MATH 60 Fall 2008 Series ad a Famous Usolved Problem
Problems = + + + + = (2- )(2+ ) 3 3 5 5 7 7 9 2-Nov-2008 MATH 60 2
Problems ( 4) = + 25 48 2-Nov-2008 MATH 60 3
Problems ( )! = + 2-Nov-2008 MATH 60 4
Problems 3 4 5 + 2 + + + + + + 3 3 3 3 2 3 4 2-Nov-2008 MATH 60 5
Problems + + + + = 2 3 4 2 3 4 = 2-Nov-2008 MATH 60 6
Problems =! =! 2-Nov-2008 MATH 60 7
Problems ( + -) =! (! ) 2 2-Nov-2008 MATH 60 8
Problems (2!) 4 = 2-Nov-2008 MATH 60 9
Problems æ ö ç çè + ø = 0 2-Nov-2008 MATH 60 0
Problems = r, r < 2-Nov-2008 MATH 60
Problems Show r r = 2 = (-r) 2-Nov-2008 MATH 60 2
Problems Fid = 2 2-Nov-2008 MATH 60 3
Prime Numbers Statemet: There are a ifiite umber of prime umbers. Proof A (Euclid): Assume ot, that is assume that there are a fiite umber of prime umbers: p, p 2, p 3,,p. p p p 2 p 3 p Let M = (p p 2 p 3 p ) +. Note that M is ot divisible by p, p 2, p 3, or p, sice oe of 3 these divide. Thus, M is divisible oly by itself ad. Therefore, M is prime ad ot i the list a cotradictio. ti 2-Nov-2008 MATH 60 4
Prime Numbers Proof B (Euler): (Usig series) Assume that there are a fiite umber of prime umbers: 2, 3, 5,,p, where p is the largest prime. = 0 æ ö ç = = 2 çè2 ø - 2 æ ö 3 = = ç è3 ø - 3 2 = 0 æö p = = ç = 0èp ø - p p- 2-Nov-2008 MATH 60 5
Prime Numbers Thus the product is fiite: æ ö æ ö æ ö æ ö æ ö æ ö æ 3 ö æ p ö ( 2) = ç ç = 0è2 ø ç = 0 3 ç = 0 p çè2 ø çp è ø ç è ø ç è ø è ø è - è ø ø What does the product o the left had side really look like? æ ö æ ö æ ö æ ö æ ö æ ö = ç = 0è ç2ø ç = 0èç 3ø ç = 0 p ç è ø è ø è ø è ø æ ö æ ö æ ö + + + + ç 2 3 + + + + 2 3 + + + + 2 3 èç 2 2 2 øè ç 3 3 3 ø èç p p p ø 2-Nov-2008 MATH 60 6
Prime Numbers If we distribute the multiplicatio through the additio we get a ifiite sum each of whose summads looks like = 2 3 5 p d d3 d 2 5 2 3 5 p 2 3 5 p d d d d d where each of the d k is a o-egative iteger. Each summad occurs exactly oce. The deomiator is a positive iteger ad each positive iteger appears exactly oce. Therefore æ æö ö æ æö ö æ ö æ ö 0 2 = = 0 3 = + + + + + + = 0 p èç çè ø ø ç ç è ø ç è ø çè è ø ø 2 3 4 5 6 2-Nov-2008 MATH 60 7 p
Prime Numbers The right had side is the harmoic series which diverges ad is ot a real umber. This is the eeded cotradictio. 2-Nov-2008 MATH 60 8
Prime Numbers This was t eough for Euler. He wated to see what size the set of primes was. He showed the sum of the reciprocals of all prime umbers diverges. Theorem: /p diverges. Proof: From above we kow that t æ ö p = - ç = p prime - çè ø 2-Nov-2008 MATH 60 9
Prime Numbers Take the log of both sides. æ ö é æ ö ù æ ö l l l l - - = = = = - - ê p prime p ú ç è ø çè - ø ç p prime è-p ë û ø p prime - ( p ) Euler ow uses the Taylor series for the logarithm. 2 3 4 l ( - x ) =-x- x - x - x - - x - 2 3 4 - - -2-3 - 4 - - l ( - p ) = p + p + p + p + + p + 2 3 4 = + + + + + + 2 3 4 p 2p 3p 4p p 2-Nov-2008 MATH 60 20
Prime Numbers Take the log of both sides. æ ö æ ö l = + + + + + + ç è ø çè ø 2 3 4 = p prime p 2 p 3 p 4 p p æö æ ö = ç + ç + + + + + 2 2-2 ç p prime èp ø p prime p çè2 3p 4p p ø æö æ ö < + + + + + + ç è ø ç è ø = 2 2 p prime p p prime p p p p æö + ç èp ø p ( p - ) p prime p prime æ ö = + C ç è p ø p prime 2-Nov-2008 MATH 60 2
Prime Numbers where C is a umber <. Sice diverges, we = kow that æ ö l ç è= ø æ ö ç èp ø diverges. Thus, diverges. p prime We do kow that if you sum the reciprocals of the twi primes that that sum is fiite ad slightly bigger tha. (Apery s costat) 2-Nov-2008 MATH 60 22
Propositio: Probability If two positive itegers are chose idepedetly ad radomly, the the probability that they are relatively l prime is 6/π 2. Proof: Let p be prime. Let be radomly chose iteger. α = Prob(p ) = /p. 2-Nov-2008 MATH 60 23
Probability If m ad are idepedetly ad radomly chose Prob(p m ad p ) = /p 2. Thus, Prob(p ot divide both m ad ) = /p 2 List the primes: p,p 2,p 3,,p k, ad let P k = Prob(p k ot divide both m ad ) = / p 2 k Claim: Divisibility by p i ad p j are idepedet. Let P be the probability that m ad are relatively prime. 2-Nov-2008 MATH 60 24
So Probability P = P P 2 P 3 P k = P PP 2 P k = - - - - 2 3 5 2 2 2 2 p k æ ö æ ö = + + + + + + + + ç 2 2 2 2 3 2 2 2 2 3 è 2 (2 ) (2 ) ø èç 3 (3 ) (3 ) ø æ ö + + + + 2 2 2 2 3 çè 5 (5 ) (5 ) ø 2-Nov-2008 MATH 60 25
Probability æ ö æ ö = + + + + 2 2 2 2 3 + + + + 2 2 2 2 3 P èç 2 (2 ) (2 ) ø ç è 3 (3 ) (3 ) ø æ ö ç + + + + 2 2 2 2 3 èçè 5 (5 ) (5 ) ø æ ö æ ö = + + + + + + + + 2 2 2 3 2 2 2 2 3 2 ç è 2 (2 ) (2 ) ø çè 3 (3 ) (3 ) ø æ ç + + + + ö 2 2 2 3 2 çè 5 (5 ) (5 ) ø 2 = + + + + + + = 2 2 2 2 2 2 3 4 5 6 6 2-Nov-2008 MATH 60 26
Riema Zeta Fuctio I 859 Reima defied a differetiable fuctio of a complex variable ζ(s) ) by ( s ) = + 2 + 3 + 4 + 5 + + + = s s s s s s = Riema kew the value of this fuctio at certai values of s. ζ(0) does ot exist. (Why?) ζ() does ot exist. (Why?) ζ(2) = π 2 /6 ζ(4) ) = π 4 /90 2-Nov-2008 MATH 60 27
Riema Zeta Fuctio Euler had computed ζ(2) for =, 2, 3,, 3. We saw last time that (2 k ) = = 2k = 2 2k- 2k B2 k (2 k)! Riema showed that ζ(s) gives a lot of iformatio about the distributio of primes. 2-Nov-2008 MATH 60 28
Riema Zeta Fuctio Questio: Where is ζ(s) = 0? It ca be show that whe it has bee exteded to all complex umbers, except Re(s) =, the it is trivially see to be zero at the egative eve itegers. Riema proved that ζ(s) ) = 0 whe s falls iside id the ifiite strip bouded by the lies x = 0 ad x =. 2-Nov-2008 MATH 60 29
Prime Number Theorem Theorem: For every real umber x let π(x) be the umber of prime umbers less tha x ad let x dt Li ( x ) = ó ô õ l( t) 2 The ( x) lim = x Li ( x ) This was prove by Hadamard ad Poussi i 896. 2-Nov-2008 MATH 60 30
Riema Zeta Hypothesis Cojecture: (Uprove) If s is a complex umber so that ζ(s) ) = 0 the Re(s) = ½. What we do kow: The lie x = ½ cotais a ifiite umber of zeroes of ζ(s). The first 70,000,000 or so lie o that t lie. 2-Nov-2008 MATH 60 3
Defiitio: Power Series If {a } is a sequece, we defie the series a x as a power series i x. For a give sequece a power series is a fuctio f (x) whose domai cosists of those values of x for which the series coverges. Power series behavior is typical to that of the geometric series. x coverges for x <, so the domai of this fuctio: f (x) = x is the ope iterval (-,). ) 2-Nov-2008 MATH 60 32
Covergece of Power Series Propositio: Suppose that the power series a x coverges for x = x ad diverges for x = x,, the a x. coverges absolutely for each x satisfyig x < x ; ; 2. diverges for each x satisfyig x > x 2-Nov-2008 MATH 60 33
Covergece of Power Series Proof: ) If x = 0 there is othig to prove. Assume x 0. a x x coverges ï {a } Ø 0 ï {a x } bouded ï$m > 0 ' a x M for all. Let x < x, the x ax = ax M x x x 2-Nov-2008 MATH 60 34
Covergece of Power Series If x/x < < ï M x/x coverges ï a x coverges ï a x coverges. 2) Assume x > x ad a x coverges. The first part ï a x coverges, cotradictig the give. 2-Nov-2008 MATH 60 35
Covergece of Power Series Propositio: Let a x be a power series, the oe of the followig must hold.. a x coverges absolutely for all x; 2. a x coverges oly at x = 0; 3. there is a umber ρ>0 so that a x coverges absolutely for x < ρ ad ddiverges for x >ρ. 2-Nov-2008 MATH 60 36
Covergece of Power Series Proof: Let S = {x œ R a x coverges}. Note: 0 œ S. S ubouded ï by previous Propositio, a x coverges absolutely for all x. Assume S bouded. Let ρ = lub{s}, which exists by the Least Upper Boud Axiom. x > ρ ï x S ï a x diverges. Let x < -ρ ad let x < x 0 < -ρ ï ρ < -x 0 < -x ï a (-x x 0 ) diverges ï a x diverges 2-Nov-2008 MATH 60 37
Covergece of Power Series Suppose $ x ' x < ρ ad a x diverges. The a x diverges for all x > x ï x is a upper boud for S. This cotradicts ρ = lub{s}. ρ is the radius of covergece for the power series. S = [ ρ ρ, ρ], ( ρ ρ, ρ], [ ρ ρ, ρ), or ( ρ, ρ) ad all ca occur. This is called the iterval of covergece. I Case, ρ = ad S = R; ; Case 2,,ρ ρ=0 ad S={0}. { } 2-Nov-2008 MATH 60 38
Ratio Test for Power Series Propositio: Let a x be a power series with radius of covergece ρ. If a 0 for all ad the. ρ = if q = 0; 2. ρ = 0 if q = ; 3. ρ = / q otherwise. {a + /a } Ø q 2-Nov-2008 MATH 60 39
Proof 2-Nov-2008 MATH 60 40
Problem x a =, a+ = = ( + 4) ( + 4) ( + )( + 5) a + ( + 4) ( + 4) = = a ( + )( + 5) ( + )( + 5) a + lim lim a ( + )( + 5) = = q ( + 4) = = = q 2-Nov-2008 MATH 60 4
x ( + 4) = + = Problem = ( + 4) coverges by compariso with / 2. =- = (-) ( + 4) coverges by Alteratig Series Test. S =- [,] 2-Nov-2008 MATH 60 42
Problem 2 = x! 0! a ( + ) =, a + =! ( + )! + a+ ( + )!! ( + ) ( + ) = = = a ( + )! ( + )! + a + lim limç a çè + + + ( + ) æ+ ö æ ö = = = + ç è ø çè ø æ ö = ç + = e= q ø = e 2-Nov-2008 MATH 60 43
Trigoometric Series Defiitio: Let {a } ad {b } be sequeces, we say that a = O(b ) if a lim exists. b 3 3 + + = O( ) 3 + 5-3 = O ( ) 4 2 4-5 + 6 2-Nov-2008 MATH 60 44
Propositio: Trigoometric Series Let {a } be a sequece so that a = O( p ) for some p <. The the series a cos(x) ad a si(x) both coverge absolutely for all x. Proof: a = O( p ) ï {a / p } coverges ï {a / p } bouded, so a / p M for all ï a cos(x) a M p ï a cos(x) coverges absolutely. 2-Nov-2008 MATH 60 45