Solutions to Problems 3, Level 4

Transcription

1 Soluios o Problems 3, Level 4 23 Improve he resul of Quesio 3 whe l. i Use log log o prove ha for real >, log ( {}log + 2 d log+ P ( + P ( d 2. Here P ( is defied i Quesio, ad parial iegraio has bee used. ii For ieger N show ha log N where C is he cosa ( N + 2 iii Deduce ha for ieger N, C N! N /2 e C ( N e ( logn N +C +O, N P ( d 2. N ( ( +O. N (You may assume ha e O(f( +O(f ( if f ( 0 as. Compare his resul wih Quesio 5 o Shee. Soluio i log ( log ( {}log + d []log {} d ( {}log + 2 log+ ( {}log log+ P ( [] d ( {} d 2 + P ( d 2

2 o iegraig by pars. ii For ieger N we have {N} 0 ad P (N 0. Complee he iegral i par i o ifiiy, wih a error i doig so of P ( d 2 P ( d d N, N N havig used par iv of Quesio. Thus we ge he error of O(/N see i par ii. iii Sarig from ( log log logn!, Epoeiae o ge N N N! N N+/2 e N e C e O(/N e C N /2 ( N e usig he assumpio i he quesio. N N ( ( +O, N 24 Improve par ii of Quesio 6 by provig ha, for all l here eiss a cosa C l such ha log l ( log l l+ logl+ +C l +O. This is a geeralisaio of Theorem 3.5. Hi Jusify log l log l ydy +E l (, where E l ( O ( log l. Do o evaluae he iegral (ha was doe i he soluio o Quesio 4 bu ow use parial summaio ad ierchage he wo iegrals you ge. The follow he proof of Theorem 3.5. Proof Lemma 3.4 gives log l log l ydy +E l (, 53

3 where E l ( O ( log l. The parial summaio gives log l log l + log l d 2 ( log l d+e l ( + The mai erms are log l d+ The errors i (32 are ( log l d+ ( d log l ydy ( y d 2 2 d log l ydy +E l ( 2. log l ydy o ierchagig he iegrals ( log l d+ y log l ydy log l y dy y l+ logl+. E l ( + The iegral here coverges absoluely sice E l ( d 2 E l ( 2 d. log l 2 d l!. Thus we complee he iegral up o, he error i doig so is by Quesio 6. Combiig log l E l ( d 2 l+ logl+ +O 54 log l 2 d logl, ( log l + E l ( d 2. (32

4 Hece C l E l ( d Use Quesio 24 o improve Quesio 8 ad show ha for every l, here eiss a polyomial Q d (y of degree d leadig coefficie /(d+ such ha log l (/ ( log l Q l+ (log+o. Soluio Use he Biomial epasio log l (/ (log log l l r0 ( l ( log r (log l r. r Thus log l (/ l r0 l r0 ( l (log l r ( r (log r r ( ( l (log l r ( r r r+ logr+ +C r +O ( log r (33 from Quesio 24. The domiae erm here is (log l+ l r0 ( l ( r r r+. I is easy o check (perhaps by wriig he biomial coefficie i erms of facorials ha ( l ( l+. r+ r l+ r+ 55

5 I which case l ( l ( r r r+ r0 l+ l+ l+ l r0 ( l+ ( r r+ l+ ( l+ k k ( k ( l+ ( l+ ( k k k0 ( ( l+ l+ l+. We ake as he required polyomial Q l+ (log l+ logl+ + l ( l ( r C r (log l r. r r0 I (33 here are l+ error erms all of he size O ( (log l /. 26 Le ζ C saisfy ζ wih ζ. i Prove ha for all N. N ζ 2 ζ ii Deduce ha coverges for all real Res > 0. ζ s iii Deduce ha as log as θ 2πk for ay k Z, he Dirichle Series siθ s 56

6 coverges for all Res > 0. Soluio i Sice he series is a fiie geomeric series we have ha Hece N N ζ ζ ζn ζ. ζ N ζ + ζ N ζ by he riagle iequaliy. Bu ζ N ζ N N ad so we have he saed resul. ii Covergece follows from Quesio 3. iii By Quesio 3 i suffices o show ha N siθ is bouded for all N. Ye N siθ Im N e iθ Im einθ e iθ, by par i wih ζ e iθ. Allowable sice θ 2πk for ay k Z implies e iθ. The sice Imz z we see ha N siθ e inθ e iθ 2 e iθ. This boud, idepede of N, meas ha he resul of Quesio 3 implies he Dirichle series coverges for Res > i Assume ha f has a coiuous derivaive ad he sum over primes f (p coverges. Prove ha he ail-ed sum saisfies p f (p π(f ( p> π(f (d. ii Assume ha f is sricly decreasig ad lim f ( 0. Deduce ha Chebyshev s iequaliy for π( leads o p> f (p log f (+ log 57 f (d,

7 for sufficiely large. iii Wha does his say of p> p(p, see i Chaper? Hi For par i use Soluio i p>f (p p> f ( π(f ( p f (d f (d. (π( π(f (d ii Use Chebyshev s boud i he form π( C log, ( p π(f (d. f (d for some cosa C. Simplify he resul i par i by replacig i by a upper boud by discardig he egaive erm π(f (, so f (p p> π(( f (d C log ( f (d, havig used he fac ha f ( > 0. For his iegral we have i which case log log ad O iegraig by pars, log f (d log f (d. f (d [ f (] + f (d. 58

8 Combie o ge p> iii For he eample give p> f (p log f (+ log p(p log ( + log f (d. Use he fac ha /2 for sufficiely large, o ge p> for sufficiely large. p(p log + log d (. d 2 log, 28 Recall from Chaper ha give N N he se N is defied by N { : p p N}. Prove ha for some κ > 0. N κlogn +O( Soluio We saw i Chaper ha N (. p p N Thus he resul follows from Theorem The oly difficuly may come wih he error. Theorem 3.22 is of he form f g+ε ad we wa a resul o /f. Ye f g g f fg ε g 2, sice f is close o g. I our case g C/log ad ε O ( /log 2 i which case ε/g 2 O ( log 2 /log 2 O(. 59

9 29 Do he followig series coverge or diverge? i for β >. ii for β >. 3 log(loglog β p 3 iii Wha are he resuls if β? Soluio i Apply Theorem 3.4 o ge log(loglog β 3 3 p(loglogp β For he iegral, chage variable y loglog o ge loglog loglog3 d log(loglog β +O(. dy y β which coverges sice β >, hece he series coverges. ii Parial Summaio gives p(loglogp β (loglog β 3 p +β 3 3 p p ( 3 p (loglog+o( (loglog β +β 3 p d (loglog +β log (loglog+o(d (loglog +β log, havig used Mere s resuls. I he iegrals agai chage variable o y loglog ad boh are see o coverge, i.e. 3 loglogd (loglog +β log 60 loglog loglog3 dy y β

10 ad, for he error erm, 3 d loglog (loglog +β log dy loglog3 y +β. iii If β he boh sums equal ad hus boh diverge. logloglog+o(, 30 (Tricky Recall a form of Möbius Iversio foud i Quesio 28 of Shee 2. This saed ha if F is a arihmeic fucio ad G( d F (d he F ( ( µ(dg. d d i Use his o show ha ( ( µ(log G F (log+ m d Λ(dF ( d. Hi sar wih log(/ log logm ad use he form of Möbius Iversio above alog wih Quesio 2. ii Geeralise Λ µ l by defiig Λ k µ l k for all k 0. Prove ha Λ k+ Λ k l+λ Λ k, (34 i.e. for all. Λ k+ ( Λ k (log+λ Λ k ( 6

11 Soluio i Sar wih log(/ log logm, ( ( µ(log G ( µ((logg m m (logf ( m ( µ((logg m µ((log d m F (d havig used Möbius Iversio from above (logf ( d (logf (+ d F (d µ((log m d ( F (dλ, d havig fiished wih (22. ii The defiiio of Λ k µ l k ca be wrie as l k Λ k. So apply par i wih F Λ k ad G l k o ge Λ k (log+ ( Λ(dΛ k ( ( µ(log log d k m d m m µ(log k+ ( m Λ k+ (. Aside O a addiioal Quesio Shee a aleraive proof of (34 is give. There are furher quesios, askig you o show ha if gcd(m, he Λ k (m ( k Λ j (Λ k j (. j 0 j k This geeralises Quesio 2, he k case, because he Λ 0 µ l 0 µ δ. Furher here are quesios showig ha Λ k is suppored o iegers wih a mos k disic prime facors, ha Λ k ( (log k ad ha Mere ype resuls hold for Λ k. 3 i a Prove ha if logy > A > 0 he 0 < < A y log < Ay. logy 62

12 b Prove ha if y/logy > /B > 0 he 0 < B y logy < y < 2 B log. ii Recall Chebyshev s Theorem, ha for some a < ad b > we have for sufficiely large. a log < π( < b log Prove ha here eis cosas D ad D 2 such ha he -h prime p saisfies D log p D 2 log, for sufficiely large. Hi Wha is π(p? I boh a ad b sar by akig he logarihm of he assumpios < Ay/logy ad By/logy < respecively, rearrage ad feed i he assumpios logy > A ad y/logy > /B Soluio i a Take he logarihm of 0 < < Ay/logy o ge log < loga+logy loglogy < logy, give he assumpio logy > A > 0, epressed as loglogy > loga. Subsiuig his lower boud o logy back i o < Ay/logy, we see ha which is he saed resul. < A y logy A y log, b Assume By/logy <. Take he logarihm of 0 < By/logy < o ge log > logb +logy loglogy 2 logy give he assumpio y/logy > /B, epressed as loglogy logb < (logy/2. Subsiuig his upper boud o logy back i o By/logy <, we see ha which is he saed resul. > B y logy B 2 y log, 63

13 ii The observaio here is ha π(p, hus Chebyshev s Theorem becomes a p < < b p. (35 logp logp The by par i b log < p < 2 a log as log as logp > b ad p /logp > /a. Sice p > hese will be saisfied if log > b ad /log > /a, i.e. if is sufficiely large i erms of a ad b. 32 Recall ha for wo fucio k( ad l(, ad cosa c, we wrie k( cl( as if, ad oly if lim k(/l( c. i Prove firs a geeral resul o limis. Assume for fucios f,g ad h ad cosas a ad b ha f ( ah( ad g( bh( as. Prove, usig resuls o limis, ha as, for all λ,µ R. λf (+µg( (λa+µbh( ii Assume he Prime Number Theorem i he form π( /log as. Prove ha for a fied c >, as. π(c π( (c log iii Recall a secod year resul ha if lim f ( l wih l > 0 he here eiss X such ha > X implies ha l/2 f ( 3l/2. Deduce from Par ii ha for ay fied c > he ierval [,c] coais a prime for all sufficiely large. Soluio i We are assumig ha f ( ah( ad g( bh( as, which ca be wrie as f ( lim h( g( a ad lim h( b. 64

14 Cosider λf (+µg( lim h( ( λf ( lim h( + µg( h( f ( g( λ lim +µ lim h( h( λa+µb feedig i our assumpios above. Thus we have show ha as. λf (+µg( (λa+µbh( ii The assumpio π( /log as meas ha π(c c logc c logc+log c log as. So we ca apply Par i wih f ( π(, g( π(c, h( /log, a ad b c o deduce, wih λ, µ, ha as. π(c π( (c log iii By he recollecio we have he eisece of a X such ha > X implies ha <p c π(c π( (c 2 log > 0 Thus here is a prime bewee ad c for all sufficiely large. 33 i Assume ha f is a icreasig fucio for which f (logf ( as. a Prove ha lim f (. b Prove ha f ( /log as. Hi for b: Wrie f (logf ( as a limi ad ake logarihms. 65

15 ii Prove ha if he -h prime p saisfies p log as he π( /log as. This is he coverse o he resul of Quesio 24. Soluio i a By coradicio, if lim f ( he here eiss M (ake o be greaer ha so logm is defied, such ha f ( < M for all. Bu his meas ha f (logf ( < M logm ad so f (logf ( lim 0 coradicig f (logf ( as. Hece lim f (. b The assumpio f (logf ( as meas ha Take logarihms o ge f (logf ( lim. Divide by logf ( o ge lim (logf (+loglogf ( log 0. lim ( loglogf ( + log logf ( logf ( 0 For he secod erm use he idea ha as y he (logy/y 0. Thus lim equivale o he saed coclusio. log logf (, 66

16 Addiioal Quesios level 4 34 Show ha if h( c as he lim log h( d c. Deduce from Quesio 2 ha IF lim ψ(/ eiss he he limi has value. (Noe, his is o a proof of he Prime Number Theorem, bu shows wha he correc saeme of he Prime Number Theorem should be. Hi You eed o verify he defiiio of limi, i.e. for all ε > 0 here eiss X such ha if > X he h( d c log < ε. To his ed wrie c c log d, subsiue i ad ry o make use of he assumpio ha h( c as. Soluio Le ε > 0 be give. The assumpio h( c as meas here eiss X 0 such ha h( c < ε/2 for all > X 0. The log h( d c log log X0 h( c d h( c d+ log h( c X 0 The firs iegral is idepede of ad is hus a cosa k, say. Hece h( d c log k log + h( c d. log X 0 There is a X such ha k log < ε 2 for > X. Therefore, if > ma(x 0,X he h( d c log ε 2 + ε/2 log 67 X 0 d ε d.

17 as required. The coclusio o lim ψ(/ follows immediaely. 35 i Differeiae log ( s wih respec o, for Res >. ii Prove ha for Res > we have logζ(s s 2 π( ( s d. HiForpariiuseQuesio2. Thisrequiresshowighaπ(Nlog( N s 0 as N. For his, recall ha a logarihm of N s is give by a power series ad for Res > his series coverges absoluely ad ca be bouded by a geomeric series ad hus summed. Soluio i Recall ha so Bu d d d d d logf ( f ( d f (, d ( d log s d d ( s ( s. ( d s d d s d e slog ( se slog d d log s Combiig we have ii Quesio 2 gives log ( p s p N d ( d log s N 2 π( s ( s. d ( ( s +π(nlog N. s +s. The log ( N s r rn rs. 68

18 As log as /N s < his sums absoluely as r rn rs r ( r N σ N σ. Simply usig π(n N we ge ( π(nlog N N s N σ 0 as N as log as σ >. O his codiio we fid ha log ( p d π( s ( s p 2 36 Recall from Problem Shee 2 ha a form of he Möbius Iversio saes ha if F is a fucio o [, ad G( F (/, he F ( (. (36 µ(g Prove he Tauzawa-Iseki Ideiy ( ( µ(log G F (log+ Λ(F m (. This ideiy is see i some elemeary proofs of he Prime Number Theorem (i.e. proofs ha do o use comple aalysis. Hi O he lef had side wrie log(/ log logm. O oe of he erms use (36, o he oher subsiue i for G(/ ad rearrage he sums. Soluio µ(log m ( ( G log m µ(g ( m µ((logg ( logf ( m µ((logg (, 69

19 by (36. For he secod erm here we have m ( µ((logg m µ((log /m ( /m F µ((log ( F N m N m N N F ( µ(logm. N m m N N Λ(NF ( N, I his ier sum he codiio m ca be dropped sice i follows from m N ad N. The ier sum is he µ(logm Λ(N by (22. m N 70