TaylorsTheorem: IntroductoryWaeSupposethatweoncewentonajourney, startingattimeaatpositionf(a)andnishingattimexatpositionf(x). Later, wedecidetorecreatethisjourney, butweveforgottenexactlywhatwedidlasttimeandsowetrytoapproximateit.Forourrstattempt,wewilltravelatconstantspeedkforthewholejourney. Wetravelatspeedkfromtimeatotimet,andsowetraveladistanceof(t a)k,andarriveatpositionf(a) + (t a)kOur aim is to arrive at position f(x) at time x, so we need to choose k so that f(x) = f(a)+(xa)k.Itsclearfromthisthatthecorrectspeedktouseis(f(x) f(a))/(x a). WemightthenrecalltheMeanValueTheorem, whichsaysthatthisfractionequalsf(c)forsomec(a, x). Inotherwords, theconstantspeedkweneedtousehappenstobeanactual speedwehadatsomepointduringouroriginaljourney.Lets seeanother waythat this result couldhavebeenreached. Sincewerestill leavingat thesame time,and we want to arrive at the same time,lets dene a function to measure the dierencebetweenourpositionsattimetinboththeoriginalandcurrentjourneys.Let(t) = f(t) f(a) (t a)k.Then(a) = 0,andourdesiredvalueofkissuchthat(x) = 0aswell. ByRollesTheoremthereissomec (a, x)suchthat(c) = 0. Since(t) = f(t) k,wehavef(c) = k. Hencef(x) = f(a) + (x a)f(c) forsomec (a, x)Andwehaveobtainedthesameresult.Wenowdecidethatconstantspeedisntaveryaccuraterepresentationofouroriginaljourney. Sowell try to mimic it more closely this time. Well start at time a at position f(a) as before, only thistimewell startotravellingatouroriginal initial speedf(a). Well nowtrytoapplyaconstantaccelerationktoadjustourspeedsothatwearriveatf(x)attimex. Thinkingofs = ut +12at2,wendthatfromtimeatotimetwetraveladistanceof(t a)f(a) +12(t a)2k,andsoattimetwearriveatpositionf(a) + (t a)f(a) +(t a)22kOuraimistoarriveatf(x)attimex. So, asbefore, well deneadierencefunction, measuringthedierencebetweenourpositionsattimetinboththeoriginalandcurrentjourneys.Let(t) = f(t) f(a) (t a)f(a) (t a)22k.Then(a) = 0,andourdesiredvalueofkissuchthat(x) = 0aswell. ByRollesTheorem,thereissomec1 (a, x)suchthat(c1) = 0.1Nowconsider(t)=f(t) f(a) (t a)k. Since(a)=(c1)=0,byasecondapplicationofRollethereissomec (a, c1)suchthat(c) = 0. Since(t) = f(t) k,weseethatk= f(c).Thuswehavef(x) = f(a) + (x a)f(a) +(x a)22f(c) forsomec (a, x)Next,wedecidetomakeourjourneyevenmorefaithfultotheoriginal: westartatf(a)attimea,travellingattheoriginalspeedf(a)andwithoriginalaccelerationf(a),andwetrytoadjustthewhatever-the-derivative-of-acceleration-issothatwearriveatf(x)attimex.Andsoon.2TaylorsTheoremwiththeLagrangeRemainderContinuingthethoughtsfromourintroductorywae,wewantksuchthatf(x) = f(a) + (x a)f(a) +(x a)22f(a) + . . . +(x a)n1(n 1)!f(n1)(a) +(x a)nn!k.Let(t) = f(t) f(a) (t a)f(a) (t a)22f(a) . . . (t a)n1(n 1)!f(n1)(a) (t a)nn!k.Our k is such that (a) = (x) = 0, and so Rolles Theorem gives us c1 (a, x) such that (c1) = 0.Now,(t) = f(t) f(a) (t a)f(a) . . . (t a)n2(n 2)!f(n1)(a) (t a)n1(n 1)!k.Since(a)=(c1)=0, asecondapplicationof RollesTheoremgivesusc2(a, c1)suchthat(c2) = 0.Now,(t) = f(t) f(a) (t a)f(a) . . . (t a)n3(n 3)!f(n1)(a) (t a)n2(n 2)!k.Since(a)=(c2)=0, athirdapplicationof RollesTheoremgivesusc3(a, c2)suchthat(c3) = 0.And so on. We reach (n)(t) = f(n)(t)k, and nd some c (a, cn1) (a, x) such that (n)(c) = 0.I.e.,suchthatk = f(n)(c).Since(x) = 0forthisk,weconcludethat,forsomec (a, x),wehavef(x) = f(a) + (x a)f(a) +(x a)22f(a) + . . . +(x a)n1(n 1)!f(n1)(a) +(x a)nn!f(n)(c).We might prefer to make the following change of notation. Let x = a+h, and since c (a, x), writec = a + hwhere (0, 1). Thenf(a + h) = f(a) + hf(a) +h22f(a) + . . . +hn1(n 1)!f(n1)(a) +hnn! f(n)(a + h).Examiningwhatwedid,weseethatfneededtoben-timesdierentiableon(a, x),butataitselfwe needed fto be only (n1)-times dierentiable,although with f(n1)continuous there. And wedidntneedf(n)tobecontinuousanywhere.ThisresultisTaylorsTheoremwiththeLagrangeremainder.TheremainderisRn(x) =(x a)nn!f(n)(c).3TaylorsTheoremwiththeCauchyRemainderOftenwhenusingtheLagrangeRemainder,wellhaveaboundonf(n),andrelyonthen!beatingthe(x a)nasn. Butif thef(n)termstartsprovidinguswithann!-shapedtermontop,suchaswithabinomialexpansion,thenwemightneedabetterexpressionthan(x a)n.WellshowthatRn(x) =(x c)n1(x a)(n 1)!f(n)(c),forsomec (a, x). (Adierentc,ofcourse.)Thiswillallowusbetterestimatesifallwehaveisageneralboundonf(n),since|x c| < |x a|.Wehavef(x) = f(a) + (x a)f(a) +(x a)22f(a) + . . . +(x a)n1(n 1)!f(n1)(a) + Rn(x)RearrangingRn(x) = f(x) f(a) (x a)f(a) (x a)22f(a) . . . (x a)n1(n 1)!f(n1)(a).Wewanttoestimatethis,soletsturnitintoafunctionofsomevariablet,bysettingF(t) = f(x) f(t) (x t)f(t) (x t)22f(t) . . . (x t)n1(n 1)!f(n1)(t).sothatF(a) = Rn(x)andF(x) = 0.Then,notingthatddt

(x t)mm!f(m)(t)

=(x t)m1(m1)!f(m)(t) (x t)mm!f(m+1)(t),wehave,byatelescopingsum,F(t) = (x t)n1(n 1)!f(n)(t)ApplyingtheMeanValueTheoremtoFontheinterval[a, x],wendsomec (a, x)suchthatF(x) F(a)x a=F(c) =(x c)n1(n 1)!f(n)(c)RearranginggivesF(a) = F(x) +(x c)n1(x a)(n 1)!f(n)(c)Thatis,asclaimed,Rn(x) =(x c)n1(x a)(n 1)!f(n)(c)ThisresultisTaylorsTheoremwiththeCauchyremainder.4TaylorsTheoremwiththeIntegral RemainderThereisanotherformoftheremainderwhichisalsouseful,undertheslightlystrongerassumptionthatf(n)iscontinuous.WellshowthatRn=

xa(x t)n1(n 1)!f(n)(t) dt.Theproofofthisisbyinduction,withthebasecasebeingtheFundamentalTheoremofCalculus.Whenn = 1,wehavef(x) = f(a) +

xaf(t) dt,soweredonebytheFTC.Assumingtheformulaaboveforthe(n 1)thcase,wehavef(x) = f(a) + (x a)f(a) +(x a)22!f(a) + . . . +(x a)n2(n 2)!f(n2)(a) + Rn1,whereRn1=

xa(x t)n2(n 2)!f(n1)(t) dt=

(x t)n1(n 1)(n 2)!f(n1)(t)

xa+

xa(x t)n1(n 1)(n 2)!f(n)(t) dt=(x t)n1(n 1)!f(n1)(a) +

xa(x t)n1(n 1)!f(n)(t) dtandsubstitutingthisintotheexpressionforf(x)givestherequiredformforRn(x).ThisresultisTaylorsTheoremwiththeintegral formoftheremainder.WecanderivetheLagrangeremainderfromthis. Sincef(n)iscontinuouson[a, x], itisboundedthereandattainsitsbounds. Saymf(n)(t)M,withtheseboundsbeingattained.m

xa(x t)n1(n 1)!dt

xa(x t)n1(n 1)!f(n)(t) dt M

xa(x t)n1(n 1)!dtSince

xa(x t)n1(n 1)!dt =(x a)nn!,wehavem(x a)nn!Rn(x) M (x a)nn!ApplyingtheIntermediateValueTheorem,wendsomec (a, x)suchthatRn(x) =(x a)nn!f(n)(c).5