Taylor s Thorm & Lagrag Error Bouds Actual Error This is th ral amout o rror, ot th rror boud (worst cas scario). It is th dirc btw th actual () ad th polyomial. Stps:. Plug -valu ito () to gt a valu. (a). Plug -valu ito th polyomial ad gt aothr valu. P(a) 3. Th absolut dirc btw th two is th rror. (a) P(a) = Actual Error Eampl Giv () =, approimat (0.) usig a d dgr Taylor polyomial ad id th rror. Eampl π What is th rror or th ourth dgr polyomial approimatio o cos wh = Eampl 3 6 Fid a ormula or th trucatio rror i w us + + + to approimat o (-,). Taylor s Formula ( ) ( a) ( ) ( ) ( )( ) ( ) ( a ) ( ) = a + a a + a + a + R ( )!! I R () 0 as, th th Taylor sris covrgs to () o th itrval I or all o I.
Lagrag Formula This mthod uss a spcial orm o th Taylor ormula to id th rror boud o a polyomial approimatio o a Taylor sris. R () = ( z)( a ) ( + )! ( + ) ( + ) = rror boud Whr: a is whr th sris is ctrd z is a valu btw a ad (z is usually a or ) ( + ) Th variabl z is a umbr btw ad a (z givig th largst valu or ( z ) ), but to id th rror boud, z ds up big qual to ithr a or. To dtrmi whthr th z valu will b th + sam as or a, you must plug ach umbr ito ( z ) to s which givs th gratst umbr. For si or cos ( + ) ( z ) =, (v i all z s giv smallr valus). Polyomial valu ± rror boud = rag o possibl valus o th sris. For ampl: I you ar tryig to id th rror o a d dgr Taylor polyomial approimatio o () =, you must irst id th 3 rd + drivativ, bcaus th ormula uss ( z ), ot ( z ). 6 () =, () =, ad () = 3 ( ) ( ) ( ) Also, or this uctio, = 0. ad a = 0. Plug ths two valus ito th 3 rd drivativ. 3 6 (0) = ( 0) 3 6 (.) = (.) = 6 = 9.5 this is biggr! Nt, plug i 9.5 or + ( z ) i th La Grag ormula: Error boud = 9.5(. 0) 3! 3 =.005
Ecptio! Wh () = si() or cos(), th valu or gratst valu o ay si or cos uctio. + ( z ) will always b qual to, bcaus that is th Eampls or La Grag Error Boud: a.) Fid th uppr boud or th rror or th 5 th dgr polyomial approimatio o. is qual to, whos sris ca b dtrmid rom th Taylor sris o 0 3 5 () () () () () () = + + + + + 0!!! 3!! 5! Th La Grag ormula is, ( z)( ) 6! (6) 6 All drivativs o ar,so 6 ( z ) = To id z, plug i th valus or a ad ito z. z. a= 0, =, 0 = = ad () 6! 6 >, so z = = = =.00377 6! 70 Th actual rror or this 5 th dgr polyomial alls somwhr btw th ral valu o +.00377. =.788 + + + + + =.7666! 3!! 5! Th rror is.788.7666, which quals 0.006. This umbr is lss tha th uppr boud or th rror, 0.00377, which shows how th La Grag ormula works.
b.) What dgr Taylor polyomial or l(.) might hav a rror lss tha 0.00? (I othr words, th uppr boud or th rror would b 0.00) First, start o with th La Grag ormula, whos valu must b lss tha 0.00: ( z)( a ) ( + )! + + Th drivativs o l() ar as ollows: < 0.00 For th uctio l(), a = ad i this cas, =. 6 ( ) = l( ), '( ) =, ''( ) =, '''( ) =, ''''( ) = 3 Sic you do t kow th valu o, a gral ormula or th + drivativ must b usd. Th ormula or th th drivativ ca b obtaid rom abov ad is as ollows: + ( ) ( )! + ( ) = To id ( ) that quatio:, simply substitut + or ito + ( ) ( )! + ( ) = + This is what you will put ito th La Grag ormula or + ( z ), chagig to z. Still, you must id th valu or z. It will b qual to ithr a or. Wh pluggig th two valus ito th abov list o drivativ or l(), you id that always producs th gratr valu, so z =. Now, th rror boud ormula looks somthig lik this: + ( ) ( )! + ( )! + ( a ) (. ) + + z = z ( + )! ( + )! + +! (.). = i = 0.00 + < ( + )! z + z + (.) < 0.00 + Nt you must simply us th cocpt o trial ad rror. Choos valus or ad kp pluggig thm i to th iquality. Wh th trm o th lt ds up big gratr tha 0.00, you kow that you hav crossd th li ad your valu or will b th prvious umbr (bor th valu cdd 0.00). 3+ (.) =.000 <.00 3+ Th valu or is 3. + (.) =.0067 >.00 +
Eampl Fid th 3 rd dgr polyomial approimatio or at, ctrd at 0. Fid th rag o possibl valus or at, ctrd at 0. Us th Lagrag rror quatio. Eampl Fid th th dgr Maclauri polyomial approimatio or cos() whr a = 0, valuatd at. Fid th Lagrag rror boud. Eampl 3 Us graphs to id a Taylor Polyomial P () or cos so that P () - cos() < 0.00 or vry i [ π, π ].