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1 Macmí36 è98-è Solutions Homework è èdue January 6, 998è Exercise Set.. èaè witèplotsè: f:=è-expèxè+*xèè3; f := 3, 3 ex + 3 x?abs plotèabsèfè,x=0..è; x By moving te mouse pointer to te extremum, we gets its coordinates to be about è0:69; 0:6è: g:=diffèf,xè; We now apply te Extreme Value Teorem èteorem.9è to jfj wic is continuous dierentiable on ë0; ë:
2 g := 3, 3 ex fsolveèg,x,0..è; evalfèsubsèx=",fèè; evalfèsubsèx=0,fèè; evalfèsubsèx=,fèè; : :60980 : Hence, by te Extreme Value Teorem, : max jfèxèj= 0:6098 to 6 digits 0çxç wit maximum occurring at x = 0:6937 è6 digits accuracyè. witèplotsè: f:=*x*cosè*xè-èx-è^; ècè plotèabsèfè,x=..è; f := x cosè xè, èx, è x
3 By moving te mouse pointer to te extremum, we gets its coordinates to be about è3.3,.99è g:=diffèf,xè; fsolveèg,x,..è; evalfèsubsèx=",fèè; g := cosè xè, x sinè xè, x + 3:30678 evalfèsubsèx=,absèfèèè; evalfèsubsèx=,absèfèèè; : :6578 Hence, by te Extreme Value Teorem, 5: max jfèxèj= 5:600 è6digits accuracyè çxç wit te maximum occurring at te rigt endpoint x =: 6. We prove by contradiction èindirect proofè. Assume, tere exist two èor moreè numbers p q bot in ëa; bë wit fèpè = fèqè=0 p6= q: Ten ëp; që ç ëa; bë we now apply Rolle's Teorem.7 to f Cëp; që to prove te existence of a c èp; qè wit f 0 ècè =0:But f 0 èxè 6= 08xëa; bë soweave a contradiction.. èabè f:=*x*cosè*xè-èx-è^; f := x cosè xè, èx, è s3:=taylorèf,x=0,è: p3:=convertès3,polynomè; Digits:=; p3 :=,+6x,x,x 3 Digits := y:=evalfèsubsèx=0.,fèè; y:=evalfèsubsèx=0.,p3èè; error:=absèy-yè; fxi:=subsèx=xi,fè; y :=,: y :=,:06 error := : r3:=diffèfxi,xi,xi,xi,xièè!*èxè^; fxi := ç cosè çè, èç, è r3 := è6 sinè çè+3çcosè çèè x r3_0.:=subsèx=0.,r3è; r3 0 := : sinè çè +: ç cosè çè 3
4 plotèr3_0.,xi=0..0.è; xi We see tat te maximum error in tis case occurs at xi=0.. r3max:=evalfèsubsèxi=0.,r3_0.èè; r3max := : Alternately, we could use trig inequalities to obtain a more conservative estimate r3max:=subsèfsinè*xiè=sinè*0.è,cosè*xiè=,xi=0.g,r3_0.è; r3max := : Or, being even more conservative, we obtain te following estimate r3max:=subsèfsinè*xiè=,cosè*xiè=,xi=0.g,r3_0.è; r3max := : NB: We can't always assume tat te remainder term attains its absolute maximum at te endpoint!. For example, if x=., ten we ave r3_.:=subsèx=.,r3è; r3 := 5: sinè çè + : ç cosè çè wrongerrorbound:=evalfèsubsèxi=.,r3_.èè; wrongerrorbound := : y:=evalfèsubsèx=.,fèè; y:=evalfèsubsèx=.,p3èè; y :=,:097973
5 error:=absèy-yè; y :=,5:5 error := : If we plot r3 from xi=0 to.5, we see tat te error term reaces a maximum absolute value near ç =0:66 wic is greater tan te actual error of.7 èas it sould be! Wy?è plotèr3_.,xi=0...è; xi fsolveèdiffèr3_.,xiè,xi,0..è; : ècdè Te degree fourt Taylor polynomial for fèxè wit x 0 =0is P èxè=,+6x,x,x 3 =P 3 èxè; i.e., te quadratic term is zero, we obtain te same approximation same actual error as in part èaè. However, we obtain a muc smaller upper bound for te error jr 5 èçèj= 0 j60 cos ç, 6ç sin çj è0:è5 ç R 5 è0è = :3653 0, wic is only sligtly greater tan te actual error of :3365 0, : Remark Did you get an upper bound of 0:79 0,? Howdoyou know tis must be wrong? If you did, see te previous page for a int onow to correct your answer. 5
6 Exercise Set.. Suppose p approximates p wit relative error at most ". In tat case jp, p j jpj ",jp,p j"p èfor p0è,,"pp,p "p, è, "è pp è + "è p: Tus for " =0,3 weave: èaè p = 50 è 9:85 p 50:5 ècè p = 500 è 98:50 p 50:50. èbè èiè 5 3 = 5 =0:6 èiiè =0:800 0:333 = 0: èiiiè =0:800 0:333=0: èivè Te relative error is te same for èiiè èiiiè 5, 0:66 5 ç :5 0,3 ècè èiè,, = 39 =0: èiiè è0:333, 0:7è + 0:50 = 0: :50 = 0: èiiiè è0:333, 0:73è + 0:50=0: :50 = 0:0 èivè èdè For èiiè , 0: èiè è è, 3 0 = =0:560 ç :87 0,3 for èiiiè , 0: èiiè è + 3 è, 3 =è0: :7è, 0:50=0:605, 0:50 = 0: èiiiè è + 3 è, 3 =è0: :73è, 0:50 = 0:606, 0:50 = 0: èivè For èiiè , 0: ç :3 0,3 for èiiiè , 0: ç :9 0,3 ç :3 0, Exercise Set.3 6. èaè Since lim sin =0 n! n sin n = n, è n è3 3! 6 + Oèè n è5 è
7 we ave jsin, 0j n ç è Te rate of convergence is Oè n è: n èbè Since we ave jsin n, 0j lim n! sin n =0 sin n = n, è n è 3 3! + Oèè n è5 è ç è Te rate of convergence is Oè n n è: ècè Since we ave sin n =ë n,è n è3 3! jsin n, 0j lim n! sin =0 n + Oèè n è5 èë = n + Oèè n è è ç è Te rate of convergence is Oè n n è: 7. èaè Since sin, cos lim =0!0 sin =, 3 3! + Oè5 è we ave terefore sin, cos cos =,! + Oè è =, 3, 6, 3 + Oè5 è =, 3 + Oè è sin, cos, 0 ç 3 è Te rate of convergence is Oè è: 7
8 èbè Since, e lim!0 =, we ave terefore,e,e x, è,è ç è e =++! + 3 3! + Oè è; =,, +Oè3 è =,, +Oè è Te rate of convergence is Oèè: ècè Since sin lim =!0 we ave terefore sin =, 3 3! + Oè5 è sin =, 6 +Oè è sin, ç 6 è Te rate of convergence is Oè è: èdè Since, cos lim =0!0 we ave terefore,cos, 0 ç è cos =,! + Oè è,cos = + Oè3 è Te rate of convergence is Oèè: 8
9 Question 8 8a Writing out te summation we ave, a b + èa b + a b è + èa 3 b + a 3 b + a 3 b 3 è + + èa n b +a n b ++a n b n, +a n b n è Tere are a total of ++3++èn,è + n terms to add. Tis total is te same as nèn+è : Terefore, te number of additions is nèn+è, wile te number of multiplications is te same as te number of terms, nèn+è : 8b Te number of multiplications can be reduced by factoring a i obtaining, from te it bracketed term tus a b + a èb + b è+a 3 èb +b +b 3 è++a n èb +b ++b n, +b n è: Tis reduces te number of multiplications to n: 9
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