10个最常见的泰勒级数展开公式common-taylor-series

CommonlyUsedTaylorSeriesserieswhenisvalid/true11x=1+x+x2+x3+x4+...notethisisthegeometricseries.justthinkofxasr=1Xn=0xnx2(1,1)ex=1+x+x22!+x33!+x44!+...so:e=1+1+12!+13!+14!+...e(17x)=P1n=0(17x)nn!=1Xn=017nxnn!=1Xn=0xnn!x2Rcosx=1x22!+x44!x66!+x88!...notey=cosxisanevenfunction(i.e.,cos(x)=+cos(x))andthetaylorserisofy=cosxhasonlyevenpowers.=1Xn=0(1)nx2n(2n)!x2Rsinx=xx33!+x55!x77!+x99!...notey=sinxisanoddfunction(i.e.,sin(x)=sin(x))andthetaylorserisofy=sinxhasonlyoddpowers.=1Xn=1(1)(n1)x2n1(2n1)!or=1Xn=0(1)nx2n+1(2n+1)!x2Rln(1+x)=xx22+x33x44+x55...question:isy=ln(1+x)even,odd,orneither?=1Xn=1(1)(n1)xnnor=1Xn=1(1)n+1xnnx2(1,1]tan1x=xx33+x55x77+x99...question:isy=arctan(x)even,odd,orneither?=1Xn=1(1)(n1)x2n12n1or=1Xn=0(1)nx2n+12n+1x2[1,1]1Math142Taylor/MaclaurinPolynomialsandSeriesProf.GirardiFixanintervalIintherealline(e.g.,Imightbe(17,19))andletx0beapointinI,i.e.,x02I.Nextconsiderafunction,whosedomainisI,f:I!Randwhosederivativesf(n):I!RexistontheintervalIforn=1,2,3,...,N.Definition1.TheNth-orderTaylorpolynomialfory=f(x)atx0is:pN(x)=f(x0)+f0(x0)(xx0)+f00(x0)2!(xx0)2+···+f(N)(x0)N!(xx0)N,(openform)whichcanalsobewrittenas(recallthat0!=1)pN(x)=f(0)(x0)0!+f(1)(x0)1!(xx0)+f(2)(x0)2!(xx0)2+···+f(N)(x0)N!(xx0)N-afinitesum,i.e.thesumstops.Formula(openform)isinopenform.Itcanalsobewritteninclosedform,byusingsigmanotation,aspN(x)=NXn=0f(n)(x0)n!(xx0)n.(closedform)Soy=pN(x)isapolynomialofdegreeatmostNandithastheformpN(x)=NXn=0cn(xx0)nwheretheconstantscn=f(n)(x0)n!arespeciallychosensothatderivativesmatchupatx0,i.e.theconstantscn’sarechosensothat:pN(x0)=f(x0)p(1)N(x0)=f(1)(x0)p(2)N(x0)=f(2)(x0)...p(N)N(x0)=f(N)(x0).TheconstantcnisthenthTaylorcoecientofy=f(x)aboutx0.TheNth-orderMaclaurinpolynomialfory=f(x)isjusttheNth-orderTaylorpolynomialfory=f(x)atx0=0andsoitispN(x)=NXn=0f(n)(0)n!xn.Definition2.1TheTaylorseriesfory=f(x)atx0isthepowerseries:P1(x)=f(x0)+f0(x0)(xx0)+f00(x0)2!(xx0)2+···+f(n)(x0)n!(xx0)n+...(openform)whichcanalsobewrittenasP1(x)=f(0)(x0)0!+f(1)(x0)1!(xx0)+f(2)(x0)2!(xx0)2+···+f(n)(x0)n!(xx0)n+...-thesumkeepsongoingandgoing.TheTaylorseriescanalsobewritteninclosedform,byusingsigmanotation,asP1(x)=1Xn=0f(n)(x0)n!(xx0)n.(closedform)TheMaclaurinseriesfory=f(x)isjusttheTaylorseriesfory=f(x)atx0=0.1Hereweareassumingthatthederivativesy=f(n)(x)existforeachxintheintervalIandforeachn2N⌘{1,2,3,4,5,...}.2BigQuestions3.Forwhatvaluesofxdoesthepower(a.k.a.Taylor)seriesP1(x)=1Xn=0f(n)(x0)n!(xx0)n(1)converge(usuallytheRootorRatiotesthelpsusoutwiththisquestion).Ifthepower/Taylorseriesinformula(1)doesindeedconvergeatapointx,doestheseriesconvergetowhatwewouldwantittoconvergeto,i.e.,doesf(x)=P1(x)?(2)Question(2)isgoingtotakesomethought.Definition4.TheNth-orderRemaindertermfory=f(x)atx0is:RN(x)def=f(x)PN(x)wherey=PN(x)istheNth-orderTaylorpolynomialfory=f(x)atx0.Sof(x)=PN(x)+RN(x)(3)thatisf(x)⇡PN(x)withinanerrorofRN(x).Weoftenthinkofallthisas:f(x)⇡NXn=0f(n)(x0)n!(xx0)n-afinitesum,thesumstopsatN.WewouldLIKETOHAVETHATf(x)??=1Xn=0f(n)(x0)n!(xx0)n-thesumkeepsongoingandgoing.Inothernotation:f(x)⇡PN(x)andthequestionisf(x)??=P1(x)wherey=P1(x)istheTaylorseriesofy=f(x)atx0.Well,let’sthinkaboutwhatneedstobeforf(x)??=P1(x),i.e.,forftoequaltoitsTaylorseries.Notice5.TakingthelimN!1ofbothsidesinequation(3),weseethatf(x)=1Xn=0f(n)(x0)n!(xx0)n-thesumkeepsongoingandgoing.ifandonlyiflimN!1RN(x)=0.Recall6.limN!1RN(x)=0ifandonlyiflimN!1|RN(x)|=0.So7.IflimN!1|RN(x)|=0(4)thenf(x)=1Xn=0f(n)(x0)n!(xx0)n.Sowebasicallywanttoshowthat(4)holdstrue.Howtodothis?Well,thisiswhereMr.Taylorcomestotherescue!22AccordingtoMr.Taylor,hisRemainderTheorem(seenextpage)wasmotivatedbyco↵eehouseconversationsaboutworksofNewtononplanetarymotionandworksofHalley(ofHalley’scomet)onrootsofpolynomials.3Taylor’sRemainderTheoremVersion1:forafixedpointx2IandafixedN2N.3Thereexistscbetweenxandx0sothatRN(x)def=f(x)PN(x)theorem=f(N+1)(c)(N+1)!(xx0)(N+1).(5)Soeitherxcx0orx0cx.Sowedonotknowexactlywhatcisbutatleastweknowthatcisbetweenxandx0andsoc2I.Remark:ThisisaBigTheorembyTaylor.Seethebookfortheproof.TheproofusestheMeanValueTheorem.Notethatformula(5)impliesthat|RN(x)|=f(N+1)(c)(N+1)!|xx0|(N+1).(6)Version2:forthewholeintervalIandafixedN2N.3AssumewecanfindMsothatthemaximumoff(N+1)(x)ontheintervalIM,i.e.,maxc2If(N+1)(c)M.Then|RN(x)|M(N+1)!|xx0|N+1(7)foreachx2I.Remark:Thisfollowsfromformula(6).Version3:forthewholeintervalIandallN2N.4Nowassumethatwecanfindasequence{MN}1N=1sothatmaxc2If(N+1)(c)MNforeachN2NandalsosothatlimN!1MN(N+1)!|xx0|N+1=0foreachx2I.Then,byformula(7)andtheSqueezeTheorem,limN!1|RN(x)|=0foreachx2I.Thus,bySo7,f(x)=1Xn=0f(n)(x0)n!(xx0)nforeachx2I.3Hereweassumethatthe(N+1)-derivativeofy=f(x),i.e.y=f(N+1)(x),existsforeachx2I.4Hereweassumethaty=f(N)(x),existsforeachx2IandeachN2N.4