Often the order of the error generated by the procedure is known In other words Kh 1 00 2 with being some known constant and K K K 00 being some other usually unknown constants For example might be the value at some 64257nal time for the solution ID: 86571
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RichardsonExtrapolationTherearemanyapproximationproceduresinwhichone rstpicksastepsizehandthengeneratesanapproximationA(h)tosomedesiredquantityA.Oftentheorderoftheerrorgeneratedbytheprocedureisknown.InotherwordsA=A(h)+Khk+K0hk+1+K00hk+2+withkbeingsomeknownconstantandK;K0;K00;beingsomeother(usuallyunknown)constants.Forexample,Amightbethevaluey(tf)atsome naltimetfforthesolutiontoaninitialvalueproblemy0=f(t;y);y(t0)=y0.ThenA(h)mightbetheapproximationtoy(tf)producedbyEuler'smethodwithstepsizeh.Inthiscasek=1.IftheimprovedEuler'smethodisusedk=2.IfRunge-Kuttaisusedk=4.ThenotationO(hk+1)isconventionallyusedtostandfor\asumoftermsoforderhk+1andhigher".SotheaboveequationmaybewrittenA=A(h)+Khk+O(hk+1)(1)Ifweweretodropthe,hopefullytiny,termO(hk+1)fromthisequation,wewouldhaveonelinearequation,A=A(h)+Khk,inthetwounknownsA;K.Butthisisreallyadi erentequationforeachdi erentvalueofh.Wecangetasecondsuchequationjustbyusingadi erentstepsize.Thenthetwoequationsmaybesolved,yieldingapproximatevaluesofAandK.ThisapproximatevalueofAconstitutesanewimprovedapproximation,B(h),fortheexactA.Wedothisnow.Taking2ktimesA=A(h=2)+K(h=2)k+O(hk+1)(2)(notethat,inequations(1)and(2),thesymbol\O(hk+1)"isusedtostandfortwodi erentsumsoftermsoforderhk+1andhigher)andsubtractingequation(1)gives 2k 1A=2kA(h=2) A(h)+O(hk+1)A=2kA(h=2) A(h) 2k 1+O(hk+1)c\rJoelFeldman.2000.Allrightsreserved.1 Henceifwede neB(h)=2kA(h=2) A(h) 2k 1(3)thenA=B(h)+O(hk+1)(4)andwehavegeneratedanapproximationwhoseerrorisoforderk+1,onebetterthanA(h)'s.Onewidelyusednumericalintegrationalgorithm,calledRombergintegration,appliesthisformularepeatedlytothetrapezoidalrule.ExampleA=y(1)=64:897803wherey(t)obeysy(0)=1;y0=1 t+4y.A(h)=approximatevaluefory(1)givenbyimprovedEulerwithstepsizeh.B(h)=2kA(h=2) A(h) 2k 1withk=2.hA(h)%#B(h)%# .1 59.9387.620 64.587.4860 .05 63.4242.340 64.856.065120 .025 64.498.6280 64.8924.0083240 .0125 64.794.04160 The\%"columngivesthepercentageerrorandthe\#"columngivesthenumberofevalu-ationsoff(t;y)used.Similarly,bysubtractingequation(2)fromequation(1),wecan ndK.0=A(h) A(h=2)+Khk 1 1 2k+O(hk+1)K=A(h=2) A(h) hk 1 1 2k+O(h)OnceweknowKwecanestimatetheerrorinA(h=2)byE(h=2)=A A(h=2)=K(h=2)k+O(hk+1)=A(h=2) A(h) 2k 1+O(hk+1)Ifthiserrorisunacceptablylarge,wecanuseE(h)=Khkc\rJoelFeldman.2000.Allrightsreserved.2 todetermineastepsizehthatwillgiveanacceptableerror.Thisisthebasisforanumberofalgorithmsthatincorporateautomaticstepsizecontrol.NotethatA(h=2) A(h) 2k 1=B(h) A(h=2).OnecannotgetastillbetterguessforAbycombiningB(h)andE(h=2).Example.SupposethatwewishedtouseimprovedEulerto ndanumericalapproximationtoA=y(1),whereyisthesolutiontotheinitialvalueproblemy0=y 2ty(0)=3Supposefurtherthatweareaimingforanerrorof10 6.IfwerunimprovedEulerwithstepsize0:2(5steps)andagainwithstepsize0:1(10steps)wegettheapproximatevaluesA(0:2)=6:70270816andA(0:1)=6:71408085.SinceimprovedEulerhask=2,TheseapproximatevaluesobeyA=A(0:2)+K(0:2)2+higherorder=6:70270816+K(0:2)2+higherorderA=A(0:1)+K(0:1)2+higherorder=6:71408085+K(0:1)2+higherorderSubtracting0=6:70270816+K(0:2)2 6:71408085 K(0:1)2+higherorder 0:01137269+0:03KsothatK0:01137269 0:03=0:38TheerrorforstepsizehisKh2+O(h3),sotoachieveanerrorof10 6weneedKh2+O(h3)=10 6)0:38h210 6)hq 10 6 0:38=0:001622=1 616:5IfwerunimprovedEulerwithstepsize1 617wegettheapproximatevalueA(1 617)=6:71828064.Inthisillustrative,andpurelyarti cal,example,wecansolvetheintialvalueproblemexactly.Thesolutionisy(t)=2+2t+et,sothattheexactvalueofy(1)=6:71828183,toeightdecimalplaces,andtheerrorinA(1 62)is0:00000119.c\rJoelFeldman.2000.Allrightsreserved.3