Sketch of Dirichlet s Theorem on Arithmetic Progressions

Transcription

1 Itroductio ad Defiitios Sketch of o Arithmetic Progressios Tom Cuchta 24 February 2012 / Aalysis Semiar, Missouri S&T

2 Outlie Itroductio ad Defiitios 1 Itroductio ad Defiitios 2 3

3 Itroductio ad Defiitios Dirichlet s theorem: Let h ad k be relatively rime ositive itegers. The there are ifitiely rime umbers i the set {k + h : Z + }. For examle, if h = 2 ad k = 3, we are focusig o these red umbers: {5, 8, 11, 14, 17, 20, 23, 26,...}, ad if h = 23 ad k = 10, we are focusig o these red umbers: {33, 56, 79, 102, 125, 148, 171, 194, 217, 240, 263,...}. We will rove rove this theorem by demostratig h(mod k) log =.

4 Theorem (Euler): Itroductio ad Defiitios 1 = Proof: For R(s) > 1, ζ(s) = 1 s = =1 log ζ(s) = log(1 s ). 1, so 1 s Taylor series: log(1 x) = x + O(x 2 ) for x ear 0; so for s > 1 ear 1, sice log(ζ(s)) = 1 2s ( s + O( 2s )) = 1 2s = 1 s + O(1), 1 2R(s) 1 2R(s). =1

5 Itroductio ad Defiitios Theorem (Euler): Proof:Let s 1 + : ad the result is 1 = 1 lim log(ζ(s)) = lim s 1 + s 1 + s + O(1), 1 = lim s 1 + s + O(1) = 1 + O(1), sice ζ(1) is the harmoic series, which diverges.

6 Itroductio ad Defiitios Defie U(k) to be the grou of iteger uits modulo k uder multilicatio; for examle U(7) = {1, 2,..., 6} Z 7, U(12) = {1, 5, 7, 11} Z 4, U(20) = {1, 3, 7, 9, 11, 13, 17, 19} Z 8. It is well kow that the cardiality of U(k) is give by the Euler totiet fuctio = k;(,k)=1 where (, k) idicates is relatively rime to k. 1,

7 Itroductio ad Defiitios We deote the Z/qZ to be the itegers modulo q. It is well kow that the th roots of uity form a multilicative subgrou of C of cardiality. We will deote the set of th roots of uity as R(). A homomorhism χ: U(k) R(k) is called a character mod k. It is well kow that there are characters mod k. We label the characters mod k as {χ 1, χ 2,..., χ }, with χ 1 called the ricial character, which seds all elemets of U(q) to 1 (all other characters are cosidered oricial).

8 Itroductio ad Defiitios Let Û(k) deote the grou of all characters mod k uder the oeratio χ a χ b ( ) = χ a ( )χ b ( ). Defie L 2 (U(k)) to be the set of comlex-valued fuctios o U(k) with resect to the ier roduct < f, g >= 1 m U(k) f (m)g(m). It is well-kow that Û(k) is a orthoormal basis of L2 (U(k)). Recall that h is relatively rime to k ad defie the characteristic fuctio δ h δ h (a) = { 1 : a h mod k 0 : otherwise.

9 Itroductio ad Defiitios Now defie ˆδ h (χ) =< δ h, χ > δ h (a)χ(a) = 1 a G = 1 χ(h). The followig reresetatio for δ h is well kow: δ h (a) = φ k r=1 ˆδ h (χ r )χ r (a) = 1 r=1 χ r (h)χ r (a).

10 Itroductio ad Defiitios We ow trivially exted all characters χ r mod k to χ r : Z/kZ R(k) {0} by the followig formula: { χr (a) : a U(k) χ r (a) = 0 : a Z/kZ \ U(k). Such a exteded character is called a Dirichlet character mod k, ad we label them {χ 1,..., χ }.

11 Itroductio ad Defiitios A Dirichlet series is a geeralizatio of the Riema zeta fuctio. Defie a Dirichlet series L χ by the formula, for R(s) > 1, Note that ζ(s) = L χ1 (s). L χ (s) = =1 χ() s. It turs out that Dirichlet s theorem boils dow to rovig that L χ (1) 0 for ay oricial character χ mod k.

12 Itroductio ad Defiitios The Dirichlet roduct of f ad g is defied to be f g ad has formula (f g)() = ( ) f (d)g. d d Let F be a comlex-valued fuctio which is defied o (0, ) with F(x) = 0 for 0 < x < 1. The geeral covolutio of F ad a arithmetic fuctio f is defied to be (f F)(x) = ( x ) f ()F. With a little algebra, ca show for arithmetical fuctios f ad g, that f (g F) = (f g) F.

13 Itroductio ad Defiitios Theorem: If h = f g, let H(x) = h(), F(x) = f (), ad G(x) = g(). The, H(x) = ( x ) f ()G = ( x ) g()f.

14 Theorem: H(x) = Itroductio ad Defiitios ( x ) f ()G = g()f ( x ). { 0 : 0 < x < 1 Proof: Defie U(x) =. The, 1 : x 1 F(x) = f () = ( ) f ()U = (f U)(x), x ad similarly, G = g U. So, ( x ) g()f = g F = g (f U) = (g f ) U = H, ( x ) f ()G = f G = f (g U) = (f g) U = H.

15 Theorem: H(x) = Itroductio ad Defiitios ( x ) f ()G = ( x ) g()f. If we set g() = 1 for all, the G(x) = [x]. The theorem yields the followig corollary: Corollary: If F(x) = f (), the f (d) = (f g)() d = H(x) = [ x ] f () = ( x ) F.

16 Itroductio ad Defiitios Euler Summatio formula: if f has cotiuous derivative f o the iterval [y, x], where 0 < y < x, the f () = y< x y f (t)dt+ x y (t [t])f (t)dt+f (x)([x] x) f (y)([y] y). Defie the Magoldt fuctio { log : = Λ() = m, rime,m 1 0 : otherwise.

17 Corollary: Itroductio ad Defiitios f (d) = [ x ] f () = F d Theorem: For x 1 we have [ x ] Λ() = log[x]!. ( x ). Proof: Use fudametal theorem of arithmetic to rereset ay k as = e i i. Aly the corollary with f () = Λ(): So i=1 [ x ] Λ() = Λ(d) = [ k ] e i log( i ) = log(). d i=1 [ x ] Λ() = log([x]!).

18 Itroductio ad Defiitios Theorem: If x 2, we have ad hece log([x]!) = x log x + O(x), x [ x ] Λ() = x log x + O(x). t [t] Proof: Realize that dt = O(log x). Take f (t) = log t 1 t ad aly Euler s summatio formula to get log([x]!) = 1< log = x 1 x log tdt + 1 = x log x x O(log x) = x log x + O(x). t [t] dt + ([x] x) log x t

19 Itroductio ad Defiitios Theorem: [ ] x log = x log x + O(x) x Proof: Sice Λ() = 0 uless = m, we have [ x ] Λ() = m=1 [ ] x m Λ( m ) = m=1 [ ] x m log, so sice m x x ad So, m=1 [ ] x m log = x [ x ] Λ() = x [ x m ] = 0 if > x, we get [ ] x log + x [ ] x log + x m=2 m=2 [ ] x m log. [ ] x m log

20 Itroductio ad Defiitios Theorem: [ ] x log = x log x + O(x) x Now the secod term i the RHS is O(x): x m=2 [ ] x m log x = x x log log x m m=2 ( 1 m=2 ( 1 2 = x log 1 1 x = x ( log x x =1 ) m ) 1 ( 1) ) log ( 1) = O(x)

21 Itroductio ad Defiitios Theorem: x Now we have [ x ] Λ() [ ] x log = x log x + O(x) = [ x x [ x = x ] log + x ] log + O(x). m=2 From earlier, we kow [ x ] Λ() = x log x + O(x). [ ] x m log

22 Itroductio ad Defiitios Theorem: x Combiig the formulas imlies x i other words [ ] x log = x log x + O(x) [ x ] Λ() = x [ ] x log + O(x) [ x ] Λ() = x log x + O(x), [ ] x log + O(x) = x log x + O(x), x [ ] x log = x log x + O(x).

23 Itroductio ad Defiitios Shairo s Tauberia Theorem: Let {a()} be a oegative sequece such that [ x ] a() = x log x + O(x), for all x 1. The for x 1 we have a() = log x + O(1). If is rime, defie Λ 1 () = log. Otherwise, defie Λ 1 () = 0. With this otatio, we ca write the theorem i the revious slide as Theorem (revious slide): [ x ] Λ 1 () = x log x + O(x). Alyig Shairo s tauberia theorem to this yields x log = Λ 1 () = log x + O(1).

24 Itroductio ad Defiitios To rove Dirichlet s theorem, it is sufficiet to rove x; h(mod k) h(mod k) log log =, = 1 log x + O(1) ( ) ad the take the limit as x, because that would yield lim x x; h(mod k) log 1 = lim log x + O(1) =. x

25 Itroductio ad Defiitios Now we ca use our reresetatio for δ h ad write x; h(mod k) log = δ h () log x χ r (h)χ r () log 1 r=1 = x = 1 χ r (h) χ r () log r=1 x = 1 log + 1 χ r (h) x r=2 x χ r () log.

26 Itroductio ad Defiitios From earlier, we kow x log so the revious slide yields = log x + O(1), x; h(mod k) If we ca rove log = log x+ 1 χ r (h) r=2 x χ r () log +O(1). 1 χ r (h) χ r () log r=2 x remais bouded, the Dirichlet s theorem would follow by takig the limit as x.

27 Itroductio ad Defiitios By a techical lemma, 1 = 1 χ r (h) r=2 x ( χ r (h) r=2 χ r () log =1 + O(1) χ r () log() ) where µ is the Möbius fuctio defied by µ()χ r () { µ( e e k ( 1) k ) = k : e 1 = e 2 =... = e k = 1 0 : otherwise. χ r () log() So we will show that coverges ad that µ()χ r () =1 = O(1). The we are doe. + O(1),

28 Itroductio ad Defiitios Theorem: Let χ is oricial character mod k, f o-egative with cotiuous egative derivative for all x x 0. The if y x x 0 we have x< y I additio, if lim x f (x) = 0, the χ()f () = χ()f () = O(f (x)). χ()f () coverges ad =1 χ()f () + O(f (x)). =1 So with f (x) = log x x we have f (x) = 1 log x ad thus the sum x 2 χ r () log() coverges. =1

29 Itroductio ad Defiitios By aother techical lemma, for r 1, L χr (1) µ()χ r () = O(1), So if we ca always divide by L χr (1), we are doe. This is why the roof of Dirichlet s theorem boils dow to showig L χ (1) 0 for ay oricial character χ mod k.

30 Itroductio ad Defiitios Theorem: For r > 1 ad real-valued χ r, L χr (1) 0. Defie A() = χ r (d), ad B(x) = A(). The it is d kow that lim x B(x) = ad for all x 1. B(x) = 2 xl χr (1) + O(1) Therefore, sice L χr (1) is the coefficiet of x ad lim x 2L χ r (1) x =, we have L χr (1) 0.

31 Itroductio ad Defiitios Theorem: For r > 1, o-real-valued χ r, L χr (1) 0. Let N(k) deote the umber of oricial o-real-valued characters mod k such that L χr (1) = 0. Note that if L χr (1) = 0, the L χr (1) = 0 as well, so N(k) is a eve umber. By a techical lemma, if L χr (1) = 0 the x; 1(modk) log = 1 N(k) log x + O(1). So if N(k) 2, the coefficiet 1 N(k) will become egative. This is a cotradictio, sice all terms i the sum i the left had side are ositive. Thus N(k) = 0.

32 Itroductio ad Defiitios I the last few slides, we have show x; h(mod k) = log x + 1 = log x + 1 = log x + 1 = log x + O(1). log χ r (h) r=2 x ( χ r (h) r=2 χ r (h) r=2 χ r () log =1 + O(1) χ r () log() ) µ()χ r () + O

33 Itroductio ad Defiitios So, takig a limit yields lim x x; h(mod k) log = h(mod k) log = lim x log x+o(1) =. Therefore there are ifiitely may rimes i the arithmetic rogressio {k + h : Z}!

34 Itroductio ad Defiitios Thak you for attedig! Refereces ANONYMOUS ARTICLE. A itroductio to aalytic umber theory. Aostol, Tom. Itroductio to Aalytic Number Theory. Sriger-Verlag 1976.