 16x

Figure 2.2. The modification of Oresme’s trick that proves that ζ (x) converges for all x > 1 (in the figure we chose x = 1.1).

So this time we are using the inequality in reverse. Next the terms are grouped as follows ζ (x) < 1 + = 1+

1 1 1 1 1 1 1 1 1 + x + x + x + x + x + x + x + x + ··· x 4  4 4 8 8  8 2  2 4

2 2x

+

4 4x

+

8 8x

+ ···

Now let us write all numerators and denominators as powers of 2. ζ (x) < 1 + = 1+

2 4 8 16 + x + x + x + ··· 2x 4 8 16 22 23 24 2 + + + + ··· 2x 22x 23x 24x

= 1 + 21−x + 22−2x + 23−3x + 24−4x + · · · 2 3 4 = 1 + 21−x + 21−x + 21−x + 21−x + · · · You should recognize the last series as a geometric series with ratio 21−x . Since x > 1 and so 0 < 21−x < 20 = 1, we see that this geometric series converges. The series for ζ (x) has only positive terms so it too must converge in this case. Indeed, each term is less than the corresponding term in the 2x−1 . geometric series. The latter sums to 1−211−x = 2x−1 −1

26

2. The zeta function

Exercise 2.2. a. Show that the graph of the zeta function has a horizontal asymptote and find it. (Hint: look again at the sum of the geometric series.) b. Also prove that the graph of the zeta function has a vertical asymptote. Where is it? Hint: What happens when x decreases to 1? c. What is the range of the function ζ (x)?

2.2 Series for well-known functions Series are not just some gadget to deal with obscure functions like the zeta function. Well-known functions such as e x , sin(x), cos(x) and ln(1 + x) can also be expressed in terms of a series, an infinite sum. Such series are indispensable tools for any mathematician. Without explaining where they come from, we will present the series of some of these functions. In any calculus course you will learn how to derive them. ex = 1 + x +

x2 x3 x4 x5 + + + + ··· 2! 3! 4! 5!

sin x = x −

x5 x7 x9 x3 + − + − ··· 3! 5! 7! 9!

cos x = 1 −

x4 x6 x8 x2 + − + − ··· 2! 4! 6! 8!

ln(1 + x) = x −

x3 x4 x5 x2 + − + − ··· 2 3 4 5

Let us have a look at the series for the exponential function e x first. We could 0 1 have written the first two terms 1 + x in the form x0! + x1! , because 0! = 1. This shows that all terms have the same appearance: a power of x over the corresponding factorial. Similarly for the other three series. The first series is an expansion of the exponential function in terms of powers of x, somewhat like an “infinite polynomial”. To show how this works, we have plotted the first five approximating polynomials of e x ; see figure 2.3. We obtain such polynomials by truncating the series. In the case of e x they are 1 1 1 1, 1 + x, 1 + x + x 2 , 1 + x + x 2 + x 3 2 2 6 and 1 1 1 1 + x + x2 + x3 + x4 2 6 24

27

2.2. Series for well-known functions 8 6 4

y 2 0

–2

–1

0

x

1

2

Figure 2.3. Graphs of the exponential function together with the first five approximating polynomials.

(here we have calculated all factorials). Close to x = 0 the approximations seem best. It is like the polynomials are folding closer and closer to the graph of the exponential function around x = 0. With some effort, everything we know about the exponential function d x e = e x. can also be read off from the series. For example we know that dx This is reflected in the series as follows: try differentiating the series as if it were a polynomial, so each term gets derived in turn. Doing this yields a surprise:

 d x2 x3 x4 x5 1+x+ + + + + ··· dx 2! 3! 4! 5! x2 x3 x4 + + + ··· 2! 3! 4! We get back exactly the same series, so differentiating does not have any effect at all, as it should be because we are dealing with the function e x . The identity e 0 = 1 also works neatly since plugging x = 0 into our series gets rid of all x-terms. In addition to things we already know, the series also pays dividends in the form of new less familiar identities. For example if you substitute x = 1 you will find e = 1 + 1!1 + 2!1 + 3!1 + 4!1 + 5!1 + · · · or 1 1 + 120 + · · · if you write out the factorials. e = 1 + 1 + 12 + 16 + 24 As is the case with the geometric series and the zeta function, we need to be careful what values of x we substitute into the above series. The series of the sine, cosine and exponential functions always converge so they can be =0+1+x+

28

2. The zeta function

Figure 2.4. Graphs of the sine (top) and cosine (bottom) together with some of their approximating polynomials.

used for any x. The series for ln(1 + x) is a different matter, though. It only converges for −1 < x ≤ 1. In the next exercise you will get to know these series a little better. It is also very useful to do some computer experiments with them; see the appendix Computer support. Exercise 2.3. a. How many terms of the series for sine do you need to get an approximating polynomial that differs less than 0.1 from the sine on every point of the domain [−7, 7]? Use a computer to find and confirm your answer.

2.3. Computation of ζ (2)

29

2 1 0

y –1 –2 –3 –4

–1

–0.5

0

x

0.5

1

Figure 2.5. A graph of the function ln(1 + x) and some of its approximating polynomials.

b. The cosine is the derivative of the sine. How does this follow by looking at their series? You can differentiate series term by term. c. Plugging in values of x into the sine and cosine series reveals a wealth of interesting identities. For example, try to find out how the series imply the following: −2 = −

π4 π6 π8 π2 + − + − ··· 2! 4! 6! 8!

Try to find some more of these identities. d. What is the series for ln(1 − x)? For what values of x does it converge? e. Differentiate the series ln(1 + x) term by term. Do you see a connection to the geometric series M(x) ? Compare your answer by directly calculating the derivative of ln(1 + x). f. Can you explain the following formula? ln 2 = 1 −

1 1 1 1 1 + − + − + ··· 2 3 4 5 6

2.3 Computation of ζ (2) By now we already know quite a bit about the zeta function. We have seen its graph, with its domain (the interval (1, ∞)) and the vertical asymptote at

30

2. The zeta function

x = 1, but can we also compute some function values, such as ζ (2) or ζ (3)? For example, the sine has some nice special values such as sin(π/6) = 12 . Special zeta values do indeed exist and are spectacular. Some of them can be computed using the series from the previous section. We will see how the 18th century mathematician Leonhard Euler (1707–1783) found the following value for ζ (2): ζ (2) =

π2 . 6

The core of Euler’s idea is that many functions are (almost) determined by their zeroes. This is a very important theme that we will return to in chapter 4 when we have identified the zeroes of the zeta function. In Euler’s case he was looking at the zeroes of the sine function (the integer multiples of π ) or . The next exercise will rather the zeroes of the modified sine S(x) = sin(πx) πx show you how to find a series for this function. Exercise 2.4. a. Using the series for the sine from the previous section show that S(x) =

π2 2 π4 4 π6 6 sin(π x) =1− x + x − x + ··· πx 3! 5! 7!

b. The value of S(0) is a bit problematic because of division by 0. However, using the series from part (a) it is still possible to assign a value to S(0). What value should that be? c. Sketch a graph of S(x) on the domain [−4, 4]. d. Determine the zeroes of S(x). To understand how functions may be determined by their zeroes, let us start as simply as possible. Suppose you know that a second degree polynomial p(x) has zeroes x = 2 and x = 3. Do you know p(x)? Not really, because this only determines the x-intercepts of the corresponding parabola, not whether it opens up or down, or where its vertex is. Now suppose that you also know that the ‘constant term’ p(0) is equal to 1. In that case the function p(x) is determined completely. It must be (1 − x2 )(1 − x3 ). Why? Well, first of all this function has the same zeroes as p(x) (try plugging in x = 2 or x = 3). Also the graph of this function intersects the y-axis at height 1 just like p(x) does. We can actually find a formula for p(x) by multiplying out the parentheses: 



1 1 1 1 x+ x2, · p(x) = 1 + − − 2 3 2 3

2.3. Computation of ζ (2)

31

so p(x) = 1 − 56 x + 16 x 2 . The upshot is that knowing three values determines a polynomial of degree two. In the same way a degree ten polynomial p(x) is completely determined by knowing that it has the following values as zeroes: 1, 2, 3, . . ., 10 and knowing that p(0) = 1. In this case the polynomial p(x) must be equal to   x x x  x 1− 1− ··· 1 − , 1− 1 2 3 10 since this polynomial has zeroes at the same locations as p(x), the same degree and the same value at 0. Again we can expand all parentheses to find an explicit formula for p(x). This takes a bit of work, so we will content ourselves with only computing the coefficient of x in the formula for p(x). This computation will be very important in what follows. Try to multiply out mentally a couple of the parentheses. What multiplications will eventually yield a term with x? These will be the products that only involve x once. This is why the coefficient of 1 . x must be equal to − 11 − 12 − 13 − 14 · · · − 19 − 10 After these preparations we are finally ready for the real deal: Euler is a polynomial of infinite degree pretended that the function S(x) = sin(πx) πx whose zeroes are 1, 2, 3, 4, . . . and −1, −2, −3, −4, . . . (all integers except 0 are zeroes, see exercise 2.4) and whose value at 0 equals 1. Arguing by analogy with the case for polynomials we discussed earlier, Euler then wrote





  x  x x  x x x 1− 1− 1− 1− 1− S(x) = 1 − 1 −1 2 −2 3 −3

  x x × 1− 1− ··· 4 −4 2 x Using that, for example, 1 − −3 1 − x3 = 1 − x32 , this may be written as    

x2 x2 x2 x2 sin(π x) 1− 2 1− 2 1 − 2 ··· = 1− 2 S(x) = πx 1 2 3 4 The big surprise is that we already know what the coefficients of the infi2 nite polynomial S(x) should be: S(x) = 1 − π3! x 2 + · · · (see the previous exercise). By comparing these coefficients with the ones that are hidden in the product formula for S(x), Euler discovered astounding new results. For 2 2 example, the coefficient of x 2 is − π3! = − π6 . By multiplying out infinitely many parentheses we can extract the very same coefficient from the product above. To get the coefficient of x 2 we only need to look at those multiplica2 tions that involve a single x 2 . These products look like − x12 times a bunch of

32

2. The zeta function 2

ones, or − x22 times infinitely many ones, and so on. It follows that the coefficient for x 2 must be − 112 − 212 − 312 − 412 − · · · But that is exactly −ζ (2). 2 Euler concluded that ζ (2) = π6 . Euler’s argument is not entirely rigorous; to validate some of his steps requires some additional mathematical technique. However there is nothing wrong with the line of his argument and the end result is indeed correct. Now that we know that the value of the zeta function is so beautiful at x = 2, it makes sense to ask about other values, such as ζ (3), ζ (4) and so on. Euler was able to compute the zeta function at all even numbers by using the above method. This is not as easy as the case x = 2, but it can be done. The first few are: ζ (2) =

π4 π6 π8 π 10 π2 , ζ (4) = , ζ (6) = , ζ (8) = , ζ (10) = . 6 90 945 9450 93555

These fractions look pretty simple, but then the next one is 691π 12 . 638512875 (this has to do with the so-called Bernoulli numbers). The zeta values at the odd integers are much more challenging. Although we can easily compute them numerically to as many decimals as we please, Euler and all mathematicians after him were unable to find a nice expression for ζ (3), ζ (5) and the other odd values. For example it took until 1978 to prove that ζ (3) is not a rational number, but that is about all we know. For ζ (5) even this is unknown. ζ (12) =

Exercise 2.5. The Wallis product formula (John Wallis, 1655) is: 2·2 4·4 6·6 8·8 π = · · · ··· 2 1·3 3·5 5·7 7·9 Use the product formula for S(x) to prove this formula. Hint: choose an appropriate value of x.

2.4 Euler’s product formula As a foretaste of the next chapter, we now present the formula that Euler used to connect the zeta function to the prime numbers. Euler’s product formula: For all x greater than 1 we have: ζ (x) =

1 1 1 1 1 × × × × × ··· . 1 1 1 1 1 − 2x 1 − 3x 1 − 5x 1 − 7x 1 − 111x

33

2.4. Euler’s product formula 8 7

ζ(x)

6 5

y 4 3 2 1 0

1

2

3

x

4

5

Figure 2.6. The graph of ζ (x) and the first ten approximations of the Euler product. This is an infinite product of terms of the form 1−1 1 , where p runs through px

all prime numbers. The next term would be 1−1 1 , then a term involving 17 13x and so on. The infinite product should be thought of as a limit. Multiplying more and more terms will get you closer and closer to the limit value, and apparently this is exactly the zeta function! Figure 2.6 shows the graph of the zeta function and also the first ten approximations to the product. Such approximations are obtained by truncating the infinite product after a finite number of terms. For the infinite series in the previous section we used similar truncations to approximate the series. For example the first product approximation is the function 1−1 1 , 2x

while the second approximation is 1−1 1 × 1−1 1 and so on. 2x 3x In the next chapter we will show you how Euler found his product formula. For Riemann this formula was the starting point of his investigation of the prime numbers. Exercise 2.6. a. In figure 2.6 compare the graph of ζ (x) to the first ten approximations of the Euler product. For what values of x are the approximations good and where do they fail? Can you explain why this is so? b. Use the computer to plot the approximations on other domains. c. For the first five approximations what is the value at x = 1? Check your answers using figure 2.6. d. Why do the approximations form an increasing sequence of functions?

34

2. The zeta function

Euler’s product formula establishes the connection between the primes and the zeta function. From chapter 1 we know that there are infinitely many prime numbers, and this fact is nicely reflected in the product formula. Indeed we know that ζ (x) has a vertical asymptote at x = 1 so if we approach 1 from the right, ζ (x) will increase and blow up. This behavior forces there to be an infinite number of primes. To see this, suppose for a moment that there were only finitely many primes. Then Euler’s product would be a finite product and this product would have a finite value at x = 1. This, however, contradicts ζ (1) = ∞. Figure 2.6 illustrates this nicely.

2.5 Looking back and a glimpse of what is to come We have already come quite close to understanding the mysterious statement of the Riemann hypothesis. It says that all non-trivial zeroes of the zeta function are on the critical line (see chapter 1). We have learned what the zeta function is and also what a zero of the zeta function should be: a number x such that ζ (x) = 0. From section 2.3, we also have got some idea why the zeroes of the zeta function are important. Indeed in that section our knowledge of the zeroes of the sine function allowed us to compute ζ (2) = π2 . The zeroes of a function often contain a lot of unexpected information. 6 The only problem is that from the series definition of the zeta function it follows that ζ (x) is always greater than 1. So it looks like the zeta function does not have zeroes at all! In the next chapter this problem will be resolved. We will learn more about the zeta function and will extend its domain to all real numbers (except 1). Later we will extend the zeta function even further, so it can also deal with complex numbers. After these extensions we will find that, indeed, there do exist zeroes and that, actually, there are two types of zeroes. First, there are easy-to-find zeroes on the negative real line (these will be called the trivial zeroes). But there are also infinitely many non-real zeroes. These are called the non-trivial zeroes. In chapter 4 we will show how the non-trivial zeroes dictate the distribution of the primes.

2.6 Additional exercises Exercise 2.7. A simple proof that ζ (2) < 2. 1 1 1 + 2·3 + 3·4 + · · · to the series for ζ (2). Compare the series R = 1 + 1·2 a. By comparing the terms of R to the corresponding terms in the series for ζ (2) show that ζ (2) < R.

35

2.6. Additional exercises

b. Prove that

1 k(k+1)

=

1 k



c. Use this to conclude 1 +

1 . k+1 1 1·2

+

1 2·3

+ ··· +

1 n(n+1)

=1+1−

1 n+1

for all n.

d. Show that this implies R = 2 and so ζ (2) < 2. e. Can you use the same technique to show that ζ (3) < 1 14 ? Exercise 2.8. Computations with the series for sine, cosine and tangent. a. According to the double angle formula we have sin(2x) = 2 cos(x) sin(x). Can you also get the left-hand side by multiplying the series for the righthand side? b. The tangent also has a series expansion. To find the coefficients of the powers of x is less straightforward. Using the series for cosine and sine find the first three terms of the series for the tangent. It is easiest to use the relation tan(x) cos(x) = sin(x). c. Show that in each of the series the sum of the first two terms is exactly the formula for the tangent line to the point x = 0 on the graph. Exercise 2.9. Computation of ζ (4). In this exercise we will compute ζ (4) in the footsteps of Euler. Just like ζ (2), the formula for ζ (4) is also hidden in the product formula for S(x). a. Find the coefficient of x 4 in the series for S(x) from section 2.3. b. Next find the same coefficient of x 4 using the product formula for S(x) by multiplying out the parentheses. c. Writing ζ (2) as an infinite sum, compare the product ζ (2)ζ (2) with the answer in part (b). d. Show that the coefficient from part (b) is a combination of ζ (2)2 and ζ (4). e. Prove that ζ (4) =

π4 90

by comparing the answers in parts (a) and (d).

Exercise 2.10. There are more primes than there are squares. Combining the Euler product and the series for the logarithm we will find a formula that not only shows that there are infinitely many primes, but also that in a certain sense they are more numerous than the squares. a. Write ln(ζ (x)) as a sum of terms of the form − ln(1 −

1 ). px

b. Using the series for the logarithm ln(1 + x) from section 2.2 show that − ln(1 − p1x ) = p1x + Rp where Rp = 2p12x + 3p13x + 4p14x + · · ·

36

2. The zeta function

c. Take x ≥ 1 and prove that Rp
1 we now have proven that ln(ζ (x)) = R +

1 1 1 1 1 + x + x + x + x + ··· x 2 3 5 7 11

Take the limit x → 1 and conclude that to infinity.

1 2

+

1 3

+

1 5

+

1 7

+

1 11

+ · · · goes

e. Show that there are more primes than squares in the sense that 112 + 1 1 + · · · converges, whereas 12 + 13 + 15 + 17 + 11 + · · · diverges. 32

1 22

+

Exercise 2.11. The cosine product formula. ). a. Determine the zeroes of the function C(x) = cos( πx 2 b. In the footsteps of Euler write down a product formula for the function C(x). c. Test your formula by computer by comparing the graph of C(x) with the graph resulting from the product of the first few terms of your formula. d. Using your formula prove that 1 1 1 1 π2 1 . + 2 + 2 + 2 + 2 + ··· = 2 1 3 5 7 9 8 e. Compare the product formulas for S(x) and C(x). Does one follow from the other?

Exercise 2.12. ζ (2) and the average sum of the divisors of a number. In this exercise we consider the behavior of the sum of the divisors of a number. It turns out to be related to ζ (2). To study this, we define σ (n) to be the sum of all divisors of n. For example, if n = 12, then its divisors are 1, 2, 3, 4, 6 and 12; so σ (12) = 28. a. Write down all divisors of the numbers 1 to 10 and compute σ (1), σ (2), . . . , σ (10).

37

2.6. Additional exercises

250

200

150

100

50

20

40

60

80

100

The above figure shows a plot of σ (n) for n = 1 . . . 100. Apparently, the graph of σ varies wildly. To gain some understanding, we work with averages instead. To this end we define A(n) to be the average of the numbers σ (1), σ (2), . . . , σ (n). Below we have plotted A(n) for n = 1 . . . 100. Now a pattern seems to emerge, which you may wish to further explore for yourself, using a computer (see the appendix Computer support). 80 70 60 50 40 30 20 10 20

40

60

80

100

b. The graph of A(n) seems close to a straight line. With the help of a computer find approximate values of its slope using A(10), A(100) and A(1000). To get some grip on the pattern in part (c), it is useful to first work with the sum S(n) of the numbers σ (1), σ (2), . . . , σ (n). Of course,

38

2. The zeta function

S(n) = nA(n). Another way to express S(n) is to say that it is the sum of all divisors of all numbers k between 1 and n. Any such divisor can be represented by a point (x, y) in the plane such that x and y are positive integers and x × y ≤ n. To get S(n), one has to add the x-coordinates of all points. c. For n = 10, plot such points (x, y) in the plane and check that adding their first coordinates does indeed yield the correct values for S(10) and A(10). Since the pattern we are looking for only concerns the behavior of the divisors for large n, we will not attempt to calculate S(n) exactly. Instead we will settle for a good estimate. To do so, we first add all points that are on the same horizontal line. d. Show that the number of points (x, y) with y = 3 lies between n3 − 1 and n3 and that the number of points on the horizontal line of height y is between yn − 1 and yn . e. Prove that the sum of the x-coordinates of the points on the line of height n2 n n2 n y lies between 2y 2 + 2y and 2y 2 − 2y . Check your answer for y = 3 and n = 10. To obtain an estimate for S(n) we have to add all contributions from all horizontal lines. Before doing so, it is useful to introduce some notation for expressing the answer. Define the harmonic numbers H (n) and the second harmonic numbers H2 (n) to be 1 1 1 + + ... + , 1 2 n 1 1 1 H2 (n) = 2 + 2 + . . . + 2 . 1 2 n H (n) =

f. Do you see how the numbers H (n) are related to the harmonic series from section 2.1? And do you see a relation between H2 (n) and ζ (2)? g. Prove that part (e) gives an upper bound for S(n) that reads: S(n) ≤

n n2 H2 (n) + H (n). 2 2

Also find a lower bound for S(n). h. We would like to get rid of the n2 H (n) part since we suspect it is relatively small. Use Oresme’s trick from section 2.1 to show that H (n) ≤ T2 (n), where T2 (n) is the counting function that counts the powers of 2 introduced in section 1.3.

39

2.6. Additional exercises

i. Use the estimate from part (h) and the bounds from part (h) to prove that n2 ζ (2), 2 i.e. that the relative error goes to zero. S(n) ∼

2

j. Conclude that A(n) ∼ n2 ζ (2) = n π12 . So on average the sum of the divisors of n is approximately equal to 2 n π12 , and, indeed, graphing A(n) yields a pattern that is approximately a 2 2 straight line with slope 12 ζ (2) = π12 . Note that π12 ≈ 0.822467. Compare this with your answers in part (b)!

3 The Riemann hypothesis In the previous chapter we encountered the zeta function and learned that the famous Riemann hypothesis has to do with its zeroes. However, the problem is that on the domain (1, ∞), where we could make sense of the series for ζ (x), there are no zeroes at all. Recall that the series is given by ζ (x) = 1 +

1 1 1 1 + x + x + x + ··· 2x 3 4 5

and since all terms are positive, we must have ζ (x) > 1 whenever the series converges. Riemann had the key idea to assign meaningful values to ζ (x) even when x is outside this domain. In other words, Riemann sought to extend the domain of the zeta function, and this is what we will do as well. As a first step we will extend the domain to include (almost) all positive real numbers, and after that even to the complex numbers. Then there are plenty of zeta zeroes to be found. After learning how to calculate with complex numbers, we will be able to fully understand the statement of the Riemann hypothesis. But let us first return to Leonhard Euler and his product formula. We have already mentioned this formula in the previous chapter, but now it is time to see where it comes from and how Euler found it.

3.1 Euler’s discovery of the product formula Looking at the series for the zeta function, it is not clear at all that it should have anything to do with the prime numbers. Euler found a surprising product formula that makes this connection clear. Euler’s product formula: For all x greater than 1 we have: ζ (x) =

1 1 1 1 1 × × × × × ··· 1 1 1 1 1 − 2x 1 − 3x 1 − 5x 1 − 7x 1 − 111x 41

42

3. The Riemann hypothesis

The next term in the product is 1−1 1 , next a term involving 17 and so on. All 13x primes make their appearance in this product expansion for the zeta function. This formula is the reason why the zeta function is of paramount importance in the study of prime numbers. According to the product formula, the zeta function is like a ‘clothesline’ to neatly hang out all the primes one after the other. The way Euler proved his formula is as striking as the formula itself. In a way it is an extension of the technique we used to treat the geometric series. So let us first briefly recall the main points of that technique. For −1 < x < 1 we were looking at M(x) = 1 + x + x 2 + x 3 + x 4 + · · · xM(x) = x + x 2 + x 3 + x 4 + x 5 · · · Subtracting and bringing M(x) in front of the parentheses we get M(x)(1 − x) = 1. Solving for M(x) would give the familiar sum of the series, but for now we prefer the formula like this. You will see why in a minute. Exactly the same technique can also be applied to the zeta function instead of M(x), and this is what Euler did for x > 1. To make things easier to read, we will first work with ζ (2) instead of ζ (x): 1 1 1 1 + 2 + 2 + 2 + ··· 22 3 4 5 1 1 1 1 1 ζ (2) = 2 + 2 + 2 + 2 + · · · 22 2 4 6 8

ζ (2) = 1 +

Subtracting and isolating ζ (2) yields:

 1 1 1 1 1 ζ (2) 1 − 2 = 1 + 2 + 2 + 2 + 2 + · · · 2 3 5 7 9 The net result is to remove all multiples of 2 in the bases of the squares in the denominators of the right-hand side. The multiples of 3 can be removed next in a similar way: 

1 1 1 1 1 ζ (2) 1 − 2 = 1 + 2 + 2 + 2 + 2 + · · · 2 3 5 7 9 

1 1 1 1 1 1 = 2 + 2 + 2 + 2 + ··· ζ (2) 1 − 2 2 3 2 3 9 15 21

3.2. Extending the domain of the zeta function

43

Subtracting and isolating ζ (2)(1 − 212 ) will yield:  

1 1 1 1 1 − 2 = 1 + 2 + 2 + ··· ζ (2) 1 − 2 2 3 5 7 At this point we have got rid not only of the multiples of 2 but also of the multiples of 3. Continuing in the same way by removing the multiples of 5, the multiples of 7, the multiples of 11 and so on for all prime numbers, only a single 1 will eventually remain on the right-hand side. What will the left-hand side look like? Every removal contributes a factor to the product on the left-hand side, so we should get:     

1 1 1 1 1 1− 2 1− 2 1− 2 1 − 2 · · · = 1. ζ (2) 1 − 2 2 3 5 7 11 This is an infinite product in which all the prime numbers appear exactly once. Solving for ζ (2) yields a neat product formula for ζ (2): ζ (2) =

1 1 1 1 1 × × × × ··· 1 1 1 1 1 − 22 1 − 32 1 − 52 1 − 72 1 − 1112 2

As an aside: in the previous chapter we have already seen that ζ (2) = π6 . This means that the present formula can be used to compute π in a new way! The entire argument that led us to the above product formula for ζ (2) also works word for word in the case of ζ (x). You can check that all we need to do is replace each square by ‘to the power x’. This then completes Euler’s proof of his product formula.

3.2 Extending the domain of the zeta function To successfully extend the domain of the zeta function to the complex numbers, Bernhard Riemann had to call on an impressively large amount of mathematical machinery. Many of the techniques he had to use were not yet fully developed or had only recently become available. We will take a more gentle route towards the zeta function and will start with a relatively simple and small extension involving real numbers only. The geometric series provides a simple model to illustrate what we would like to do. Recall that we used the notation M(x) = 1 + x + x 2 + x 3 + · · · and that we viewed M(x) as a function of x where x has to be inside the domain (−1, 1). In chapter 2 we found that this function M(x) is equal 1 on this domain. This is interesting because although M(2) = 1 + to 1−x 2 + 22 + 23 + 24 + 25 + · · · is meaningless (the series diverges) the formula 1 does make sense for x = 2. So we can extend the geometric M(x) = 1−x

44

3. The Riemann hypothesis 8 6 4 2

ζ(x) η(x)

y 0 –2 –4 –6 –8

1

2

3

x

4

5

6

Figure 3.1. A graph of ζ (x) for x > 0. Notice the vertical asymptote at x = 1. The special point x = 1 is called a pole. The graph of η(x) is also shown. 1 series function M(x) using the identity M(x) = 1−x to a domain that is vastly larger than where the series itself is convergent. More precisely we can extend the domain to include all real numbers except 1 (compare exercise 2.1 from chapter 2). The same principle can also be used for the zeta function, once we find a suitable formula for ζ (x) that still makes sense outside the present domain (1, ∞). By a simple trick we can find a formula that works for all x in (0, ∞). This formula (see below) will then allow us to extend the domain of ζ (x) to (0, ∞). In figure 3.1 you can already see how nicely the zeta function extends to its new and larger domain. Here is the trick to find a new formula for ζ (x) that is valid for all x > 0. We would like to modify the series for ζ (x) shown below so that it ‘converges more easily’. Recall that the series for the zeta function consists of positive terms only and looks like

1 1 1 1 1 + x + x + x + x + ··· x 2 3 4 5 6 By adding a minus sign to every other term we get a very similar series that defines a function we call the eta function ζ (x) = 1 +

1 1 1 1 1 + x − x + x − x + ··· 2x 3 4 5 6 The added minus signs make sure terms cancel out more, so the series does not grow so fast. It turns out that the series for η(x) actually converges for η(x) = 1 −

45

3.3. A crash course on complex numbers

all positive x, even when x is less than 1. For example in exercise 3.1 you will see that η(1) = ln(2) while ζ (1) is infinite. Using the computer we can easily approximate the value of the eta function at any positive number, for example η( 21 ) ≈ 0.6048986. The same exercise 3.1 also challenges you to find a strikingly simple connection between the zeta function and the eta function: ζ (x) =

1 η(x). 1 − 21−x

This is the formula we are looking for to extend the domain of the zeta function. Although the left-hand side is only defined for x > 1 (the domain of the zeta series) we can use the right-hand side to define the zeta function for 0 < x < 1. After all, the right-hand side only depends on η(x) which is already defined for all positive x. To see how this works, we will now compute ζ ( 12 ). We cannot use the series for the zeta function because that will diverge, but using η( 21 ) ≈ 0.6048986 and the above formula we see ζ ( 12 ) = 1−1√2 η( 21 ) ≈ −1.4603545. Only x = 1 remains problematic. Our new formula is of no help here because the denominator becomes 0. This has to do with the fact that we already know that the zeta function has a vertical asymptote at x = 1. Such points where the function has a vertical asymptote are called poles of the function. Exercise 3.1. a. Show that η(1) = ln 2 using the series for ln(1 + x) from chapter 2. b. Compute η(2) exactly. Hint: what is ζ (2)? c. Write down the series for ζ (x) − η(x). d. Show that ζ (x) − η(x) = 21−x ζ (x) so that ζ (x) =

1 1−21−x

η(x).

3.3 A crash course on complex numbers To make further progress towards the Riemann hypothesis and the connection between prime numbers and the zeta function, it is essential to get to know the complex numbers. It is well-known that the function x 2 + 1 does not have any zeroes. Such a zero x would be absurd, since it solves the equation x 2 = −1, in other √ words x = −1, and the square root of √ −1 does not exist. From now on, however, we will choose to work with √ −1, even though it does not make sense as a real number. By carrying −1 along in all our calculations and pretending it behaves just like an ordinary square root, we get numbers called

46

3. The Riemann hypothesis

√ √ √ the complex numbers. For example: 2 + −1 + 3 + 4 −1 = 5 + 5 −1 √ √ √ √ and (1 + −1)(1 + 2 −1) = 1 − 2 + 3 −1 = −1 + 3 −1. In the last example we have worked out the parentheses as we would do for ordinary √ numbers, using ( −1)2 = −1. In terms of complex numbers −4 now also has a square to the ordinary rules for square roots we have √ √root: According √ √ −4 = 4 −1 = 2 −1. We see that any negative real √ number now has a complex square root that can be expressed in terms of√ −1. a real This all sounds fantastic but can we make sense of −1? It is not √ number so it has no place on the number line. How can we visualize −1, let alone √ the other complex numbers? A good way to do exactly this is to give −1 its very own number line, perpendicular to the ordinary line of real numbers. √ If we think of the real numbers as the x-axis, then we want −1 on the y-axis. All other complex numbers are combinations to place √ of −1 and real numbers so they will all neatly fit in the plane. So we get a number plane √ instead of the usual number line. For example the complex number 1 + 2 √ −1 is assigned the point in the plane with coordinates (1, 2) −3). Another way to describe the last and 2 − 3 −1 corresponds to (2,√ number is 2 times 1 plus −3 times −1 so two√unit steps on the horizontal axis and three unit steps downwards on the new −1 axis. The upshot is that complex numbers are simply points in a plane (the complex plane) just like real numbers are points on a line (the real line). Computing with complex numbers is not different from what we are used to, as long as we carry √ −1 as if it were some sort of variable, keeping in mind that along √ the ( −1)2 = −1. √ At this point it is usual to introduce a new notation for −1, since writing it all the time soon gets cumbersome. The abbreviation that is√commonly √ used for −1 is i . So from now on the complex number x + y −1 will be written x + y i (or x + i y). In the complex plane this number corresponds to the point (x, y). The special numbers x + 0 i and 0 + y i are usually abbreviated to x and y i . The number i itself (also known as 0 + 1 i ) sits at the point (0, 1) in the plane, see figure 3.2. Now we introduce one more piece of terminology: the absolute value |z| of a complex number z = x + y i . It is the distance in the plane from the origin to the point (x, y) corresponding to z. According to thePythagorean theorem there is a nice formula for the absolute value: |z| = x 2 + y 2 . The miracle about complex numbers is that we somehow get extremely lucky: by construction, the equation x 2 + 1 = 0 has a solution in terms of complex numbers, namely i (the number − i is another solution, check this!), but the miracle is that every other quadratic equation has a solution, too! The reason is that the quadratic formula now always gives an answer, since we

47

3.4. Complex functions and powers 6i

2+5i

5i

z = x + yi

yi

−4 + 2 i

2i

|z |

i −6

−5

−4

−3

−2

−1

−i

−2 i

−5 - 3 i

0

1

2

3

x

5

6

3-2i

−3 i

Figure 3.2. The complex plane. know how to interpret square roots of negative real numbers. This may not seem so very special, but it gets better: all cubic (i.e., degree three) equations also have solutions, and all fourth and higher degree equations as well! So even hopelessly complicated equations such as x 198 + 15x 44 + 3x 2 + 3 = 0 now always have solutions in the complex numbers. By the way, this equation does not have any real solutions, since the left-hand side is always positive for real values of x. Exercise 3.2. a. Compute the (complex) zeroes of x 2 + 2x + 2 using the quadratic formula. Hint: do not get discouraged by the negative discriminant; the usual rules for the square root will deal with it now! b. Check your answer by plugging the found zeroes into the formula.

3.4 Complex functions and powers Think of all new things we can do with complex numbers. All functions you know take on a richer and more exciting form when you plug in complex numbers. Let us start with the simple function f (x) = x 2 + 1. Plugging in a complex number for x yields a complex number, for example f (1 + i ) = (1 + i )2 + 1 = 1 + 2 i . To remind ourselves that we are dealing with complex numbers we will use the variable z instead of the usual x. For example f (z) = z2 + 1. Functions that take complex numbers as input are called complex functions. Following Riemann, we can do the same for the

48

3. The Riemann hypothesis

zeta function. Writing the familiar series formula in terms of the complex variable z, we have 1 1 1 1 + z + z + z + ··· z 2 3 4 5 Plugging in complex numbers such as z = 2 − i , we run into a slight problem. What do complex powers mean? For example what is 22− i ? To find the answer we go back to the series from chapter 2. Recall that in chapter 2 we grappled with the following series ζ (z) = 1 +

ex = 1 + x + sin x = x −

x2 x3 x4 x5 + + + + ··· 2! 3! 4! 5!

x3 x5 x7 x9 + − + − ··· 3! 5! 7! 9!

x2 x4 x6 x8 + − + − ··· 2! 4! 6! 8! One of the nice things we noticed about them, was that they converge for all real numbers x. Now that we know how to do simple calculations with complex numbers we can attempt to plug in a complex number z = x + i y. It will turn out that these series still converge, no matter what numbers we plug in. This is important because now the all important exponential, sine and cosine functions have become complex functions e z , sin z, cos z. All the known equations for these functions still hold, for example e z1 +z2 = e z1 e z2 . The same goes for any of the trigonometric formulas involving sine and cosine. So far, we only have changed letters in known equations. To see that using complex numbers actually does show the old equations from a fresh new perspective, let us concentrate on a special case of the exponential function. Instead of substituting a general complex number, let us choose z = 0 + i y and consider the series for e z . Before continuing, recall that ( i y)2 = i 2 y 2 = −y 2 , ( i y)3 = i 3 y 3 = − i y 3 , ( i y)4 = i 4 y 4 = y 4 , ( i y)5 = i 5 y 5 = i y 5 , and so on. Now writing out the series for e z again yields one of Euler’s greatest hits: cos x = 1 −

y2 y3 y4 y5 −i + +i − ··· 2! 3! 4! 5! 



y4 y3 y5 y2 + − ··· + i y − + − ··· = 1− 2! 4! 3! 5!

ei y = 1 + i y −

= cos y + i sin y.

49

3.4. Complex functions and powers

i

ey i

i sin y

y 0

cos y

1

Figure 3.3. The point e y i on the unit circle in the complex plane. In the last step we simply recognized the terms involving i as the series of the sine, and the other terms as the series of the cosine. So Euler’s formula says that the complex exponential function actually includes both the sine and the cosine functions! e i y = cos y + i sin y.

(3.1)

Choosing the special value y = π , we get a famous and very elegant formula e i π = −1 (recall that cos π = −1 and sin π = 0). This formula combines three of the most important constants in mathematics: e , π and 1, and the new kid on the block i . Formula (3.1) can also be interpreted geometrically if we recall that we assigned to each complex number a point in the plane. The right-hand side then represents the point (cos y, sin y). Such points always are on the unit circle and this particular point corresponds to the point at angle y (in radians); see figure 3.3 on page 49. Since y can be interpreted as an angle (the angle between the positive x-axis and the radius), it is commonly replaced by a Greek letter such as ϕ (phi). Euler’s formula then looks like e i ϕ = cos ϕ + i sin ϕ. To avoid confusion with Euler’s product formula we will name it Euler’s circle formula. Note that one often writes eϕ i instead of e i ϕ .

50

3. The Riemann hypothesis

Exercise 3.3. a. Try to prove the following: e −π i = −1,

e 3π i = −1,

e

π 2

i

= i,

e 2π i = 1,

e 2kπ i = 1

for all integers k. b. Show that e− i ϕ = cos ϕ − i sin ϕ. c. Convince yourself that cos ϕ =

ei ϕ +e − i ϕ 2

and sin ϕ =

ei ϕ −e − i ϕ . 2i

d. Prove: if z = x + i y (where x and are y real), then e z = e x cos y + i e x sin y.

3.5 The complex zeta function Now that we know what is meant by ez we can go back to the zeta function and try once more to make sense of its series when we plug in complex numbers. The problem was that we did not know what to make of powers like 2z and 3z and so on, but now that we have the exponential function we can simply say 2 = eln 2 and therefore 2z = e(ln 2)z . Recall that for any positive real number eln x = x (the logarithm and the exponential are each other’s inverse). For example, if z = 2 − 3 i , then we can compute (see also exercise 3.3.b): 22−3 i = 22 2−3 i = 22 e −3 i ln 2 = 4 cos(3 ln 2) − i 4 sin(3 ln 2). Notice how Euler’s circle formula was used in the last step. The above computation contains all we need to know to treat the complex zeta series itself. Here it is once more 1 1 1 1 ζ (z) = 1 + z + z + z + z + · · · 2 3 4 5 The series turns out to converge for all complex numbers z = x + i y satisfying x > 1. In other words, for all numbers in the complex plane that are to the right of the vertical line x = 1. Similarly, the eta function can also be turned into a complex function using its series. In this case the series converges for z in the half plane to the right of the y-axis. As we did before we can again use the formula 1 η(z), 1 − 21−z this time to extend the domain of the complex zeta function to all complex numbers z = x + i y where x is positive. Except the pole z = 1 of course because there the formula does not apply. There is an important relationship between the zeta values ζ (x + i y) and ζ (x − i y). The complex numbers x + i y and x − i y are mirror images ζ (z) =

3.6. The zeroes of the zeta function

51

in the x-axis and this relationship continues to hold for their zeta values ζ (x + i y) and ζ (x − i y): they are mirror images in the x-axis also. In exercise 3.4 you will see a neat application of this fact. Why this mirror property persists is easier to explain in a concrete example. A general proof would follow exactly the same steps. Let us take the complex numbers 5 + 3 i and 5 − 3 i . Their corresponding points in the plain (5, ±3) are mirror images in the x-axis. Now consider plugging them into the series for the zeta function. The first term of that series is 1. The next term is 21z . What happens if you substitute 5 + 3 i or 5 − 3 i ? Using Euler’s circle formula we get: 1 = 2−5−3 i = 2−5 2−3 i = 2−5 e −3 i ln 2 25+3 i = 2−5 cos(3 ln 2) − i 2−5 sin(3 ln 2) and 1 25−3 i

= 2−5+3 i = 2−5 23 i = 2−5 e 3 i ln 2 = 2−5 cos(3 ln 2) + i 2−5 sin(3 ln 2).

These numbers are again mirror images in the x-axis. The same goes for all other terms in the series. It must also hold for the series as a whole because the mirror image of the sum is the sum of the mirror images (why?).

3.6 The zeroes of the zeta function So far the zeta function ζ (z) has been defined for all complex numbers z = x + i y with x > 0, i.e. for all z in the right half plane. Using more advanced techniques from the theory of complex functions, Riemann was able to extend the domain even further to include all complex numbers except 1. In the next chapter we will see a glimpse of how he did that. For now we will only say that the zeta function does indeed have infinitely many zeroes, so there are infinitely many complex numbers z satisfying ζ (z) = 0. These zeroes turn out to come in two families: the trivial zeroes and the non-trivial zeroes. The first got their name because they are relatively easy to find. They are precisely all negative even integers −2, −4, −6, . . . The second family of zeroes is much more interesting. All we know about these so called non-trivial zeroes is that they must all lie inside the critical strip 0 ≤ x ≤ 1, see figure 3.4. Riemann discovered that these elusive nontrivial zeroes play the main part in his connection between the zeta function

52

3. The Riemann hypothesis

critical strip

x+yi

yi

pole

trivial zeroes i −6

−4

−2

0

1

x

critical line

Figure 3.4. In the complex plane we have drawn the critical strip {0 ≤ x ≤ 1}, the critical line {x = 12 }, the pole z = 1 and the trivial zeta zeroes z = −2, z = −4 and z = −6. The non-trivial zeroes are all located inside the critical strip but a bit further away. The Riemann hypothesis states that they are actually all on the critical line.

and prime counting functions. In chapter 1 we already mentioned that the non-trivial zeta zeroes are at the core of the explicit formula for Chebyshev’s logarithmic prime counting function ψ(x). The Riemann hypothesis states that the non-trivial zeroes must all lie on the critical line. Finally we are able to understand what this means: the critical line is the vertical line precisely in the middle of the critical strip. It is the line x = 12 . Riemann hypothesis (1859): All non-trivial zeroes of the zeta function are located on the critical line. So far nobody has been able to prove that the Riemann hypothesis is true, nor has it been refuted, and not for lack of trying. However, we do know quite a bit about the non-trivial zeta zeroes. For example, they always come in mirror pairs: if x + y i is a zero, then so is its mirror image x − i y (see the remark at the end of section 3.5). By this observation we only need to look for zeroes in the upper half plane. Finding a single zero off the critical line will refute the Riemann hypothesis, but so far this has not happened. Riemann himself computed the first few zeroes and found that they are all on the critical line. Figures 3.5 and 3.6 show the first ten pairs of non-trivial

53

3.6. The zeroes of the zeta function

zeta zeroes, written in the form 12 ± y i . It is usual to order the pairs of zeroes so that the values of y increase. _1 2 + 50 i 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2

_1 2 + 40 i

± 14.134725 i

_1 2 + 30 i

± 21.022040 i

_1 2 + 20 i

± 25.010856 i

_1 2 + 10 i

± 30.424878 i

_1 2

± 32.935057 i ± 37.586176 i

_1 2 - 10 i

± 40.918720 i

_1 2 - 20 i

± 43.327073 i

_1 2 - 30 i

± 48.005150 i

_1 2 - 40 i

± 49.773832 i

_1 2 - 50 i

Figure 3.5. The first ten pairs of non-trivial zeroes of the zeta function. The value of y is correct to 6 decimals.

Figure

3.6. The critical line together with the smallest twenty nontrival zeroes of the zeta function.

Exercise 3.4. a. Using a computer, check for a few negative even integers z that indeed ζ (z) = 0. b. Also check that the zeta function vanishes at the non-trivial zeroes given in figure 3.5. Since the y-values are approximate, you will not get exactly zero. c. To get a better view of the zeroes, it is useful to plot the absolute value |ζ ( 21 + i y)|, see figure 3.7 on page 54. Use your computer to check that there are exactly ten zeroes on the critical line between 0 and 50. d. Can you find the next zero by exploring further? Try first plotting from 0 to some number M > 50 to find the rough location, and then try zooming in. Find the next zero to at least two digits accuracy. e. Draw a graph of |ζ ( 21 + i y)| for −20 ≤ y ≤ 20. Do you notice anything special? Explain what you see, using the remark at the end of section 3.5.

54

3. The Riemann hypothesis

2.5 2 1.5 1 0.5 0

5

10

y

15

20

25

Figure 3.7. The graph of f (y) = |ζ ( 12 + i y)| for 0 ≤ y ≤ 29. This function describes the absolute value of the zeta function on the critical line. Notice that f (0) = |ζ ( 21 )| ≈ 1.4603545 (see section 3.2). The first three non-trivial zeroes of ζ (z) are visible.

3.7 The hunt for zeta zeroes The title of Riemann’s 1859 article was Ueber die Anzahl der Primzahlen unter einer gegebenen Gr¨osse (On the number of primes below a certain bound). On page four he remarks, in passing, that it is probable that all zeroes of the zeta function are located on the critical line. This statement of course has become known as the famous Riemann hypothesis. Next he writes: Hiervon w¨are allerdings ein strenger Beweis zu w¨unschen; ich habe indess die Aufsuchung desselben nach einigen fl¨uchtigen vergeblichen Versuchen vorl¨aufig bei Seite gelassen, da es f¨ur den n¨achsten Zweck meiner Untersuchung entbehrlich schien. (It is certainly preferable to find a mathematical proof of this statement; however, after a few unsuccessful attempts I have given up for now, because it did not seem necessary for the present work.) More than one hundred and fifty years later the proof still has not been found, even though many of the most talented mathematicians have tried. It has become clear that the Riemann hypothesis is one of the central questions in all of mathematics and its resolution would give the whole subject a huge impulse. There are long lists of theorems that would automatically be true once we knew that the Riemann hypothesis holds true. However, if a zero is found off the critical line, not only the Riemann hypothesis fails, but large swathes of mathematical theories will be destroyed with it.

3.8. Additional exercises

55

Even verifying the Riemann hypothesis experimentally is not an easy task. After numerically finding a zero one needs to show that the found zero is exactly on the critical line, so it starts with exactly 12 and not just approximately with 0.5000000 . . .. To get such exact results, advanced techniques are needed. So far the first one hundred billion pairs of zeroes have been found to sit exactly on the critical line. None of them deviates the slightest bit from Riemann’s prediction! And yet the search seems hopeless, since there will be ever more zeroes to check. For now the Riemann hypothesis seems a good bet, but any sort of proof is lacking. However, in the last decades of the twentieth century, Hugh Montgomery and Andrew Odlyzko found fascinating patterns in the relative spacing of the zeta zeroes. These patterns are reminiscent of patterns in particle physics that may lead to new insight, but so far the analogy is merely experimental. To learn more about this, we refer to chapter 18 of the book Prime Obsession by John Derbyshire (see Appendix C). The importance of proof in mathematics is not just in knowing for certain that something is or is not the case. Rather it is a test of our mastery so far, and the proof of an important theorem invariably results in unexpected connections and insights. This was certainly the case, for example, when Andrew Wiles proved Fermat’s last theorem in 1995. A proof of the Riemann hypothesis is expected to provide even more interesting spin-offs.

A look ahead We have seen that Euler’s product formula translates our knowledge of the zeta function into information about the prime numbers. To extract this information we started looking at the zeroes of the zeta function. Riemann’s hypothesis states that all non-trivial zeroes are on the critical line. In the next chapter we will use these zeroes to find out more about the prime numbers. For example, we will return to the prime number theorem and to the explicit formula for the prime counting function ψ(x). It will become apparent that the Riemann hypothesis neatly summarizes much that we would like to know about the prime numbers.

3.8 Additional exercises Exercise 3.5. A variation on the zeta function. In this exercise we will study the following variation on the zeta function: 1 1 1 1 1 1 1 1 D(x) = x − x + x − x + x − x + x − x + · · · 1 3 5 7 9 11 13 15

56

3. The Riemann hypothesis

As it turns out, this function also has a product formula in the style of Euler. a. Choose the minus signs in the formula below so the equation holds true:      1 1 1 1 ··· D(x) = 1 ± 31x 1 ± 51x 1 ± 71x 1 ± 111x b. Using the method from section 3.1. give a proof of the formula you found in part (a). c. Explore the behavior of D(x) as x decreases to 1. Exercise 3.6. Extending the domain of a series. Consider the following two series: (x + 1)3 1 (x + 1) (x + 1)2 + + + + ··· 2 4 8 16



  1 1 3 1 2 N(x) = 2 + 4 x − + 16 x − + ··· +8 x− 2 2 2 M(x) =

a. Find the values of x such that the series M(x) and N (x) converge. b. Prove: if both M(x) and N (x) converge, then M(x) = N (x). This implies that one can be used to extend the domain of the other. c. Can you find another series that will help to extend the domain even further? d. Do you see a relation between M, N and the geometric series? Exercise 3.7. Surprises with the complex sine and cosine functions. We are used to the fact that the sine and cosine functions are bounded. For example, −1 ≤ cos(x) ≤ 1. a. Using the formulas from exercise 3.3.c, sketch the graphs of f (x) = cos( i x) and g(x) = i sin( i x) for real values of x. b. Show that for complex z the functions cos(z) and sin(z) can grow arbitrarily large. c. Can you find complex numbers z such that cos2 z + sin2 z = 2? Exercise 3.8. Complex powers a. Compute i i . More precisely: rewrite this number in the standard form x + i y.

57

3.8. Additional exercises

b. Show that ( i i ) i = − i . c. Do you see a pattern? What would (( i i ) i ) i be? Hint: Write i = e

πi 2

.

Exercise 3.9. Evaluating the zeta function. a. Find an approximation of ζ (5 + 3i) by hand, using the computation at the end of section 3.5. b. Check your answer of part (a) by using a computer. c. Why does the series for ζ (5 + 3i) converge? Exercise 3.10. The argument of the zeta function. To better understand the behavior of the zeta function along the critical line, we will study how its argument changes there. By argument we mean the following. A complex number z can be thought of as an arrow starting at the origin and ending at the point z in the complex plane. The arrow has length equal to the absolute value |z| and a direction that can be expressed in terms of the angle (in radians) between the arrow and the positive x-axis. This angle is called the argument of the complex number z, notation: arg(z). a. Using the computer, plot the argument of the zeta function along the critical line, i.e., investigate the graph of g(y) = arg(ζ ( 21 + i y)) for real values of y. b. Notice that the graph exhibits certain jumps. When do they occur? 2 1 0 –1 –2 –3 5

10

15

y

20

25

30

Figure 3.8. The graph of g(y) = arg ζ ( 12 + i y) for 0 ≤ y ≤ 32.

58

3. The Riemann hypothesis

c. How big are the jumps? d. Can you explain this jumping behavior? Exercise 3.11. All algebraic equations have solutions. a. Show that any complex number z can be written as z = |z|e ϕ i for some real angle ϕ. This angle is the argument arg(z) we introduced in the previous exercise. b. Show that multiplying two complex numbers multiplies their absolute values and adds their arguments. c. What is the image of the complex function f (z) = z10 if z runs once around the circle with center 0 and radius R? Now we can show that, for example, g(z) = z10 + z4 + 3 has at least one zero in the complex plane. To this end, we first investigate the curve that is described by g(z) if z runs once around the circle with center 0 and radius R. d. Comparing g(z) to f (z), show that for very large R the image of g(z) is a curve that winds ten times around the origin. Why? And how large do we need R to be for this to be true? e. For very small R, the image of g(z) is a curve that doesn’t wind around the origin at all. Why? How small is small in this case? f. Conclude that there must be at least one value of R for which the image of g(z) is a curve that passes through the origin. Why does this imply that g(z) has a zero? g. Replace g(z) by any polynomial g(z) = zn + an−1 zn−1 + · · · + a0 and f (z) by zn (with n ≥ 1). Show that g(z) must have at least one zero in the complex plane.

4 Primes and the Riemann hypothesis In the past chapters we have worked hard at understanding the zeta function and the Riemann hypothesis, hoping that it would help us to get a better grip on our main question: how are the prime numbers distributed among the other numbers? Everything will finally come together in this last chapter and you will see that the so-called explicit formula fulfills our promises. It is a formula for the logarithmic prime counting function ψ(x) from chapter 1 in terms of the zeroes of the zeta function. Here is the formula once more (it is valid for all x > 1): ψ(x) = x − ln(2π) −

 xρ ρ

ρ

.

The complex numbers ρ in the last term are the zeroes of the zeta function, both the trivial and the non-trivial ones. Also recall that ψ(x) counts the primes and their powers less than x, giving each power of p weight ln p. This unusual way of counting turns out to be the most natural one and results in the simplest formulas. Actually, ψ(x) contains the same information as the more common prime counting function π(x), see chapter 1. In this chapter we will have a closer look at the formula for ψ(x) and find out how the locations of the zeroes makes all the difference. In particular the prime number theorem from chapter 1 turns out to be equivalent to the statement that all non-trivial zeros are inside the critical strip. But first we will complete the extension of the domain of the zeta function by invoking one of Riemann’s most remarkable formulas, the so-called functional equation.

59

60

4. Primes and the Riemann hypothesis

4.1 Riemann’s functional equation In this first section we delve deeper into the theory of the zeta function and its zeroes. The Riemann hypothesis claims that all non-trivial zeroes should be on the critical line x = 12 . But why this line? Why not the y-axis? And where do those trivial zeroes come from? The answer to both questions and much more is given by Riemann’s functional equation. This was his main tool to further extend the domain of the zeta function. Before showing you his equation, let us first review to what extent we have enlarged the domain so far. In chapter 3 we have defined the complex zeta function by the usual series that is valid on the half plane x > 1. Next, the trick with the eta function allowed us to enlarge the domain slightly to the half plane x > 0. To make further progress we will need Riemann’s functional equation. Even now, more than 150 years after Riemann, the complex integration techniques required to prove it would take us much too far afield, so we just present it without proof. Riemann’s functional equation: ζ (−z) =

πz −2 · z! sin ζ (z + 1). (2π)z+1 2

Notice how this formula expresses ζ (−z) in terms of ζ (z + 1). If you know one zeta value the equation will tell you the other and vice versa. Suppose z = x + i y is a complex number in the right half plane (so x ≥ 0). Then −z is in the left half plane where we do not yet know the zeta function. However, since z + 1 is in the right half plane Riemann’s functional equation allows us to indirectly define ζ (−z). For example, take z = 2 + 3 i. The point z + 1 = 3 + 3 i is located in the half plane x ≥ 1 so we know how to compute ζ (z + 1) using the zeta series (the computer will tell you that ζ (3 + 3 i ) ≈ 0.94 + 0.08 i ). The value of ζ (−z) = ζ (−2 − 3 i ), however, is problematic since both the zeta series and the eta series diverge in the left half plane. Using the functional equation we can overcome this difficulty and find ζ (−2 − 3 i ) ≈ 0.13 + 0.12 i. In this way the domain of the zeta function extends to all complex numbers except the pole z = 1. Looking at bit more carefully at the functional equation we run into a problem: πz −2 · z! sin ζ (z + 1) ζ (−z) = (2π)z+1 2

61

4.1. Riemann’s functional equation

20 10

y 0 –10 –20 –3

–2

–1

0

x

1

2

3

4

Figure 4.1. A plot of the factorial function x! for −4 ≤ x ≤ 4. Note the poles at x = −1, −2, −3, −4. The function x! turns out to have poles at all negative integers x.

what does z! mean? The factorial function z! only makes sense for positive integers. For example 4! = 1 × 2 × 3 × 4 = 24. To compute ζ (−2 − 3 i ) we need to know what is meant by the factorial (2 + 3 i )!. The solution is that before using the functional equation we first need to extend the domain of the factorial function z! just like we have been doing for the zeta function. First we extend it to all real numbers (see figure 4.1) and next to the complex numbers. How exactly this is done is treated in any introductory course on complex analysis. Figure 4.1 shows a graph of the factorial function for real values of x. You should be able to recognize 0! = 1! = 1, 2! = 2, 3! = 6 and 4! = 24. There turn out to be poles (vertical asymptotes) at all negative integers. Strictly speaking such values are outside the domain of the function. The factorial function does not have any zeros at all. Although we will not attempt proving or even properly sketching a proof of Riemann’s functional equation, some of the main ideas are too wonderful to leave unsaid. Riemann was not just interested in prime numbers, he also studied more physical questions such as the propagation of heat through a metal wire. For such processes he was able to write down equations that he could solve in special cases. The functional equation for the zeta function comes from two such solutions, one for the left-hand side and one for the right-hand side. Since heat can propagate in only one way, these two solutions must be equal. More details can be found in chapter 9 of the book A primer

62

4. Primes and the Riemann hypothesis

of analytic number theory, from Pythagoras to Riemann by Jeffrey Stopple (see Appendix C on page 99). Riemann’s functional equation immediately shows the special role of z = 12 . Looking at complex numbers as points in the complex plane, you notice that the point 12 is exactly in between the points z + 1 and −z, no matter which z is taken (in exercise 4.2 we ask you to verify this). Geometrically viewed the points z + 1 and −z are connected by a rotation of 180 degrees with rotation center the point 12 . z

z+1

180o 0 1_ 2

1

−z

Riemann’s formula describes the relation between zeta values of such point pairs. For z = 2 + 3 i the points z + 1 = 3 + 3 i and −z = −2 − 3 i are such a pair. And for z = − 12 they coincide with the rotation center, since then z + 1 = −z = 12 . Perhaps you have already noticed there is another problem at z = 0 in the functional equation. In that case we would have ζ (z + 1) = ζ (1) but there is a pole at z = 1 so we cannot use this value to determine the left-hand side is there to save us. This factor goes to 0 as z ζ (0). Luckily the factor sin πz 2 goes to 0, thus compensating the ζ (z + 1) that blows up. It turns out to be possible to take the limit as z goes to 0 and define ζ (0) by this limit. Compare the situation in chapter 2 where we considered the function at x = 0. For this value S(x) was not really defined either, S(x) = sin(πx) πx since both numerator and denominator are 0 and yet using a limit we were

63

4.2. The zeroes of the zeta function

able to see that S(0) should be 1. In the case of the zeta function the right value turns out to be ζ (0) = − 12 : try it with the computer. Also refer to the graph in figure 4.2 on page 64, showing the zeta function on the real line −16.5 ≤ x ≤ 0. Exercise 4.1. a. Compute ζ (−1). Hint: Do you remember ζ (2)? b. In one of his famous notebooks the Indian genius Srinivasa Ramanujan (1887–1920) writes: 1 + 2 + 3 + 4 + 5 + 6 + ··· = −

1 . 12

Explain why this is not as crazy as it looks. c. Compute (− 21 )! using Riemann’s functional equation. d. By plugging z = −1 into the functional equation, deduce that the factorial function z! must have a pole at z = −1.

4.2 The zeroes of the zeta function In the previous chapter we have met the two families of zeta zeroes, the trivial ones and the non-trivial ones. Let us now have a closer look at them, starting with the trivial zeroes. Knowing Riemann’s formula makes it easy (almost trivial!) to find these zeroes. Try substituting z = 2 into the functional equation. The equation relates ζ (−z) = ζ (−2) to ζ (z + 1) = ζ (3), more precisely:

 −2 · 2! 2π ζ (−2) = sin ζ (3). 3 (2π) 2 Since sin π = 0, we must also have ζ (−2) = 0. We have found our first zeta zero! In the same way the sine function creates infinitely many zeta zeroes ζ (−4) = 0, ζ (−6) = 0 and so on at all the even negative integers. These zeroes are called the trivial zeroes of the zeta function. All other zeroes of the zeta function are called non-trivial zeroes. Although it is very difficult to say where exactly they are located, finding out regions where they cannot be is rather easy. For example, Euler’s product formula from chapter 3 tells us that there cannot be any complex zeroes z = x + i y with x > 1. Here is the product formula again: ζ (z) =

1 1−

1 2z

·

1 1 1 1 · · · ··· 1 1 1 1 − 3z 1 − 5z 1 − 7z 1 − 111 z

64

4. Primes and the Riemann hypothesis 0.6 0.4 0.2 0 –0.2 –0.4 –0.6

–16

–14

–12

–10

–8

x

–6

–4

–2

0

Figure 4.2. A plot of ζ (x) for the real numbers −16.5 ≤ x ≤ 0. Notice the zeroes ζ (x) = 0 when x = −2, x = −4, . . . , x = −16. All terms are non-zero and the terms get closer and closer to 1 as p grows (see exercise 4.2.d). Roughly speaking, this prevents the product from ever coming close to zero. (Actually proving this statement requires a bit more technique and is a good exercise for undergraduate mathematics students.) Except for the trivial zeroes the zeta function has no zeroes to the left of the y-axis. This follows easily from the previous remark and another application of Riemann’s functional equation. The argument goes as follows: suppose you had a zero in the left half plane and call it −z. According to the functional equation we have πz −2 · z! sin ζ (z + 1). 0 = ζ (−z) = (2π)z+1 2 Since the left-hand side is zero, so is the right-hand side. But z + 1 is to the right of the line x = 1 and for such numbers we just showed that the zeta function cannot be zero. The factorial function is non-zero as well (it has no zeroes!) so the sine function must be zero here. But the sine function can only be zero if z is an even integer so −z must have been a trivial zero all along. There are no other zeroes to the left of the y-axis. So far we have narrowed down our search for the non-trivial zeroes to the so called critical strip 0 ≤ x ≤ 1. The line x = 12 exactly in the middle of the strip passes through the point of symmetry z = 12 of the functional equation. This line is called the critical line, see figure 4.3. Remember that the Riemann hypothesis says that the critical line contains all non-trivial zeta zeroes.

65

4.2. The zeroes of the zeta function

critical strip

x+yi

yi

pole

trivial zeroes i −6

−4

−2

0

1

x

critical line

Figure 4.3. A picture of the complex plane showing the critical strip {0 ≤ x ≤ 1}, the critical line {x = 12 }, the pole z = 1 and the trivial zeta zeroes z = −2, z = −4 and z = −6. The non-trivial zeta zeroes are located inside the critical strip but are out of range of the picture. According to the Riemann hypothesis they should actually all lie on the critical line. The functional equation implies that if the Riemann hypothesis is true, the zeta function has the least possible number of zeroes. To see why, suppose that the Riemann hypothesis is false. 1-x+iy x +iy Then there exists a non-trivial zero z = x + i y off the critical line. Applying the functional equation will immediately give us a zero 1 − z on the other side of the critical line as well (here we are making use of the facts that neither the factorial function nor the sine have zeroes in the critical strip). Notice how the zeroes z 1_ and 1 − z are related by a 180 degree rotation 2 1 around the point 2 . From chapter 3 we already know that the mirror image of a non-trivial zero is again a zero. The upshot is that every nontrivial zero outside the critical line actually gives rise to four such renegade zeroes: x ± i y and 1 − x ± i y. 1-x-iy x -iy Exercise 4.2. a. Show that the point 12 in the complex plane is exactly in between the points −3 − 2 i and 4 + 2 i .

66

4. Primes and the Riemann hypothesis

b. Prove that for arbitrary complex numbers z, the point middle between z + 1 and −z.

1 2

is exactly in the

c. Suppose a > 0 is a positive real number and z = x + i y is an arbitrary complex number. Prove that |a z | = a x . Hint: write a z = e z ln a . d. Suppose z0 = x0 + i y0 is some fixed complex number with x0 > 1. Prove that the terms 1−1 1 in the Euler product for ζ (z0 ) converge to 1 as p goes p z0 to infinity. Hint: Consider the denominator 1 − chapter 3.

1 pz0

and use exercise 3.3d from

Exercise 4.3. In this section we have claimed several times that the only zeroes of the sine function are the integer multiples of π . Why can the sine not have any ‘non-trivial’ zeroes somewhere in the complex plane? a. Prove that sin z = 0 implies that z is real by writing the sine in terms of two exponential functions. b. Same question for the cosine and the tangent. c. Does the exponential function itself have any zeroes?

4.3 The explicit formula for ψ(x) We started our journey in chapter 1 asking how the primes are distributed. To find the answer, we counted the number of primes less than x using π (x) and plotted this prime counting function. This revealed a striking pattern: x . The approximation is the number of primes less than x is roughly ln(x) fairly accurate in the sense that the relative error decreases to 0 as x goes x . This last result is the famous prime to infinity. In other words π(x) ∼ ln(x) number theorem. In that chapter we promised to explain more about this theorem later. Next we found a more suitable prime counting function in ψ(x). The logarithmic ψ(x) may look more complicated at first, in that it counts prime powers as well, giving all the powers of p weight ln p. However, this function works very well to offset the growth of the prime numbers, and its use results in more manageable formulas. Most importantly, the connection with the zeta function is more clearly visible in this case. To recall how ψ(x) works, here is a sample computation: ψ(10) = 3 × ln(2) + 2 × ln(3) + 1 × ln(5) + 1 × ln(7) ≈ 7.8320 since below 10 there are three powers of 2, two powers of 3 and the primes 5 and 7. The pattern for ψ(x) is even more striking than

4.3. The explicit formula for ψ(x)

67

it is for π(x). We noticed that ψ(x) ∼ x, i.e., the relative error between ψ(x) and x goes to 0. At the end of the first chapter we showed that our estimates for the two prime counting functions actually come down to the same thing. If ψ(x) ∼ x, x . The reason is that ψ(x) then the prime number theorem is true, π(x) ∼ ln(x) closely resembles π (x) ln(x). In the next two chapters we spent a lot of time on the zeta function and its complex zeroes. All that hard work will now pay off handsomely as our ideas about the prime numbers and the zeta function come together in the explicit formula. This formula tells us how much the prime counting function ψ(x) deviates from x. The zeroes of the zeta function control exactly how much the deviation will be. Here is the explicit formula again:  xρ . ψ(x) = x − ln(2π) − ρ ρ The first term x is already familiar and it also is the dominant one, since we noticed that ψ(x) is roughly equal to x. The other terms are corrections to get the exact answer. For example, the function y = x − ln(2π ) is a slightly better approximation to ψ(x). You get even closer by subtracting more and ρ more terms xρ where ρ denotes a zero of the zeta function. To get the exact answer you need to make this subtraction for all zeta zeroes. That is the content of the explicit formula. Where does this strange constant ln(2π) come form? We cannot tell you the whole story here, but we will say that it is equal to ζ (0)/ζ (0). Not that it makes much of a difference, since ln(2π ) ≈ 1.837877067. It merely shifts the diagonal y = x downwards a bit.  ρ Now let us have a closer look at the term − ρ xρ . Every ρ denotes a zero of the zeta function and every zero is counted once. Since we already know that the zeroes come in two families, trivial and non-trivial, we can  ρ split the sum ρ xρ to see how much each of the two families contributes:  xρ ρ

ρ

=

 xt  xu + . t u t trivial u non-trivial

We have denoted the trivial zeroes −2, −4, −6, . . . by t, while u runs over the non-trivial zeroes. Surprisingly we can simplify the first sum, the one over the trivial zeroes. The answer turns out to be

  xt 1 1 = ln 1 − 2 . t 2 x t

68

4. Primes and the Riemann hypothesis

Using the series expansion for ln(1 + y) from chapter 2, it is not too hard to see why. We want to apply it to the left-hand side of the above formula, setting y = − x12 . Here is the expansion again: ln(1 + y) = y −

y2 y3 y4 y5 + − + − ··· 2 3 4 5

Plugging in y = − x12 yields: 

 1 1 1 1 1 1 1 ln 1 − 2 = − 2 − 4 − 6 − 8 − ··· 2 x 2 x 2x 3x 4x x −2 x −4 x −6 x −8 + + + + ··· −2 −4 −6 −8  xt . = t t

=

In the last step we recognized the trivial zeroes t = −2, −4, −6, −8, . . . The series converges because we always assume that x > 1 when working with the prime counting function, and in this case y = − x12 satisfies the required −1 < y < 0. The upshot is that we can make the explicit formula even more explicit as follows:

  u 1 x 1 ψ(x) = x − ln(2π) − ln 1 − 2 − . 2 x u u The contribution of the term − 12 ln(1 − x12 ) coming from the trivial zeroes is negligible for large x, since the function f (x) = − 12 ln(1 − x12 ) goes to zero rather fast as x goes to infinity: see figure 4.4. To get ψ(x) exactly it is necessary to include this term, but the main contributions come from x and the non-trivial zeroes u. So roughly speaking we may simplify the formula as follows:  xu ψ(x) ≈ x − . u u Refer to the next exercise for the exact meaning of ≈. Exercise 4.4. a. Let us have a closer look at the contribution f (x) = − 12 ln(1 − x12 ) of the trivial zeroes. Compute f (x) and simplify the expression you find as much as possible. In this way prove that f (x) < 0 for all x > 1. b. Suppose y = f (x) with x > 1. Express x in terms of y.

69

4.4. Pairing up the non-trivial zeroes 0.025 0.02 0.015 0.01 0.005 0

10

20

30

x

40

50

Figure 4.4. A plot of the ‘trivial’ contribution f (x) = − 12 ln(1 − val 5 ≤ x ≤ 55.

1 ) x2

on the inter-

c. Find x0 > 0 such that f (x0 ) = 0.001. Make sure your answer is accurate to at least three digits. d. Show that for all x > x0 we have     x u   1.8368 < ψ(x) − x +  < 1.8379.  u u This confirms our remark that only the non-trivial zeroes contribute significantly to the explicit formula, so for all practical purposes we can write the explicit formula as:  xu ψ(x) ≈ x − u u (the absolute error is less than two!)

4.4 Pairing up the non-trivial zeroes One of the mysterious aspects of the explicit formula

  u x 1 1 . ψ(x) = x − ln(2π ) − ln 1 − 2 − 2 x u u is that ψ(x) is clearly real (it counts things!) whereas the right-hand side involves complex zeroes u. How can something real result from adding all u these terms xu ? The key is pairing up mirror image zeroes. Recall that in

70

4. Primes and the Riemann hypothesis

chapter 3 we found that if u = v + i w is a zero, then so is its mirror image u = v − i w. The notation u is used to denote the mirror image of a complex number u. Combining the contributions of u and u, we get a real contribution for the pair (we will carefully derive this result below; see also page 51 for similar results): −

xu 2 v xu − =− x cos(w ln(x) − α). u u |u|

This solves the mystery, because after pairing up we see that the right-hand side is a real function of x. Why spoil the simple and elegant form of the individual terms by writing them in terms of ugly cosines and powers? This  u form not only verifies that the sum u xu is real, but also leads to a better understanding of the relation between ψ(x) and the non-trivial zeroes. For example, it will allow us to express the prime number theorem in terms of a condition on the zeroes explicitly. Another reason is that these real terms are easier to interpret. Some mathematicians refer to them as the ‘music of the primes’, and we will show you why. Before going into the music of the primes we need to take a closer look at the notation in the right-hand side of the formula above. In the figure below, we have drawn the non-trivial zero u = v + i w and its twin u = v − i w, together with the relevant angles. Since u is located inside the critical strip, we must have 0 ≤ v ≤ 1. Let us assume that u is the ‘upper twin’ in the sense that w > 0. Recall from chapter 3 that |u| denotes the absolute value of the complex number u, its distance from the origin. u √ According to the wi Pythagorean theorem |u| = v 2 + w2 . Next, α (in radians) measures the angle between the line connecting u to the origin and the positive real axis. We |u| can express both u and u in terms of these quantities as follows (see also chapter 3, exercise 3.3): u = v + i w = |u| cos α + i |u| sin α = |u| e i α , u = v − i w = |u| cos α − i |u| sin α = |u| e − i α . The following computations, verifying the above formula for combining the contribution of the zeta pair as a cosine, are a little technical. Upon first reading we suggest skimming through to get a general idea. If desired the details can be verified later, but skipping over them is possible too.

α 0

−w i

v

_ u

4.4. Pairing up the non-trivial zeroes

71

From the formulas for u and u it follows that 1 1 −iα = u−1 = (|u| e i α )−1 = |u|−1 e − i α = e u |u| and 1 iα 1 = (u)−1 = (|u| e − i α )−1 = |u|−1 e i α = e . u |u| Combining the contributions of u and u (recall that x u = e u ln x = ): x e v w i ln x



xu 1 i α v − i w ln x xu 1 − i α v w i ln x x e − − =− e e x e u u |u| |u| =−

e v ln x i (w ln(x)−α) + e − i (w ln(x)−α) e |u|

=−

2 v x cos(w ln(x) − α). |u|

In the last step we used Euler’s circle formula cos ϕ = 12 (e i ϕ + e − i ϕ ) (chapter 3, exercise 3.3). This completes our computation. We enumerate the non-trivial zeroes in the upper half plane starting from below and call them u1 , u2 , u3 , . . . where uk = vk + i wk and 0 < w1 ≤ w2 ≤ w3 ≤ . . . (compare our enumeration of the first ten such zeroes in chapter 3). The explicit formula then becomes:

  ∞ 2 vk 1 1 x cos(wk ln(x) − αk ). ψ(x) = x − ln(2π ) − ln 1 − 2 − 2 x |u k| k=1 Exercise 4.5. In the table in figure 3.5 on page 53 the first ten non-trivial zeroes in the upper half plane u1 , u2 , . . ., u10 are given. a. For each of these uk compute its absolute value |uk | to three decimals and compare your answer to wk . b. Also compute the corresponding angle αk for every uk . Use radians and round to three digits. How close are your answers to π2 ? c. Plot some ‘cosine contributions’ to the explicit formula and compare them to the graphs in figure 4.5 on page 77.

72

4. Primes and the Riemann hypothesis

4.5 The prime number theorem As promised, we will find a surprisingly direct connection between the prime number theorem and the location of the non-trivial zeroes of the zeta function. In terms of the logarithmic prime counting function ψ(x) the prime number theorem states ψ(x) ∼ x. In other words the relative error goes to = 0. Using our explicit formula, let us write out what zero limx→∞ ψ(x)−x x this limit says in terms of the zeta zeroes. In the limit we can safely ignore the small terms ln(2π ) and the trivial contribution (see the previous exercise), so we get: ∞  ψ(x) − x 2 vk −1 cos(wk ln(x) − αk ). = − lim x x→∞ x→∞ x |u k| k=1

lim

Notice how the division by x has been combined with the xkv terms. If the prime number theorem is to hold, then the last limit should be 0. This will fail if at least one of the vk is equal to 1, because then the corresponding power of x becomes 0 and we are left with a cosine that will not die out. More precisely, the contribution of uk then is |u2k | cos(wk ln(x) − αk ). Since the cosine varies between −1 and 1, this contribution oscillates between − |u2k | and |u2k | . Because the cosines in all terms in the series have different frequencies, the terms with vk = 1 will not be able to cancel out each other, so these terms will cause the entire sum to stay away from 0 as x grows. (An actual proof of this fact requires some standard mathematical techniques.) Since vk = 1 means that the corresponding zeta zero uk = vk + i wk sits on the boundary of the critical strip, our conclusion so far is that if the prime number theorem is true, then there can be no zeta zero on the boundary of the critical strip. (Why is the left boundary forbidden, too?) Conversely, if there are no zeta zeroes on the boundary of the critical strip, then the prime number theorem must be true. This is not as easy to prove. We do know that in this case all cosine terms will damp out and go to zero because the power of x in front is vk − 1 < 0. However, the fact that the individual terms go to zero does not imply that their sum does the same. After all, we are summing infinitely many terms; see exercise 4.6. The fact that in this case the sum of the cosine terms does converges to zero if x goes to infinity, was the key to the proof of the prime number theorem by Hadamard and De la Vall´ee Poussin. Curiously they independently discovered two very similar proofs of the prime number theorem in 1896. Their proofs have been vastly simplified in the past hundred years, but the details still are too difficult for this book.

73

4.6. A proof of the prime number theorem

Summarizing, the prime number theorem now can be phrased entirely in terms of the location of the zeta zeroes: If the prime number theorem holds, then the boundary of the critical strip cannot contain any zeroes of the zeta function. Conversely, if the latter is true, so is the prime number theorem. In the next, rather technical section we will sketch a proof of the prime number theorem in the sense that we will prove that there are no zeta zeroes on the boundary of the critical strip. Exercise 4.6. In this exercise we explore some of the more subtle points mentioned in the above paragraph. a. We have indicated why a zeta zero u = v + iw on the right boundary of the critical strip (so v = 1) is in contradiction with the prime number theorem. Why is the same true for a zeta zero on the left boundary of the critical strip (v = 0)? b. In our discussion of the cosine terms above, we mentioned that although each of the terms goes to zero as x goes to infinity, this does not necessarily imply that their sum goes to zero as x goes to infinity as well. Here we will see an example. Instead of cosine terms we have the terms e −x fk (x) = . x k

Show that for any fixed integer k ≥ 0 the function fk (x) converges to 0 as x → ∞. c. Now we consider the sum F (x) of all terms fk (x). By using the geometric 1 series with ratio e − x prove that F (x) = f0 (x) + f1 (x) + f2 (x) + · · · =

∞  k=0

fk (x) =

1 1 × 1 . x 1 − e −x

d. Finally notice that even though its terms converge to 0 as x goes to ∞, the sum F (x) goes to 1 as x goes to ∞.

4.6 A proof of the prime number theorem In this section we will sketch a proof of the fact that there are no zeroes on the right boundary of the critical strip, i.e., on the line x = 1. Given the discussion in the previous section, this boils down to a proof of the prime

74

4. Primes and the Riemann hypothesis

number theorem. The proof is somewhat tricky and may be skipped without disrupting the flow of the chapter. The proof (by contradiction) proceeds along the following lines: suppose that there exists a zero z = 1 + i w on the border of the critical strip. We will show how this leads to a contradiction. To do so, we compare our zero z = 1 + i w to the pole z = 1 and the value of zeta at the auxiliary point 1 + 2 i w. It can be proved that the pole z = 1 is compensated by the zero z = 1 + w i in the sense that ζ (1)ζ (1 + w i ) is a finite complex number (on page 62 a similar trick was used to show that ζ (0) = − 12 ). Also z(1 + 2w i ) is finite, and it follows that ζ (1)3 ζ (1 + i w)4 ζ (1 + 2 i w) = (ζ (1)ζ (1 + i w))3 ζ (1 + i w)ζ (1 + 2 i w) = 0. This implies that lim ζ (a)3 ζ (a + i w)4 ζ (a + 2 i w) = 0.

a→1

By a contrived argument we will show that, on the other hand, Euler’s product formula implies that for any real a with a > 1 |ζ (a)3 ζ (a + i w)4 ζ (a + 2 i w)| ≥ 1. It follows that the limit for a → 1 from above also must be greater than or equal to 1. This is the required contradiction. We will now focus on proving the above inequality for a > 1. It is easier to take the natural logarithm of both sides, and prove the resulting inequality instead: 3 ln |ζ (a)| + 4 ln |ζ (a + i w)| + ln |ζ (a + 2 i w)| ≥ 0. (Here we have used familiar rules for the logarithm.) Our strategy is to prove that the above combination of logarithms is greater than 0 by expanding each of the logarithms using Euler’s product formula. Recall that Euler’s formula splits the zeta function into an infinite product of factors, one for each prime. Taking logarithms, we get an infinite sum over the primes:          1 1 1 ln |ζ (z)| = − ln 1 − z  − ln 1 − z  − ln 1 − z  − . . . 2 3 5 Here we have only shown the first three terms: the contributions of the primes 2, 3 and 5 to ln |ζ (z)|. We will prove our inequality prime by prime.

4.6. A proof of the prime number theorem

75

For example for the prime 2 we will show that:          1  1  1     3 ln 1 − a  + 4 ln 1 − a+ i w  + ln 1 − a+2 i w  ≤ 0. 2 2 2 Next we will prove the same statement for the prime 3:          1  1  1     3 ln 1 − a  + 4 ln 1 − a+ i w  + ln 1 − a+2 i w  ≤ 0, 3 3 3 then for the prime 5, and so on. The argument will be exactly the same for every prime, so we will focus on the case of the prime 2. So if we can prove that          1  1  1 3 ln 1 − a  + 4 ln 1 − a+ i w  + ln 1 − a+2 i w  ≤ 0, 2 2 2 then we are done. The key inequality that we will use is an easy consequence of the well-known double angle formula cos 2ϕ = 2 cos2 ϕ − 1 for real angles ϕ: 3 cos 0 + 4 cos ϕ + cos 2ϕ ≥ 0 (try to derive this inequality using the double angle formula). Notice how familiar our combination of terms looks: you already see the coefficients 3, 4 and 1 of each of the three terms in the inequality we want to prove. Also note how the part involving i in the exponents somehow matches the angle ϕ (this is why we write cos 0 instead of 1). What is missing are the cosines. To make progress, we would like to expand the three logarithms using the familiar series 1 1 ln(1 − y) = −y − y 2 − y 3 − . . . 2 3 from chapter 2. However, before doing so we need to deal with the absolute value signs. For this we need the following fact (that you might like to prove for yourself, writing z = |z|e i ϕ ): if ln z = a + i b, then ln |z| = a. So our strategy will be to compute for each of the three values of z the logarithm ln(1 − 21z ) using the series above, and throw away all terms involving i . This is what we get: 

1 1 1 ln 1 − z = −2−z − 2−2z − 2−3z − . . . 2 2 3 To get rid of everything involving i , let us write z = x + i y and use the rules for logarithm and Euler’s circle formula to get: 2−z = e −z ln 2 = 2−x e − i y ln 2 = 2−x cos(y ln 2) − i 2−x sin(y ln 2).

76

4. Primes and the Riemann hypothesis

Remember that we were allowed to throw out the terms with i , so the sine term will be cast out. Doing the same for the other powers of 2 we find:    1 1 ln 1 − z  = −2−x cos(y ln 2) − 2−2x cos(2y ln 2) 2 2 1 − 2−3x cos(3y ln 2) − . . . 3 These are the cosines we were looking for! Now we are very close to proving our inequality:          1  1  1 3 ln 1 − a  + 4 ln 1 − a+ i w  + ln 1 + a+2 i w  ≤ 0. 2 2 2 Recall that we supposed that a > 1, so indeed we have 0 < 21a < 1 and our series for the logarithm may be used. Also note that |2a+ i w | = |2a+2 i w | = 2a . Plugging in the values z = a, z = a + i w and z = a + 2 i w we get the three equations:    1 1 + 2−2a 3 cos(0) + ... −3 ln 1 − a  = 2−a 3 cos(0) 2 2    1 1  −4 ln 1 − a+ i w  = 2−a 4 cos(w ln 2) + 2−2a 4 cos(2w ln 2) + . . . 2 2    1 1  − ln 1 − a+2 i w  = 2−a cos(2w ln 2) + 2−2a cos(4w ln 2) + . . . 2 2 Our claim is that adding up these three equations gives something that is greater than or equal to 0. The cosine inequality can be applied to each triple of corresponding terms to show that adding those three together yields a sum that is ≥ 0. For example the first three are 2−a 3 cos(0) + 2−a 4 cos(w ln 2) + 2−a cos(2w ln 2) ≥ 0. This concludes our proof!

4.7 The music of the primes We have seen that the explicit formula for ψ(x) can be written as

  ∞ 1 2 vk 1 ψ(x) = x − ln(2π ) − ln 1 − 2 − x cos(wk ln(x) − αk ). 2 x |uk | k=1

4.7. The music of the primes

77

Figure 4.5. Graphs of the first (top) and the tenth (bottom) cosine contribution to the explicit formula for ψ(x).

The summation contains infinitely many terms |u2k | x vk cos wk ln(x) − αk . What does the graph of such a term look like? Figure 4.5 shows a couple of plots corresponding to u1 = 12 + 14.134725 . . . i and u10 = 12 + 49.773832 . . . i. They look like sine waves and since any sound can be expanded into a number of sine waves we could say that the same is happening here. Every zeta zero produces a ‘tone’ and it is as if together they generate the ‘music of the primes’. Unlike in ordinary music here we are not dealing with ordinary cosine waves. In our case the ‘amplitude’ of the terms is not constant but rather a

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function of x. It is |u2k | x vk and since vk > 0 the amplitude increases. Continuing the analogy, the music of the primes gets louder and louder. For the first hundred billion such terms vk actually is 12 and if Riemann’s hypothesis is true this pattern will persist. In this case the amplitude is always of the form 2 √ x. It increases as the square root of x, as is clearly visible in figure 4.5. |uk | If the Riemann hypothesis fails, then a counter example must include four special zeroes that are related by symmetry: vk ± wk i and 1 − vk ± wk i (see page 65) where 12 < vk ≤ 1. This will disrupt the music of the primes, since one tone will now be much louder than the others. The result is a dramatic loss of symmetry that causes all kinds of related arguments to fail. In this sense we may rephrase the Riemann hypothesis as the statement that the music of the primes is as pleasant as possible. Coming back to the ‘cosine terms’, the ‘frequency’ of such a term decreases slowly because of the wk ln x instead of the more usual wk x. This effect is also visible in the plots in figure 4.5. Finally we can remark that the ‘phase’ αk converges to π2 as k goes to infinity. This is because the zeroes uk move up higher inside the critical strip as k grows. (For the same reason |uk | is well approximated by wk for large k). The explicit formula contains infinitely many ‘overtones’. Truncating the series after the first n terms gives an interesting sequence of approximations to ψ(x). Figure 4.6 shows the graph of two such approximation of ψ(x), given by the formula:

  n 2 vk 1 1 x cos(wk ln(x) − αk ) x − ln(2π) − ln 1 − 2 − 2 x |uk | k=1 for n = 10 and n = 100. In other words, we have included the first n pairs of zeta zeroes to approximate the step function ψ(x) (also shown in the figure). It is amazing how the wavy line is able to follow and in the end become the step function. On page 80 we present a similar figure, this time on the bigger domain 0 ≤ x ≤ 500 and with n = 50. Again, it is stunning to see how the ‘music of the primes’ manages to follow the seemingly erratic behavior of the logarithmic prime counting function ψ(x).

4.8 Looking back We have come a long way in understanding the prime numbers. From naively counting them, to writing down an explicit formula for their logarithmic counting function in terms of the zeroes of the zeta function. We now understand the significance of the Riemann hypothesis: it describes

4.8. Looking back

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Figure 4.6. Graph of ψ(x) and its approximation using the first n pairs of nontrivial zeroes of the zeta function for n = 10 (top) and n = 100 (bottom) on the domain 0 ≤ x ≤ 100.

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4. Primes and the Riemann hypothesis

the optimal, most symmetric situation possible for the prime numbers. In terms of the explicit formula, this means that the estimate given by the prime number theorem is as accurate as possible. We have merely scratched the surface of a vast world of amazing results, questions, conjectures and theorems on prime numbers, the zeta function and the Riemann hypothesis. Of course there are many topics that we were not able to treat properly. We would have loved to tell you how the explicit formula can be derived from Euler’s product and the intuition (see chapter 2) that every function is (almost) determined by its zeroes. But we were able to show you the gist of a the proof of the prime number theorem and the truly stunning beauty of the ideas behind Riemann’s functional equation. The function ψ(x) at first may have seemed a strange choice for a prime counting function, but this too had a wonderful, yet more difficult explanation. 500

400

300

200

100

100

200

300

400

500

Figure 4.7. Graph of ψ(x) and its approximation using the first 50 pairs of nontrivial zeroes of the zeta function on the domain 0 ≤ x ≤ 500. Every book has an end but we do hope that it will also be a beginning for a fruitful study of more advanced topics. We hope you have enjoyed this first encounter with the subject of analytic number theory (as it is known to mathematicians). On many occasions we have merely indicated how things work, without going into much detail. Once you pursue mathematics further, in college or otherwise, you will learn about these details. One of the glories

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of the subject, after all, is that nothing needs to be accepted on faith. On the other hand, we hope that you have seen a glimpse of what mathematics is all about: gaining profound insight into patterns and structures that at first sight look impregnable. Through the ages mathematicians have shed light on many intricate structures and have solved many a difficult problem. Every solution, however, leads to fresh directions for research. Many such questions have to do with prime numbers, and perhaps after reading this book you now will understand what the mathematician Paul Erd˝os meant when he paraphrased Einstein and said: “God may not play dice with the Universe, but there’s something strange going on with the prime numbers!”

4.9 Additional exercises Exercise 4.7. In this exercise you will prove that the term ln(2π ), that (0) . occurs in the explicit formula, is equal to ζζ (0) a. Differentiate the series for the eta function η(x) term by term. b. Take the logarithm of the Wallis-product (see exercise 2.5) and show that this equals 2η (0). c. Differentiate the relation η(x) = (1 − 21−x )ζ (x) to prove that ln(2π ) = ζ (0) . ζ (0) Hint: The value of ζ (0) is mentioned in section 4.1. Exercise 4.8. The eta function. a. Does the eta function have zeroes too? Where are they located? b. Can you find a functional equation for the eta function similar to that of Riemann for the zeta function? Exercise 4.9. The existence of non-trivial zeroes of the zeta function. a. Show that the function f (x) = x − ln(2π) − 12 ln(1 − x12 ) is continuous for x > 1. In particular prove that the graph of f (x) does not have any jumps. b. Suppose that the zeta function had only trivial zeroes. Explain how this is in contradiction with the explicit formula. c. In the same way prove that the zeta function must have infinitely many non-trivial zeroes.

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Exercise 4.10. Infinitely many primes of the form 4n + 1. The function D(x) from exercise 3.5 is very similar to the zeta function. For example, in that exercise we found it has an Euler type product formula. What we have not yet seen is that taking the logarithm of this product actually tells us that there must be infinitely many primes of the form 4n + 1. In exercise 1.9, where we proved the same for the family of primes of the form 4n + 3, we were unable to do this. a. Suppose A is the sum of all terms p1 where p is a prime of the form 4n + 1 and suppose B is the sum of all terms p1 where p is a prime of the form 4n + 3. Show that A + B = ∞. b. Taking the logarithm of D(x) show that A − B = ln π4 . c. Conclude that there are both infinitely many primes of the form 4n + 1 and infinitely primes of the form 4n + 3. d. Using the computer, can you find any zeroes of the function D(x)? Exercise 4.11. A simple step function. It is very difficult to imagine how step functions such as ψ(x) can be expressed in terms of a combination of waves. And yet this is exactly what happened in the explicit formula. To get a feel for this we are going to prove a similar formula for the vastly simpler step function g(t). ⎧ π ⎪ ⎪ ⎨ 4 if 2kπ < t < (2k + 1)π , g(t) = 0 if t = kπ, ⎪ ⎪ ⎩− π if (2k − 1)π < t < 2kπ . 4

Here k is an integer. Step by step we will show that g(t) = sin t +

1 1 sin 3t + sin 5t + · · · 3 5

The graphs of g(t) and sin t + 13 sin 3t + · · · +

−π

0

1 11

sin 11t are shown below.

π

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4.9. Additional exercises

To do this exercise you need to know two facts about the complex natural logarithm function ln z. 1. If z = re i ϕ , then ln z = ln r + i ϕ. However, since the value of ϕ is only determined up to integer multiples of 2π, so is the value of the logarithm. 2. Just like in the real case (see page 26) the complex function ln(1 + z) has a series expansion. This series converges for all z satisfying |z| ≤ 1, z = −1. In this case we have: ln(1 + z) = z −

z2 z3 z4 z5 + − + + ··· 2 3 4 5

a. Using the computer compare the graph of g(t) to that of the sum of the first few terms in the above series of sine functions. b. Show that the function F (z) = 1+z has the property that for all z with 1−z |z| ≤ 1, z = 1, −1 

z5 z3 + + ··· . ln F (z) = 2 z + 3 5 c. Show that F (z) = −F ( 1z ). d. Prove that for all z satisfying |z| ≥ 1, z = 1, −1

 

1 1 1 1 ln F =2 + 3 + 5 + ··· . z z 3z 5z e. Now take 0 < t < π and set z = e i t . In other words, take a z on the upper half of the unit circle. Show that in that case we have 1z = e − i t . This is a point on the lower half of the unit circle. f. Prove that for z = e i t and 0 < t < π we have 

  

1 1 1 3 1 5 1 + z − 3 + z − 5 + ··· . ln(−1) = 2 z− z 3 z 5 z Conclude that the right-hand side is constant for 0 < t < π . After all it is one of the values of ln(−1) = π i + 2kπi. Below we will find out which value of ln(−1) is required here. g. First show that z = e i t and 0 < t < π together imply that the right-hand side of the above equation equals

 1 1 4 i sin t + sin 3t + sin 5t + · · · . 3 5

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h. Now take t = 12 π. Prove that in this case we must have 

1 1 4i 1 − + − ··· 3 5 and that the expression between parentheses is positive and less than 1. i. Conclude that we need to choose the value ln(−1) = π i in part (f). Explain that this implies that for 0 < t < π we must have sin t + 1 sin 3t + 15 sin 5t + · · · = π4 . 3 j. Define the function h(t) by h(t) = sin t +

1 1 sin 3t + sin 5t + · · · 3 5

Show that for all t we have h(−t) = −h(t) and conclude that for −π < t < 0 we have h(t) = − π4 . k. Prove: h(t) = 0 for all t = kπ (k is an integer) and that h(t) is a periodic function of period 2π. Finally conclude that for all t we have g(t) = h(t), as desired. Exercise 4.12. Complex prime numbers. In this this exercise we will not study ordinary integers but the so-called complex integers (or Gaussian integers). These are numbers of the form a + b i where a and b are integers (both positive and negative are allowed). The product of two complex integers is again a complex integer. Just like we did for ordinary integers we can once again ask for the elementary building blocks for multiplication. In other words, what are the ‘complex prime numbers’? All ordinary integers can be written as a product of (positive) primes. Except for the ordering of the prime factors this factorization is unique. In the case of the complex integers the role of ±1 is played by the four numbers ±1 and ± i . These four special integers are called units and we will not count them as complex primes, in the same way that +1 and −1 are not considered ordinary primes. This said, let us explore the world of the complex primes. a. Prove the remarkable factorization 2 = (1 + i )(1 − i ). The number 2 may be a prime in the ordinary sense of the word but it is definitely not a complex prime! b. The so-called norm N(a + i b) = |a + i b|2 = a 2 + b2 of a complex integer is an ordinary integer and it is positive unless both a = b = 0.

4.9. Additional exercises

85

Show that the norm is multiplicative: N((a + i b)(c + i d)) = N (a + i b)N (c + i d). c. Using the norm, prove that the integer 3 (viewed as a complex integer) is a complex prime. Are 1 + i and 1 − i also complex primes? What about −1 + 5 i ? d. Are there infinitely many complex primes? e. Plot all complex primes of norm less than 6 in the plane. For every complex prime you have found, compute its norm. Do you see a pattern? f. Define a zeta-type function called Z(x) that fits well with our new notion 1 of complex integers. Z(x) is the infinite series with terms (a 2 +b 2 )x where a + i b = 0. Write down the first few terms in the series for Z(x). Do you notice something special? g. Find an Euler type product for our zeta-ish function Z(x) in which all complex primes a + b i satisfying a > 0 and b ≥ 0 occur. h. Using the product formulas, show that the three zeta-type functions ζ (x), D(x) (see exercise 4.8) and Z(x) we have defined so far, are related by Z(x) = 4 ζ (x)D(x).

A Why big primes are useful Bernhard Riemann and other mathematicians from his time (the nineteenth century) investigated prime numbers only because they were intrigued by their mathematical properties. But at the end of the twentieth century it turned out that big prime numbers had completely unexpected applications in cryptography, the science of secret writing. Using modern cryptographic methods, digital messages can be enciphered and provided with digital signatures in a virtually unbreakable way. Anyone banking electronically or paying bills via the internet uses this technique. In many cryptosystems primes of more than a hundred digits are used. One such system is RSA, dating from 1976. RSA is an acronym of the surnames of its inventors Rivest, Shamir and Adleman. Their system rests, among others, on the following mathematical pillars: 1. Big primes exist in an incredible amount. For instance, the number of primes with one hundred digits is greater than the number of elementary particles in the universe (this is estimated to be a number of about 80 digits). 2. There are very fast methods (prime tests) to check whether big numbers (say, of a few hundred digits) are prime. 3. Therefore it is easy to find in a very fast way such big primes at random. 4. If a prime test for a big number establishes that it is a composite number (so not a prime), such a test does not yield a factorization. Whether or not it is a prime is determined by advanced, indirect mathematical methods and not by factoring. 5. So for big numbers there is an essential difference between prime testing and factorization. The former is practically easy, while the latter is much more difficult. 87

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6. The best known factorization methods don’t get farther than numbers of at most 180 digits. Even using the most powerful computers, such a factorization may take months or even years. For the factorization of numbers of more than 200 digits (without easy to find small factors), using the most powerful computer methods will take longer than the age of the universe. So anyone with a laptop computer with a suitable computer algebra package may provide himself or herself with an unbreakable personal secret: a. Find two big primes p and q, both of at least 100 digits. b. Compute their product m = p × q. Publish m, for instance via the internet, but keep p and q secret. Then nobody else will be able to factor m into its prime factors p and q. For example: using our computer we found in less than one second the following prime numbers: p =3229246017998554007515224836513361914702373052\\ 0271018550194525717774432256934607386472425891419\\ 14189510779823172959 and q =7290296671268297309960532140532240494628547914\\ 0585148545026416949232978455741748983083120534503\\ 031988954803507 These are primes of 114 and 109 digits, respectively. Their product m = p × q is a 223-digit number: m =2354216149572126238363172480353074730958039028\\ 0888693756566192326869447063642517995941919547776\\ 3155988025576735691295786995025765022671705998887\\ 6696311244500434825826341239488571885867496423578\\ 76735037159043223082873120767213 In less than one second, using a prime test our computer verified that m is not a prime number, but nobody not knowing p and q will be able to conjure up them from m alone! To tell how such a personal secret may be enlarged to a complete cryptosystem like RSA would lead us too far afield, but you may find out for yourself more details, for instance using Wikipedia. You will be amazed by the amount of advanced mathematics that is necessary. We can safely say

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that cryptography still is big business and that all cryptographers are professional mathematicians. Since progress in this field is very fast – time and again new cryptosystems are invented, weak points in existing systems are repaired, broken codes have to be replaced – cryptographers must keep their knowledge up-to-date continuously.

B Computer support Using a computer for solving some of the exercises For solving some of the exercises in this book you should use a computer. There are many suitable commercial packages, such as Maple or Mathematica, but also free packages like Sage (see below). In this appendix we have chosen a free, accessible package for which you only need an internet browser. We use the web site Wolfram Alpha (abbreviated WA): http://www.wolframalpha.com WA is a mix of a search machine, a calculator and a computer algebra package. Although it is based upon Mathematica, there is no obligatory standard format or grammar to enter commands. A drawback might be that sometimes you get more results than you expect. The big advantage is that everybody can immediately start using it and that the site in time probably becomes better and better. While WA is suitable for solving all the exercises in this book, the reader who wants to do heavier computations may be interested in the free program Sage. http://www.sagemath.org/. At the end of this appendix we will give a short introduction to this program. We now will show which WA commands can be used in the exercises.

Chapter 1 Exercise 1.2. To plot a graph of the prime counting function, you enter: plot primepi[x] x = 0 to 100

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primepi is the name of the prime counting function and the numbers 0 and 100 specify the domain on which the function should be plotted. Of course, you may vary the bounds 0 and 100 at will. Exercise 1.3. To compute π (40000) you enter primepi[40000]. In Wolfram Alpha two graphs can be plotted together if you put the functions between braces, e.g.: plot {2x, primepi[x]} x = 0 to 100 Exercise 1.4. The functions f (x) and T2 (x) are written in WA as Log[2,x] and Floor[Log[2,x]]. They can be plotted together by entering: plot {Log[2,x], Floor[Log[2,x]]} x = 1 to 100 Exercise 1.5. To plot T2 (x) together with a power function, e.g., x 0.3 , you may enter in WA: plot {xˆ0.3, Floor[Log[2,x]]} x = 0 to 100 As always, the domain may be adapted by changing the bounds 0 and 100. Exercise 1.6. To plot the two functions, you may enter: plot {x/ln[x], primepi[x]} x = 2 to 100 The relative error is plotted as follows: plot (primepi[x]-x/ln[x])/primepi[x] x = 2 to 100 Exercise 1.7. In WA you may compute π (100) by entering primepi[100]. Exercise 1.8. The two functions are plotted as follows: plot {primepi[x]ln[x], x} x = 2 to 100 A plot of the relative error results from: plot (primepi[x]ln[x]- x)/x x = 2 to 100 Exercise 1.14. You may use the following command: plot {Li[x],primepi[x],x/Log[x]} x = 2 to 10000 Remark: most books and software packages use a different lower bound in the integral to define the function Li(x): sometimes 0, sometimes 2. For

Computer support

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convenience, we have chosen e as the lower bound. The difference is just a small constant, so it is not essential.

Chapter 2 Numerically computing the first 100 terms of the geometric series as mentioned in paragraph 2.1 is done in WA as follows: Sum[0.5ˆk,{k,0,99}] Here we have taken x = 0.5. Exercise 2.3. Using WA you find the first 6 terms of the series for the sine function by entering: Series[Sin[x],{x,0,11}] You may also plot the power series approximation, for instance up to the term with x 11 , on the interval [−7, 7] by typing: plot sum[(-1)ˆk xˆ(1+2 k)/(1+2 k)!,{k,0,5}] x = -7 to 7 Exercise 2.6. In WA you may plot the tenth product approximation as follows: plot product[1/(1-1/prime[n]ˆx), n = 1 to 10] x = 1 to 5 The zeta function itself may be plotted by typing: plot zeta[x] x = 1 to 10 Exercise 2.7. In WA an infinite sum may be computed as follows: Sum[1/((k-1)k(k+1)),{k,2,Infinity}] A complex fraction may be written as the sum of simpler fractions by using the command Apart. Just try: Apart[1/((k-1)k(k+1))] Exercise 2.8. More terms of the tangent series can be found using: Series tan Exercise 2.11. The product of the first 16 terms of C(x) can be plotted as follows: plot Product[1-xˆ2/(2k+1)ˆ2,{k,0,15}] x = -Pi to Pi

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Exercise 2.12. The sum of the divisors σ (n) of a number n can be computed by typing: DivisorSigma[1,n] It can be plotted for n = 1 . . . 40 by writing Plot[DivisorSigma[1,n],{n,1,40}] The average of the first 10 values of σ (n) can be computed as follows: 1/10 Sum[DivisorSigma[1,n],{n,1,10}]

Chapter 3 Exercise 3.1. In WA the function η(x) is called after its inventor Gustav Lejeune Dirichlet: plot DirichletEta[x] x = 0 to 6 This should be the same as figure 3.1. Exercise 3.4. For any given z you may compute ζ (z) in WA by entering, e.g., zeta[5+2I]. Note that Wolfram Alpha uses the capital I for i = √ −1. The absolute value of the zeta function on the critical line may be plotted by entering: plot |zeta[1/2+x*I]| x = 0 to 30 Exercise 3.7. The hyperbolic cosine and the hyperbolic sine are known in WA as cosh and sinh. For instance: plot sinh[x] x = -2 to 2 Exercise 3.9. Values of the zeta function, complex or not, may be found using, e.g., zeta[5+2*I] Exercise 3.10. The argument of the zeta function on the critical line may be plotted as follows: plot arg[zeta[1/2+x*I]] x = 0 to 30 Note that the argument is determined up to integer multiples of 2π . In WA the argument is always taken between −π and π .

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Exercise 3.11. To plot the image of the function g(z) = z10 + z4 + 3 as z runs through the circle with radius R around the origin, we note that g(Re i t ) = R 10 cos(10t) + R 4 cos(4t) + 3 + i (R 10 sin(10t) + R 4 sin(4t)), where 0 ≤ t ≤ 2π . In WA you may plot the image for, e.g., R = 2 by typing parametric plot (2ˆ(10)*cos(10t)+2ˆ(4)*cos(4t)+3, 2ˆ(10)*sin(10t)+2ˆ(4)*sin(4t)), t from 0 to 2*pi

Chapter 4 Exercise 4.5. You may plot a cosine term using the following command, in which u, v, w and a stand for the numbers |uk |, vk , wk and αk , respectively. For each zero these values have to be computed separately (but according to Riemann’s hypothesis always vk = 12 should hold): plot 2xˆv/u*cos[w ln[x] - a] x = 0 to 100

Exercise 4.8. In WA the eta function may be computed using DirichletEta[z].

Exercise 4.9. The function values of D(x) may be computed using DirichletL[4,2,x].

Exercise 4.11. The approximation of the first 6 sines may be plotted as follows: plot Sum[Sin[(2k+1)x]/(2k+1),{k,0,5}] x = -5 to 5

Using Sage Sage is a free open source alternative to heavy duty computer algebra packages such as Mathematica and Maple. Due to a growing and very active community Sage can actually compete with its commercial competitors and sometimes it is even faster than them. An additional advantage of Sage is that it is built on top of the popular and easy to learn programming language Python. This makes Sage programs more readable and more versatile.

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Sage is freely available for Macintosh and Linux and can also be run on Windows using an emulator such as virtual box. For more information on how to download Sage, see http://www.sagemath.org/ Here we will give a small sample of Sage code for plotting and computing functions we have met in this book. The prime counting function π is known in Sage as prime_pi(x) and plotting π (x) in the range (50, 100) is done as follows: plot(prime_pi, 50,100) Other useful functions with almost self explanatory titles are: The list of ‘all’ prime numbers: Primes() next_prime(x) yields the next prime greater than x, is_prime(x) tests whether x is prime, is_prime_power(x) tests whether x is a prime power, prime_powers(x,y) yields all prime powers in the range between x and y. The Chebyshev ψ(x) function is not built in, but can be programmed as follows: def psi(x): sum =0 for q in prime_powers(2,floor(x)): sum += log(factor(q)[0][0]) return sum Usually one plots a function f in Sage by writing plot(f, a, b), here a and b are the begin and end points. However doing so for the psi function will be quite slow. A faster way to draw the same plot on the interval [0, x] is given by the function plot_psi(x) below: def plot_psi(x): points = [(0,0)] sum = 0; for q in prime_powers(2,floor(x)): sum += log(factor(q)[0][0]) points += [(q,sum)] return plot_step_function(points,aspect_ratio=1)

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The zeta function can be computed using the command zeta(x) A list of the first 100000 zeroes of the zeta function is available within Sage using the command zeta_zeros() More Sage code will be available on the web site of this book (or the web site of the first author).

C Further reading and internet surfing The interested reader may find lots of information on the Riemann hypothesis both in the literature and on the internet. We especially recommend the following books and internet sites.

Books and articles John Derbyshire, Prime Obsession, Bernhard Riemann and the Greatest Unsolved Problem in Mathematics, London, 2003, ISBN 0-452-28525-9. This book alternates biographical chapters on Riemann’s life with chapters in which the mathematics of the primes is explained in a lucid and very accessible way. Marcus du Sautoy, The Music of the Primes, why an unsolved problem in mathematics matters, London, 2003, ISBN 1-84115-580-2. Brilliantly written with a keen eye for human aspects of the mathematicians who worked on the quest for Riemann’s so far unsolved hypothesis. Jeffrey Stopple, A primer of analytic number theory, from Pythagoras to Riemann, Cambridge, 2003, ISBN 0-521-81309-3. A very accessible text for readers who would like to delve a little deeper into the mathematics. Don Zagier, The first 50 million prime numbers, The Mathematical Intelligencer, 0 (1977), 7–19. An inspiring and concrete introduction to the Riemann hypothesis. It also can be found on the internet (use Google).

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Further reading and internet surfing

Internet sites The book Primes by Barry Mazur and William Stein is an excellent introductory text on the Riemann hypothesis, and while the book remains unfinished at the time of this writing, it is already available on their web site together with many useful programs written in the free mathematics software package Sage (see the appendix Computer support): http://www.wstein.org/rh/ On the site http://www.maths.ex.ac.uk/˜mwatkins/zeta/ encoding2.htm of the British mathematician Matthew Watkins you will find a beautiful animation, devised by Raymond Manzoni, of the approximations of ψ(x) of the form

 n  2 vk 1 1 x cos wk ln(x) − αk x − ln(2π ) − ln 1 − 2 − 2 x |uk | k=1 in which n runs from 1 to 250 (scroll to the lower end of the webpage). Another site containing a beautiful animation is by Glen Pugh: http://web.mala.bc.ca/pughg/psi/ The following prime pages are also worthwhile: http://primes.utm.edu/ On the home page of the Clay Mathematics Institute you can find the exact conditions to earn the one million dollar prize for solving the Riemann hypothesis. In addition they link to a couple of interesting articles and lectures on the subject. http://www.claymath.org/millennium/ Riemann_Hypothesis/ Lists of accurately computed zeta zeroes can be found on the site of Andrew Odlyzko: http://www.dtc.umn.edu/˜odlyzko/zeta_tables/ If you want to know more about Riemann, Euler, Chebyshev and all the other great mathematicians mentioned in this book, you should visit: http://www-history.mcs.st-and.ac.uk/ Of course, there are many more interesting sites on the Riemann hypothesis. Also many so-called “proofs” are posted. In general, unfortunately, these cannot be taken seriously!

D Solutions to the exercises Below we present solutions to the exercises. We do not include the computer assignments in these solutions. To the Additional exercises not all solutions are fully worked; sometimes we only give hints. To facilitate the reading, we always repeat the exercise, sometimes in a condensed form, before giving the solution.

Chapter 1: Prime numbers Exercise 1.1. a. Using table 1.1 find all primes smaller than 100. b. What numbers smaller than 100 can you construct by multiplication using only the numbers 3 and 8 as building blocks? c. Is it possible to construct 103 = 2 × 3 × 17 + 1 using only 2, 3 and 17? What about 104? d. Is the number 2 × 3 × 5 × 7 × 11 + 1 prime? And what about 2 × 3 × 5 × 7 × 11 × 13 + 1? a. There are 25 prime numbers less than 100, namely 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. b. 3, 8, 9, 24, 27, 64, 72, 81. c. 103 is not divisible by one of the building blocks 2, 3 and 7, so it cannot be constructed. 104 is divisible by the building block 2, but 104 = 2 × 2 × 2 × 13 and the building block 13 (or 26, or 52, or 104) is missing. Both numbers thus cannot be constructed. d. 2 × 3 × 5 × 7 × 11 + 1 = 2311 is a prime number, but 2 × 3 × 5 × 7 × 11 × 13 + 1 = 30031 = 59 × 509 is a composite number. 101

102

Solutions to the exercises

Exercise 1.2. a. Using the computer examine graphs of π (x) on various domains. First take 0 ≤ x ≤ 100, as in figure 1.1. This graph should also be in line with the outcome of exercise 1.1.a. Check this! b. Also try the domains 0 ≤ x ≤ 1 000, 0 ≤ x ≤ 10 000, . . . , 0 ≤ x ≤ 1 000 000. c. Find the smallest prime number greater than 1 000 000. d. What is the ten-thousandth prime number? c. Since π (1 000 001) = 78498 and π (1 000 003) = 78499 the number 1 000 003 is the smallest prime number greater than 1 000 000. d. Some trying yields π(104727) = 9999 and π (104729) = 10000 so 104729 is the ten-thousandth prime number. Exercise 1.3. a. Using the computer you can check that π (40 000) = 4203. Find functions f (x) = cx and g(x) = x a that satisfy f (40 000) = g(40 000) = 4203. Find the constants c and a up to three decimals accuracy. b. On the interval [0, 40 000] the functions f (x) and g(x) approximate π (x) reasonably well, but they will fail on a larger interval. Check this by comparing f (100 000) and g(100 000) with π (100 000) = 9592. a. c = 0.105, a = 0.787 b. f (100 000) = 10500 (or 10507.5 if you rounded c to more than three decimals). g(100 000) ≈ 8610. Exercise 1.4. a. Using a computer draw the graphs of T2 (x) and f (x) = log2 x on various domains. b. Use f (x) to give an estimate for the number T2 (1050 ). How good is your estimate? b. Since 0 ≤ f (x) − T2 (x) < 1 for all x, T2 (1050 ) ≈ f (1050 ). Using wellknown properties of the logarithm, we deduce f (1050 ) = log2 (1050 ) = 50 log2 10 ≈ 166.096 so 166 is a reasonable estimate. Since the difference between f (x) and T2 (x) is always between 0 and 1 and since T2 (x) is an integer less than or equal to f (x), we conclude T2 (x) = 166. Exercise 1.6. a. Plot the functions π(x) and L(x) and the relative error on various domains.

103

Solutions to the exercises

b. What grows faster eventually? Sort the functions below, first the function that grows slowest as x → ∞, and last the function that grows fastest as x → ∞: √ x ln x, x/(ln x), x/(ln x), x 1.001 , x 0.4 . √ b. You should use that x = x 0.5 and that the natural logarithm grows slower than every positive power. Precisely formulated, using a limit: lim

x→∞

ln x =0 xa

for any a > 0.

The correct order from ‘slow’ to ‘fast’ therefore is √ x x , , x ln x, x 1.001 . x 0.4 , ln x ln x The quotient of each pair of consecutive terms goes to zero as x goes to infinity. Only for the first and the last pair is this not clear at once. For the first pair notice that lim √

x→∞

x 0.4 x 0.4 ln x ln x = lim = lim 0.1 = 0, x→∞ x x/(ln x) x→∞ x 0.5

and for the last pair use that lim

x→∞

x ln x ln x = lim 0.001 = 0. x→∞ x x 1.001

Do you understand why for limits you cannot always trust your intuition (or a calculator)? Sometimes you need a proof ! Exercise 1.7. a. Use the function L(x) from the prime number theorem to find approximate answers to the following questions: How many 5 digit primes are there? How many 100 digit primes? Is the number of primes with 100 digits greater or less than the number of elementary particles in the universe? (The latter is estimated to be an 80 digit number) b. For the first question in part (a) compare your answer with the exact answer that can be found by computer using the π (x) function. (Warning: do not try to get an exact value for π (100) by computer!) a. The number of primes of 5 digits equals π(99999) − π (9999). Replacing this by L(99999) − L(9999) with L(x) = lnxx we get 7600.17 . . . while the computer yields the exact value 8363.

104

Solutions to the exercises

For the number of primes of 100 digits, it is impossible to find the exact value by computer. However, L(10100 ) =

10100 10100 ≈ ln 10100 230.2585093

which is a number of 98 digits. Similarly, L(1099 ) is a number of 97 digits, so the difference again is a 98-digit number (you may verify that the first two digits of the difference are 39). The number of primes of 100 digits should be in the same order, which means that there are way more 100 digit primes than there are elementary particles in the universe! Exercise 1.8. a. Compute ψ(10) and ψ(15) to three decimals accuracy. b. Use the computer to experiment with the graph of π (x) ln x and the relative error in the approximation x. a. ψ(10) = 3 × ln 2 + 2 × ln 3 + ln 5 + ln 7 ≈ 7.832. ψ(15) = 3 × ln 2 + 2 × ln 3 + ln 5 + ln 7 + ln(11) + ln(13) ≈ 12.795.

Additional exercises: Exercise 1.9. Two families of prime numbers. a. Suppose M = 4n + 3. Show that M must have at least one prime divisor of the form 4n + 3. b. Suppose p1 , . . . , pN are prime numbers of the form 4n + 3. Show that the numbers M1 = p1 × p2 × · · · × pN + 2 and M2 = p1 × p2 × · · · × pN + 4 are not divisible by the primes p1 , . . . , pN . c. Prove: if N is even, then M1 = p1 × p2 × · · · × pN + 2 is of the form 4n + 3 for some n. Conclude that in that case M1 is divisible by a prime of the form 4n + 3 that is not included in the original list p1 , . . . , pN . d. Prove: if N is odd, then M2 = p1 × p2 × · · · × pN + 4 is of the form 4n + 3. Deduce that in this case M2 is divisible by a prime of the form 4n + 3 that is not included in the original list p1 , . . . , pN . e. Conclude that there are infinitely many prime numbers of the form 4n + 3. What does this say about the number of primes in the other family, i.e., those of the form 4n + 1? a. Remark that the product of two numbers of the form 4n + 1 is again of this form: (4a + 1)(4b + 1) = 4(4ab + a + b) + 1.

Solutions to the exercises

105

b. Dividing M1 by one of the primes pi yields remainder 2. Dividing M2 by one of the primes pi yields remainder 4 unless pi = 3. In that case the remainder is 1. In all cases M1 and M2 thus are not divisible by any of the the primes pi . c. Multiplying two numbers of the form 4n + 3 yields a number of the form 4m + 1. So for even N , the number p1 × · · · × pN is the product of N/2 numbers of the form 4n + 1, and that again is of the form 4n + 1. Finally add 2 and apply part (b). d. Same reasoning as in part (c). e. The above does not say anything about the number of primes of the form 4n + 1. Exercise 1.10. The n-th prime number. a. Write a computer program that converts the input of a number n into an output that is the n-th prime number. b. Estimate how the size of the n-th prime number depends on n. a. We present the most simple way to construct a list of all primes below a certain number N. In the programming language Python, for instance, it looks as follows: N = 100 Primes = range(2,N) for j in range(2,N): for k in range(2,N): if j*k in Primes: Primes.remove(j*k) print Primes In this program we took N = 100. We start with a list of all numbers from 2 to N (that is the meaning of range) and call it Primes. Next we remove from the list Primes all numbers of the form j × k, with 2 ≤ j, k < N. Finally we print the list Primes of all prime numbers less than or equal to N. The above program is simple, but certainly not fast. Using the so-called Sieve of Erathosthenes you probably can write a much faster program! b. Referring to the prime number theorem we know that π (x) ≈ lnxx . Denoting the n-th prime by p(n), it is the inverse function of π (x) in the sense

106

Solutions to the exercises

that π (p(n)) = n. So we are looking for the inverse function of lnxx . An exact formula is hard to find, but a reasonable approximation is x ln x. Using the computer, do some experiments to verify this. Exercise 1.11. Counting functions. a. The square counting function K(x) is defined as follows. K(x) is the √ number of squares less than or equal to x. Show that f (x) = x is a good approximation of K(x) in the sense that the relative error goes to zero for x → ∞. b. For any increasing sequence of numbers {a} = a1 < a2 < · · · we can define a counting function T{a} (x). To do so, define T{a} (x) to be the number of a1 , . . . an that are less than or equal to x. How are the graph of T{a} (x) and the graph of {a} related? By the graph of {a} we mean the graph of the function that is equal to an for all n ≤ x < n + 1 √ √ √ a. Clearly, 0 ≤ x − K(x) ≤ 1. Therefore, the relative error x−K(x) goes x to 0 as x goes to infinity. b. The graphs are each other’s reflection in the line y = x since they correspond to mutually inverse functions. Exercise 1.12. Rate of growth. a. Come up with a function f (x) that grows faster than x but slower than all functions cx a for a > 1 and c > 0 as x → ∞. In other words, find a function f (x) such that lim

x→∞

x =0 f (x)

and

lim

x→∞

f (x) = 0. cx a

b. Come up with a function g(x) that goes to infinity faster than x but slower than the function f (x) you found in part (a). c. Find a function h(x) that goes to infinity faster than x but slower than the function g(x) you found in part (b). Is it possible to keep going on like this? a. The function f (x) = x ln x is a possible candidate. b. Take for instance g(x) = x(ln x)1/2 . c. Find out for yourself, using the following hint: ln(ln(x)) goes to infinity, but much slower than ln(x).

107

Solutions to the exercises

Exercise 1.13. The function ψ(x). a. Show that T2 (x) ≤ log2 x. b. Show that ψ(x) ≤ π(x) ln x. c. Is it also true that for every x > 2 we have ψ(x) < π (x) ln x ? d. Prove that ψ(x) is equal to the natural logarithm of the least common multiple of the numbers 1, 2, 3, . . . , x . Here x means that we need to round down x to the nearest integer. a. This follows from the fact that T2 (x) is the greatest integer less than or equal to log2 x. b. Write ψ(x) = T2 (x) ln 2 + T3 (x) ln 3 + T5 (x) ln 5 + · · · , in which sum only the first π(x) primes are included. Now use part (a) to show that ψ(x) ≤ log2 (x) ln 2 + log3 (x) ln 3 + · · · = π (x) ln(x). See also section 1.6. c. Yes, for x > 2 there is always a prime number p < x such that Tp (x) < logp (x). d. The least common multiple of m and n can be computed by noting for each prime p, the power to which it occurs in the prime factorizations of m and n, respectively. Taking for each p the maximum of these two powers yields the power of p in the factorization of the least common multiple of m and n. In the least common multiple of 1, 2, 3, . . . , x , each prime p occurs with maximum power k(p) for which pk(p) ≤ x. The natural logarithm ln(pk(p) ) = k(p) ln p of this power exactly is the term for p in the sum that defines ψ(x), so the natural logarithm of the least common multiple indeed is equal to ψ(x). Exercise 1.14. A better approximation for the prime counting function. a. We can try to improve the approximation L(x) by looking at how fast the graph grows. To do this first show that L (x) =

1 1 . − ln(x) (ln x)2

b. L(x) grows a bit too slowly. If we remove the term − (ln1x)2 from L (x) we will get a function that grows a bit faster. The resulting function is called the logarithmic integral Li(x) and satisfies Li (x) = ln1x . To fix matters

108

Solutions to the exercises

we set Li(e) = 0 and write

 Li(x) = e

x

dt . ln t

Using the computer, test whether Li(x) actually approximates the prime counting function. Make sure you check a couple of domains. See also figure 1.8 on page 19. c. For the following exercises you need to know the technique of partial intergration. Part (b) suggests that Li(x) and π (x) agree to some extent, but what about the relative error? We will see that the relative error, as expected, converges to zero. First show by partial integration that  x x dt Li(x) = . −e + 2 ln x e (ln t) d. Next prove, again by partial integration, that  x dt x x + 2 − 2e . + Li(x) = ln x (ln x)2 (ln x)3 e e. Below we will show that the integral from part (d) is less than √

x+

Cx (ln x)2

for certain C > 0. First prove that this implies that indeed the relative error between Li(x) and L(x) goes to zero. f. Does this automatically imply that the relative error between Li(x) and π (x) also goes to zero? x g. We will now show that the integral e (lndtx)3 from part (d) is less than √ √ x + (lnCxx)2 . Divide the integral into two pieces: the part from e to x √ √ and the part from x to x. Prove that the integral from e to x is less √ than x. h. For the second integral we will argue as follows. Show that  x  x dt dt 1 (see the text preceding the exercise) and since limx→∞ 21−x = 0, we have limx→∞ 1−211−x = 1. So the graph of the zeta function is in between the horizontal line y = 1 and the graph of the function 1−211−x , which has this line as an asymptote. Therefore the graph of zeta function also has this line as an asymptote. b. The vertical asymptote must be the line x = 1 since ζ (1) = +∞ and ζ (x) is finite for all x > 1. To prove that this line really is a vertical asymptote, we must show that ζ (x) can be made arbitrarily large by choosing x sufficiently close to 1. Imagine a very big number M, for instance one million. Since ζ (1) = 1 + 12 + 13 + 14 + . . . = ∞, by adding a sufficiently large number of terms, we can get higher than M. Then we have

1+

1 1 1 + + ··· + > M, 2 3 N

111

Solutions to the exercises

in which N is that number of terms. For x > 1 and sufficiently close to 1, then also 1+

1 1 1 + x + ··· + x > M 2x 3 N

and, consequently, ζ (x) = 1 +

1 1 1 1 1 + x + · · · > 1 + x + x + · · · + x > M. x 2 3 2 3 N

So, indeed, it is possible to get ζ (x) > M by choosing x sufficiently close to 1. This reasoning holds for any M, so the line x = 1 is a vertical asymptote. c. The range of the zeta function therefore is the open interval (1, ∞). A graph of ζ (x) is shown in figure 2.6 on page 33.

Exercise 2.3. a. How many terms of the series for sine do you need to get an approximating polynomial that differs less than 0.1 from the sine on every point of the domain [−7, 7]? Use a computer to find and confirm your answer. b. The cosine is the derivative of the sine. How does this follow by looking at their series? You can differentiate series term by term. c. Plugging in values of x into the sine and cosine series reveals a wealth of interesting identities. For example, try to find out how the series imply the following: −2 = −

π2 π4 π6 π8 + − + − ··· 2! 4! 6! 8!

Try to find some more of these identities. d. What is the series for ln(1 − x)? For what values of x does it converge? e. Differentiate the series ln(1 + x) term by term. Do you see a connection to the geometric series M(x) ? Compare your answer by directly calculating the derivative of ln(1 + x). f. Can you explain the following formula? ln 2 = 1 −

1 1 1 1 1 + − + − + ··· 2 3 4 5 6

112

Solutions to the exercises

a. Up to the 15-th power the difference still is too big, but from power 17 on it is all right. Then the approximation is sin x ≈ x −

x5 x7 x 17 x3 + − + ··· + . 3! 5! 7! 17!

(For x = 7 the difference from the sine function is approximately 0.08.) 3

2

2

b. The derivative of x is 1, the derivative of − x3! is − 3x3! = − x2! , et cetera. c. Choose x = π in the cosine series and use that cos(π ) = −1. d. For −1 ≤ x < 1 we have ln(1 − x) = −x −

x3 x4 x5 x2 − − − − ··· 2 3 4 5

e. Differentiating term by term yields the series 1 − x + x 2 − x 3 + x 4 − x 5 + · · · = M(−x). 1 Since M(x) = 1−x for −1 < x < 1, we have on that domain M(−x) = 1 and this is exactly the derivative of ln(1 + x). 1+x

f. Take x = 1 in the series for ln(1 + x). Exercise 2.4. a. Using the series for the sine from the previous section show that S(x) =

π2 2 π4 4 π6 6 sin(π x) =1− x + x − x + ··· πx 3! 5! 7!

b. The value of S(0) is a bit problematic because of division by 0. However, using the series you found in part (a) it is still possible to assign a value to S(0). What value should that be? c. Sketch a graph of S(x) on the domain [−4, 4]. d. Determine the zeroes of S(x). a. Substituting π x for x in the sine series yields sin π x = π x − Divide each term by π x.

π 5x5 π 7x7 π 9x9 π 3x3 + − + − ··· 3! 5! 7! 9!

113

Solutions to the exercises

b. S(0) = 1. This is in accordance both with the series and with limx→0 sin(πx) = 1. πx 1

−4

−3

−2

−1

0

1

2

3

4

c. The precise locations of all local maxima and minima are difficult to find, but a global sketch is easily drawn (see the figure and also part (d)). d. The zeroes occur where the numerator is 0, except for x = 0. So the zeroes are the integers ±1, ±2, ±3, . . . Exercise 2.5. a. The Wallis product formula (John Wallis, 1655) is: 2·2 4·4 6·6 8·8 π = · · · ··· 2 1·3 3·5 5·7 7·9 Use the product formula for S(x) to prove this formula. Taking x = 12 in the product formula for S(x) yields

   2 1 1 1 1 = = 1− 2 1− 2 1 − 2 ··· π/2 π 2 4 6 22 − 1 42 − 1 62 − 1 · · ··· 22 42 62 1·3 3·5 5·7 = · · ··· 2·2 4·4 6·6

=

(Recall that k 2 − 1 = (k − 1)(k + 1) for all k.) Reversing the entire formula yields the Wallis product. Exercise 2.6. a. In figure 2.6 compare the graph of ζ (x) to the first ten approximations of the Euler product. For what values of x are the approximations good and where do they fail? Can you explain why? b. Use the computer to plot the approximations on other domains.

114

Solutions to the exercises

c. For the first five approximations, what is the value at x = 1? Check your answers using figure 2.6. d. Why do the approximations form an increasing sequence of functions? a. The approximations are good for large x and bad for x near to 1. At x = 1 the product approximations always are finite while ζ (1) = ∞. Therefore the approximations for x near to 1 never can be good. b. The first five approximating values at x = 1 are 2, 3, respectively.

15 35 , 8 4

and

77 , 16

c. Each new term is greater than 1 (since each denominator is less than 1), so the product approximations form an increasing sequence of functions.

Additional exercises Exercise 2.7. A simple proof that ζ (2) < 2. 1 1 1 + 2·3 + 3·4 + · · · to the series for ζ (2). Compare the series R = 1 + 1·2 a. By comparing the terms of R to the corresponding terms in the series for ζ (2) show that ζ (2) < R. b. Prove that

1 k(k+1)

=

1 k



1 . k+1

c. Use the previous part to conclude 1 + 1 for all n. n+1

1 1·2

+

1 2·3

+ ··· +

1 n(n+1)

=1+1−

d. Show that this implies R = 2 and so ζ (2) < 2. e. Can you use the same technique to show that ζ (3) < 1 14 ? a. Apart from the first term (equal to 1) every term of the series R looks like 1 1 . This is greater than (n+1) 2 , the corresponding term in the series for n(n+1) ζ (2). b. Put the right-hand side over a common denominator. c. Substituting part (b) in each term of the sum (for k = 1 . . . n), each time the second fraction cancels with the first fraction in the next term. 1 1 e. Yes, use the infinite sum 1 + 1·2·3 + 2·3·4 + · · · This sum can be calculated by splitting each fraction into three partial fractions. See also the appendix Computer support. As a bonus, try to tackle ζ (4) in a similar way.

115

Solutions to the exercises

Exercise 2.8. Computations with the series for sine, cosine and tangent. a. According to the double angle formula we have sin(2x) = 2 cos(x) sin(x). Can you also get the left-hand side by multiplying the series for the right hand side? b. The tangent also has a series expansion. To find the coefficients of the powers of x is less straightforward. Using the series for cosine and sine find the first three terms of the series for the tangent. It is easiest to use the relation tan(x) cos(x) = sin(x) for this. c. Show that in each of the series the sum of the first two terms is exactly the formula for the tangent line to the point x = 0 on the graph. 2

4

3

a. The series for 2 cos(x) sin(x) is 2(1 − x2! + x4! − · · · )(x − x3! + · · · ). Expanding term by term:  

1 1 1 1 1 + x3 + 2 + + x5 + · · · = 2x + 2 2! 3! 4! 2!3! 5!

x5 5!



Simplifying the coefficient of x 3 yields: −2( 2!1 + 3!1 )x 3 = −2 3+1 x3 = 3! (2x)3 5 − 3! . Simplifying the coefficient of x (with common denominator 5!) 5

yields in a similar way (2x) . These terms agree with the series for sin(2x). 5! And so on. If you want to do all terms in a general way, you may use Newton’s binomial formula. b. Suppose tan x = a0 + a1 x + a2 x 2 + a3 x 3 + · · · . We will determine the unknown coefficients a0 , a1 , . . . step by step from the relation tan(x) cos(x) = sin(x). Write the cosine and the sine as series, expand parentheses and compare the coefficients. First the left-hand side: 

x4 x2 + − ··· (a0 + a1 x + a2 x 2 + a3 x 3 + . . .) 1 − 2! 4!  

1 1 x 2 + a3 − a1 x3 + · · · = a0 + a1 x + a2 − a0 2! 2! Comparing this with the sine series, we first see that a0 = 0 and a1 = 1. Next, it follows that a2 = 0 and a3 − 12 = − 3!1 so a3 = 13 . The next coefficient is 0. And so on. See also the appendix Computer support. c. The equation of the tangent line to the graph of the function f in 0 is given by y = f (0) + f (0) x. Remark: the general formula for the n-th term in the series for f is in which f (n) denotes the n-th derivative.

f (n) (0) , n!

116

Solutions to the exercises

Exercise 2.9. Computation of ζ (4). In this exercise we will compute ζ (4) in the footsteps of Euler. Just like ζ (2), the formula for ζ (4) is also hidden in the product formula for S(x). a. Find the coefficient of x 4 in the series for S(x) from section 2.3. b. Next find the same coefficient of x 4 using the product formula for S(x) by multiplying out the parentheses. c. Writing ζ (2) as an infinite sum, compare the product ζ (2)ζ (2) with the answer in part (b). d. Show that the coefficient from part (b) is a combination of ζ (2)2 and ζ (4) using part (c). e. Prove that ζ (4) =

π4 90

by comparing the answers in parts (a) and (d).

a. The coefficient of x 4 in the series for S(x) is

π4 5!

b. Expanding parentheses in the product formula yields the coefficient of x 4 as the sum of all terms a 21b2 with a < b and positive integers a and b. c. We take for ζ (2) the infinite series expansion. Expanding parentheses in ζ (2)2 then yields ζ (4) plus two times the sum of all terms a 21b2 with a < b. d. The coefficient of x 4 equals 12 (ζ (2)2 − ζ (4)). e. In the equation 12 (ζ (2)2 − ζ (4)) =

π4 5!

the only unknown is ζ (4).

Remark: In principle, all the other positive even zeta values ζ (2n) may be found in a similar way. Exercise 2.10. There are more primes than there are squares. a. Write ln(ζ (x)) as a sum of terms of the form − ln(1 −

1 ). px

b. Using the series for the logarithm ln(1 + x) from section 2.2 show that − ln(1 − p1x ) = p1x + Rp where Rp = 2p12x + 3p13x + 4p14x + · · · c. Take x ≥ 1 and prove that Rp
1 we now have proven that ln(ζ (x)) = R +

1 1 1 1 1 + x + x + x + x + ··· x 2 3 5 7 11

Take the limit x → 1 and conclude that to infinity.

1 2

+

1 3

+

1 5

+

1 7

+

1 11

+ · · · goes

117

Solutions to the exercises

e. Show that there are more primes than there are squares in the sense that the series 112 + 212 + 312 + · · · converges, whereas the series 12 + 13 + 15 + 1 1 + 11 + · · · diverges. 7 a. The logarithm of the Euler product yields a sum of terms ln( 1−1 1 ) in px

which p is a prime. Use the rules for the logarithm. b. Substitute

1 px

for x in the series for ln(1 + x).

c. In the formula for Rp , first factor out p12 . Since p is a prime, p ≥ 2 holds. Therefore we can compare the expression between parentheses with the geometric series with ratio p1 . Finally remark that 1−1 1 ≤ 2. p

d. As x tends to 1 from above, ζ (x) goes to infinity and so does ln(ζ (x)). This means that the sum of all terms p1 evidently must also go to infinity, since R is a finite number and thus may be ignored. e. The series with the reversed squares converges to ζ (2) < ∞. Exercise 2.11. The cosine product formula. ). a. Determine the zeroes of the function C(x) = cos( πx 2 b. In the footsteps of Euler write down a product formula for the function C(x). c. Test your formula by computer by comparing the graph of C(x) with the graph resulting from the product of first few terms of your formula. d. Using your formula prove that 1 1 1 1 π2 1 . + 2 + 2 + 2 + 2 + ··· = 2 1 3 5 7 9 8 e. Compare the product formulas for S(x) and C(x). Does one follow from the other? a. The zeroes are precisely the odd integers. b. C(x) = (1 − 2

x x x )(1 − x1 )(1 − −3 )(1 − x3 )(1 − −5 )(1 − x5 ) · · · −1 2 2 − x32 )(1 − x52 ) · · · So we have written C(x) as

= (1 − x12 )(1 uct over all positive odd integers.

a prod-

d. Comparing coefficients of x 2 in the product formula and the series formula for C(x) yields the desired result, exactly as in the computation of ζ (2). Remark: it may also be derived directly from the formula for ζ (2).

118

Solutions to the exercises

e. Remark that S( x2 )C(x) = S(x). Compare this with the double angle formula. Exercise 2.12. ζ (2) and the average sum of the divisors of a number. Let σ (n) be the sum of the divisors of n. a. Write down all divisors of the numbers 1 to 10 and compute σ (1), σ (2), . . . , σ (10). Define A(n) to be the average of the numbers σ (1), σ (2), . . . , σ (n). b. The graph of A(n) seems close to a straight line. With the help of a computer find approximate values of its slope using A(10), A(100) and A(1000). Let S(n) = nA(n). Then S(n) is the sum of all divisors of all numbers k between 1 and n. Any such divisor can be represented by a point (x, y) in the plane such that x and y are positive integers and x × y ≤ n. To get S(n), one has to add the x-coordinates of all points. c. For n = 10, plot such points (x, y) in the plane and check that adding their first coordinates does indeed yield the correct values for S(10) and A(10). d. Show that the number of points (x, y) with y = 3 lies between n3 − 1 and n and that the number of points on the horizontal line of height y is 3 between yn − 1 and yn . e. Prove that the sum of the x-coordinates of the points on the line of height n2 n n2 n y lies between 2y 2 + 2y and 2y 2 − 2y . Check your answer for y = 3 and n = 10. To obtain an estimate for S(n) we have to add all contributions from all horizontal lines. Define 1 1 1 + + ··· + , 1 2 n 1 1 1 H2 (n) = 2 + 2 + · · · + 2 . 1 2 n H (n) =

f. Do you see how the numbers H (n) are related to the harmonic series from section 2.1? And do you see a relation between H2 (n) and ζ (2)? g. Prove that part (f) gives an upper bound for S(n) that reads: S(n) ≤

n n2 H2 (n) + H (n). 2 2

Also find a lower bound for S(n).

119

Solutions to the exercises

h. Use Oresme’s trick from section 2.1 to show that H (n) ≤ T2 (n), where T2 (n) is the counting function that counts the powers of 2 introduced in section 1.3. i. Use the estimate from part (g) and the bounds from part (h) to prove that S(n) ∼ j. Conclude that A(n) ∼

n2 ζ (2). 2

π2 n ζ (2) = n . 2 12

10

8

6

4

2

2

4

6

8

10

a. 1 only has a single divisor, 1 itself, so σ (1) = 1. The prime numbers p have divisors 1 and p so σ (p) = p + 1. Next 4 has divisors 1, 2 and 4 so σ (4) = 7. The number 6 has divisors 1, 2, 3 and 6 so σ (6) = 12. The number 8 has divisors 1, 2, 4 and 8 so σ (8) = 15. The number 9 is divisible by 1, 3 and 9 so σ (9) = 13 and finally 10 has divisors 1, 2, 5 and 10 so that σ (10) = 18. b. Using Wolfram Alpha we found that A(10)/10 = 0.87, A(100)/100 ≈ 0.8299 and A(1000)/1000 ≈ 0.8231. c. For n = 10 we have drawn all points (x, y) with positive integer coordinates satisfying xy ≤ 10. Adding the x coordinates of all points in each horizontal line and finally adding those together we get 55 + 15 + 6 + 3 + 3 + 1 + 1 + 1 + 1 + 1 = 87 = S(10) and A(10) = 8.7.

120

Solutions to the exercises

This agrees with the numbers found in part (a): σ (1) + · · · + σ (10) = 1 + 3 + 4 + 7 + 6 + 12 + 8 + 15 + 13 + 18 = 87. d. The number of points (x, 3) such that 3x ≤ n is the number positive multiples of 3 less than or equal to n. If n is divisible by 3, then there are exactly n3 such multiples and otherwise this number is rounded down to the nearest integer. Likewise the number of points (x, y) for some fixed number y ≤ n, is the number of positive multiples of y that are less than or equal to n. There can never be more than yn such multiples and if y does not divide n, then yn is rounded down to an integer that is always more than yn − 1. e. The sum of the x-coordinates of the points along a horizontal line at height y looks like 1 + 2 + 3 + · · · + k where the number k is either yn or the sum is between more than yn − 1. Since 1 + 2 + 3 + · · · + k = k(k+1) 2 n2 n n2 n − 2y and 2y 2 + 2y . For example for n = 10 the line at height y = 3 2y 2 contains 3 such points and the sum of their x-coordinates is 6. This is n2 n n2 n consistent with our estimates since 2y 2 − 2y ≈ 3.9 and 2y 2 + y ≈ 7.2. f. The harmonic numbers H (n) are a truncated version of the harmonic series defined in section 2.1. It was shown there that limn→∞ H (n) = ∞. In other words the harmonic series diverges. Similarly the second harmonic numbers H2 (n) are a truncated version of the series for ζ (2). In particular 2 limn→∞ H2 (n) = ζ (2) = π6 . g. Recall that S(n) is the sum of the x-coordinates of all points on the horizontal lines. In part (e) we showed that the sum of the points on the n2 n height y line is at most 2y 2 + 2y . Since the heights y of the lines run from y = 1 to y = n we will get an upper bound from S(n) if we sum

n2 2y 2

+

n 2y

2

for y = 1 · · · n. The quadratic parts add up to n2 H2 (n) and the rest adds up 2 to n2 H (n). This gives the upper bound S(n) ≤ n2 H2 (n) + n2 H (n). To find a lower bound we use the lower bound for the contribution of the points n2 n in a single line from part (f). This is 2y 2 − 2y so by the same reasoning we find the lower bound

n2 H2 (n) 2

− n2 H (n) ≤ S(n).

h. The trick is explained in figure 2.2 except that now we set x = 1 and the series is truncated at the number n. Still the idea is the same. We notice 1 where 2a(k) is the biggest power that for any number k we have k1 ≤ 2a(k) of 2 less than or equal to k. If in the sum H (n) we now replace each term k1 1 by 2a(k) , then we will count every power of 2 less than n with total weight 1 except the last one, so H (n) ≤ T2 (n).

121

Solutions to the exercises

we need to show that 2S(n) goes to ζ (2) as n goes i. To show S(n) ∼ n2 ζ (2) 2 n2 to infinity. We do this in two steps. First we will prove an upper bound 2S(n) ≤ ζ (2) + f (n) where f (n) goes to zero as n → ∞ and then we will n2 where again g(n) goes prove a similar lower bound ζ (2) + g(n) ≤ 2S(n) n2 to zero. To prove the upper bound we start by multiplying the upper bound ≤ H2 (n) + n1 H (n). According to part (h), from part (g) by n22 to get 2S(n) n2 H (n) ≤ T2 (n). Recall that T2 (n) ≤ log2 (n) and that limn→∞ (log2 n)/n = 0 (see section 1.3). It follows that n1 H (n) goes to 0 as n → ∞. Moreover, H2 (n) goes to ζ (2), so our upper bound is complete with f (n) = −ζ (2) + H2 (n) + n1 H (n). The lower bound is proven in exactly the same way. j. If we remember that A(n) = n1 S(n) and directly from part (i).

ζ (2) 2

=

π2 , 12

then this follows

Chapter 3: The Riemann hypothesis Exercise 3.1. a. Show that η(1) = ln 2 using the series for ln(1 + x) from chapter 2. b. Compute η(2) exactly. c. Write down the series for ζ (x) − η(x). d. Show that ζ (x) − η(x) = 21−x ζ (x) so that ζ (x) = a. η(1) = 1 −

1 2

+

1 3



1 4

+

1 5



1 6

1 η(x). 1 − 21−x

+ · · · = ln 2.

b. Using the formula on page 45 we find η(2) = (1 − 21−2 )ζ (2) = 2 1 π2 = π12 . 2 6 c. From the series of ζ (x) and η(x) follows that ζ (x) − η(x) = d. Factor out

2 2x

2 2 2 2 + x + x + x + ··· x 2 4 6 8

= 21−x from each term:

ζ (x) − η(x) = 2

1−x

1 1 1 1 + x + x + x + ··· 2 3 4

 = 21−x ζ (x),

hence ζ (x)(1 − 21−x ) = η(x). Division by 1 − 21−x yields the desired result.

122

Solutions to the exercises

Exercise 3.2. a. Compute the (complex) zeroes of x 2 + 2x + 2 using the quadratic formula. b. Check your answer by plugging the found zeroes into the formula. √ √ √ a. x = −2±2 −4 = −1 ± i since −4 = 2 −1 = 2 i . b. (−1 ± i )2 + 2(−1 ± i ) + 2 = 1 − 1 − (±2 i ) − 2 ± 2 i + 2 = 0. Exercise 3.3. a. Try to prove the following: π e −π i = −1, e 3π i = −1, e 2 i = i , e 2π i = 1, e 2kπ i = 1 for all integers k. b. Show that e − i ϕ = cos ϕ − i sin ϕ. c. Convince yourself that cos ϕ =

e i ϕ +e − i ϕ 2

and sin ϕ =

e i ϕ −e − i ϕ . 2i

d. Prove: if z = x + i y (where x and y are real), then e z = e x cos y + i e x sin y. Using Euler’s circle formula, we get a. e −π i = cos(−π) + i sin(−π) = −1, e 3π i = cos(3π ) + i sin(3π) = −1, e

π 2

i

= cos π2 + i sin π2 = i ,

e 2π i = cos(2π ) + i sin(2π) = 1, e 2kπ i = cos(2kπ) + i sin(2kπ) = 1. b. e − i ϕ = e i (−ϕ) = cos(−ϕ) + i sin(−ϕ) = cos(ϕ) − i sin(ϕ). c. e i ϕ + e − i ϕ = cos ϕ + i sin ϕ + cos ϕ − i sin ϕ = 2 cos ϕ. Similarly e i ϕ − e − i ϕ = 2 i sin ϕ. Next, divide by 2 or 2 i , respectively. d. e z = e x+ i y = e x e i y = e x (cos y + i sin y) = e x cos y + i e x sin y. Exercise 3.4. a. Using a computer, check for a few negative even integers z that indeed ζ (z) = 0. b. Also check that the zeta function vanishes at the non-trivial zeroes given in figure 3.5. Since the y-values are approximate, you will not get exactly zero. c. To get a better view of the zeroes, it is useful to plot the absolute value |ζ ( 21 + i y)|, see figure 3.7 on page 54. Use your computer to check that there are exactly ten zeroes on the critical line between 0 and 50.

Solutions to the exercises

123

d. Can you find the next zero by exploring further? Try first plotting from 0 to some number M > 50 to find the rough location, and then try zooming in. Find the next zero to at least two digits accuracy. e. Draw a graph of |ζ ( 21 + i y)| for −20 ≤ y ≤ 20. Do you notice anything special? Explain what you see, using the remark at the end of section 3.5. d.

+ 52.97 i is the next zero. First find out that it is somewhere in the neighborhood of 12 + 53 i and then zoom in. As a matter of fact, in more decimals accuracy this zero is 12 + 52.9703214777 i . For much more decimals, see also the following web site: http://www.dtc.umn. edu/˜odlyzko/zeta_tables/zeros2. Furthermore, it should be stressed that this is no proof, but rather an ‘experimental’ verification. To really prove that there is a zero at this point, more advanced mathematical techniques must be used. 1 2

e. The graph is symmetric in the vertical axis since |ζ ( 21 + i y)| = |ζ ( 21 − i y)| for all real y. Indeed, for all z we know that ζ (z) = ζ (z) so |ζ (z)| = |ζ (z)|.

Additional exercises Exercise 3.5. A variation on the zeta function. In this exercise we will study the following variation on the zeta function: 1 1 1 1 1 1 1 1 − x + x − x + x − x + x − x + ··· 1x 3 5 7 9 11 13 15 As it turns out, this function also has a product formula in the style of Euler. D(x) =

a. Choose the minus signs in the formula below so the equation holds true:      1 1 1 1 D(x) = ··· 1 ± 31x 1 ± 51x 1 ± 71x 1 ± 111x b. Using the method from section 3.1. give a proof of the formula you found in part (a). c. Also explore the behavior of D(x) as x decreases to 1. a. The term

1 1± p1x

gets a minus sign if p is a prime of the form 4n + 1 and a

plus sign if p is of the form 4n + 3. b. First compute (1 + 31x )D(x). The result is exactly equal to D(x) from which all terms with a base that is a multiple of 3 have been removed.

124

Solutions to the exercises

Going on in this manner, always using the correct plus or minus signs, all terms are removed until at last only 1 remains. c. Unlike for the zeta function, D(1) turns out to be a finite number. It is possible to show that D(1) = π4 , e.g., by plugging in x = i in the series for ln(1 + x). Exercise 3.6. Extending the domain of a series. Consider the following two series: (x + 1)3 1 (x + 1) (x + 1)2 + + + + ··· 2 4 8 16 

 

1 2 1 1 3 +8 x− N(x) = 2 + 4 x − + 16 x − + ··· 2 2 2

M(x) =

a. Find the values of x such that the series M(x) and N (x) converge. b. Prove: if both M(x) and N (x) converge, then M(x) = N (x). This implies that one can be used to extend the domain of the other. c. Can you find another series that will help to extend the domain even further? d. Do you see a relation between M, N and the geometric series? a. M(x) converges for all x with |x + 1| < 2, N (x) converges for all x with |x − 12 | < 12 . b. On their domains M and N both are equal to a geometric series with sum 1/(1 − x). c. The series

1 3

+

(x+2) 9

+

(x+2)2 27

+ · · · extends the domain to (−5, 1).

1 . d. In this exercise we have seen examples of series for the function 1−x Given a number a = 1 we always can construct such a series in powers of (x − a) as follows:

1 1 1 1 = = 1−x 1 − a − (x − a) 1 − a 1 − x−a 1−a =

x−a 1 (x − a)2 + + + ··· 1−a (1 − a)2 (1 − a)3

Above, we presented the cases a = 12 , −1 and −2. Exercise 3.7. Surprises with the complex sine and cosine functions. a. Using the formulas from exercise 3.3.c, sketch the graphs of f (x) = cos( i x) and g(x) = i sin( i x) for real values of x.

125

Solutions to the exercises

b. Show that for complex z the functions cos(z) and sin(z) can grow arbitrarily large. c. Can you find complex numbers z such that cos2 z + sin2 z = 2? −x

−x

and i sin( i x) = sinh(x) = e −e . These a. cos( i x) = cosh(x) = e +e 2 2 functions are called the hyperbolic cosine and the hyperbolic sine, respectively. x

x

b. For large real x > 0 we have cosh(x) = e x −e −x ≈ 12 e x , which is very large. 2

e x +e −x 2

≈ 12 e x and sinh(x) =

c. No. From the series formulas it is clear that ez = cos z + i sin z for all complex numbers z. From this follows that cos2 z + sin2 z = (cos z + i sin z)(cos z − i sin z) = ez e−z = 1. The rule of thumb is that in going from real to complex, inequalities loose their validity, but equalities still hold. Exercise 3.8. Complex powers a. Compute i i . More precisely: rewrite this number in the standard form x + i y. b. Show that ( i i ) i = − i . c. Do you see a pattern? What would (( i i ) i ) i be? a. i = e 2 i so i i = e 2 i = e− 2 . π

π

b. ( i i ) i = (e− 2 ) i = e− π

2

πi 2

π

= −i.

c. The pattern repeats: ((( i i ) i ) i ) i = i . Exercise 3.9. Evaluating the zeta function. a. Find an approximation of ζ (5 + 3i) by hand, using the computation at the end of section 3.5. b. Check your answer of part (a) by using a computer. c. Why does the series for ζ (5 + 3i) converge? a. ζ (5 + 3 i ) = 2−5 cos(3 ln 2) − i 2−5 sin(3 ln 2) + 3−5 cos(3 ln 3) − i 3−5 sin(3 ln 3) + 5−5 cos(3 ln 5) − i 5−5 sin(3 ln 5) + 7−5 cos(3 ln 7) − i 7−5 sin(3 ln 7) + · · · b. ζ (5 + 3 i ) ≈ 0.9804 − 0.0254 i .

126

Solutions to the exercises

1 1 1 1 1 1 c. ζ (5 + 3 i ) = 15+3 i + 25+3 i + 35+3 i + · · · ≤ | 15+3 i | + | 25+3 i | + | 35+3 i | + 1 1 1 · · · = 15 + 25 + 35 + · · · = ζ (5) < ∞.

Exercise 3.10. The argument of the zeta function. a. Using the computer, plot the argument of the zeta function along the critical line. In other words study the graph of the function g(y) = arg(ζ ( 21 + i y)) for real values of y. b. Notice that the graph exhibits certain jumps. When do they occur? c. How big are the jumps? d. Can you explain this jumping behavior? b. The graph seems to jump precisely at each zero of the zeta function. c. The jump always is equal to exactly π . A difference in argument of π hints to a change of sign; just find the difference in arguments of z and −z. d. We only can explain this if we suppose that the zeroes of the zeta function on the critical line are simple zeroes. Unfortunately, as yet, nobody has succeeded in proving that this is true. By a simple zero we mean that the derivative at a zero is not also zero. If 12 + i p is a simple zero of ζ , then close to 12 + i p the function ζ (z) is well approximated by a first degree polynomial:

 



1 1 1 + ip + ζ + ip z− + ip ζ (z) ≈ ζ 2 2 2



 1 1 + ip z− + ip . = ζ 2 2 As soon as z = 12 + i y on the critical line passes the zero 12 + i p, the y − p changes sign, and consequently also z − ( 21 + i p) = ( 21 + i y) − ( 12 + i p) = i (y − p) exhibits a sign change. If ζ ( 12 + i p) = 0 this means a jump of π in the argument of ζ (z). Exercise 3.11. All algebraic equations have solutions. a. Show that any complex number z can be written as z = |z|e ϕ i for some real angle ϕ. This angle is the argument arg(z) we introduced in the previous exercise. b. Show that multiplying two complex numbers multiplies their absolute values and adds their arguments.

127

Solutions to the exercises

c. What is the image of the complex function f (z) = z10 if z runs once around the circle with center 0 and radius R? Now we can show that, for example, g(z) = z10 + z4 + 3 has at least one zero in the complex plane. To this end, we first investigate the curve that is described by g(z) if z runs once around the circle with center 0 and radius R. d. Comparing g(z) to f (z), show that for very large R the image of g(z) is a curve that winds ten times around the origin. Why? And how large do we need R to be for this to be true? e. For very small R, the image of g(z) is a curve that doesn’t wind around the origin at all. Why? How small is small in this case? f. Conclude that there must be at least one value of R for which the image of g(z) is a curve that passes through the origin. Why does this imply that g(z) has a zero? g. Replace g(z) by any polynomial g(z) = zn + an−1 zn−1 + · · · + a0 and f (z) by zn (with n ≥ 1). Show that g(z) must have at least one zero in the complex plane.  a. If z = x + i y, then |z| = x 2 + y 2 . For z = 0 we take arg(z) = ϕ as the angle between the positive x-axis and the arrow connecting the origin to the point z. The argument arg(z) is determined up to integer multiples of 2π . b. z1 z2 = |z1 |e ϕ1 i |z2 |e ϕ2 i = (|z1 ||z2 |)e (ϕ1 +ϕ2 ) i . c. The circle of radius R around the origin is mapped by f (z) = z10 onto the circle with radius R 10 around the origin. If arg(z) increases from 0 to 2π , then arg(z10 ) = 10 arg(z) increases from 0 to 10 × 2π , so the image of the circle |z| = R is a circle that winds 10 times around the origin. d. g(z) = z10 + z4 + 3 = z10 (1 + (z4 )/(z10 ) + 3/(z10 )). For big values of |z| = R, the absolute value of (z4 )/(z10 ) + 3/z10 can be made arbitrarily small. For instance, taking R > 2 we have      g(z) − f (z)   z4 1 3  3 =    z10 + z10  ≤ |z|6 + |z|10  f (z) =

3 1 1 3 1 + < . + 10 < 6 R R 64 1024 32

1 Then |g(z) − f (z)| < 32 |f (z)|, and since f (z) = z10 winds ten times around the origin if z runs once around the circle |z| = R, the same holds for g(z) = f (z) + (g(z) − f (z)).

128

Solutions to the exercises

e. For small values of R, the image of the circle stays close to 3 since g(0) = 3. Take, e.g., R < 12 . Then |z10 | < 1/1024 and |z4 | < 1/16, so |g(z) − 3| = |z10 + z4 | < 2/16 = 1/8. Hence g(z) is too close to 3 to wind around the origin. In the figure below we have drawn the curve for R = 2 (top) and R = 12 (bottom). In the bottom figure, the center is the complex number z = 3, in the top figure the center is the origin. Along the border xy-coordinates are indicated.

129

Solutions to the exercises

f. Letting R increase from 12 to 2, by continuity there must be values of R where the image of the circle |z| = R passes through the origin. Every passing through the origin yields a zero of g(z). To see how one may use the computer to actually plot the image of the circle with radius R for various values of R; see the appendix Computer support. g. In the general case, a similar reasoning applied to g(z) = zn + an−1 zn−1 + · · · + a0 yields the proof that g(z) has at least one zero. Actually, degree n equations (with n ≥ 1) have exactly n complex solutions if you count each solution with its proper multiplicity. The idea is that once you have found a zero z = a of a polynomial g(z) of degree n, you can factor it out by writing g(z) = (z − a)h(z), where h(z) is a polynomial of degree n − 1. Going on in this way, the polynomial g(z) can be factored completely into n linear factors.

Chapter 4: Primes and the Riemann hypothesis Exercise 4.1. a. Compute ζ (−1). b. In one of his famous notebooks the Indian genius Srinivasa Ramanujan (1887–1920) writes: 1 + 2 + 3 + 4 + 5 + 6 + ··· = −

1 . 12

Explain why this is not as crazy as it looks. c. Compute (− 21 )! using Riemann’s functional equation. d. By plugging z = −1 into the functional equation deduce that the factorial function z! must have a pole at z = −1. a. Take z = 1 in Riemann’s functional equation. Since ζ (2) = 1 . and 1! = 1 we have ζ (−1) = − 12

π2 , sin π2 6

=1

b. Ramanujan, not knowing the work of Euler or Riemann when he wrote this, investigated the series 1 1 1 1 + z + z + z + ··· z 2 3 4 5 (that we know as the series of the zeta function) for several values of z. Apparently he had ideas about extending the function defined by the series also for values of z for which the series does not converge. Such a z value is z = −1. Just plugging this in the series yields 1+

1 + 2 + 3 + 4 + 5 + ···

130

Solutions to the exercises

Ramanujan apparently had discovered that this ‘extended function’ takes 1 in this point, in accordance with Riemann’s functional the value − 12 equation (see part (a)). c. For z = − 12 we have −z = 1 + z = 12 . Then in Riemann’s equation on both sides ζ ( 21 ) appears, which is a number unequal to zero. Dividing both sides by ζ ( 21 ) yields

  π  −2 − 12 ! −π 1 2 1= sin = − ! . sin √ (2π)1/2 4 2 4 2π Since sin

π 4

=

1 2

√ √ 2 it follows that − 12 ! = π .

d. We already know that ζ (0) = − 12 . Furthermore, sin(− π2 ) = −1 and (2π)0 = 1. If (−1)! were finite, the right-hand side of Riemann’s equation would be finite, too. But the left-hand side is ζ (1), and for z = 1 the zeta function has a pole. The function z! therefore must have a pole at z = −1. For those who want to know more about the factorial function, which in higher mathematics appears as the Gamma function (z), where

(z) = (z − 1)!, we refer to http://en.wikipedia.org/wiki/ Gamma_function.

Exercise 4.2. a. Show that the point 12 in the complex plane is exactly in between the points −3 − 2 i and 4 + 2 i . b. Prove that for arbitrary complex numbers z, the point middle between z + 1 and −z.

1 2

is exactly in the

c. Suppose a > 0 is a positive real number and z = x + i y is an arbitrary complex number. Prove that |a z | = a x . d. Suppose z0 = x0 + i y0 is some fixed complex number with x0 > 1. Prove that the terms 1−1 1 in the Euler product for ζ (z0 ) converge to 1 as p goes p z0 to infinity. a. The x-coordinate of the point halfway between −3 − 2 i and 4 + 2 i is 1 , while the y-coordinate is 0. 2 b. If z = x + y i , then z + 1 = 1 + x + y i and −z = −x − y i . Just as in part (a) it follows that the midpoint of z + 1 and −z is 12 + 0 i = 12 . c. |a z | = |e z ln a | = |e (x+ i y) ln a | = |e x ln a ||e i y ln a | = e x ln a = a x .

131

Solutions to the exercises

d. For each number p > 0 we have 1 = p−z0 = e −z0 ln p = e (−x0 −y0 i ) ln p = e −x0 ln p · e (−y0 ln p) i pz0 = e −x0 ln p cos(y0 ln p) − i e −x0 ln p sin(y0 ln p). From x0 > 1 and p → ∞ it follows that e x0 ln p → ∞ so e −x0 ln p = 1 → 0. Since −1 ≤ cos(y0 ln p) ≤ 1 and −1 ≤ sin(y0 ln p) ≤ 1, e x0 ln p from this it follows that e −x0 ln p cos(y0 ln p) → 0 and e −x0 ln p sin(y0 ln p) → 0. Conclusion: p1z0 → 0 so 1− p1z0 → 1 and also 1−1 1 → 1. p z0

Note that in this derivation we did not use that x0 > 1. In fact, the condition x0 > 0 is already sufficient. We only considered the individual terms in the Euler product. They all converge to 1 for p → ∞ if x0 > 0, so if z0 is in the right half plane. But in order to prove that the Euler product itself, i.e., the infinite product of those terms, converges to a finite value, and, moreover, that this value is not 0, it is necessary to suppose that x0 > 1. To prove this, however, is outside the scope of this book. Exercise 4.3. a. Prove that sin z = 0 implies that z is real by writing the sine in terms of two exponential functions. b. Same question for the cosine and the tangent. c. Does the exponential function itself have any zeroes? a. sin z = 21i (e i z − e − i z ) so sin z = 0 is equivalent to e i z − e − i z = 0, i.e., e 2 i z = 1. Write z = x + i y; then e 2 i z = e 2 i x e −2y . It follows from e z = 1 that y = 0 and 2 i x = 2kπ i , so indeed all zeroes of the sine function are the real numbers x = kπ . b. and c. are proven in a similar way (the answer to (c) is “no”). Exercise 4.4. a. Let us have a closer look at the contribution f (x) = − 12 ln(1 − x12 ) of the trivial zeroes. Compute f (x) and simplify the expression you find as much as possible. In this way prove that f (x) < 0 for all x > 1. b. Suppose y = f (x) with x > 1. Express x in terms of y. c. Find x0 > 0 such that f (x0 ) = 0.001. Make sure your answer is accurate to at least three digits. d. Show that for all x > x0 we have     x u   1.8368 < ψ(x) − x +  < 1.8379.  u u

132

Solutions to the exercises

a. The chain rule yields f (x) = −

1 1 −2 1 1 (−1) 3 = − 3 =− 2 1 − x12 x x −x x(x 2 − 1)

and this, indeed, is negative for all x > 1. b. From y = − 12 ln(1 − x12 ) it follows that −2y = ln(1 − x12 ), so e −2y = 1 − x12 , hence x12 = 1 − e −2y . Since x > 1 we may conclude that  1 . x = 1−e1−2y = √1−e −2y c. x0 = 22.372 d. ψ(x) − x +



xu u u

= − ln(2π ) + f (x), so for x > x0 we have     x u   1.8368 < ln(2π ) − 0.001 < ln(2π ) − f (x) = ψ(x) − x +   u u < ln(2π ) < 1.8379.

Exercise 4.5. In the table in figure 3.5 on page 53 the first ten non-trivial zeroes in the upper half plane u1 , u2 , . . . , u10 are given. a. For each of these uk compute its absolute value |uk | to three decimals and compare your answer to wk . b. Also compute the corresponding angle αk for every uk . Use radians and round to three digits. How close are your answers to π2 ? c. Plot some ‘cosine contributions’ to the explicit formula and compare them to the graphs in figure 4.5 on page 77. k

wk

|uk |

αk

1 2 3 4 5 6 7 8 9 10

14.134725 21.022040 25.010856 30.424878 32.935057 37.586176 40.918720 43.327073 48.005150 49.773832

14.144 21.028 25.016 30.429 32.939 37.590 40.922 43.330 48.008 49.776

1.535 1.547 1.551 1.554 1.556 1.557 1.559 1.559 1.560 1.561

Compare π/2 ≈ 1.5707963. Explanation  of part (a):

|uk | = vk2 + wk2 (see section 4.4). For the given zeroes we always have vk = 12 , so, e.g., |u1 | =  (0.5)2 + (14.134725)2 ≈ 14.144. Explanation of part (b): tan αk = wvkk so in these cases, where vk = 12 , we have tan αk = 2wk , and using a computer (inverse tangent function) you find the values of αk in the last column.

133

Solutions to the exercises 1.5

1.5

k=2

1 0.5

0.5

0

0

–0.5

–0.5

–1

–1

–1.5

20

40

1.5

60

x

80

100

–1.5

0

0 –0.5

–1

–1 20

40

1.5

60

x

80

100

80

100

80

100

80

100

80

100

k=5

–1.5

20

40

1.5

k=6

1

60

x

k=7

1

0.5

0.5

0

0

–0.5

–0.5

–1

–1 20

40

1.5

60

x

80

100

–1.5

0.5

0

0

–0.5

–0.5

–1

–1 40

60

x

40

80

100

–1.5

60

x

k=9

1

0.5

20

20

1.5

k=8

1

–1.5

60

x

0.5

–0.5

–1.5

40

1

0.5

–1.5

20

1.5

k=4

1

k=3

1

20

40

60

x

134

Solutions to the exercises

On page 79 a graph is given of the approximating function x − ln(2π ) −

 

10 2 vk 1 1 ln 1 − 2 − x cos(wk ln(x) − αk ) 2 x |uk | k=1

in which the first two times ten non-trivial zeroes are included. Exercise 4.6. In this exercise we explore the difficulties of summing the infinitely many cosine contributions of the non-trivial zero pairs. ∞  2 vk −1 cos(wk ln(x) − αk ). x |u k| k=1

a. We have indicated why a zeta zero u = v + iw on the right boundary of the critical strip (so v = 1) is in contradiction with the prime number theorem. Why is the same true for a zeta zero on the left boundary of the critical strip (v = 0)? b. In our discussion of the cosine terms above, we mentioned that although each of the terms goes to zero as x goes to infinity, this does not necessarily imply that their sum goes to zero as x goes to infinity as well. Here we will see an example. Instead of cosine terms we have the terms e −x . x k

fk (x) =

Show that for any fixed integer k ≥ 0 the function fk (x) converges to 0 as x → ∞. c. Now we consider the sum F (x) of all terms fk (x). By using the geometric 1 series with ratio e − x prove that F (x) = f0 (x) + f1 (x) + f2 (x) + · · · =

∞  k=0

fk (x) =

1 1 × 1 . x 1 − e −x

d. Finally notice that even though its terms converge to 0 as x goes to ∞, the sum F (x) goes to 1 as x goes to ∞. a. Applying Riemann’s functional equation to any possible zero z = i y (with y real) on the left boundary of the critical strip, would yield a zero 1 − z = 1 − i y on the right boundary, which contradicts the prime number theorem. b. For k = 0 we have limx→∞ f0 (x) = limx→∞ 1/x = 0. For k > 0 we have limx→∞ e −k/x = e 0 = 1, so then also limx→∞ fk (x) = 0.

135

Solutions to the exercises

c. Apply the summation formula for a geometric series (see page 22). d. Consider G(x) = 1/F (x) and put y = 1/x. Then x → ∞ implies y → 0, so 1 1 − e −y (1 − e −y ) = lim . y→0 y y→0 y

lim G(x) = lim x(1 − e −1/x ) = lim

x→∞

x→∞

Using the series for the exponential function e −y = 1 − y + we get

1 2 y 2!

− . . .,

1 − (1 − y + 12 y 2 − . . .) 1 − e −y = lim = 1, y→0 y→0 y y

lim G(x) = lim

x→∞

so also limx→∞ F (x) = 1.

Additional exercises Exercise 4.7. In this exercise you will prove that the term ln(2π ), that (0) . occurs in the explicit formula, is equal to ζζ (0) a. Differentiate the series for the eta function η(x) term by term. b. Take the logarithm of the Wallis product (see exercise 2.5) and show that this equals 2η (0). c. Differentiate the relation η(x) = (1 − 21−x )ζ (x) to prove that ln(2π ) = ζ (0) . ζ (0) a. η (x) = − ln1x1 +

ln 2 2x



ln 3 3x

+ ···

b. The Wallis product yields ln( π2 ) = 2 ln 2 − 2 ln 3 + 2 ln 4 − · · · = 2η (0). c. Differentiating the relation between eta and zeta yields η (x) = (1 − 21−x )ζ (x) + ln(2)21−x ζ (x). Plug in x = 0 and use that ζ (0) = − 12 . Exercise 4.8. The eta function. a. Does the eta function have zeroes too? Where are they located? b. Can you find a functional equation for the eta function similar to that of Riemann for the zeta function? a. From η(x) = (1 − 21−x )ζ (x) it follows that the zeroes of ζ (x) are also zeroes of η(x). Moreover, η(z) is also zero when 1 − 21−z = 0, with the

136

Solutions to the exercises

exception of z = 1. The value z = 1 is not a zero of η(z) since ζ (z) has a pole at that point. The zeroes of 1 − 21−z are the complex numbers i + 1. z = − 2kπ ln 2 b. The same relation between the eta and the zeta function yields  π z  1 − 21+z −2 · z! η(−z) = sin η(1 + z). z+1 (2π) 2 1 − 2−z Exercise 4.9. The existence of non-trivial zeroes of the zeta function. a. Show that the function f (x) = x − ln(2π) − 12 ln(1 − x12 ) is continuous for x > 1. In particular prove that the graph of f (x) does not have any jumps. b. Suppose that the zeta function had only trivial zeroes. Explain how this is in contradiction with the explicit formula. c. In the same way prove that the zeta function must have infinitely many non-trivial zeroes. a. The function f (x) is composed of a finite number of continuous functions, so it is also a continuous function. b. Suppose that the zeta function had only trivial zeroes. Then the explicit formula for ψ(x) would be:

 1 1 ψ(x) = x − ln 2π − ln 1 − 2 . 2 x This is impossible since ψ(x) is not continuous (it has jumps at al prime powers), while the right-hand side would be a continuous function. c. The same argument is valid if there were only finitely many non-trivial zeroes, since then also the right-hand side would be a continuous function. Exercise 4.10. Infinitely many primes of the form 4n + 1. a. Suppose A is the sum of all terms p1 where p is a prime of the form 4n + 1 and suppose B is the sum of all terms p1 where p is a prime of the form 4n + 3. Show that A + B = ∞. b. Taking the logarithm of D(x) from exercise 3.5 on page 55 show that A − B = ln π4 . c. Conclude that there are both infinitely many primes of the form 4n + 1 and infinitely primes of the form 4n + 3.

137

Solutions to the exercises

d. Using the computer can you find any zeroes of the function D(x)? a. The series A + B + 12 is the series of all fractions p1 with p prime. In exercise 2.10 we have proven that the sum of this series is infinity. b. Just like in exercise 2.10 on page 35 we take the logarithm of the Euler product, but this time the Euler product for the function D(x). Letting x go down to x = 1, we get the required expression for A − B. c. From A + B and A − B we find A and B. d. There is also a kind of Riemann hypothesis for the function D(x). It says that also the non-trivial zeroes of D(x) all are on the critical line. The first non-trivial zero of D(x) is approximately 12 + 6.020899 i . Exercise 4.11. A simple step function. It is very difficult to imagine how step functions such as ψ(x) can be expressed in terms of a combination of waves. And yet this is exactly what happened in the explicit formula. To get a feel for this we are going to prove a similar formula for the vastly simpler step function g(t). ⎧ π ⎪ ⎪ ⎨ 4 if 2kπ < t < (2k + 1)π , g(t) = 0 if t = kπ , ⎪ ⎪ ⎩− π if (2k − 1)π < t < 2kπ . 4

Here k is an integer. Step by step we will show that g(t) = sin t +

1 1 sin 3t + sin 5t + · · · 3 5

a. Using the computer compare the graph of g(t) to that of the sum of the first few terms in the above series of sine functions. b. Show that the function F (z) = 1+z has the property that for all z with 1−z |z| ≤ 1, z = 1, −1 it satisfies:

 z3 z5 ln F (z) = 2 z + + + ··· . 3 5 c. Show that F (z) = −F ( 1z ). d. Prove that for all z satisfying |z| ≥ 1, z = 1, −1 it holds that:

 

1 1 1 1 ln F + 5 + ··· . =2 + z z 3z3 5z

138

Solutions to the exercises

e. Now take 0 < t < π and set z = e i t . In other words, take a z on the upper half of the unit circle. Show that in that case we have 1z = e − i t . This is a point on the lower half of the unit circle. f. Prove that for z = e i t and 0 < t < π we have 



   1 3 1 1 5 1 1 + z − 3 + z − 5 + ··· . ln(−1) = 2 z− z 3 z 5 z Conclude that the right-hand side is constant for 0 < t < π . After all it is one of the values of ln(−1) = π i + 2kπ i. Below we will find out which value of ln(−1) is required here. g. First show that z = e i t and 0 < t < π together imply that the right-hand side of the above equation equals 

1 1 4 i sin t + sin 3t + sin 5t + · · · . 3 5 h. Now take t = 12 π . Prove that in this case we must have 

1 1 4i 1 − + − ··· 3 5 and that the expression between parentheses is positive and less than 1. i. Conclude that we need to choose the value ln(−1) = π i in part (f). Explain that this implies that for 0 < t < π we must have sin t + 1 sin 3t + 15 sin 5t + · · · = π4 . 3 j. Define the function h(t) by h(t) = sin t +

1 1 sin 3t + sin 5t + · · · 3 5

Show that for all t we have h(−t) = −h(t) and conclude that for −π < t < 0 we have h(t) = − π4 . k. Prove: h(t) = 0 for all t = kπ (k is an integer) and that h(t) is a periodic function of period 2π . Finally conclude that for all t we have g(t) = h(t), as we set out to prove. a. See the appendix Computer support. b. Use that ln F (z) = ln(1 + z) − ln(1 − z) and subtract the two series. c. Just plug in. d. If |z| ≥ 1, then | 1z | ≤ 1. Substitute

1 z

for z in the series of part (b).

139

Solutions to the exercises

f. In part (c) take the natural logarithm of both sides and transform the resulting equation. g. z − 1z = e i t − e − i t = 2 i sin t, et cetera.

z3 −

1 z3

= e 3 i t − e −3 i t = 2 i sin 3t,

h. sin(2k + 1) π2 = 1 for even k and −1 for odd k. Furthermore,  

 

1 1 1 1 1 1 1 + − + ··· = 1 − − − − − · · · < 1. 0< 1− 3 5 7 3 5 7 9 i. Just plug in. As a matter of fact, we already have seen in the solution to exercise 3.5 that the sum of this series equals π4 . j. For all k we have sin k(−t) = − sin kt. k. For all integer n and k we have sin(2n + 1)(kπ ) = 0 and for all t we have sin k(t + 2π ) = sin kt. Exercise 4.12. Complex prime numbers. a. Prove the remarkable factorization 2 = (1 + i )(1 − i ). The number 2 may be a prime in the ordinary sense of the word but it is definitely not a complex prime! b. The so-called norm N (a + i b) = |a + i b|2 = a 2 + b2 of a complex integer is an ordinary integer and it is positive unless both a = b = 0. Show that the norm is multiplicative: N ((a + i b)(c + i d)) = N (a + i b)N (c + i d). c. Using the norm, prove that the integer 3 (viewed as a complex integer) is a complex prime. Are 1 + i and 1 − i also complex primes? What about −1 + 5 i ? d. Are there infinitely many complex primes? e. Plot all complex primes of norm less than 6 in the plane. For every complex prime you have found, compute its norm. Do you see a pattern? f. Define a zeta-type function called Z(x) that fits well with our new notion 1 of complex integers. Z(x) is the infinite series with terms (a 2 +b 2 )x where a + i b = 0. Write down the first few terms in the series for Z(x). Do you notice something special? g. Find an Euler type product for our zeta-ish function Z(x) in which all the complex primes a + b i satisfying a > 0 and b ≥ 0 occur.

140

Solutions to the exercises

h. Using the product formulas, show that the three zeta-type functions ζ (x), D(x) (see exercise 4.8) and Z(x) we have defined so far, are related by Z(x) = 4 ζ (x)D(x). c. First remark that the units ±1 and ± i are the only complex integers with norm 1. All other complex integers unequal to 0 have norm greater than 1. If z = uv is a composed complex integer and u and v are not units, it follows that N (u) < N(z) and N(v) < N(z). Now suppose that 3 = uv. Since N(3) = 9, then N (u) = N (v) = 3 should hold. But the sum of two squares cannot be 3. The same reasoning holds for any (ordinary) prime p of the form p = 4n + 3 since the sum of two squares cannot be a multiple of four +3 (a square in fact always is a multiple of four or a multiple of four +1). All these ordinary primes therefore are also complex primes. We already know that there are infinitely many of those primes, so this also answers part (d). Convince yourself that 1 ± i are complex primes. However, since −1 + 5 i = (1 + i )(2 + 3 i ) this is not a complex prime. e. They are 1 + i , 1 + 2 i , 2 + i , 3, 2 + 3 i , 3 + 2 i , 1 + 4 i and 4 + i with norms 2, 5, 5, 9, 13, 13, 17 and 17, respectively. Perhaps you have noticed that the norm of a complex prime always is a prime of the form 4n + 1, unless it is on one of the axes. For positive primes of the form p = 4n + 3 the norm is equal to p2 . There is a famous theorem by Fermat saying that each prime p = 4n + 1 can be written in exactly one way as the sum of two squares. Using this theorem we can find all complex primes. f. We combine terms with the same denominator, and order them by ascending norm. All numerators then are integer multiples of 4. 4 8 4 4 8 8 4 + x + x + x + x + x + x x 2 4 5 8 9 10 13 4 8 4 8 12 + x + x + x + x + x + ··· 16 17 18 20 25

Z(x) = 4 +

Notice that each numerator gives the number of ways to write the denominator as the sum of two squares a 2 + b2 , taking into account the order of a and b. Explain this. As an example, verify that 25 indeed can be written in exactly 12 ways as the sum of two squares a 2 + b2 .

141

Solutions to the exercises

g. The original proof of Euler’s product formula also works in this case. The underlying principle is that each complex integer unequal to 0 can be written as a unit times the product of complex primes of the form a + b i with a > 0 and b ≥ 0, and that this notation, up to the order of the complex prime factors, is unique. With Euler’s method, step by step for each complex prime factor, all multiples are removed from the series until only the first term 4 remains. This yields the formula  4 , Z(x) = 1 1 − (a 2 +b 2 )x where the product is taken over all complex primes a + b i with a > 0 and b ≥ 0. h. We write each of the three zeta-like functions Z(x), ζ (x) and D(x) as Z(x) = 4 since all terms in the an Euler product. Next we show that ζ (x)D(x) products cancel. First we take the special complex prime 1 + i . The accompanying Euler factor for Z(x) is 1−1 1 . This factor is canceled by 2x the first factor from ζ (x). Now we look at the primes in the first quadrant not on the horizontal or vertical axis. Their norms are ordinary primes of the form p = 4n + 1. In complex terms these correspond to two distinct primes a + i b and b + i a in the first quadrant with a 2 + b2 = p. Their Euler factors in Z(x) are both 1− 1 1 = 1−1 1 . One of them matches (a 2 +b2 )x

px

the corresponding factor in ζ (x) and the other matches the same factor in D(x). The remaining Euler factors correspond to primes of the form 4n + 3, and in a similar way these factors can be matched. Keep in mind that the Euler factor in Z(x) of such a prime p is equal to 1 1 1 = . 1 1 1 − p2x 1 − px 1 + p1x Remark: The complex integers also are known as the Gaussian integers.

Index i , 46 ζ (2), 30, 34, 114 absolute value, 46, 53, 57, 122 algebraic equations, 58 approximating polynomial, 26 argument, 57 Bays, Carter, 20 Bernoulli numbers, 32 Chebyshev, Pafnuti, 12 complex functions, 48 complex integers, 84 complex natural logarithm, 83 complex numbers, 46 complexe prime numbers, 84 convergence, 22 cosine product formula, 36, 117 counting functions, 17 critical line, 52, 64 critical strip, 51, 64 cryptography, 87 de la Vall´ee Poussin, Charles, 11, 72 Derbyshire, John, 99 divergence, 23 domain, 23 du Sautoy, Marcus, 99 Erd˝os, Paul, 81 error absolute, 10 relative, 8, 9 eta function, 44, 135 Euler, Leonhard, 30 circle formula, 49 computation of ζ (2), 31 computation of ζ (4), 35, 116 product formula for ζ (x), 32, 35, 41 factorial function, 61

factorization, 87 family of prime numbers, 16 functional equation, 59 Gamma function, 130 Gaussian integers, 84, 141 geometric series, 22 convergence, 22 divergence, 23 ratio, 22 sum formula, 23 Hadamard, Jacques, 11, 72 harmonic series, 23 Hudson, Richard, 20 hyperbolic cosine, 94 hyperbolic sine, 94

Littlewood, John E., 20 logarithm, 7 logarithmic integral, 107 logarithmic integral, Li(x), 18 logarithmic prime counting function ψ(x), 12, 14 explicit formula, 59, 67, 69, 76 Mazur, Barry, 100 mirror images, 51 Montgomery, Hugh, 55 music of the primes, 77 natural logarithm, 10 norm of a complex number, 84, 139 Odlyzko, Andrew, 55 Oresme, Nicholas, 23

partial integration, 18, 108 pole, 44 power counting function Tp , 9

143

144 power counting functions, 7 prime, 2 prime counting function π (x), 3 prime number, 1, 2 prime number theorem, 11, 72 prime powers, 11 prime test, 87 Python, 95 Ramanujan, Srinivasa, 63, 129 Riemann, Bernhard, ix, 1, 41, 43, 54 Riemann hypothesis, ix, 14, 16, 34, 41, 52, 54, 55, 60, 64, 65 rotation, 62 RSA, 87 Sage, 91, 95, 100 series, 22 for cos x, 26 for e x , 26 for ln(1 + x), 26 for sin x, 26 for tan x, 35, 115 sine product formula, 31 Skewes, Stanley, 20 Stein, William, 100

Index square counting function, 17, 106 step function, 82, 137 Stopple, Jeffrey, 99 sum of the divisors of a number, 36, 118 theorem of Fermat, 140 variation on the zeta function, D(x), 56, 123 Wallis, product formula, 32, 113 Wolfram Alpha, 91 Zagier, Don, 99 zeroes, 30 zeta function complex series, 50 non-trivial zeroes, 51 trivial zeroes, 51 argument, 57 extension of domain, 43 non-trivial zeroes, 59, 63, 81, 136 product formula, 41 series, 44 summation formula, 21 trivial zeroes, 59, 63, 67