Worksheet: Taylor Series, Lagrange Error Bound ilearnmath.net

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1 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.

2 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

3 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 () () () () () () = !!! 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 = = = = ! 70 Th actual rror or this 5 th dgr polyomial alls somwhr btw th ral valu o = =.7666! 3!! 5! Th rror is , which quals This umbr is lss tha th uppr boud or th rror, , which shows how th La Grag ormula works.

4 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 = < ( + )! z + z + (.) < 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 < Th valu or is 3. + (.) =.0067 >.00 +

5 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 ].