How to Prove the Riemann Hypothesis Author: Fayez Fok Al Adeh.


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1 How o Prove he Riemnn Hohesis Auhor: Fez Fok Al Adeh. Presiden of he Srin Cosmologicl Socie P.O.Bo,387,Dmscus,Sri Tels: ,735 Emil:hf@scsne.org Commens: 3 ges SubjClss: Funcionl nlsis, comle vribles, generl mhemics. Absrc: I hve lred discovered simle roof of he Riemnn Hohesis. The hohesis ses h he nonrivil zeros of he Riemnn ze funcion hve rel r equl o.5. I ssume h n such zero is s =+ bi.i use inegrl clculus in he firs r of he roof. In he second r I emlo vriionl clculus. Through equions (5) o (59) I consider () s fied eonen, nd verif h =.5.From equion (6) on wrd I view () s rmeer ( <.5 ) nd rrive conrdicion. A he end of he roof (from equion (73)) nd hrough he ssumion h () is rmeer, I verif gin h =.5 Noe: This er hs been ublished in he Journl, Sceime & Subsnce, No,6, P. The Riemnn ze funcion is he funcion of he comle vrible s = + bi (i = ), defined in he hlf lne > b he bsolue convergen series () ζ ( s) = n s nd in he whole comle lne b nlic coninuion. The funcion ζ (s) hs zeros he negive even inegers , 4, nd one refers o hem s he rivil zeros. The Riemnn hohesis ses h he nonrivil zeros of ζ (s) hve rel r equl o.5. We begin wih he equion () ζ (s) = And wih (3) s= +bi (4) ζ ( + bi ) = I is known h he nonrivil zeros of ζ (s) re ll comle. Their rel rs lie beween zero nd one. If < < hen [ ] (5) ζ (s) = s d ( < < ) s + [] is he ineger funcion Hence [ ] (6) d = s +
2 Therefore (7) ([ ] ) bi d = (8) ([ ] ) bi d = (9) ([ ] )(cos( b log ) i sin( b log )) d = Sering he rel nd imginr rs we ge () ([ ] ) cos( b log ) d = () ([ ] ) sin( b log ) d = According o he funcionl equion, if ζ (s) = hen ζ ( s) =. Hence we ge besides equion () () + ([ ] )sin( b log ) d = In equion () relce he dumm vrible b he dumm vrible (3) ([ ] ) sin( b log ) d = We form he roduc of he inegrls ()nd (3).This is jusified b he fc h boh inegrls () nd (3) re bsoluel convergen.as o inegrl () we noice h + ([ ] )sin( b log ) d + (( )) d ( where ((z)) is he frcionl r of z, ((z))<) + ([]) sin ( ) d = lim( ) + d + lim ( ) + (())d + ( is ver smll osiive number) (since (()) = whenever <) = + lim ( < + lim ( ) + ) + And s o inegrl (3) + (())d + ([ ] )sin( ) (( )) d d = + ([ ] )sin( b log ) d = lim ( ) d + lim( ) (( )) d + ( is ver smll osiive number) (since (()) = whenever <) d
3 = < + lim( ) d = (( )) d + 3 Since he limis of inegrion do no involve or, he roduc cn be eressed s he double inegrl (4) + ([ ] )([ ] )sin( )sin( ) dd = Thus (5) + ([ ] )([ ] )(cos( b log + b log ) cos( b log b log )) dd = (6) + ([ ] )([ ] )(cos( ) cos( )) dd = Th is (7) + ([ ] )([ ] )cos( ) dd = + ([ ] )([ ] )cos( ) dd Consider he inegrl on he righhnd side of equion (7) (8) + ([ ] )([ ] )cos( ) dd In his inegrl mke he subsiuion = z The inegrl becomes dz d = z dz (9) z ([ ] )([ ] )cos( z) d z z z Th is ()  z ([ ] )([ ] )cos( z) dzd z z This is equivlen o () z ([ ] )([ ] )cos( b log z) dzd z z If we relce he dumm vrible z b he dumm vrible, he inegrl kes he form () ([ ] )([ ] ) cos( b log ) dd Rewrie his inegrl in he equivlen form (3) + ( [ ] )([ ] )cos( ) dd Thus equion 7 becomes
4 (4) + ([ ] )([ ] )cos( ) dd = + ( [ ] )([ ] )cos( ) dd Wrie he ls equion in he form (5) + ([ ] ) cos( b log ){ ( 4 [ ] ) ([])}d d= Le < be n rbirr smll osiive number.we consider he following regions in he lne. (6) The region of inegrion I = [, ) [, ) (7) The lrge region I =[, ) [, ) (8) The nrrow sri I =[, ) [, ] (9) The nrrow sri I 3 = [,] [, ) Noe h (3) I = I U I U I 3 Denoe he inegrnd in he lef hnd side of equion (5) b (3) F (,) = + ([ ] ) cos( b log ){ ( [ ] Le us find he limi of F (,) s nd. This limi is given b (3) Lim [  (( )) ] cos ( ) [  (( )) + (()) ] ((z)) is he frcionl r of he number z, ((z)) < )([])} The bove limi vnishes,since ll he funcions [  (()) ],cos ( ),  (( )), nd (()) remin bounded s nd Noe h he funcion F (,) is defined nd bounded in he region I. We cn rove h he inegrl (33) F(,) d d is bounded s follows I (34) F(,) d d = [  (( )) ] cos ( ) [  (( )) + (()) I I ] d d [  (( )) ] cos ( ) [  (( )) + (()) ] d d = I ( cos ( ) [  (( )) + (()) ] d) [  (( )) ]d ( cos ( ) [  (( )) + (()) ] d) [  (( )) ] d ( cos ( ) [  (( )) + (()) ] d) [  (( )) ] d
5 < = 5 [ (( )) + (()) ] d = { lim( ) + [ (( )) + (()) ] d [ (( )) + (()) ] d } [ (( )) + (()) ] d + lim( ) where is ver smll rbirr osiive. number.since he inegrl lim( ) show h lim( ) + [ (( )) + (()) ] d is bounded, i remins o [ (( )) + (()) ] d is bounded. Since >,hen (( )) = nd we hve lim( ] d ) + = lim( ) + = lim( ) + < lim( = ( ) [ + (()) ] d [ + (()) ] d [ + ] d Hence he boundedness of he inegrl I ) + F(,) d d is roved. [ (( )) + (()) Consider he region (35) I4=I U I3 We know h (36) = F(,) d d = F(,) d d + I I I4 nd h (37) F(,) d d is bounded I From which we deduce h he inegrl F(,) d d (38) I4 F(,) d d is bounded Remember h (39) F(,) d d = F(,) d d + F(,) d d
6 6 I4 I I3 Consider he inegrl (4) F(,) d d F(,) d d = I ( I { (( )) (()) } cos( b log ) d) d ( { (( )) (()) } cos( b log ) d) d ( { (( )) (()) } cos( b log ) d) d { (( )) (()) } d d (This is becuse in his region (()) = ). I is eviden h he inegrl )) (()) inegrl d { (( } d is bounded,his ws roved in he course of roving h he F(,) d d is bounded.also i is eviden h he inegrl I is bounded. Thus we deduce h he inegrl (4) F(,) d d is bounded I Hence,ccording o equion(39),he inegrl F(,) d d is bounded. Now consider he inegrl (4) F(,) d d I3 We wrie i in he form (4) F(,) d d = ( I3 I3 {(( )) } (()) cos (b log) d ) d ( This is becuse in his region (()) = ) {(( )) } (()) cos (b log) d ) d ( ( (()) cos (b log) d ) {(( )) } d
7 7 {(( )) ( (()) d ) Now we consider he inegrl wih resec o (43) (()) d } = (lim ) d+ (lim ) (()) d + ( where is ver smll rbirr osiive number).( Noe h (())= whenever <). ) + Thus we hve (lim nd (lim ) Hence he inegrl (43) Since inegrl be H(). Thus we hve d= ) + (()) d < (lim (()) d is bounded. (()) cos (b log)d (()) cos (b log)d d d= (()) d, we conclude h he is bounded funcion of. Le his funcion (44) (()) cos (b log)d = H () K ( K is osiive number ) Now equion (44) gives us (45) K (()) cos (b log)d K According o equion (4) we hve (46) F(,) d d = ( I3 {(( )) } ( K) d = K {(( )) } (()) cos (b log) d ) d {(( )) Since F(,) d d is bounded, hen I3 he inegrl {(( )) } (47) G = d is bounded We denoe he inegrnd of (47) b (48) F = { (( )) } } d {(( )) } d is lso bounded. Therefore
8 8 Le δ G [F] be he vriion of he inegrl G due o he vriion of he inegrnd δ F. Since (49) G [F] = F d (he inegrl (49) is indefinie ) ( here we do no consider s rmeer, rher we consider i s given eonen) δg[ F] We deduce h = δf( ) h is (5) δ G [F] = δ F () Bu we hve (5) δ G[F] = d δg[ F] δ F() ( he inegrl (5) is indefinie) δf( ) Using equion (5) we deduce h (5) δ G[F] = d δ F() ( he inegrl (5) is indefinie) Since G[F] is bounded cross he elemenr inervl [,], we mus hve h (53) δ G[F] is bounded cross his inervl From (5) we conclude h (54) G δ = d F() δ = df d δ = [ F δ ] ( = ) [ F δ ] ( = ) d Since he vlue of [ F δ ] ( = )is bounded, we deduce from equion (54) h (55) lim ( ) F δ mus remin bounded. Thus we mus hve h (56) (lim ) [ δ { (( )) }] is bounded. Firs we comue δ (57) (lim ) Aling L 'Hosil ' rule we ge δ d( δ) (58) (lim ) = (lim ) = d We conclude from (56) h he roduc (59) (lim ) { (( )) }mus remin bounded. Assume h =.5.( remember h we considered s given eonen )This vlue =.5 will gurnee h he quni { (( )) } will remin bounded in he limi s ( ).Therefore, in his cse (=.5) (56) will roch zero s ( ) nd hence remin bounded. Now suose h <.5.In his cse we consider s rmeer.hence we hve (6) G []= d F (, ) (he inegrl (6) is indefinie ) Thus δ G (6) [ ] F (, ) = δ Bu we hve h (6) δ G [ ] = d δ G [ ] δ ( he inegrl (6) is indefinie ) δ Subsiuing from (6) we ge (63) δ G [ ] = d F (, ) δ ( he inegrl (63) is indefinie )
9 We reurn o equion (49) nd wrie (64) G = lim ( ) 9 Fd ( is ver smll osiive number <<) = { F ( ) lim ( ) F ( ) }  lim ( ) Le us comue (65) lim ( ) F ( ) = lim ( ) Thus equion (64) reduces o (66) G F ( ) =  lim ( ) df df (( )) = Noe h he lef hnd side of equion (66) is bounded. Equion (63) gives us F (67) δ G =lim ( ) d δ ( is he sme smll osiive number <<) We cn esil rove h he wo inegrls df nd d F δ re bsoluel convergen.since he limis of inegrion do no involve n vrible, we form he roduc of (66) nd (67) (68) K = lim( ) df d F δ = lim( ) FdF δ d ( K is bounded quni ) Th is F F (69) K = lim( ) [ ( )  ( ) ] [ δ ( )  δ ( ) ] We conclude from his equion h F F (7) { [ ( )  lim( ) ( ) ] [δ ( ) ] } is bounded. ( since lim( )δ =, which is he sme hing s lim( ) δ = ) F F Since ( ) is bounded, we deduce once h mus remin bounded in he limi s ( ), which is he sme hing s sing h F mus remin bounded in he limi s ( ). Therefore. (( )) (7) lim ( ) mus remin bounded Bu (( )) (( )) (7) lim ( ) = lim( ) (( )) = lim( ) = lim( ) I is eviden h his ls limi is unbounded. This conrdics our conclusion (7) h
10 (( )) lim ( ) mus remin bounded (for <.5 ) Therefore he cse <.5 is rejeced.we verif here h,for =.5 (7)remins bounded s ( ). We hve h (73) (( )) <  Therefore )) (( (74) lim(.5) ( ) We consider he limi (75) lim(.5) ( ) We wrie (76) = (lim ) (.5 + ) Hence we ge (77) lim(.5) ( ) = lim ( ) < lim(.5) ( ) (.5+) =lim ( ) = (Since lim( ) = ) Therefore we mus l L 'Hosil ' rule wih resec o in he limiing rocess (75) ( ) (78) lim(.5) ( ) = lim(.5) ( ) ( ) = lim(.5) ( ) Now we wrie gin (79) = (lim ) (.5 + ) Thus he limi (78) becomes ( ) (8) lim(.5) ( ) (.5 + ) (.5 + ) = lim ( ) = lim ( ).5.5 (.5 + ) = lim ( ) ( Since lim ( ) = ).5 We mus l L 'Hosil ' rule.5 (.5 + ) (.5 + ) (8) lim ( ) = lim ( ) = lim ( ) = (.5 + ) Thus we hve verified here h,for =.5 (7) roches zero s ( ) nd hence remins bounded. We consider he cse >.5. This cse is lso rejeced, since ccording o he funcionl equion, if ( ζ (s) =) (s = + bi) hs roo wih >.5,hen i mus hve noher roo wih noher vlue of <.5. Bu we hve lred rejeced his ls cse wih <.5 Thus we re lef wih he onl ossible vlue of which is =.5 Therefore =.5 This roves he Riemnn Hohesis.
11 References  Tich mrsh,e.c.(999) The Theor of he Riemnn ze funcion,london : oford universi ress, Aosol,om M.(974) Mhemicl Anlsis, Reding,Msschuses ;Addison wesle Publishing Comn : P.76, P.77, P.39,P.39,P Edwrds,H.M. (974) Riemnn "s ze funcion, New York : Acdemic ress, Inc., P.37 4 Aosol,Tom M. (976) Inroducion o Anlic Number Theor, New York: Sringer Verlg, P.49 P _ Koblis, Nel (984) P dic Numbers, P dic Anlsis, nd Ze Funcions,New York : Sringer Verlg, P.9 P.8. 6_ Geiner,Wler ; Reinhrd,Jochim (996) Field Qunizion,Berlin :Sringer Verlg,37 39.
How to prove the Riemann Hypothesis
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