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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 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.
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.
The Riemann Hypothesis
A Great Unsolved Mathematical Problem
presented by
John Derbyshire
author of Prime Obsession (2003) and Unknown Quantity (2006)
Mr. Derbyshire’s Math Books
Mr. Derbyshire’s Math Books (cont.)
Prime Obsession
(2003)
Unknown Quantity
(2006)
• 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.
• A history of algebra for
non-mathematicians
• Published in paperback
May 2007
• Includes references to
Riemann’s work in
function theory and
topology
Mr. Derbyshire’s Most Recent Book
We Are Doomed: Reclaiming Conservative Pessimism (2009)
The Riemann Hypothesis
All nontrivial
zeros of the zeta
function have real
part one-half.
Q: What is the RH about ?
A: It is about prime numbers.
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,
47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101,…
More Prime Numbers
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.
Unsolved Problems About Prime
Numbers
• Goldbach’s Conjecture
Every even number after 2 is the sum of two
prime numbers (e.g. 98 = 19 + 79).
• The Prime Pair Conjecture
There are infinitely many pairs of primes just two
apart (like 41 and 43).
• The Riemann Hypothesis
Topic 1
*
The Sieve [Решето] of
Eratosthenes
Eratosthenes of Cyrene
• Greek, 276-194 B.C.
• Cyrene is in today’s
Libya, then part of postAlexander 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.
The Sieve of Eratosthenes
10 11
20 21
30 31
40 41
50 51
60 61
70 71
80 81
90 91
100 101
2
3
4
5
6
7
8
9
12 13 14 15 16 17 18 19
22 23 24 25 26 27 28 29
32 33 34 35 36 37 38 39
42 43 44 45 46 47 48 49
52 53 54 55 56 57 58 59
62 63 64 65 66 67 68 69
72 73 74 75 76 77 78 79
82 83 84 85 86 87 88 89
92 93 94 95 96 97 98 99
102 103 104 105 106 107 108 109
Sieving out the 2’s
10 11
20 21
30 31
40 41
50 51
60 61
70 71
80 81
90 91
100 101
2
3
4
5
6
7
8
9
12 13 14 15 16 17 18 19
22 23 24 25 26 27 28 29
32 33 34 35 36 37 38 39
42 43 44 45 46 47 48 49
52 53 54 55 56 57 58 59
62 63 64 65 66 67 68 69
72 73 74 75 76 77 78 79
82 83 84 85 86 87 88 89
92 93 94 95 96 97 98 99
102 103 104 105 106 107 108 109
Sieving out the 3’s
10 11
20 21
30 31
40 41
50 51
60 61
70 71
80 81
90 91
100 101
2
3
4
5
6
7
8
9
12 13 14 15 16 17 18 19
22 23 24 25 26 27 28 29
32 33 34 35 36 37 38 39
42 43 44 45 46 47 48 49
52 53 54 55 56 57 58 59
62 63 64 65 66 67 68 69
72 73 74 75 76 77 78 79
82 83 84 85 86 87 88 89
92 93 94 95 96 97 98 99
102 103 104 105 106 107 108 109
Sieving out the 5’s
10 11
20 21
30 31
40 41
50 51
60 61
70 71
80 81
90 91
100 101
2
3
4
5
6
7
8
9
12 13 14 15 16 17 18 19
22 23 24 25 26 27 28 29
32 33 34 35 36 37 38 39
42 43 44 45 46 47 48 49
52 53 54 55 56 57 58 59
62 63 64 65 66 67 68 69
72 73 74 75 76 77 78 79
82 83 84 85 86 87 88 89
92 93 94 95 96 97 98 99
102 103 104 105 106 107 108 109
Sieving out the 7’s
10 11
20 21
30 31
40 41
50 51
60 61
70 71
80 81
90 91
100 101
2
3
4
5
6
7
8
9
12 13 14 15 16 17 18 19
22 23 24 25 26 27 28 29
32 33 34 35 36 37 38 39
42 43 44 45 46 47 48 49
52 53 54 55 56 57 58 59
62 63 64 65 66 67 68 69
72 73 74 75 76 77 78 79
82 83 84 85 86 87 88 89
92 93 94 95 96 97 98 99
102 103 104 105 106 107 108 109
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 . . .
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.
Topic 2
*
The Basel Problem
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.
0.3
0.03
0.003
0.0003
0.00003
0.000003
......
+ ________________
0.3333333333333 . . .
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
0.00003
0.000003
0.0000003
......
+ ________________
0.3333333333333 . . .
1 = 1
1/2 = 0.5
1/3 = 0.3333333333333
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1666666666666
1/7 = 0.1428571428571
......
+ ____________________
∞
The Basel Problem (cont.)
Consider the Harmonic Series:
1 1 1 1 1 1
1      
2 3 4 5 6 7
(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.
The Bernoulli Brothers
Jakob (1654-1705)
Johann (1667-1748)
The Basel Problem (cont.)
What about this sum?
1 1 1
1 1
1
1 2  2  2  2  2  2 
2 3 4 5 6
7
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).
Leonhard Euler
• Swiss, 1707-1783
• Ranked 1st among
mathematicians in Murray’s
Human Accomplishment.
• Studied under Johann
Bernoulli at Basel.
• Prolific, pious (Calvinist),
domestic.
• St. Petersburg Academy of
Sciences, 1727-41 and
1766-83.
• Court of Frederick the
Great, 1741-1766.
Charles Murray’s 2003 book
Human Accomplishment
Mathematicians ranked
The Basel Problem – Solved!
Euler proved that
He then pursued the matter further and
came up with:
The Golden Key
Topic 3
*
The Golden Key
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:
1 1 1 1 1 1
1 s  s  s  s  s  s 
2 3 4 5 6 7
. . . where ‘s’ is any number at all ?”
When Does the Sum Exist?
1
1
1
1
1
1
1 s  s  s  s  s  s 
2
3
4
5
6
7
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.
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.
1 1 1
 s   1  s  s  s  
2 3 4
Here is its graph.
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 . . .
Sieving out the 2’s
Here’s the zeta function. Call this “Expression 2a.”
1 1 1 1 1 1
 s   1  s  s  s  s  s  s  
2 3 4 5 6 7
Multiply both sides by 1/2s:
1
1 1 1 1
1
1
1



s








s
s
s
s
s
s
s
s
2
2 4 6 8 10 12 14
Call that “Expression 2b.”
Now subtract Expression 2b from Expression 2a:
1
1 1 1 1
1
1

1  s   s   1  s  s  s  s  s  s  
3 5 7 9 11 13
 2 
Look! – I sieved out the 2’s!
(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.)
Sieving out the 3’s
Call that last result “Expression 3a.”
1
1
1
1
1
1
1



1  s   s  1  s  s  s  s  s  s  
2 
3
5
7
9
11 13

Multiply both sides by 1/3s:
1
3s
1

1


s
 2
1
1
1
1
1




s







s
s
s
s
s
9 15
21
27
33

Call that “Expression 3b.”
Now subtract Expression 3b from Expression 3a:
1 
1

1

1


s 
3
2s


1
1
1
1
1




s

1







s
s
s
s
s
5
7
11 13
17

Look! – I sieved out the 3’s!
Sieving out the 5’s
Call that last result “Expression 5a.”
1 
1 
1
1
1
1
1

1  s 1  s   s   1  s  s  s  s  s  
3 
2 
5
7
11
13
17

Multiply both sides by 1/5s:
1
5s
1 
1
1
1
1
1
1



1

1


s







s 
s 
s
s
s
s
s
2 
5
25
35
55
65
 3 
Call that “Expression 5b.”
Now subtract Expression 5b from Expression 5a:
1 
1 
1
1
1
1
1
1



1

1

1


s

1







s 
s 
s 
s
s
s
s
s
7 11 13 17 19
 5  3  2 
Look! – I sieved out the 5’s!
The Golden Key – At Last!
Just as with the original sieve, I can go on sieving for ever.
My final result will be:
1 
1 
1 
1  1 
1

 1  s  1  s 1  s 1  s 1  s 1  s   s   1
 13   11  7  5  3  2 
Or, to rearrange slightly:
 s  
1
1
1 s
2

1
1
1 s
3

1
1
1 s
5

1
1
1 s
7

1
1
1 s
11

1
1
1 s
13

. . . With zeta on the left and all the primes represented
on the right.
*** That is The Golden Key! ***
Why a “Key”? Why “Golden”?
More formally:
 s    1  p
s

1
p
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.
Topic 4
*
Zeta’s Hidden Depths
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.
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
S x   1  x  x 2  x 3  x 4  x 5  
A Slight Detour (cont.)
However: If
S x   1  x  x  x  x  x  
2
Then
Subtracting
3
4
5
x S x   x  x 2  x 3  x 4  x 5  x 6  
1  x S x  1
So
1
S x  
1 x
Well, that’s nice. But . . .
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.
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.”
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 →
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?”
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)?
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
 1  s   2
1 s
1 s 
 sin 
  s  1!  s 
 2

. . . 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).
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:
 s  
n
B2 k  s  2k  2 
1
1


 
s  1 2 k 1 2k  2k  1 
 s  2n   B2 n1 x  x
 
 
dx
s

2
n

1
1
2
n

1
x


Topic 5
*
Zeros of a Function
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.
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 xaxis at several points.
• Its value then is zero.
• I have found some
zeros of the zeta
function!
The Riemann Hypothesis
All nontrivial
zeros of the zeta
function have real
part one-half.
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?
Topic 6
*
Complex Numbers
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.
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.
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
Topic 7
*
Complex Functions
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.
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:
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.
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.
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)
Topic 8
*
The Distribution of
Primes
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.
N
Number
of primes
less than 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
Percentage
error
-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 mathematicians, Jacques Hadamard and Charles de la Vallée Poussin.
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.
The Primes “Thin Out” (cont.)
Start with one thousand and count forward in blocks of 100. 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.
Topic 9
*
Riemann’s 1859 Paper
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!
Riemann’s Answer
• How many primes are there between 1
and x?
• Riemann’s answer:
 n 
n n J
 x
n
where
 
J x   Li( x)   Li x



dt
 log 2   2
x t t  1 log t


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 μfunction, the log function,
summation and
integration.
• Only Li, the logarithmicintegral function, is not
elementary. Here’s its
graph for real arguments.
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 1.04516378011749278…
The “American” Li(x)
The Möbius μ-function
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(1000000000000), 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.
The Möbius μ-function (cont.)
N
μ(N)
N
μ(N)
N
μ(N)
N
μ(N)
N
μ(N)
1
1
8
0
15
1
22
1
29
-1
2
-1
9
0
16
0
23
-1
30
-1
3
-1
10
1
17
-1
24
0
31
-1
4
0
11
-1
18
0
25
0
32
0
5
-1
12
0
19
-1
26
1
33
1
6
1
13
-1
20
0
27
0
34
1
7
-1
14
1
21
1
28
0
35
1
The Heart of the Matter
• The sensational part of Riemann’s result is
Li x 



• 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.
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.
0.5 + 14.134725142i
0.5 + 21.022039639i
0.5 + 25.010857580i
0.5 + 30.424876126i
0.5 + 32.935061588i
0.5 + 37.586178159i
0.5 + 40.918719012i
0.5 + 43.327073281i
0.5 + 48.005150881i
0.5 + 49.773832478i
0.5 + 52.970321478i
0.5 + 56.446247697i
0.5 + 59.347044003i
0.5 + 60.831778525i
0.5 + 65.112544048i
0.5 + 67.079810529i
0.5 + 69.546401711i
0.5 + 72.067157674i
0.5 + 75.704690699i
0.5 + 77.144840069i
. . . . . . .
. . . . . . .
The Riemann Hypothesis
All nontrivial
zeros of the zeta
function have real
part one-half.
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.”
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.
Footnote 1
*
The Chebyshev Bias
Pafnuty Lvovich Chebyshev
• Russian (1821-94)
• 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).
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.
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.
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…….
Footnote 2
*
The Riemann Hypothesis
Song
The Riemann Hypothesis Song
The Riemann Hypothesis Song
(continued)
The Riemann Hypothesis Song
(concluded)