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Introduction I have been told that this Lectorium of the Polytechnic Museum has been a venue for many great names in modern Russian culture, including.

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Presentation on theme: "Introduction I have been told that this Lectorium of the Polytechnic Museum has been a venue for many great names in modern Russian culture, including."— Presentation transcript:

1 Introduction I have been told that this Lectorium of the Polytechnic Museum has been a venue for many great names in modern Russian culture, including poets: Mayakovskiy, Blok, Yevtushenko, Voznesensky, and others. It is a great honor to speak at a place associated with such immortals; but I have the superstitious feeling that I should try to win a little favor with their ghosts before proceeding with my talk. In a spirit of humility, therefore, I offer a few apt lines from Boris Pasternak. I beg you, and any ghosts who may be listening, to excuse my poor Russian.

2 Есть в опыте больших поэтов
Черты естественности той, Что невозможно, их изведав, Не кончить полной немотой. В родстве со всем, что есть, уверясь И знаясь с будущим в быту, Нельзя не впасть к концу, как в ересь, В неслыханную простоту. Но мы пощажены не будем, Когда ее не утаим. Она всего нужнее людям, Но сложное понятней им.

3 The Riemann Hypothesis
A Great Unsolved Mathematical Problem presented by John Derbyshire author of Prime Obsession (2003) and Unknown Quantity (2006)

4 Mr. Derbyshire’s Math Books

5 Mr. Derbyshire’s Math Books (cont.)
Prime Obsession (2003) All about the Riemann Hypothesis Mixes math with history and biography Awarded the 2007 Euler Prize (“for popular writing on a mathematical topic”) by the Mathematical Association of America. Unknown Quantity (2006) A history of algebra for non-mathematicians Published in paperback May 2007 Includes references to Riemann’s work in function theory and topology

6 Mr. Derbyshire’s Most Recent Book
We Are Doomed: Reclaiming Conservative Pessimism (2009)

7 All nontrivial zeros of the zeta function have real part one-half.
The Riemann Hypothesis All nontrivial zeros of the zeta function have real part one-half.

8 Q: What is the RH about ? A: It is about prime numbers.

9 Prime Numbers A prime number doesn’t divide by anything (except itself and 1). 63 divides by 3, 7, 9, and 21, so 63 is not a prime number. 29 divides by … nothing (except 29 and 1), so 29 is a prime number. First few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, , 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101,…

10 More Prime Numbers

11 There is an Infinity of Primes
Do the primes go on for ever? Yes: the Greeks proved it. Proof : Suppose there were a biggest prime, P. Form this number: (12345 P) + 1. Plainly it is bigger than P. Yet it is not divisible by P, nor by any smaller number. If you try, you always get remainder 1! Therefore either it is not divisible by anything at all, or the smallest number that divides it is bigger than P. In the first case it’s a prime bigger than P; in the second, its smallest factor is a prime bigger than P. Since these both contradict my original “suppose,” my original “suppose” must be wrong. There is no biggest prime.

12 Unsolved Problems About Prime Numbers
Goldbach’s Conjecture Every even number after 2 is the sum of two prime numbers (e.g. 98 = ). The Prime Pair Conjecture There are infinitely many pairs of primes just two apart (like 41 and 43). The Riemann Hypothesis

13 The Sieve [Решето] of Eratosthenes
Topic 1 * The Sieve [Решето] of Eratosthenes

14 Eratosthenes of Cyrene
Greek, b.c. Cyrene is in today’s Libya, then part of post-Alexander Greek Egypt under Ptolemy II. All-round intellectual: good achievements in philosophy, astronomy, geography, drama, ethics. Accurately measured the Earth’s circumference, and the tilt of Earth’s axis.

15 The Sieve of Eratosthenes
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109

16 Sieving out the 2’s 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109

17 Sieving out the 3’s 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109

18 Sieving out the 5’s 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109

19 Sieving out the 7’s 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109

20 Result of Sieving By the time I have sieved out the 7’s, a number that escaped sieving would have to NOT be divisible by 2, 3, 5, or 7. Un-sieved numbers would therefore be divisible ONLY by 11 or numbers bigger than 11. The smallest such number (not counting 11 itself) is 11 × 11, which is 121. So . . .

21 Result of sieving (cont.)
By sieving up to 7 (which is the 4th prime), I found all primes less than 1111. In general: By sieving up to the n-th prime, I can find all primes less than the square of the (n+1)-th prime.

22 Topic 2 * The Basel Problem

23 The Basel Problem — Preamble
Math allows you to add up an infinity of numbers and get a definite, finite sum. Example → So this infinity of numbers adds up to one-third. The trick is for the numbers to get smaller and smaller fast enough. ________________

24 Convergence and Divergence
The numbers in the yellow box get smaller very fast. You can keep adding them for ever: the sum is finite. The numbers in the pink box get smaller, but not fast enough: If you add them for ever, the sum is infinite. 0.3 0.03 0.003 0.0003 + ________________ 1 = 1 1/2 = 0.5 1/3 = 1/4 = 0.25 1/5 = 0.2 1/6 = 1/7 = + ____________________

25 The Basel Problem (cont.)
Consider the Harmonic Series: (This was the pink box on the previous slide.) The numbers being added get smaller, but not fast enough. The sum is not finite. It “adds up to infinity.” The Harmonic Series diverges. This was proved by Nicole d’Oresme around 1370. It was proved again by the Bernoulli brothers, Jakob (1682) and Johann (1695). The brothers were successive Professors of Mathematics at the University of Basel, in Switzerland.

26 The Bernoulli Brothers
Jakob ( ) Johann ( )

27 The Basel Problem (cont.)
What about this sum? Does it also diverge? Or does it add to a finite number? If the latter, what is the number? This was the Basel Problem. Both Bernoulli brothers tackled it without success. It was solved at last in 1735 by Leonhard Euler, a native of Basel (though he was living in St. Petersburg, Russia at the time).

28 Leonhard Euler Swiss, Ranked 1st among mathematicians in Murray’s Human Accomplishment. Studied under Johann Bernoulli at Basel. Prolific, pious (Calvinist), domestic. St. Petersburg Academy of Sciences, and Court of Frederick the Great,

29 Charles Murray’s 2003 book Human Accomplishment
Mathematicians ranked

30 The Basel Problem – Solved!
Euler proved that He then pursued the matter further and came up with: The Golden Key

31 Topic 3 * The Golden Key

32 From Particular to General
Fruitful math is often done by starting from a particular result and generalizing it. Having solved the Basel problem, Euler asked: “What if, instead of squares in the denominators, I had some other power ? What can we say about this infinite sum: . . . where ‘s’ is any number at all ?”

33 When Does the Sum Exist? Well, when s = 1, this is just the Harmonic Series, which diverges, because the terms don’t get smaller fast enough. When s is less than 1, the terms get smaller even more slowly. In fact, when s is less than zero, they don’t get smaller at all – they get bigger! So when s = 1 or less, the sum diverges “to infinity.” When s = 2 this is the Basel problem. Euler showed that it then adds up to π2/6 . In fact there is a definite sum for any value of s greater than 1.

34 The Zeta Function So long as s > 1, when I put in a number s (the argument), I will get out a definite sum (the value). This means I have a function. It is called the zeta function. Here is its graph.

35 Euler’s great stroke of genius
Euler’s great stroke of genius was to apply the sieve of Eratosthenes to the zeta function. Why not? In both cases you start off with a list of all the whole numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, . . . You end up with only the prime numbers. Like this . . .

36 Sieving out the 2’s Here’s the zeta function. Call this “Expression 2a.” Multiply both sides by 1/2s: Call that “Expression 2b.” Now subtract Expression 2b from Expression 2a: Look! – I sieved out the 2’s!

37 (A Slight Difference in the Sieving)
(Note a slight difference in my sieving technique. Working the original Sieve of Eratosthenes, I left the first instance of each prime standing. Here I sieve out that first instance with all the rest.)

38 Sieving out the 3’s Call that last result “Expression 3a.”
Multiply both sides by 1/3s: Call that “Expression 3b.” Now subtract Expression 3b from Expression 3a: Look! – I sieved out the 3’s!

39 Sieving out the 5’s Call that last result “Expression 5a.”
Multiply both sides by 1/5s: Call that “Expression 5b.” Now subtract Expression 5b from Expression 5a: Look! – I sieved out the 5’s!

40 *** That is The Golden Key! ***
The Golden Key – At Last! Just as with the original sieve, I can go on sieving for ever. My final result will be: Or, to rearrange slightly: . . . With zeta on the left and all the primes represented on the right. *** That is The Golden Key! ***

41 Why a “Key”? Why “Golden”?
More formally: The Greek capital pi says “Multiply together all such expressions, for all (in this case) primes p.” (What Σ does for addition, Π does for multiplication.) This is a key because it unlocks a door between Number Theory (“the higher arithmetic”) and Function Theory (graphs, smoothness, calculus, etc.) It is golden because all the power of function theory can now be brought to bear on problems about primes.

42 Topic 4 * Zeta’s Hidden Depths

43 What Use is Zeta? The Golden Key tells us that all the properties of the prime numbers are contained somehow in the zeta function. But what use is that? Zeta looks really boring. Remember its graph → Ah, but zeta has hidden depths! First, a slight detour.

44 A Slight Detour Here is a different function, also defined by an infinite sum. Like zeta, it exists only in a limited range. If x is 1 or more, the sum diverges “to infinity.” Likewise if x is -1 or less. Graph of the function

45 A Slight Detour (cont.) However: If Then Subtracting So
Well, that’s nice. But . . .

46 A Slight Detour (cont.) . . . S(x) and 1/(1-x) have different graphs. The graphs are identical between -1 and +1, but outside those bounds S(x) has no values (because the infinite sum diverges), while 1/(1-x) has perfectly good values: e.g. 1/(1-3) = -½.

47 A Slight Detour – The Moral
The moral of the detour is: Using an infinite sum to express a function may “work” over only a part of the function’s range. The function may have values even where the infinite sum “doesn’t work.”

48 The Hidden Zeta Is that the case with zeta? Of course it is!
Here is a graph of zeta between s = –4 and s = 1 → Here’s a graph between s = –14 and s = 0 →

49 A Common Point of Bafflement
“OK, I understand the business with S(x). There’s that honest, well-defined function 1/(1-x). There’s also the infinite sum, but it only works over part of the function’s range. Got that.” “With zeta, though, the infinite sum is all we have. There’s nothing equivalent to that 1/(1-x). Wha? Huh?”

50 The Infinite Sum for ζ(s)
Point of bafflement: The sum only ‘works’ when s is greater than 1. So how can I talk about ζ(½) or ζ(-1)?

51 Relief for the Baffled There are ways to define zeta outside the range where the infinite sum works. Unfortunately they involve heavy-duty math. Best known is the functional equation . . . which gives ζ(1 – s) in terms of ζ(s), using standard mathematical functions (sine, factorial). So if you know the value of ζ(4) , you can work out a value for ζ(-3).

52 More Relief for the Baffled
There are other expressions for zeta outside the range where the infinite sum works, the range s > 1. For any positive whole number n, for example, the following thing is true for any s bigger than –2n:

53 Topic 5 * Zeros of a Function

54 Function Argument, Function Value
Remember how functions work. You put a number in (the argument); you get a number out (the function value). If you put argument 7 into the function x2 – 1, you get out the function value 48. If an argument produces function value zero, that argument is a zero of the function. The zeros of the function x2 – 1 are –1 and 1. Either argument will yield function value zero.

55 Zeros of the Zeta Function
Recall the graph I showed a few frames ago, of the values of ζ between –14 and 0. The graph crosses the x-axis at several points. Its value then is zero. I have found some zeros of the zeta function!

56 All nontrivial zeros of the zeta function have real part one-half.
The Riemann Hypothesis All nontrivial zeros of the zeta function have real part one-half.

57 Trivial and Nontrivial
Yes, we have spotted some zeros of the zeta function. In fact ζ is zero for every negative even whole number argument: –2, –4, –6, –8, –10, –12, . . . Unfortunately these are trivial zeros, of no great interest. The Riemann Hypothesis concerns nontrivial zeros. So Where the heck are they?

58 Topic 6 * Complex Numbers

59 Imaginary Numbers By the rule of signs, the square of a number must be positive. E.g. 2  2 = 4, –2  –2 = 4. There’s no way to get “–4” by squaring something. Negative numbers have no square roots! By the late 16th century this restriction was holding up mathematical progress. Some bold mathematicians began to allow square roots of negative numbers. They figured that you only need a square root for –1. Call it i. Then the square root of –4 is 2i. Now every negative number has a square root. The square root of a negative number is called an imaginary number. Regular numbers are real numbers.

60 Complex Numbers If you multiply two imaginary numbers you get a real number: (–2i)(3i) = 6. If you add two imaginary numbers you get another imaginary number: (–2i)+(3i) = i. If you multiply an imaginary number by a real number, you get an imaginary number: (–2i)(3) = – 6i. If you add a real number to an imaginary number, they “don’t mix”: –2+3i. These “don’t mix” numbers are terrifically useful in math, though. We call them complex numbers. A complex number has a real part and an imaginary part. For –2+3i the real part is –2, the imaginary part is 3i.

61 Complex Arithmetic Arithmetic with complex numbers is very easy.
You just have to remember that i 2 = –1. Example: multiply 1–i by –2+3i. By the ordinary rules of algebra: (1–i) (–2+3i) = –2 + 3i + 2i – 3i 2 = –2 + 3i + 2i + 3 = 1 + 5i

62 Topic 7 * Complex Functions

63 Complex Functions Once complex numbers are allowed, why not use them as arguments for functions? Example: Use the number –2+3i as an argument for the function x 2 – 1. The square of –2+3i, by ordinary algebra, is –5–12i. Subtract 1 for the function value –6–12i. Function theory using complex numbers as arguments and function values – The Theory of Functions of a Complex Variable – is a rich and rewarding field of mathematics.

64 More Advanced Functions of a Complex Variable
More advanced mathematical functions can often be defined by power series – that is, infinite sums of powers. The exponential function, for instance:

65 Complex Functions in History
Complex Function Theory was a huge growth point in early 19th-century math. It was sexy. By 1850 there was a big, solid body of theory. Bernhard Riemann used that theory to investigate ζ as a function of a complex variable. The powerful tools of complex function theory could then be applied to problems about prime numbers.

66 Bernhard Riemann German, 1826-1866
Tied 9th (with Pascal) among mathematicians in Murray’s Human Accomplishment. Introverted, poor, ill, pious (Lutheran). Studied under Gauss at Göttingen. Brilliant imaginative mathematician.

67 Carl Friedrich Gauss German, 1777-1855
Ranked 4th among mathematicians in Murray’s Human Accomplishment. Supervised Riemann’s doctoral thesis (1851). “Parva sed matura.” First stated the Prime Number Theorem (1792)

68 The Distribution of Primes
Topic 8 * The Distribution of Primes

69 The Prime Number Theorem (PNT)
If N is a whole number and π(N) is the number of prime numbers less than N, then π(N) approaches ever more closely to N / log(N) as N gets larger. Number of primes less than N Percentage error N N / log(N) 1,000 1,000,000 1,000,000,000 1,000,000,000,000 168 78,498 50,847,534 37,607,912,018 145 72,382 48,254,942 36,191,206,825 -16.05 -8.45 -5.37 -3.91 To prove the PNT was a great challenge in 19th-century math. The theorem was proved at last in 1896 by two French mathe-maticians, Jacques Hadamard and Charles de la Vallée Poussin.

70 Prime Numbers “Thin Out”
The PNT was the first good idea about the distribution of primes. The prime numbers are scattered among the other numbers in a way that is at the same time random (you never know when the next one will appear) yet orderly (there are rules for how many you can expect to find). This is the fascination. The main feature of the distribution of primes is that they “thin out.” In higher ranges of numbers, there are fewer primes.

71 The Primes “Thin Out” (cont.)
Start with one thousand and count forward in blocks of So the first block goes from 1,001 to 1,100. The second block goes from 1,101 to 1,200,… and so on. The numbers of primes in the first ten blocks are: 16,12,14,11,17, 12, 15, 12, 12, 13 – average 13.4 per hundred numbers. If, instead of starting with one thousand, you start with one million, then the numbers of primes in the first ten blocks are: 6, 10, 8, 8, 7, 7, 10, 5, 6, 8 – average 7.5 per hundred. If, instead of starting with one thousand, you start with one trillion, the numbers are: 4, 6, 2, 4, 2, 4, 3, 5, 1, 6 – average 3.7. This is the “thinning out” of the primes. The PNT was the first mathematical expression for this observed phenomenon.

72 Topic 9 * Riemann’s 1859 Paper

73 Riemann’s 1859 Paper Title: “On the Number of Primes Less Than a Given Quantity” Asks: How many primes are there between 1 and x ? Begins with The Golden Key. Applies complex function theory to the ζ function. Arrives at an exact answer!

74 Riemann’s Answer How many primes are there between 1 and x?
where

75 Good Grief! If you’re not a seasoned mathematician, that looks scary.
Yet in fact it’s made up of standard functions and methods: the Möbius μ-function, the log function, summation and integration. Only Li, the logarithmic-integral function, is not elementary. Here’s its graph for real arguments.

76 Note on the Log-integral Function
The log-integral function Li(x) is got by integrating 1/log(t) from t = 0 to t = x (“American” definition), or from t = 2 to t = x (“European” definition). When t = 1, log(t)=0, so that 1/log(t) is infinite. This creates issues with the “American” definition! The issues can be finessed, however; and Riemann used the “American” definition. Most math software packages also use it (e.g. Mathematica), so I use it too. For any x, the two definitions differ by … The “American” Li(x)

77 The Möbius μ-function Definition
The Möbius μ-function is an arithmetic function. That means that only positive whole-number arguments are allowed. You are allowed to discuss μ(5) or μ( ), but you may NOT discuss μ(½) or μ(-3). Definition μ(n) has the value 1 when n = 1, and also when n is the product of an even number of different primes: for example 2170=2x5x7x31, so μ(2170)=1. μ(n) has the value -1 when n is the product of an odd number of different primes: for example 561=3x11x17, so μ(561)=-1. μ(n) is 0 otherwise: for example 20=2x2x5, so μ(20)=0.

78 The Möbius μ-function (cont.)
1 2 -1 3 4 5 6 7 N μ(N) 8 9 10 1 11 -1 12 13 14 N μ(N) 15 1 16 17 -1 18 19 20 21 N μ(N) 22 1 23 -1 24 25 26 27 28 N μ(N) 29 -1 30 31 32 33 1 34 35

79 The Heart of the Matter The sensational part of Riemann’s result is
This says: “For all possible values of ρ (Greek ‘rho’), add up Li(xρ).” What are these ρ? Why, they are the nontrivial zeros of the ζ function! They emerged from Riemann’s attack on The Golden Key via complex function theory. They are – no surprise – complex numbers. They lie at the heart of Riemann’s topic: the distribution of the prime numbers.

80 The Nontrivial Zeros There are infinitely many of them.
The first 20 are listed here → Riemann himself computed the first and third (not very accurately). For each nontrivial zero a + bi there is also one equal to a – bi. J.P. Gram computed the first 15 nontrivial zeros in 1903. Logician Alan Turing in 1953 computed 1,104 of the little devils. In 2004, Gourdon and Demichel computed ten trillion! Every single one computed to date has real part one-half. i i i i i i i i i i i i i i i i i i i i

81 All nontrivial zeros of the zeta function have real part one-half.
The Riemann Hypothesis All nontrivial zeros of the zeta function have real part one-half.

82 Riemann’s Statement of the Hypothesis
Riemann guessed that all nontrivial zeros of the ζ function have real part one-half. In the 1859 paper “On the Number of Primes Less Than a Given Magnitude” he states an equivalent fact, then adds → “One would of course like to have a rigorous proof of this, but I have put aside the search for such a proof after some fleeting vain attempts (einigen flüchtigen vergeblichen Versuchen) because it is not necessary for the immediate objective of my investigation.”

83 150 Years On We still do not know whether all nontrivial zeros of the zeta function have real part one-half. There is now a great mass of theory about prime numbers. A large part of it consists of theorems whose statement begins: “Assuming the truth of the Riemann Hypothesis . . .” A resolution of the RH – a proof that it is true, or a proof that it is false – would revolutionize this branch of mathematics.

84 Footnote 1 * The Chebyshev Bias

85 Pafnuty Lvovich Chebyshev
Russian ( ) Born to a prosperous military family (his father fought against Napoleon). Studied at Moscow University. Taught at St. Petersburg. Did important work on prime numbers. Came close to proving the PNT (1850).

86 The Chebyshev Bias -- Examples
Divide a prime number (other than 2) by 4. The remainder must be either 1 or 3. For the primes up to 101 (which are: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101) the remainders are 3, 1, 3, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 3, 1, 3, 3, 1,3, 3, 1, 1, 1. That’s 12 remainder-1 primes and 13 remainder-3 primes – a slight bias in favor of remainder-3 primes. For the primes up to 1,009 the counts are 81 and 87 – a stronger bias in favor of remainder-3s. Up to 10,007 the counts are 609 and 620 – still favoring remainder-3s! This bias is only violated – and very briefly – at 26,861, when reminder-1 primes take the lead. In the first 5.8 million primes, remainder-1 primes hold the lead at only 1,939 places. This is a Chebyshev bias.

87 The Chebyshev Bias (cont.)
Instead of dividing by 4, divide by 3. Now the remainder (ignore p = 3) is either 1 or 2. The bias is to 2. This bias is not violated until p = 608,981,813,029! (Discovered in 1978.) Chebyshev biases express deep properties of the whole numbers. Even though you may wait a long time for the first violation of a Chebyshev bias, each bias is violated infinitely many times.

88 Chebyshev Biases (cont.)
Chebyshev biases give us a good informal reason to think that the RH may not be true. Similar – though mathematically deeper – biases are found all over prime number theory. The pattern is: A certain theorem (“Remainder-3 primes will always outnumber remainder-1 primes”) is true for all the numbers we check, into the thousands or even millions. Then, at some very high value, a violation occurs! The RH has been found true for trillions of nontrivial zeroes, but…….

89 Footnote 2 * The Riemann Hypothesis Song

90 The Riemann Hypothesis Song

91 The Riemann Hypothesis Song (continued)

92 The Riemann Hypothesis Song (concluded)


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