Numerical differentiation (Finite difference)

Transcription

1 MA umercl Alyss Week 4 To tk d wrte portle progrms we eed strct model of ter computtol evromet. Ftful models do est ut tey revel tt evromet to e too dverse forcg portle progrmmers to lot eve te smplest cocrete tsks to strct mosters. We eed sometg smple or f ot so smple ot so cprcously comple Wllm M. K (or Jue 5 9 Toroto Cd Wllm K s mtemtc d computer scetst wo receved te Turg Awrd 989 for s fudmetl cotrutos to umercl lyss. K ws te prmry rctect ed te IEEE stdrd for flotg-pot computto. He s ee clled Te Fter of Flotg Pot. Possle project topc: Go to Prof. K s persol wepge ttp:// d red s ppers d gve lecture o s results. Te terestg oes re MATLAB s loss s oody s g Bestly umers etc. Questo: How to clculte f ( we te lytc formul s ot vlle? umercl dfferetto (Fte dfferece Gol: to clculte f (. A frst order metod for f ( f ( + f ( lm = f 0 ( f( + f ( Tt s for smll we c use f( + f More precsely we c wrte f ( for smll f( + f ( = f ( + umercl ppromto e( 0 s 0 to ppromte f (. e Dscretzto error - -

2 MA umercl Alyss Let us fd out te order of e(. Tylor epso of f( + roud : f( += f e + f( + f f ( + f = f ( + f + O( ( = f ( + O( = O( f( + f + O( s frst order metod (forwrd dfferece for clcultg f (. Bg O d smll o otto: Wrte f = O ( g f 0 f ( d s ouded for suffcetly lrge. Wrte f g ( o( g g > for R suffcetly lrge = f f g goes to zero s goes to f +. Also wrte f g (red f s symptotc to g f s. For emple g + = O + e l = o Te proof goes lke ts. Sce Te trd oe s lttle rder to prove. We ote tt e e e e e l l l l l = = = l Sce l s For lrge eoug + = + s te frst two follow. l 0 s (Apply L Hoptl s Rule to see ts. l l l. Hece for lrge l / l l / e e e 0 = = = 0 s. By te squeeze teorem e l 0 s. ote: ( For frst order metod f we reduce y fctor of te e s ppromtely reduced y fctor of. - -

3 MA umercl Alyss ( Te metod s ccurte (ect we f ( = or f ( =. A secod order metod for f ( f( + f Tylor epso roud f( += f + = f ( + e( f ( + f + f 6 f( += f ( f ( + f f 6 f( + f ( = f ( + f ( 6 e = f ( f( + f clcultg f (. 6 + O( = O( + O( 4 + O( 4 + O( s secod order metod (cetrl dfferece ppromto for ote: ( For secod order metod f we reduce y fctor of te e reduced y fctor of = 4. ( Te metod s ccurte (ect for f ( =. Let us ceck f ( ( ( + ( f + f 4 = = = f = = A fourt order metod for f ( f ( ++ 8 f ( + 8 f = f ( 5 ( e 0 + f ( 4 + O( 5 = O( 4 = f ( + e( s ppromtely = cse. ote: ( For fourt order metod f we reduce y fctor of te e( s ppromtely reduced y fctor of 6. f =. ( Te metod s ccurte for 4 Oter fte dfferece metods: - -

4 ( 4 MA umercl Alyss f + + f + f f ( = + f ( + O( ( 4 f + + f + f f ( = + O( (o-cetrl fte dfferece ppromto For ger order dervtves we ve ( + + ( f f f f ( = + O ( + ( + + ( ( f f f f f ( = + O ( + 4 ( ( + ( f f f f f f ( = + O 4 4 Totl error umercl dfferetto Cosder te frst order metod f( + f ( = f ( + umercl oted wt fte precso e ( Dscretzto error e( = O( f( + f ( s OT te umercl we get from computer! Te umercl we get from computer s fl( f ( + fl( f ( were fl( f ( + = f ( + ( +ε ε ~ 0 6 (IEEE doule precso represetto: mce precso fl( f ( = f ε ~ 0 6 +ε Tus we ve fl( f ( + fl( f ( = f ( + ( +ε f +ε - 4 -

5 Te dscretzto error: MA umercl Alyss = f + + f + ε f ( ε ( f ( + + f ( ε = f ( + e( + ε ~ 0 6 Dscretzto error e( C 0 s 0 Te effect of roud-off error: f ( + + f ( ε Effect of roud-off error 6 ~ f ( + + f ( 0 s 0 Te totl error s defed s fl( f ( + fl( f ( E T ( = f ( umercl oted wt fte precso ( = e( + f ( + + f ( ε C + f ( + + f ( ε Cosder te smple stuto were te error s gve y ET ( = + E T ( = = 0 = 0 = 0 At =0 8 E T ( tts te mmum. m E T ( =0 8 (Drw te grp of E T ( = ote: m E T s te mmum totl error we c ceve usg te umercl metod. For te secod order metod we ve E T ( C + f ( + + f ( Cosder te smple stuto tt ll coeffcets re oe d ε = 0 6. ε - 5 -

6 MA umercl Alyss 0 0 ET ( = + E ( = = = = / 5. T At =0 5. E T ( tts te mmum m E T ( =0 0.7 For te fourt order metod ET ( = + E T ( = 4 = 0 =0. At =0. E T ( tts te mmum m E T ( =0.8 ote: frst_ord.m: secod_ord.m: fourt_ord.m: 8 m 0 = m E T 5. = m E T m 0. = m E T m 0 ( =0 8 ( =0 0.7 ( =0.8 For ger order metod te mmum totl error s smller d te mmum totl error s ceved t lrger of (prctclly ceper. Ts s wy ger order metods re more desrle prctce. See Codes/Totl_error/frst_ord.m secod_ord.m d fourt_ord.m. umercl tegrto Te defte tegrl I = f d s defed Clculus s lmt of wt re clled Rem sums. It s te proved tt I = f ( d = F ( F ( were f F s y tdervtve of ; ts s te Fudmetl Teorem of Clculus. My tegrls c e evluted y usg ts formul d sgfct porto of Clculus ooks s devoted to ts pproc. oeteless most tegrls cot e evluted y ts formul ecuse most tegrds f ( do ot ve tdervtves epressle terms of elemetry fuctos. Emples of suc tegrls re π 0 0 π e d s d - 6 -

7 MA umercl Alyss Oter metods re eeded for evlutg suc tegrls. Gol: To clculte (ppromte f d We wll troduce two of te oldest d most populr umercl metods: te trpezodl rule d Smpso s rule. Strtegy: Dvde [ ] to sutervls of te sze =. Let = + = 0 (te pots 0 re clled te umercl tegrto ode pots (Drw te rel s to sow f ( d = f ( d = [ ] d = + = 0 To ppromte f ( d we oly eed to ppromte f ( d. Trpezodl rule: ( f + f = f ( d + e ( Error umercl ppromto f = f ( ( = O( e We wll derve te trpezodl rule lter. Composte trpezodl rule: Sum from = to = we ot f + f = f ( d + e = = = 0 f + f + f = f d + E = Error umercl ppromto T ( f - 7 -

8 MA umercl Alyss = e ( E = O( = O( = O = =. We s douled s lved d te term = were we ve used te fct tt decreses y fctor of 4. To llustrte ow fuctos s re reused we s douled cosder f 0 f T ( f = + f ( + = [ f0 + f + f] + = 0 = = = Also T f T f : 4 f 0 f 4 T4 ( f = + f ( + f ( + f ( + = 0 4 ( f + f + f + f + f = 0 = = = = 4 = T T we c see tt oly Comprg 4 fucto s re kow from T. f f eed to e evluted for 4 T s te oter Te cetrl de ed most formuls for ppromtg f ( d s to replce f ppromtg fucto wose tegrl c e evluted. If we ppromte f ( y te ler polyoml P = ( f ( + ( f ( wc terpoltes P ( over [ ] f t d (.e. P ( f ( P ( f ( y = =. Te tegrl of s te re of te trpezod (drw fgure d t s gve y f + f f + f P ( d d ( f s lmost ler o [ ]. = = (trpezodl rule. Ts s good ppromto f - 8 -

9 MA umercl Alyss To mprove te ove pproc use qudrtc terpolto to ppromte f ( o [ ] Let P ( e te qudrtc polyoml tt terpoltes. Usg ts to ppromte f f d P d d we get f t. / / = f ( + f ( / + f ( d / / / / Te complete evluto of ts tegrl yelds f ( d f ( 4 f ( / f ( (Smpso s rule Smpso s rule: ( 6 f + 4 f + f / = f ( d + e ( Error umercl ppromto f = f ( f / = f ( = + / / e ( = O 5 We wll derve te Smpso s rule lter. Composte Smpso s rule: Sum from = to = we ot 6 f + 4 f / + f = f ( d + e ( = = = 6 f 0 + f + f + 4 f / = = = f ( d + E ( Error umercl ppromto = e ( E = O( 5 = O( 5 = O 4 = = - 9 -

10 MA umercl Alyss Smpso s rule s ee mog te most populr umercl tegrto metods for more t two cetures. Go troug Codes\um_tegrto\trpezod.m wc uses te composte trpezodl rule to 5 compute f ( d wt from clculus: = 0.5 = 0.5 = = = f = s. Te ect soluto s kow f d = s d = cos = cos 0.5 cos = = 0.5 Te solute error from ts computto s out You c try te codes wt dfferet s of d see ow te error s reduced. Go troug Code\um_tegrto\smpso.m wc uses te composte Smpso rule to solve te ove prolem. Te solute error ow s reduced to Tlk out omework ssgmet. Prolem : use te ect of s d to compute te error. You c plot te error vs or. Prolem : compute rule. g s = + e d cos( s.5 for [ 0 4] 0 s wt te Composte Smpso s ote: MATLAB s ult- fuctos for umercl tegrto. TRAPZ Trpezodl umercl tegrto. QUAD umerclly evlute tegrl dptve Smpso qudrture. QUADL umerclly evlute tegrl dptve Lotto qudrture. QUADD umerclly evlute doule tegrl over plr rego. DBLQUAD umerclly evlute doule tegrl over rectgle. TRIPLEQUAD umerclly evlute trple tegrl. For te model prolem we c try te followg: >>=[0.5:0.0:]; >>y=s(; - 0 -

11 MA umercl Alyss >>trpz(y Or Dervto of trpezodl rule: For geerl tervl procedure s s follows. [ ] we c mke cge of vrles to mp t to ɶ = + = + [ ] mdpot of te tervl = + + dɶ = d / f ( d = f + + d ɶ ɶ / So we oly eed to cosder specl tervl.. Te def We use I[ f ]== f + f to ppromte f [ ] I f umercl ppromto / = f ( d + e( / Error Questo: How to determe d? / / / ( d. We ve Aswer: We requre tt I[ f ]= f ( d for f( = d f( = / / I[ ]= d ( + = + = / - -

12 MA umercl Alyss / I[ ]= d + = 0 = 0 / + = = 0 I [ f ]= f + f Questo: Wt s te order of e(? = = Aswer: Let us do error lyss. We use te Tylor epso of f( f( = f ( 0+ f ( 0 + f ( 0 otce tt ot I f / [ ] d f ( / / / e( = I[ f ] f ( d + d re ler terms of f(. / = I f ( 0 + f ( 0 + f ( 0 + f ( 0 + f ( 0 + f ( 0 + d / = f ( 0 I[ ] d / + f 0 / / / I[ ] d + f ( 0 I[ ] / d + / Te frst two terms o te rgt d sde re zero. Te trd term s / I[ ] d = / e( = f ( 0 + = O Tus we derved + / / = 6 f + f = f ( d + e( / Error umercl ppromto e( = O Apply ts to [ ] we ve / - -

13 MA umercl Alyss ( f + f = f ( d + e ( Error umercl ppromto f = f ( ( = O( e Ts s te trpezodl rule. A possle wekess te trpezodl rule c e ferred from te T f ove lyss. If f ( does ot ve two cotuous dervtves o [ ] te does coverge more slowly? Te swer s yes for some fuctos especlly f te frst dervtve s ot cotuous. Sme sttemet s true for Smpsos rule s we wll see te smple code (Codes\Estmte_order\emple_4f.m Dervto of Smpso s rule: def We use I[ f ]== f + f ( 0+ f 0 to ppromte f [ ] I f umercl ppromto / = f ( d + e( / Error Questo: How to determe 0 d? / - - / / ( d. We ve Aswer: we requre tt I[ f ]= f ( d for f( = f( = d f( = / / I[ ]= d ( = = / / I[ ]= d + = 0 = 0 / / I[ ]= d / = = 0 + = + = = 6 0 = 4 6 = 6 + =

14 MA umercl Alyss I [ f ]= 6 f + 4 f( 0 + f Let us do error lyss. / e( = I[ f ] f ( d / / = I f ( 0 + f ( 0 + f ( 0 + f ( 0 + f ( 0 + f ( 0 + d / = f ( 0 I[ ] d / + f 0 / + f ( ( 0 I[ ] 6 d / / / / I[ ] d + f ( 0 I[ ] / d / + f ( 4 ( 0 4 / I[ 4 ] 4 d + / Te frst tree terms o te rgt d sde re zero. Te fourt term s lso zero ecuse s odd fucto. Te fft term s / I[ 4 ] 4 d = 6 / e( = f ( 4 ( = O 5 Tus we derved / 5 / = f + 4 f ( 0+ f = f ( d + e( / Error e( = O 5 umercl ppromto Applyg ts to [ ] we ve / ( 6 f + 4 f + f / = f ( d + e ( Error umercl ppromto f = f ( f / = f ( = + / / e ( = O 5 Ts s te Smpso s rule

15 MA umercl Alyss Te trpezodl rule s te preferred tegrto rule we we re delg wt smoot perodc tegrds. Guss umercl Itegrto Guss umercl tegrto metod s umercl metod tt s sed o te ect tegrto of polyomls of cresg degree; o sudvso of te tegrto tervl s used. To llustrte te dervtve of suc tegrto formuls we restrct our tted to te tegrl I ( f = f ( d. Its relto to tegrls over [ ] For tegrl over [ ] s s follows. I f = f d troduce te ler cge of + + t vrle = t trsformg te tegrl to + + t ( I ( f = f dt ew tegrd fucto So we cosder I f w f j j j= I f = f d. Te tegrto formul s to ve te geerl form = d we requre tt te odes [ ] d wegts [ w w w ] so cose tt I ( f = I ( f for ll polyomls f Cse = : Te tegrto formul s te form f ( d w f ( [cse] We eed to fd w. It s to e ect for polyomls of s lrge degree s possle. Usg f ( d forcg equlty [cse] gve us = w ow use f ( = d g force equlty (cse. Te 0 = w = 0 sce w = 0. Tus [cse] ecomes f ( d f ( 0 = I ( f [cse_] e of s lrge degree s possle

16 MA umercl Alyss Ts s te mdpot formul wc s ect for ll ler polyomls. To see tt [cse] s ot ect for qudrtcs let f ( =. Te te error [cse_] s gve y d ( 0 = 0. Cse : = Te tegrto formul s + f d w f w f eed to determe four uspecfed quttes: w w. To determe tese we requre t to e ect for te four moomls: = f Ts leds to te four equtos = w + w 0 = w + w 0 = w + w = w + w Ts s oler system four ukows; ts soluto c e sow to e w = w = = = log wt oe sed o reversg te sgs of d. Ts yelds te tegrto formul: f d f f I f + = From eg ect for te moomls oe c sow ts formul wll e ect for ll polyomls of degree. It c lso e sow y drect clculto to ot e ect for te 4 degree 4 polyoml f ( =. Emple: Appromte = = I e d e e Usg cse formul we get I e e / / = I I Te error s qute smll for usg suc smll umer of ode pots

17 MA umercl Alyss Cse > : We seek te formul I ( f wj f ( j = (Guss umercl tegrto j= metod wc s uspecfed prmeters w w y forcg te tegrto formul to e ect for te polyomls = f I tur ts forces I ( f I ( f = for ll polyomls f of degree. Ts leds to te followg system of oler equtos ukows: = w + w + + w 0 = w + w + + w 0 = w + w + + w = w + w + + w = w + + w 0 = w + + w Te resultg formul s of te order (-. Solvg ts system s formdle prolem. w ve ee clculted d collected tles for te Tkfully te odes { } most commoly used s of. d wegts { } Tere s lso oter pproc to te developmet of te umercl tegrto formul I f = w f usg te teory of ortogol polyomls. From tt teory t c e j j j= sow tt te odes [ ] re te zeros of te Legedre polyoml P o te tervl [ ] d ts wegts re gve y w = ( P Sce te Legedre polyomls re well-kow te odes [ ] of degree c e foud wtout y recourse to te oler system. We lst odes d wegts for severl cses elow

18 MA umercl Alyss umer of pots Pots Wegts w 0 ± ± ± 6 / 5 / 7 ± + 6 / 5 / / ± 5 0 / ± / Legedre polyomls: P0 ( = P ( = ( + P ( = ( + P ( P ( + Te frst few Legedre polyomls re: 0 P ( ( ( + 5 ( + (Boet s recurso formul Te grps of tese polyomls (up to = 5 re sow elow: - 8 -

19 MA umercl Alyss Guss Lotto rules Also kow s Lotto qudrture med fter Dutc mtemtc Reuel Lotto. It s smlr to Guss qudrture wt te followg dffereces:. Te tegrto pots clude te ed pots of te tegrto tervl.. It s ccurte for polyomls up to degree were s te umer of tegrto pots. Lotto qudrture of fucto f( o tervl [ +]: Ascsss: s te ( st zero of P' (. Wegts: Remder: - 9 -