Queen Mary University of London Riemann s Hypothesis, Rewley House, 15 September 2013
Symbols Z = {..., 3, 2, 1,0,1,2,3,4,...} the integers R = the real numbers integers, fractions, irrationals C = the complex numbers more later e = 2.7182818284... Euler s number γ = 0.5772156649... EulerMascheroni (is it irrational??) τ = 6.2831853071... circle constant (= 2π) = the size of Z (for our purposes) i = 1 imaginary more later
Powers 2 0 = 1 2 1 = 2 2 2 2 3 = 2 2+3 = 2 5 = 32 ( 2 2 ) 3 = 2 2 3 = 2 6 = 64( 2 23 = 2 8 = 256) 2 1 = 1 2 2 1/2 = 2 = 1.41421356237... e τ = 535.4916555247... (e τ ) 1 = e τ 1 = e τi = 1 A cautionary tale: 1 = 1 = 1 1 = 1 1 = ( 1 ) 2 = 1
Capital Sigma Σ 1+2+3+...+n abbr. 1 2 +2 2 +3 2 +...+n 2 abbr. 1 r +2 r +3 r +...+n r abbr. r 1 +r 2 +r 3 +...+r n abbr. n k = 1 2 n(n+1) n k 2 = 1 6 n(n+1)(2n+1) n k r = S r (n) n r k = r (1 rn ) 1 r (sum of geometric progression) 1 1 2 + 1 3 1 4 +... abbr. ( 1) k+1 = 0.6931471805... k
Capital Pi Π 1 2 3... n abbr. 1 2 2 2 3 2... n 2 abbr. 2 1 2 2 2 3... 2 n abbr. n k = n! (n factorial) 0! = 1! = 1,2! = 2,3! = 6,4! = 24,... n k 2 = (n!) 2 n 2 k = 2 1+2+3+...+n = 2 n k = 2 1 2 n(n+1) = 2 2 1.3. 42 3.5. 62 5.7... abbr. (2k) 2 (2k 1)(2k +1) = 1 4 τ (Wallis s Product) (( 2 ) n ) n+1
Pascal s Triangular Cornucopia ( n ) k or n C k ( 0 ) ( 0 = 1, 7 ) ( 2 = 7 ) 5 = 21, etc
The Power of Pascal I Recall S r (n) denotes the sum 1 r +2 r +...n r = n kr. In the early 18th century, Jakob Bernoulli and Sansei Takekazu-Kowa Seki independently discovered: S r (n) = 1 r +1 r ( ) r +1 ( 1) k B k n r+1 k k k=0 where B 0 = 1, B 1 = 1 2, B2 = 1 6, B 3 = 0, B 4 = 1 30, B 5 = 0,... E.g. S 2 (n) = (( ) ( ) ( ) 1 3 3 3 B 0 n 3 B 1 n 2 + )B 2 n 1 2+1 0 1 2 (1 n 3 3 12 n2 +3 16 ) n = 1 3 = 1 ( 2n 3 +3n 2 +n ) = 1 6 6 n(n+1)(2n+1)
The Power of Pascal II The Bernoulli numbers: B 0 = 1, B 1 = 1 2, B2 = 1 6, B 3 = 0, B 4 = 1 30, B 5 = 0, B 6 = 1 42, B 7 = 0,... How can we find B 100? Are the odd B s (except B 1 ) always zero? Pascal to the rescue: n ( ) n+1 B k = 0, for n 1 k k=0 which means if we know B 0,B 1,...B n then we can get B n+1. 8 ( ) 9 E.g. B k = k k=0 = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 9 9 9 9 9 9 9 9 9 B 0 + B 1 + B 2 + B 3 + B 4 + B 5 + B 6 + B 7 + B 8 0 1 2 3 4 5 6 7 8 ( ) ( ) 9 9 1 + 1 ( ) 9 0 1 2 + 1 ( ) 9 2 6 + 1 ( ) 9 4 30 + 1 ( ) 9 6 42 + B 8 8 Solve for B 8 : B 8 = = 1 + 9 1 2 + 36 1 6 + 126 1 30 + 84 1 42 + 9B 8 = 0 1 9 ( 1 + 9 2 36 6 + 126 30 84 ) = 1 42 30
Convergent to Roughly, a sequence of numbers a 1,a 2,a 3,... converges to limit L if the difference between a k and L gets progressively closer to zero. E.g. (1) the sequence of fractions 1 1, 2 1, 3 2, 5 3, 8 5, 13,..., consisting 8 of ratios of successive Fibonacci numbers, converges to the golden ratio ϕ = 1 ( ) 2 1+ 5 = 1.6180339887... For short we can write F n+1 lim = ϕ. n F n (2) the sequence of fractions 1 1, 3 2, 7 5, 17 12, 41,..., defined by 29 a 1 = n 1 d 1 = 1 1, a 2 = n 2 d 2 = n 1 +2d 1 n 1 +d 1, a 3 = n 3 d 3 = n 2 +2d 2 n 2 +d 2,..., converges to limit 2. That is, lim n a n = 2.
Asymptotic to Roughly, a sequence of numbers a 1,a 2,a 3,... is asymptotic to M if the ratio of a k to M gets progressively closer to one (i.e. % error goes to zero). This may involve two sequences a 1,a 2,a 3,..., and b 1,b 2,b 3,... Their difference may get bigger and bigger: lim a n b n =. n a n But this still allows their ratios to converge: lim = 1. We n b n write a n b n, for short ( a n is asymptotic to b n ). E.g. The factorial function n! is asymptotic to τnn n e n. n! τnn n e n : the upper red line plots n! (extended continuously); the lower blue line is Stirling s approximation
An asymptotic for the Bernoulli numbers The sequence of even Bernoulli numbers B 0,B 2,B 4,... is asymptotic, in absolute value, to the sequence 2(2k)! τ 2k. ( 1) k+1 B 2k 2(2k)! τ 2k B 2k (thin red line) plotted against 2(2k)! τ 2k (wide blue line) in the range 11 n 20. Plots of B 2k 2(2k)! τ 2k for k = 11,...,20 (note difference in vertical scales!)
The (natural) logarithm What power of 2 gives me 16? The answer is written as a function: log 2 (16) log to base 2 of 16. The function has value 4 at argument 16 because 16 = 2 4. What power of e = 2.71828... gives me 16? The answer is log e (16). This is calculated as ln(16) on your calculator ( ln for natural log ) but mathematicians write it as log(16) (base e is assumed unless otherwise stated). The value of log(16) is 2.77258... meaning (2.71828...) 2.77258... = 16. Plots of log a (x) for a = 2,e and 10 (green, red and blue, respectively). For base e the slope as the curve passes the horizontal axis is precisely 1. Rules: log((a b) c ) = c log(ab) = c(loga+logb) = loga c +logb c.
How to calculate logs On your calculator, every scientific function is calculated by adding enough terms in a suitable series: For natural logs, the series is specified as log(1 x) = x x2 2 x3 3 x4 4... This is valid provided 1 x < 1. E.g. log2 = log(1 ( 1)) = ( 1) ( 1)2 2 ( 1)3 3 ( 1)4 4 = 1 1 2 + 1 3 1 4 + = 0.6931471805......
The area integral The area between a curve y = f(x) and the horizontal axis between x = a and x = b is abbreviated to b a f(x)dx. 1 0 ( logx)dx = 1 1 0 ( logx)2 dx = 2 1 0 ( logx)3 dx = 6 1 0 ( logx)4 dx = 24
Our star turn: the primes Euclid s Book IX, Proposition 20 They form an infinity. The Fundamental Theorem of Arithmetic They construct any positive integer uniquely: n = p a 1 r (r, primes p i and 1 pa 2 2 par powers a i determined uniquely n). They are unpredictable: 11 is prime, 1111111111111111111 is prime, i.e. 18 k=0 10k. What about 10 10 k=0 10k?
A digression... In a single week in April 2013 two massive breakthroughs in analytic number theory were announced: Week Goldbach: any odd number greater than 7isasumofthreeoddprimes(HaraldHelfgott). E.g. 35 = 11+11+13. Previously known for all large odd numbers and (Tao, 2012) for five primes. Small prime gaps: there are infinitely many pairs of primes separated by 70 million or less (Yitang Zhang). Quickly improved to approx. 5000 or less. Previous best: smallest prime gap grows slower than primes themselves(goldston Pintz Yıldırım, 2005).