MATH 6101 Fall Problems. Problems 11/9/2008. Series and a Famous Unsolved Problem (2-1)(2 + 1) ( 4) 12-Nov-2008 MATH

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1 /9/008 MATH 60 Fall 008 Series ad a Famous Usolved Problem = = ( - )( + ) Nov-008 MATH 60 ( 4) = Nov-008 MATH 60 3

2 /9/008 ( )! = + -Nov-008 MATH Nov-008 MATH = 3 4 = -Nov-008 MATH 60 6

3 /9/008 =! =! -Nov-008 MATH 60 7 = - ( ) + (! ) -Nov-008 MATH 60 8 = (!) 4 -Nov-008 MATH

4 /9/008 = 0 æ ö è + -Nov-008 MATH 60 0 = r, r < -Nov-008 MATH 60 Show = r r = ( - r) -Nov-008 MATH 60 4

5 /9/008 Fid = -Nov-008 MATH 60 3 Statemet: There are a ifiite umber of prime umbers. Proof A (Euclid): Assume ot, o, that is assume that there are a fiite umber of prime umbers: p, p, p 3,,p. Let M = (p p p 3 p ) +. Note that M is ot divisible by p, p, p 3, or p, sice oe of these divide. Thus, M is divisible oly by itself ad. Therefore, M is prime ad ot i the list a cotradictio. -Nov-008 MATH 60 4 Proof B (Euler): (Usig series) Assume that there are a fiite umber of prime umbers:, 3, 5,,p, where p is the largest prime. æö = = = 0 èø - æö 3 = = è3-3 = 0 = 0 æö p = = èp - p p- -Nov-008 MATH

6 /9/008 Thus the product is fiite: æ ö æ ö æ ö æ ö æ ö æ ö æ3ö æ p ö ( ) = = 0 = 0 3 = 0 p è ø è ø è ø è ø èp-ø è ø è ø è 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 = = 0 = 0 3 = 0 p è è ø è è ø è è ø æ ö æ ö æ ö è øè è p p p -Nov-008 MATH 60 6 If we distribute the multiplicatio through the additio we get a ifiite sum each of whose summads looks like = 3 5 p 3 5 d d3 d5 dp d d 3 d5 dp p 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 ö æ ö æ ö æ ö æ ö æ ö = = 0 = 0 3 = 0 p è è ø ø è è ø ø è è ø ø Nov-008 MATH 60 7 The right had side is the harmoic series which diverges ad is ot a real umber. This is the eeded cotradictio. -Nov-008 MATH

7 /9/008 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 æ ö = - p prime è - p = -Nov-008 MATH 60 9 Take the log of both sides. æ ö é æ öù æ ö l = l = l = -l -p - - = ê p prime p ú è ø è - ø p prime è-p ë û p prime Euler ow uses the Taylor series for the logarithm. - ( ) 3 4 l( - x) =-x- x - x - x -- x l( - p ) = p + p + p + p + + p = p p 3p 4p p -Nov-008 MATH 60 0 Take the log of both sides. æ ö æ ö l = è ø èp p 3p 4p p = p prime æ ö æ ö = p prime èp p prime p è 3p 4p p ø æ ö æ ö < p è ø p è p p p ø = p prime p prime p prime p prime p prime æö + p è ø p( p-) æ ö = + C èp -Nov-008 MATH 60 7

8 /9/008 where C is a umber <. Sice = kow that æ ö l è= æ ö diverges. Thus, diverges. prime p p è ø diverges, we 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) -Nov-008 MATH 60 Probability Propositio: If two positive itegers are chose idepedetly ad radomly, the the probability that they are relatively prime is 6/π. Proof: Let p be prime. Let be radomly chose iteger. α = Prob(p ) = /p. -Nov-008 MATH 60 3 Probability If m ad are idepedetly ad radomly chose Prob(p m ad p ) = /p. Thus, Prob(p ot divide both m ad ) = /p List the primes: p,p,p 3,,p k, ad let P k = Prob(p k ot divide both m ad ) = / p k Claim: Divisibility by p i ad p j are idepedet. Let P be the probability that m ad are relatively prime. -Nov-008 MATH

9 /9/008 Probability P = P P P 3 P k So = P PP P k = p k æ ö æ ö = è ( ) ( ) ø è 3 (3 ) (3 ) ø æ ö è 5 (5 ) (5 ) -Nov-008 MATH 60 5 Probability æ ö æ ö = P è ( ) ( ) è 3 (3 ) (3 ) æ ö è 5 (5 ) (5 ) æ ö æ ö = è ( ) ( ) è 3 (3 ) (3 ) æ ö è 5 (5 ) (5 ) = = Nov-008 MATH 60 6 Riema Zeta Fuctio I 859 Reima defied a differetiable fuctio of a complex variable ζ(s) by ( s) = = 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() = π /6 ζ(4) = π 4 /90 -Nov-008 MATH

10 /9/008 Riema Zeta Fuctio Euler had computed ζ() for =,, 3,, 3. We saw last time that ( k) = = k = B k- k k ( k)! Riema showed that ζ(s) gives a lot of iformatio about the distributio of primes. -Nov-008 MATH 60 8 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 the ifiite strip bouded by the lies x = 0 ad x =. -Nov-008 MATH 60 9 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) The ( x) lim = x Li( x) This was prove by Hadamard ad Poussi i Nov-008 MATH

11 /9/008 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 lie. -Nov-008 MATH 60 3 Power Series Defiitio: 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 (-,). -Nov-008 MATH 60 3 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 ;. diverges for each x satisfyig x > x -Nov-008 MATH 60 33

12 /9/008 Covergece of Power Series Proof: ) If x = 0 there is othig to prove. Assume x 0. a x coverges ï {a x } Ø 0 ï {a x x } bouded ï $ M > 0 ' a M for all. Let x < x, the x ax = ax M x x x -Nov-008 MATH Covergece of Power Series If x/x < ï M x/x coverges ï a x coverges ï a x coverges. ) Assume x > x ad a x coverges. The first part ï a x coverges, cotradictig the give. -Nov-008 MATH 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;. a x coverges oly at x = 0; 3. there is a umber ρ>0 so that a x coverges absolutely for x < ρ ad diverges for x >ρ. -Nov-008 MATH 60 36

13 /9/008 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 0 ) diverges ï a x diverges -Nov-008 MATH 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), or ( ρ, ρ) ad all ca occur. This is called the iterval of covergece. I Case, ρ = ad S = R; Case, ρ=0 ad S={0}. -Nov-008 MATH Ratio Test for Power Series Propositio: Let a x be a power series with radius of covergece ρ. If a 0 for all ad {a + /a } Ø q the. ρ = if q = 0;. ρ = 0 if q = ; 3. ρ = / q otherwise. -Nov-008 MATH

14 /9/008 Proof -Nov-008 MATH x ( 4) Problem = = ( + 4) ( + )( + 5) a, a+ = + a+ ( + 4) ( + 4) = = a ( + )( + 5) ( + )( + 5) a+ ( + 4) lim = lim = = q a ( + )( + 5) = = q -Nov-008 MATH 60 4 x ( 4) = + = = Problem ( + 4) coverges by compariso with / /. =- = (-) ( + 4) coverges by Alteratig Series Test. S =- [,] -Nov-008 MATH

15 /9/008 x = 0! Problem ( + ) a =, a+ =! ( + )! + a+ ( + )!! ( + ) ( + ) = = = a ( + )! ( + )! + a + lim lim a è ( + ) æ+ ö æ ö = = = + è ø è ø æ ö = + = e= q ø = e -Nov-008 MATH Trigoometric Series Defiitio: Let {a } ad {b } be sequeces, we say that a = O(b ) if a lim exists. b + + = O( ) = O -3 ( ) -Nov-008 MATH Trigoometric Series Propositio: 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. -Nov-008 MATH