就是证明π的超越性的一篇文章。。很简短的

TheTranscendenceofπ

SteveMayer

November2006

Abstract

Theproofthatπistranscendentalisnotwell-knowndespitethefactthatitisn’ttoodi cultforauniversitymathematicsstudenttofollow.Thepurposeofthispaperistomaketheproofmorewidelyavailable.Abonusisthattheproofalsoshowsthateistranscendentalaswell.

Thematerialinthesenotesarenotmine;itistakenfromasupplementissuedbyIanStewartasanadjuncttoaRingsandFieldscoursein1970attheUniversityofWarwick.

De nition.AcomplexnumberisalgebraicoverQifitisarootofapolynomialequationwithrationalcoe cients.

Thusaisalgebraiciftherearerationalnumbersα0,α1,...,αnnotall0,suchthatα0an+α1an 1+...+αn 1a+αn=0.

De nition.Acomplexnumberistranscendentalifitisnotalgebraic,soitisnottherootofanypolynomialequationwithrationalcoe cients.

Inprovingthatitisimpossibleto’squarethecircle’byaruler-and-compassconstructionwehavetoappealtothetheorem:

TherealnumberπistranscendentaloverQ

Thepurposeofthissupplementistoindicate,forthosewhomaybeinterested,howthistheoremmaybeproved.

Itispossibletoprovethatthereexisttranscendentalrealnumbersbyusingin nitecardinals,aswas rstdonebyCantorin1874.EarlierLiouville(1844)had∞ actuallyconstructedtranscendentals,forexample10 n!istranscendental.1

n=1

1Aproofcanbefoundathttp://rutherglen.ics.mq.edu.au/math334s106/m2334.Dioph.Liouville.pdf

就是证明π的超越性的一篇文章。。很简短的

However,nonaturallyoccurringrealnumber(suchaseorπ)wasprovedtran-scendentaluntilHermite(1873)disposedofe.πheldoutuntil1882whenLinde-mann,usingmethodsrelatedtothoseofHermite,disposedofthat.In1900DavidHilbertproposedtheproblem:

Ifa,barerealnumbersalgebraicoverQ,ifa=0or1andbisirrational,proveabistranscendental.

Thiswassolvedindependentlyin1934bytheRussian,Gelfond,andaGerman,Schneider.

Beforeprovingtranscendenceofπweshallproveanumberofsimilartheorems,usingsimplerversionsofthe nalmethod,asanaidtocomprehension.Thetoolsneededare rst-yearanalysis.2

Theorem1.πisirrational

+1nProof.LetIn(x)= 1(1 x2)cos(αx)dx

Integratingbypartswehave

α2In=2n(2n 1)In 1 4n(n 1)In 2

whichimpliesthat

α2n+1In=n!(Pnsin(α)+Qncos(α))

wherePn,Qnarepolynomialsofdegree<2n+1inαwithintegercoe cients.Remark.degPn=n,degQn=n 1

bπ,andassumeπisrational,sothatπ=,a,b∈Z2a2n+1bInFrom(*)wededucethatJn=isaninteger.OntheotherhandJn→0n!

asn→∞sincebis xedandInisboundedby

+1 πx dxcos2 1Putα=Jnisaninteger,→0.ThusJn=0forsomen.Butthisintegrandiscontinuous,andis>0inmostoftherange( 1,+1),soJn=0.Contradiction. (*)(n≥2)2Thiswastruein1970.Isitstilltruetoday?

就是证明π的超越性的一篇文章。。很简短的

Theorem2.π2isirrational(soπdoesnotlieinanyquadraticextensionofQ)

aProof.Assumeπ2=,a,b∈Z.bDe ne

xn(1 x)n

,f(x)=n! 2n 2n 2 n0(2n)nf(x)+...+( 1)πf(x)G(x)=bπf(x) π

(superscriptsindicatingdi erentiations).Weseethatthevalueofanyderivativeoffat0or1iseither0oraninteger.AlsoG(0)andG(1)areintegers.Now d[G (x)sin(πx) πG(x)cos(πx)]=G (x)+π2G(x)sin(πx)dx

=bnπ2n+2f(x)sin(πx)sincef(2n+2)(x)=0

=π2ansin(πx)f(x)

sothat

π

01 1 G(x)sin(πx) G(x)cos(πx)ansin(πx)f(x)dx=π0

=0+G(0)+G(1)

=integer.

Butagaintheintegralisnon-zeroand→0asn→∞.Thusagainwehaveacontradiction. Gettingmoreinvolved,now:

Theorem3(Hermite).eistranscendentaloverQ

Proof.Supposeamem+...+a1e+a0=0(ai∈Z).WLOGa0=0

xp 1(x 1)p(x 2)p...(x m)p

De nef(x)=(p 1)!whereforthemomentpisarbitraryandprime.

De neF(x)=f(x)+f (x)+...+f(mp+p 1)(x).

Nowif0<x<m,

mp 1mmp

|f(x)|≤(p 1)!

mmp+p 1

=(p 1)!

就是证明π的超越性的一篇文章。。很简短的

Alsosothatd x(eF(x))=e x[F (x) F(x)]= e xf(x)dx

0jaj je xf(x)dx=aj e xF(x)0

=ajF(0) aje jF(j).

Multiplyingbyejandsummingoverj=0,1,...mweget

m

j=0ajej 0je xf(x)dx=F(0).0

= m j=0ajF(j)ajf(i)(j).+p 1mmp

j=0i=0

Weclaimthateachf(i)(j)isaninteger,divisiblebypexceptwhenj=0andi=p 1.Foronlynon-zerotermsarisefromtermswherethefactor(x j)phasbeendi erentiatedptimes,andthenp!cancels(p 1)!andleavesp,exceptintheexceptionalcase.

WeshowthatintheexceptionalcasethetermisNOTdivisiblebyp.Clearlyf(p 1)(0)=( 1)p...( m)p.WeCHOOSEplargerthanm,whenthisproductcannothaveaprimefactorp.Hencetheright-handsideoftheaboveequationisaninteger=0.Butasp→∞theleft-handsidetendsto0,usingtheaboveestimatefor|f(x)|.Thisisacontradiction. Theorem4(Lindemann).πistranscendentaloverQ

Proof.Ifπsatis esanalgebraicequationwithcoe centsinQ,sodoesiπ(i=√Letthisequationbeθ1(x)=0,withrootsiπ=α1,...,αn.Noweiπ+1=0so

(eα1+1)...(eαn+1)=0

Wenowconstructanalgebraicequationwithintegercoe cientswhoserootsaretheexponentsofeintheexpansionoftheaboveproduct.Forexample,theexponentsinpairsareα1+α2,α1+α3,...,αn 1+αn.TheαssatisfyapolynomialequationoverQsotheirelementarysymmetricfunctionsarerational.Hencetheelementarysymmetricfunctionsofthesumsofpairsaresymmetricfunctionsoftheαsandarealsorational.Thusthepairsarerootsoftheequationθ2(x)=0withrationalcoe cients.Similarlysumsof3αsarerootsofθ3(x)=0,etc.Thentheequation

θ1(x)θ2(x)...θn(x)=0

就是证明π的超越性的一篇文章。。很简短的

isapolynomialequationoverQwhoserootsareallsumsofαs.Deletingzerorootsfromthis,ifany,weget

θ(x)=0

θ(x)=cxr+c1xr 1+...cr

andcr=0sincewehavedeletedzeroroots.Therootsofthisequationarethenon-zeroexponentsofeintheproductwhenexpanded.Calltheseβ1,...βr.Theoriginalequationbecomes

eβ1+...eβr+e0+...e0=0

ie

wherekisaninteger>0

Nowde ne eβi+k=0(=0sincetheterm1...1exists)

f(x)=cxsp 1[θ(x)]p

(p 1)!

wheres=rp 1andpwillbedeterminedlater.

De ne

F(x)=f(x)+f (x)+...+f(s+p)(x). d xeF(x)= e xf(x)asbefore.dx

Hencewehave x

e xF(x) F(0)= e yf(y)dy

Puttingy=λxweget

F(x) eF(0)= x

Letxrangeovertheβiandsum.Since

r

j=1x 01e(1 λ)xf(λx)dλ. eβi+k=0wegetβj 01F(βj)+kF(0)= r j=1e(1 λ)βjf(λβj)dλ.

CLAIM.ForlargeenoughptheLHSisanon-zerointeger.

r f(t)(βj)=0(0<t<p)byde nitionoff.Eachderivativeoforderpormorehasafactorpandafactorcs,sincewemustdi erentiate[θ(x)]penoughtimestoget=0.Andf(t)(βj)isapolynomialinβjofdegreeatmosts.Thesumis

j=1

就是证明π的超越性的一篇文章。。很简短的

symmetric,andsoisanintegerprovidedeachcoe cientisdivisiblebycs,whichciofitis.(symmetricfunctionsarepolynomialsincoe cients=polynomialsincdegree≤s).Thuswehave

r

j=1f(t)(βj)=pktt=p,...,p+s.

ThusLHS=(integer)+kF(0).

f(t)(0)=0

f(p 1)(0)=cscprWhatisF(0)?t=0,...,p 2.(cr=0)

t=p,p+1,....f(t)(0)=p(someinteger)

SotheLHSisanintegermultipleofp+cscprk.Thisisnotdivisiblebypifp>k,c,cr.Soitisanon-zerointeger.ButtheRHS→0asp→∞andwegettheusualcontradiction.