American Journal of Multidisciplinary Research & Development (AJMRD)Volume 05, Issue 04 (May - 2023), PP 22-25ISSN: 2360-821Xwww.ajmrd.com
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Research Paper Open AccessThe proof of the Riemann conjectureLiaoTengTianzheng International Mathematical Research Institute, Xiamen, ChinaAbstract:In order to strictly prove the hypothesis and conjectures in Riemann''s 1859 paper on the Number ofPrime Numbers Not greater than x from a pure mathematical point of view, this paper uses Euler''s
formulaandprovesthatRiemann''shypothesisandRiemann''sconjecturearecompletelycorrect.Key words:Euler''s formula, Riemann ζ(s) function, Riemann function ξ(t), Riemann hypothesis, Riemannconjecture . I. 1 IntroductionRiemann hypothesis and Riemann conjecture are an important and famous mathematical problemleft by Riemann in his paper "On the Number of prime Numbers not greater than x"
[1], which is ofgreat significance for the study of prime number distribution and known as the biggest unsolvedmystery in mathematics.After years of hard work, I have solved this problem. The research showsthat the Riemann hypothesis and the Riemann conjecture are valid.which is of great significance forthe study of prime number distribution and known as the biggest unsolved mystery in mathematics.After years of hard work, I have solved this problem. The research shows that the RiemannhypothesisandtheRiemannconjecturearevalid.II. ReasoningIf s is any complex number,and Re(s) is anyreal number,then the expressions for both Euler''sZeta() function andRiemann''s Zeta() function canbe written as ζ(s) =
nn?s (n∈ Z+,s ∈C，ntraverses a finite number of natural numbers whose value size is ordered, without repetition oromission), According to Mr. Riemann''s paper "On the Number of prime Numbers Not greaterthan the given number x", we get the expression ζ(1-s)=21?sπ ?sCos(π s2 )Γ(s)ζ(s) for Riemann''sZeta()function.Let''sprovethat ζ(s)and ζ((s)arecomplexconjugationsofeachother.Because e=2.718281828459045... ,e is a natural constant, I use for Multiplication, ^ formultiplication,thenbasedoneuler''s eix=cosx+isin(x)(x∈R),
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get(e^(3i))^(2)=(cos(3)+isin(3))^(2)=cos(23)+isin(23)=cos(6)+isin(6)，becausee^(6i)=cos(6)+isin(6),so(e^(3i))^(2)=e^(6i)，Ingeneral,(e^(bi))^(c)=e^(bci)(b∈R，c∈R)isestablished.Forx>0(x∈R),supposeej=x(e=2.718281828459045…， x∈Randx>0，j∈R)，thenj=ln(x),basedoneuler''s e
ix=cos(x)+isin(x)(x∈R),willget eji = eln(x)i=cos(lnx)+isin(lnx)(x∈Randx>0).supposey∈Rand y ≠ 0, now let''s figure out expression for xyi(x∈R,andx>0,y∈Rand y ≠ 0)isxyi=(ej)yi=(eji)y=(cos(lnx) + isin(lnx))y.Suppose s is any complex number, and s=ρ+yi (ρ∈R,y∈R and y ≠ 0,s∈C) ,then let''s find theexpressionof xs(x∈Randx>0,s∈C),Youputs=ρ+yi (ρ∈R,y∈Rand y ≠ 0,s∈C) and xyi=(ej)yi=(eji)y=(cos lnx + isin(lnx))y into xsandyouwillgetxs = x(ρ+yi) = xρxyi = xρ(cos(lnx) +isin(lnx))y = xρ(cos ylnx + isin(ylnx)),Base on xyi =(ej)yi = (eji)y = (cos lnx + isin(lnx))y and s = ρ ? yi(ρ ∈ R,y ∈ R and y ≠ 0,s ∈ C),Plugthemintothefollowingequation,andwegetx
s =x(ρ?yi) =xρx?yi =xρ(xyi)?1 = xρ(cos(lnx) +isin(lnx))?y =xρ(cos ?ylnx + isin ?ylnx )=xρ cos ylnx ? isin ylnx .Thenζ s = 1xs = 1xρ+yi = ( 1xρ ? 1xyi ) = x?ρ 1(cos lnx +isin(lnx))y= (x?ρ(cos lnx + isin(lnx))?y) = (x?ρ(cos ylnx ? isin(ylnx)) ,and ζ s = 1n
s = 1xρ?yi = ( 1xρ ? 1x?yi ) = ( 1xρ ? 1(cos(lnx) + isin(lnx))?y )= (x?ρ(cos?(lnx) + isin(lnx))y) = (x?ρ(cos ylnx + isin(ylnx)) ,andζ 1 ? s = 1x1?s = 1x1?ρ?yi = ( 1x1?ρ ? 1x?yi ) = xρ?1 1(cos lnx + isin(lnx))?y= xρ?1 (cos lnx + isin(lnx))y) = xρ?1 (cos ylnx + isin(ylnx)) ,SoX=x?ρ(cos ylnx ? isin(ylnx)),Y=x
?ρ(cos ylnx +isin(ylnx)),XandYarecomplexconjugatesofeachother,thatisX=Y, soζ s = 1xs = X and ζ s = 1xs = Y ,
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thenζ(s)=ζ(s) [2] .As Riemann said in his paper,n traverses a finite number of natural numbers whose value size isordered,withoutrepetitionoromission,sox=1,2,3...n-1,n...,let''sjustpluginζ s = nn?s (n∈ Z+,s ∈C，ntraversesafinitenumberofnaturalnumberswhosevaluesizeisordered,withoutrepetitionoromission).Obviously,ζ s =ζ(ρ+yi)=
1xs = X =[1?ρCos(yln1)+2?ρCos(yln2)+3?ρCos(yln3)+4?ρCos(yln4)+...]-i[1?ρSin(yln1)+2?ρsin(yln2)+3?ρsin(yln3)+4?ρsin(yln4)+...]=U-Vi,U=[1?ρCos(yln1)+2?ρCos(yln2)+3?ρCos(yln3)+4?ρCos(yln4)+...],V=[1?ρSin(yln1)+2?ρsin(yln2)+3?ρsin(yln3)+4?ρsin(yln4)+...],Thenζ(s)=ζ(ρ-yi)= 1xs = Y =[1?ρCos(yln1)+2?ρCos(yln2)+3?ρCos(yln3)+ 4?ρCos(yln4)+...]+i[1?ρSin(yln1)+2?ρsin(yln2)+3?ρsin(yln3)+4?ρsin(yln4)+...]=U+Vi,U=[1
?ρCos(yln1)+2?ρCos(yln2)+3?ρCos(yln3)+4?ρCos(yln4)+...],V=[1?ρSin(yln1)+2?ρsin(yln2)+3?ρsin(yln3)+4?ρsin(yln4))+...],ζ 1 ? s = xρ?1 (cos ylnx + isin(ylnx)) =[ 1ρ?1Cos(yln1)+ 2ρ?1Cos(yln2)+3ρ?1Cos(yln3)+4ρ?1Cos(yln4)+...]+i[1ρ?1Sin(yln1)+ 2ρ?1sin(yln2)+ 3ρ?1sin(yln3)+ 4ρ?1sin(yln4)+...],ζ(s)andζ(s)arecomplexconjugatesofeachother,thatisζ(s)=ζ(s)[2].Accordingtheequationζ(1-s)=21?sπ ?sCos(π s2 )Γ(s)ζ(s)obtainedbyRiemann,sinceRiemannhasshownthatthe Riemann ζ(s) function has zero,that is,inζ(1-s)=21?sπ ?sCos(πs2 )Γ(s)ζ(s),if ζ(s)=0holds. Sowhen ζ(s)=0 ,then ζ(1-s)=0, and it must be true that ζ(1-s)=ζ(s)=0, and it must be true that ζ(s)=ζ(s)=0anditmustbetruethatζ(1-s)=ζ((s)=0.So1-s=sor1-s=s ,then s=
12 or s=12+yi (y∈Rand y ≠ 0,s∈C),soonlywhen ρ=12 is true.Becauseζ(12)>0,so1-s=siswrong,and1-s=s is true ,so s=12+yi (y∈Rand y ≠0,s∈C).And only when ρ=12 and ζ(s)=0, the next three equations, ζ(ρ+yi)=0, ζ(1 ? ρ ?yi)=0, andζ(ρ -yi)=0 are all true,so only s = 12 +yi (y ∈ R and y ≠ 0,s ∈ C) is true,or sayonly s = 12 +ti (t ∈ R and t ≠ 0,s ∈ C）is true.Riemann got
s2 (s-1)π?s2ζ(s)=ξ(t) and ξ(t)=12 - (t2 +14) 1∞Ψ(x) x?34cos(12tlnx)dx in his paper,ors2 (s-1)π?s2ζ(s)=ξ(t)andξ(t)=4 1∞d(x32Ψ’(x))dx x?14cos(12tlnx)dx,Becasueζ(12+ti )=0(t∈Rand t ≠ 0,s∈C）isture,so s2 (s-1)π?s2ζ(12+ti )=ξ(t)=0(t∈Rand t ≠ 0,s∈C)ands2 (s-1)π?s2ζ(12+ti )=4 1∞d(x32Ψ’(x))dx x?14cos(12tlnx)dx=ξ(t)=0,and
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ξ(t)=12 -(t2 +14) 1∞Ψ(x) x?34cos(12tlnx)=0 [1],so the roots of equations s2 (s-1)π?s2ζ(12+ti )=ξ(t)=0 and4 1∞d(x32Ψ’(x))dx x?14cos(12tlnx)dx=ξ(t)=0 and ξ(t)=12 - (t2 +14) 1∞Ψ(x) x?34cos(12tlnx)=0 must all be realnumbers.When ζ(s)=0 and ξ(t)=0, the real part of the equation ξ(t)=0 must be real between 0 and T.Because the real part of the equation ξ(t)=0 has the number of complex roots between 0 and Tapproximately equal to T2πln T2π ? T2π[1] ,This result of Riemann''s estimate of the number of zeros wasrigorously proved by Mangoldt in 1895. Then,when ζ(s)=0 and ξ(t)=0, the number of real roots of thereal part of the equation ξ(t)=0 between 0 and T must be approximately equal to
T2πln T2π ? T2π[1] ,So,when ζ(s)=0,theRiemannhypothesisandtheRiemannconjectureareperfectlyvalid.III. ConclusionTheRiemannhypothesisandtheRiemannconjecturearecompletelycorrect.IV. Significance and prospectAfter the Riemann hypothesis and the Riemann conjecture are proved to be completely valid, theresearch on the distribution of prime numbers and other studies related to the Riemann hypothesis andtheRiemannconjecturewillplayadrivingrole.Readerscandoalotinthisrespect.V. Thanks
Thankyouforreadingthispaper. VI. ContributionThesoleauthor,posestheresearchquestion,demonstratesandprovesthequestion.VII. AuthorName:TengLiao(1509135693@139.com),SoleauthorSetting:TianzhengInternationalInstituteofMathematicsandPhysics,Xiamen,ChinaWorkunitaddress:237AirportRoad,WeiliCommunity,HuliDistrict,XiamenCityZipCode:361022 References
[1] Riemann:《OntheNumberofPrimeNumbersLessthanaGivenValue》；[2] John Derbyshire(America):《PRIME OBSESSION》P218,BERHARD RIEMANNAND THE GREATEST UNSOlVED PROBLEM IN MATHMATICS,Translated by ChenWeifeng, Shanghai Science and Technology Education Press,China,https://www.doc88.com/p-54887013707687.html;
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[3] XieGuofang:Onthenumberofprimenumberslessthanagivenvalue-NotestoRiemann''soriginalpaperproposingtheRiemannconjecture,https://max.book118.com/html/2021/0519/8016007070003102.shtm
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