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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.
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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)
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 Mr. Derbyshire’s Math Books (cont.)
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 byanything (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,…
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: (12345 . . . P) + 1. Plainly it is bigger thanP. 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 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.
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 1111. 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 infinityof 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: (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? 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 bookHuman 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: . . . where ‘s’ is any number at all ?”
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.
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.
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.” Multiply both sides by 1/2s: Call that “Expression 2b.” Now subtract Expression 2b from Expression 2a: 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.” Multiply both sides by 1/3s: Call that “Expression 3b.” Now subtract Expression 3b from Expression 3a: Look! – I sieved out the 3’s!
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!
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! ***
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.
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
A Slight Detour (cont.) However: If Then Subtracting So 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 partof 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)?