5.1 Evaluating a polynomial

Transcription

1 5 Evluting polynomil We will now look t evluting polynomil t point Your first tougt my be: tis is esy: just substitute te vlues, nd presto Unfortuntely, tis is not te cse 5 Execution time First, tere is te cost of performing tis evlution: ow do we efficiently evlute polynomil? Horrible evlution Suppose tt te coefficients of polynomil re stored in n rry coeffs of lengt n suc tt coeffs() is te coefficient of t n nd coeffs(n + ) is te constnt coefficient In tis cse, we could implement function [result] = polyvl( coeffs, t ) result = 0; n = lengt(coeffs); end for k = :lengt(coeffs) result = result + coeffs(k)*pow( t, n - k ); end If you were to implement powering s follows: function [result] = pow( t, n ) if n < 0 result = 0/pow( t, -n ); else result = 0; end end for i = :n result = result*t; end ten tere is specil plce in ell reserved for you: evluting polynomil of degree n would ve run-time of (n )

2 Less orrible evlution If you were to implement powering s follows: function [result] = pow( t, n ) if n < 0 result = 0/pow( t, -n ); elseif n == 0 result = 0; elseif n == result = t; else result = pow( t, n/ ); result = result*result; end end if (n % ) == result = result*t; end In tis cse, evluting polynomil of degree n would ve run-time of (n ln(n)) but would lso require (ln(n)) dditionl memory due to te size of te cll stck Getting tere Suppose, insted, we used: function [result] = polyvl( coeffs, t ) result = 0; term = 0; n = lengt(coeffs); end for k = :lengt(coeffs) result = result + coeffs(n - k + )*term; term = term*t; end Te run time is now (n), but tere re still mny unnecessry clcultions, for we ve n multiplictions Fortuntely, te memory required is only ()

3 Horner s rule Suppose, insted, we used Horner s rule for evluting polynomils: p t t t t t n n n n 0 t t t t n n n n 0 t t t t n n3 n n 0 n t t t t t n 0 For exmple, te polynomil 3t t 9t + 80 cn be written s (3t + 47)t 9)t + 80 From descriptive point-of-view, tis looks ugly; owever, from n implementtion point-of-view it is elegnt: function [result] = polyvl( coeffs, t ) result = 00; end for k = :lengt(coeffs) result = result*t + coeffs(k); end Te run time is now (n) nd tere re only n multiplictions nd n dditions Prctice questions Evlute te polynomil x 3x + t te points x = 0 nd x = 0 using Horner s rule 3

4 5 Numericl stbility Second, ow do we evlute te polynomil p t t t 0 so s to void numericl instbility? For exmple, t is lrge reltive to t, but is comprble to 0, ten unfortuntely, te following my occur if we re using four significnt digits: Suppose t = 000 nd = but = 0, so t + = 000 Suppose now tt 0 = In tis cse, our polynomil, evlutes to 0000; owever, te correct nswer is 00, so te result is not so ccurte Te solution to tis is to find te coefficients reltive to point close to were we intend to evlute our polynomil For exmple, suppose tt we were evluting our polynomil in te vicinity of t = 0000 In tis cse, we could rewrite te polynomil s In tis cse, we cn rewrite tis s p t t t 0 p t 000 t t Tis new polynomil, wen evluted t t = 000 does indeed evlute to 000, but not only tt, it does resonbly well for ll vlues round t = 000 Now Horner s rule my be written s: function [result] = polyvl( coeffs, t, t0 ) result = 00; dt = t - t0; end for k = :lengt(coeffs) result = result*dt + coeffs(k); end Note tt if we re evluting results close to te vlue of t 0, ten p t t t t t, ten t t 0 is smll, so we re lwys multiplying te previous result by smll vlue nd only ten dding on te next coefficient Tus, ssuming tt t t 0 is smll reltive to te coefficients, tis will often be quite resonble In generl, rewriting polynomil in te form were we re multiplying by t to one were we multiply by t t 0 is noting more tn solving system of liner equtions Tis is not te most difficult tsk, so it cn be done quite simply Insted, more likely, we re going to be sked: given set of points (t 0, y 0 ), (t, y ), (t, y ),, find polynomil tt psses troug tese points Prctice questions Evlute te polynomil (x 5) 3(x 5) + t te points x = 5 nd x = 49 using Horner s rule 4

5 53 Differentition We will look t number of mens of pproximting te derivtive, including: simple centered divided difference formul, bckwrd divided difference formul, nd 3 using interpolting polynomils to get better pproximtions Centered divided difference Suppose we wnt to evlute te derivtive of function f t point x One wy of doing tis is to smple te function s follows: Tis is bsed on te limit f f x x f x f x f x f x lim, 0 so if is smll enoug, tis sould be resonble pproximtion Alterntively, suppose tt we ve signl y t some time t, so we don t ctully know wt y is in te future; tt is, t time t + In tis cse, we must use te pproximtion nd so our pproximtion is y t y t y t lim, 0 y t y t y t Now, ow good is ec of tese pproximtions? Using Tylor series, we ve Subtrcting tese two, we ve Dividing bot sides by, we get f x f x f x f x f 6 f x f x f x f x f f x f x f x f 3 f f x f x f x f 3 f 3 5

6 Becuse x x nd x x, by te intermedite vlue teorem, tere must be point x x suc tt f f f, so we will simply write Now, ssuming tt te tird derivtive is not too wild, ten tis is pproximtely for some vlue x x f x f x f x f 6 3 Note tt if we reduce by lf, te error drops by pproximtely one qurter If we reduce by fctor of 0, te error drops by fctor of 00 We cn see tis wit tis exmple: sin 00 sin 00 sin , 00 nd subtrcting te pproximtion from te exct solution, we get Now we ve to find f We don t know te exct vlue of, but te cosine function on tt intervl is monotonic, so we need only look t te endpoints nd ensure tt our error ppers in tt intervl: 3 3 f nd f nd te error does fll between tese two vlues We lso see tt sin 000 sin 000 sin nd subtrcting te pproximtion from te exct solution, we get Tus, you cn see tt by dividing by 0, te error drops by fctor of 00 Additionlly, te two end-points of te error bounds re nd f 6 f Now, tis is importnt: wile we see ere tt yes, te estimte of te error is very successful, we will not know tis in generl However, wt is criticl ere is tt te formul does work Prctice questions Estimte te derivtive of f(x) = xe x t x = using = 0 nd ten = 00 using tis formul Demonstrte tt te estimtor of te error is resonble by estimting te verge vlue of te pproprite derivtive by evluting tt derivtive t x = 9 nd x = nd gin t x = 99 nd x = 0 6

7 Sow tt te error of f x f x f x is O( ) 3 You re prticipting in project were you re ttempting to detect te source of pollution long river Your sensors re plced 05 km prt, nd te most recent reding give concentrtions of HCO 3 t tree of te sensors s being 89 mg/l, 095 mg/l nd 97 mg/l, wit te lst reding being te furtest upstrem Wt is n estimte s to te grdient of te concentrtion (mg/l/m) t te middle sensor in te downstrem direction? ( 007 mg/l/m) Bckwrd divided difference On te oter nd, te one-sided derivtive is not so successful: we ve te Tylor series nd solving tis for te derivtive yields yt yt y t y y t y t y t y Tis is n interesting contrst to te previous: now te error depends on te verge vlue of te second derivtives multiplied by : divide by nd te error drops by, divide by 0, te error drops by 0: sin sin 00 sin , 00 wic, if we subtrct tis from te exct nswer, s n error of , weres sin sin 000 sin wic, if we subtrct tis from te exct nswer, s n error of Tus, you cn see tt by dividing by 0, te error drops by fctor of 0 Now, to see tt te error fll witin te expected intervls, we ve nd f nd f f nd f Consequently, te first formul is better tn te second; owever, te first only works if we ve informtion bout te system bot before nd fter te point in question: someting we simply do not ve in rel-time system 7

8 Implementtion Suppose you ve n embedded system were periodiclly, sensor returns reding nd you my spordiclly need n pproximtion of te derivtive Tis mens you would require dt structure tt cn estimte te rte of cnge of te given dt Suc dt structure would only ve to store te lst tree entries, so n rry of cpcity tree would be pproprite In clculting te derivtive, te formul lso requires te vlue between te smples, or te t We could pss tis vlue wit te constructor As for te required member functions: Te interrupt ndling routine would ve to dd te next dtum to tis dt structure, so we would require some sort of function tt would pss tt new dtum We will use circulr rry to ensure tt ec insertion is () We will need function tt clcultes te rte of cnge bsed on te given dt clss Dt_colltion { privte: double delt_t; double buffer[3]; size_t lst; ; public: Dt_collection( double dt ); void store_dtum( double x ); double derivtive() const; Dt_collection::Dt_collection( double dt ): delt_t{dt, buffer{0, 0, 0, lst{0 { // Empty constructor void Dt_collection::store_dtum( double x ) { lst = (lst + ) % 3; buffer[lst] = x; // 3 y(t) - 4 y(t - ) + y(t ) // Return // double Dt_collection::derivtive() const { return ( 30*buffer[lst] - 40*buffer[(lst + ) % 3] + buffer[(lst + ) % 3] )/(0*delt_t); 8

9 Prctice questions Estimte te derivtive of y(t) = te t t t = using = 0 nd ten = 00 using tis formul Demonstrte tt te estimtor of te error is resonble by estimting te verge vlue of te pproprite derivtive by evluting tt derivtive t x = nd x = nd gin t x = nd x = 0 Sow tt te error of y t y t y t is O() 3 You wnt to estimte te derivtive t time t but only by using istoric dt (t time t, t, etc) Tus, you could use y t y t y t previous pproximtion? How does te size of te coefficient of te error term differ from te 4 You ve sensor tt probes voltge t rte of 0 Hz Its lst tree redings re 735 V, 733 V nd 7339 V, wit te lst being te most recent Estimte te rte of cnge of voltge t te current time (008 V/s) 5 Bsed on te result, wt is te best estimtor of te voltge in te next lf second? (7339 V V/s 05 s = 7397 V) Using interpolting polynomils Insted, let s look t different wy of finding tese formuls using polynomil interpoltion If we interpolte tree points (x 0, f ), (x 0, f 0 ) nd (x 0 +, f ) we get polynomil in x of degree two, if we differentite tis wit respect to x, we get polynomil of degree, nd if we evlute tis t te point x 0, we get te formul f f wic is te sme formul we previously sw Similrly, if we interpolte te two points (t 0, y ) nd (t 0, y 0 ), differentite te interpolting polynomil t t, nd evlute te resulting liner polynomil t = t 0, we get te formul, y y 0 Tis is te formul from te pproximtion of te derivtive Suppose now we find te interpolting polynomil of te tree points two points (t 0, y ), (t 0, y ) nd (t 0, y 0 ), differentite te interpolting polynomil t t, nd evlute te resulting liner polynomil t = t 0, we get te formul 3y 4y y 0 9

10 If you wis to see tis, te following Mple code finds te interpolting polynomil, differentites it, nd ten evlutes tt derivtive t te point t 0 : > curve := CurveFitting:-PolynomilInterpoltion( [t[0], t[0] -, t[0] - *], [y[0], y[-], y[-] ], tu ); > dcurve := diff( curve, tu ); # Differentite te curve wrt tu > evl( dcurve, tu = t[0] ); # Evlute tt derivtive t t[0] > simplify( %, size ); # Simplify te previous result mking it compct 0

11 Tt is, 3y t 4y t y t y t Question: wt is te error? In tis eqution, we ve y(t ) nd y(t ), so let us write tese two Tylor series down: 3 3 y t y t y t y t y 6 y t y t y t y t y y t y t y t y Now, we will multiply te first by four nd subtrct from tt te second to get Now, if we look t y t y t 3y t y t y y y 4 y, we note tt tere must be some vlue t t suc tt y y y nd dividing y by llows us to rewriting tis eqution nd get te simplified form were t t Compre te errors: were x x 3y t 4y t y t y t y 3 3 f x f x f x f 6 3 Tus, te centered pproximtion is twice s ccurte s our bckwrd pproximtion To demonstrte tis, let us gin tke our exmple of 3sin 4sin 00 sin 00 sin , 00 wic, if we subtrct tis from te exct nswer, s n error of , weres 3sin 4sin 000 sin 000 sin wic, if we subtrct tis from te exct nswer, s n error of

12 Notice gin tt our pproximtions of te errors re indeed very close: nd f nd f f nd f Prctice questions Estimte te derivtive of y(t) = te t t t = using = 0 nd ten = 00 using tis formul Demonstrte tt te estimtor of te error is resonble by estimting te verge vlue of te pproprite derivtive by evluting tt derivtive t x = nd x = nd gin t x = nd x = 0 Sow tt te error of 3y t 4 y t y( t ) y t is O( ) 3 You wnt to estimte te derivtive t time t but only by using istoric dt (t time t, t, etc) Tus, you could use 3y t 4y t y( t 3 ) y t Sow tt now te error is O() 4 Disppointed wit te O() error in te previous question, you find te interpolting polynomil of te points (t 3, y(t 3)), (t, y(t )) nd (t, y(t )) You differentite tis interpolting polynomil nd evlute it t t to get te formul 5y t 8y t 3 y( t 3 ) y t Sow tt now te error is O( ) 5 You experiment nd find tt te formul Prove tis Be sure to note te y(t 3) 8y t 9 y t y( t 3 ) y t s n error equl to O( ) 6 4 You ve sensor tt probes voltge t rte of 0 Hz Its lst tree redings re 735 V, 733 V nd 7339 V, wit te lst being te most recent Estimte te rte of cnge of voltge t te current time (009 V/s) 5 Bsed on te result, wt is te best estimtor of te voltge in te next lf second? (7339 V V/s 05 s = 7384 V)

13 Higer derivtives We will look t bot centered divided difference nd bckwrd divided difference pproximtions, in bot cses, using te sme pproc: finding interpolting polynomils, differentiting tose twice, nd ten evluting tem t te pproprite point Centered divided-difference pproximtions Suppose we wnt to estimte iger derivtive Agin, if we find te interpolting qudrtic pssing troug te tree points x, x nd x +, we get te pproximtion f x f x f x f x weres if we find te interpolting qudrtic pssing troug te tree points t, t nd t, we get te pproximtion y t y t y t y t Note tt tese re te sme formuls, only te first is centered wile te second is not Using Tylor series, we cn once gin estimte te error Adding tese two, we get f x f x f x f x f x f 6 4 f x f x f x f x f x f f x f x f x f x f f 4 4 nd solving for te second derivtive, we ve were x x f x f x f x f x f 4 3

14 To demonstrte tis, let us gin tke our exmple of sin 099 sin sin 0 sin , 00 wic, if we subtrct tis from te exct nswer, s n error of , weres sin 0999 sin sin 00 sin wic, if we subtrct tis from te exct nswer, s n error of Notice gin tt our pproximtions of te errors re indeed very close: nd 4 3 f nd f f nd f Prctice questions Estimte te second derivtive of y(t) = te t t t = using = 0 nd ten = 00 using tis formul Demonstrte tt te estimtor of te error is resonble by estimting te verge vlue of te pproprite derivtive by evluting tt derivtive t x = 9 nd x = nd gin t x = 99 nd x = 0 Sow tt te error of f x f x f x f x is O( ) 3 Suppose tt te concentrtion of gs long tube re 053 g/l, 05 g/l, 057 g/l nd 06 g/l Ec of te redings re tken 0 cm prt Clculte te concentrtion pressure t te two interior points (g/l/m ) Te locl rte of cnge in te concentrtion is proportionl to tis concentrtion pressure Do you expect te concentrtion t ec of te interior points to increse, sty te sme, or decrese in te next few seconds? (6 g/l/m nd 0 g/l/m, nd tus te concentrtion t te first interior point sould increse, wile te concentrtion t te second sould sty pproximtely uncnged) 4

15 4 Here re four potogrps of te Flcon Hevy booster rockets lnding tken from te YouTube cnnel of SpceX Te potogrps re tken one second prt Assuming te booster rockets re 45 m tll, estimte te verticl speed nd ccelertion of te booster in te two middle frmes (Tis utor estimtes te verticl speed nd ccelertion in te second frme s 9 m/s nd 5 m/s, respectively; wile in te tird frme s 77 m/s nd 33 m/s, respectively) 5

16 Bckwrd divided-difference pproximtions Similrly, but considering t nd t, we ve: wic now becomes 3 3 y t y t y t y t y 6 y t y t y t y t 4 y y t y t y t y t y 6 4 y t y t y t y t y Now we must be more creful: we wnt te formul for te second derivtive, so we will subtrct twice te first from te second: y t y t y t y t y y 3 3 Solving for te second derivtive nd estimting wit te verge vlue of te tird derivtive in te vicinity of t, we ve y t y t y t 3 y t y Tus, we see tt tis formul s n error tt is O( ) for te pproximting te second derivtive t te center, but only O() for pproximting te derivtive t eiter end-point 6

17 To demonstrte tis, let us gin tke our exmple of sin 098 sin 099 sin sin , 00 wic, if we subtrct tis from te exct nswer, s n error of , weres sin 0999 sin sin 00 sin wic, if we subtrct tis from te exct nswer, s n error of Notice gin tt our pproximtions of te errors re indeed very close: nd f f 3 3 n f d n f d Tus, despite te fct te formul is te sme, only sifted, te error is significntly iger As n exmple of finding cubic interpolting polynomil (interpolting te points t, t, t nd t 3), if we find suc polynomil, differentite it, nd evlute t it t te point t, we get te pproximtion y t 5y t 4y t y t 3 4 y t y Implementtion We need good pproximtion of te second derivtive, but tis requires four tps nd not just tree Consequently, we must cnge our buffer cpcity of 3 As you my ve noted, te buffer cpcity ws rd-coded Tis is described s mgic number, s it is not entirely obvious were te 3 comes from Insted, we will now ssign to te constnt member vrible BUFFER_CAPACITY te cpcity of te internl buffer (now modified to 4), updte te oter functions, nd ten dd function for clculting te second derivtive clss Dt_colltion { privte: size_t const BUFFER_CAPACITY{4; double delt_t; double buffer[buffer_capacity]; size_t lst; ; public: Dt_collection( double dt ); void store_dtum( double x ); double derivtive() const; double second_derivtive() const; Dt_collection::Dt_collection( double dt ): delt_t{dt, buffer{, lst{0 { 7

18 // Empty constructor void Dt_collection::store_dtum( double x ) { lst = (lst + ) % BUFFER_CAPACITY; buffer[lst] = x; // 3 y(t) - 4 y(t - ) + y(t ) // Return // double Dt_collection::derivtive() const { size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; return (30*buffer[lst] - 40*buffer[lst] + buffer[lst])/(0*delt_t); // y(t) - 5 y(t - ) + 4y(t ) y(t - 3) // Return // // double Dt_collection::second_derivtive() const { size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst3{(lst + (BUFFER_CAPACITY - 3)) % BUFFER_CAPACITY; return ( 0*buffer[lst] - 50*buffer[lst] + 40*buffer[lst] - buffer[lst3] )/(delt_t* delt_t); 8

19 Prctice questions Estimte te second derivtive of y(t) = te t t t = using = 0 nd ten = 00 using tis formul Demonstrte tt te estimtor of te error is resonble by estimting te verge vlue of te pproprite derivtive by evluting tt derivtive t x = nd x = nd gin t x = nd x = 0 Sow tt te error of y t y t y t y t is O() 3 You wnt to estimte te second derivtive t time t but only by using istoric dt (t time t, t, etc) Tus, you could use y t 3 y t y t y t How does te error coefficient cnge? 4 Suppose you ve progrmmed n pp to estimte ow quickly runner is bot running nd ccelerting Your ppliction tkes tree potogrps in rpid succession (05 s) It ten mesures te leding edge of te cest, nd bsed on te eigt of te individul ( vlue tt must be entered seprtely), it estimtes te reltive distnces Te distnces trvelled te second nd tird potogrps reltive to te first re 7 m nd m Wt is resonble estimte of te ccelertion of te individul t te moment te lst potogrp is tken? ( 35 m/s ) 9

20 5 Here re tree potogrps of te Flcon Hevy tken from te YouTube cnnel of SpceX Te tree potogrps re one second prt Is te ccelertion positive or negtive, nd wy? If te Flcon Hevy is 70 m tll, estimte te verticl speed nd ccelertion of te Flcon Hevy in te tird frme (Tis utor estimtes te verticl speed nd ccelertion s 890 m/s nd 38 m/s, respectively) 6 Wile trusters on lunc rockets my be on te order of megnewtons (te Flcon 9 trusters produce 7607 MN of force), on trip to Mrs, smller trusters will be used to correct te direction of trvel, nd s suc will only produce force on te order of newtons Te force of truster cn be clculted s function of te fuel tt is injected into te nozzle, nd wile suc truster requires rmp-up time to go to mximum trust, tese engines cn be cut off lmost instntneously by cutting off te fuel supply Suppose tt te following re estimtes of te trust (newtons) tken t 0 Hz: Te redings re: Use te tree-vlue bckwrd divided-difference formuls for bot te rte of cnge nd ccelertion to determine te first nd second derivtives of te force t time t = 08 nd t = 7 Wy is te pproximtion t te first time resonble, but entirely erroneous t te second? 0

21 54 Integrtion We will look t two pproces to integrtion: clcultion of n integrl troug polynomil interpoltion, nd composite pplictions of te bove formuls 54 Use of polynomil interpoltion Suppose we ve n intervl (, b) on wic function f is defined, nd we d like to pproximte te integrl b f xdx Suppose we cn only smple tis function t finite number of points in pproximting tis integrl We could coose te endpoints s our two smple points: x 0 = nd x = b wit f 0 = f(x 0 ) nd f = f(x ) Trpezoidl rule In tis cse, given te informtion we ve, te best pproximtion is to use te trpezoidl rule, wic determines te integrl by clculting te re of te trpezoid Tus, te pproximtion is b f xdx f0 fb Tus, we re multiplying weigted verge of te vlue of te function by te widt of te intervl Integrting tis from to b yields f f x f x x f x b b b b f dx f x dx f x x dx f x dx

22 Simplifying tis, we ve: If we rewrite tis, we ve b b b d d d f b f x x f x x x f x x Dividing bot sides by two yields: Let s integrte function we know: 09 b b b b f x dx f x dx xf x dx f x dx b b b d d f x x f b f xf x f x x f b 6 b f x dx f b f bf b f f x dx f b 6 b b f f b f x dx f b 6 b b 3 f xdx f f b b f b 3 sin xdx sin 09 sin Wen we subtrct tis from te exct solution, we get n error of Wen we look t te errors sin nd sin , we see tt te error does fll between tese two vlues Now, wit tis formul, we cnnot reduce b, becuse we wnt to specificlly clculte tt integrl In order to reduce te error, we will look t oter tecniques 3 3

23 If we wnted to pproximte n integrl over longer intervl wit one dditionl smple, we cn use te trpezoid rule twice: Now te pproximtion is x x0 f xdx f0 f f f f0 f f Te error is simply te sum of te errors pplied to ec sub-intervl; fter simplifiction, tis becomes: Let s pply tis to te integrl bove: f b 3 sin xdx sin 09 sin sin Wen we subtrct tis from te exct solution, we get n error of Wen we look t te errors sin nd sin , we see tt te error does fll between tese two vlues 3

24 Simpson s rule We could do better, owever, s tis integrl lmost certin underestimtes te ctul ccumulted signl, for we expect te system to be differentible For exmple, suppose we re periodiclly smpling te ngulr speed t wic n ntenn is rotting In tt cse, te integrl is te cnge in position of te ntenn, nd te ntenn s inerti, so its ngulr velocity cnnot cnge bruptly Consequently, we could find n interpolting qudrtic function nd integrte it: Rter tn working tis out, Tis is n pproprite point to demonstrte te power of symbolic computtion Here is n exmple of Mple code were we: find n interpolting polynomil between points (, f 0 ), ( +, f ), ( +, f ) wit respect to te vrible x, integrte tt interpolting polynomil from x = to x = +, nd 3 simplify te resulting expression > curve := CurveFitting:-PolynomilInterpoltion( [, +, + *], [f0, f, f], x ); > int( curve, x = + * ); > simplify( %, size ); Tus, we see te pproximtion of te integrl is 4, 0 6 f f f b nd tis is weigted verge of te vlues of te polynomil multiplied by te widt of te intervl 4

25 Beyond te scope of tis course, te error for Simpson s rule is Let s pply tis to te integrl bove: f b sin xdxsin 094sin sin Wen we subtrct tis from te exct solution, we get n error of Wen we look t te errors 4 4 sin nd sin , we see tt te error does fll between tese two vlues Simpson s rule over one time intervl Now, suppose you reding dt from sensor nd summing tt dt For exmple, to clculte te totl energy used, you would integrte smple of te power usge On te oter nd, to find te totl distnce trvelled, you could integrte te speed Given te most recent reding, you would like to estimte te integrl over te lst intervl, from t to t Te most strigt-forwrd is to use te trpezoidl rule Te issue ere is tt it is not very good pproximtion, but Simpson s rule ssumes you re pproximting te integrl over two time steps Suppose you ve lredy pproximted te previous integrl up to time t In tis cse, Simpson s rule won t elp Suppose, owever, you find te interpolting qudrtic polynomil over te pst two time intervls, but integrted it only over te lst time intervl, tis my indeed give better pproximtion of te integrl: Using Mple, we cn find tt te pproximtion of te integrl to be t 5y t 8y t y t ytdt t Te error will still be on te sme mgnitude of tt of Simpson s rule, but wt is interesting ere is tt tis is one 5 8 of our first pproximtions tt contins negtive weigt coefficient in our weigted verge: 5

26 You my wonder wy coefficient would ve suc weigted verge Tis is perps best explined wit simple cse of looking t te devitions from were te y vlues re in strigt line If te point y(t ) increses, te interpolting qudrtic polynomil goes down on te rigt-nd sub-intervl, so positive increse in y(t ) results in decrese in te estimted integrl Alterntively, if te point y(t ) decreses, te interpolting polynomil increses on te rigt-nd sub-intervl, tus incresing te pproximtion of te integrl 6

27 Implementtion Adding to our dt collection dt structure, we cn now estimte te integrl; owever, tis will require keeping trck of te integrl clculted to tis point clss Dt_colltion { privte: size_t const BUFFER_CAPACITY{4; double delt_t; double buffer[buffer_capacity]; size_t lst; double running_sum; ; public: Dt_collection( double dt ); void store_dtum( double x ); double derivtive() const; double second_derivtive() const; double integrl() const; Dt_collection::Dt_collection( double dt ): delt_t{dt, buffer{, lst{0, integrl{00 { // Empty constructor void Dt_collection::store_dtum( double x ) { lst = (lst + ) % BUFFER_CAPACITY; buffer[lst] = x; // Te integrl must be clculted regrdless s to weter or not te // vlue of te integrl is queried // 5 y(t) + 8 y(t - ) - y(t - *) // running_sum += // size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; running_sum += (5*buffer[lst] + 8*buffer[lst] - buffer[lst])/0*delt_t; 7

28 // 3 y(t) - 4 y(t - ) + y(t ) // Return // double Dt_collection::derivtive() const { size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; return (30*buffer[lst] - 40*buffer[lst] + buffer[lst])/(0*delt_t); // y(t) - 5 y(t - ) + 4y(t ) y(t - 3) // Return // // double Dt_collection::second_derivtive() const { size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst3{(lst + (BUFFER_CAPACITY - 3)) % BUFFER_CAPACITY; return ( 0*buffer[lst] - 50*buffer[lst] + 40*buffer[lst] - buffer[lst3] )/(delt_t* delt_t); // Return te stored vlue of te integrl // - tis vlue will be updted ec time new reding is provided double Dt_collection::integrl() const { return running_sum; 8

29 Simpson s 3 / 8 t rule Te finl pproximtion we will use is termed Simpson s 3 / 8 t rule, wic finds n interpolting cubic nd integrtes it We will sow tis gin using Mple: > curve := CurveFitting:-PolynomilInterpoltion( [, +, + *, + 3*], [f0, f, f, f3], x ); > int( curve, x = + 3* ); > simplify( %, size ); Tus, we see te pproximtion of te integrl is 3 3, f f f f b nd tis is weigted verge of te vlues of te polynomil multiplied by te widt of te intervl 9

30 Beyond te scope of tis course, te error for Simpson s 3 / 8 t rule is Te coefficient is 4 / 9 t te error of Simpson s rule Let s pply tis to te integrl bove: f b sin xdx sin 093sin 0963sin 03 sin Wen we subtrct tis from te exct solution, we get n error of , wic is very close to 4/9 t te error of Simpson s rule (te rtio is 04444) Wen we look t te errors 4 4 sin nd sin , we see tt te error does fll between tese two vlues Simpson s 3 / 8 t rule over one time intervl As wit Simpson s rule, we could find te interpolting cubic polynomil over tree time intervls, but only integrte tt in te most recent intervl: Using Mple, we cn find tis pproximtion using t 9y t 9y t 5y t y t 3 ytdt 4 t Wit no oter informtion, te error would be tird te error of Simpson s 3 / 8 t rule 30

31 Implementtion As wit trcking te integrl using Simpson s rule, we would now updte te integrl wit Simpson s 3 / 8 t rule clss Dt_colltion { privte: size_t const BUFFER_CAPACITY{4; double delt_t; double buffer[buffer_capacity]; size_t lst; double running_sum; ; public: Dt_collection( double dt ); void store_dtum( double x ); double derivtive() const; double second_derivtive() const; double integrl() const; Dt_collection::Dt_collection( double dt ): delt_t{dt, buffer{, lst{0, integrl{00 { // Empty constructor void Dt_collection::store_dtum( double x ) { lst = (lst + ) % BUFFER_CAPACITY; buffer[lst] = x; // Te integrl must be clculted regrdless s to weter or not te // vlue of te integrl is queried // 9 y(t) + 9 y(t - ) - 5 y(t - *) + y(t - 3) // running_sum += // 4 size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst{(lst + (BUFFER_CAPACITY - )) % BUFFER_CAPACITY; size_t const lst3{(lst + (BUFFER_CAPACITY - 3)) % BUFFER_CAPACITY; running_sum += ( 9*buffer[lst] + 9*buffer[lst] - 5*buffer[lst] +buffer[lst3] )/40*delt_t; As before, te integrl function would simply return te integrl member vrible 3

32 54 Composition rules Unlike differentition, one cnnot cnge te bounds of n integrl, for tt is exctly wt we re trying to pproximte Unfortuntely, we cnnot continue using iger degree polynomils, s tey become more sensitive to noise Insted, owever, we cn sub-divide te intervl [, b] into n sub-intervls, nd ten pply ny of te previous rules to ec sub-intervl nd ten dd tem To demonstrte tese tecniques, we will sow integrting te following generic function from to b: Composite trpezoidl rule Suppose we divide te intervl into n sub-intervls, nd pply te trpezoidl rule to ec In tis cse, let us denote b so te intervl is divided into te points x0, x, x,, xn b were xk k If we denote n f k f x k, ten te pproximtion is f0 f f f fn fn Rerrnging tis, we get b f xdx f0 f f fn fn Question: wt is te error? It must be te sum of te individul errors: f f f 3 f n Recll, tt ec sub-intervl is of widt Collecting similr terms, we ve 3 n 3 f f f 3 f n f k k 3 were x k k x Te problem is tt tt wile te coefficient k my go to zero s gets smller, simultneously, n becomes lrger, nd so te sum n f k grows To contrst te two, we observe tt k n k k f 3

33 is sum of points evluted t n, nd by ssuming te second derivtive is continuous, tere must be point b suc tt it equls te verge of tese vlues so Substituting tis into our formul, we get f nf n f k n k k n f k k 3 nf 3 n f k nd becuse b, it follows tt n b, nd tus we ve n b n d b f x x f0 fk fn f k or n 0 k n b k b f xdx f f f f If you wnted to be explicit wit respect to te function evlutions (s f f k n b b f x dx f f k f b f k k ), you could write tis s You my note tt b, so tis is weigted verge of tese evlutions of te n times function on te intervl [, b] multiplied by te widt of te intervl Te components of b f re eiter fixed or bounded (ssuming te second derivtive is not too wild) Tus, lve nd te error goes down by fctor of four; increse te number of intervls by ten nd te error goes down by fctor of 00 33

34 As n exmple, let us pproximte n integrl sligtly lrger tn te one previously given We know, for exmple, 3 tt te integrl sin x dx cos cos Te minimum nd mximum vlues of te second derivtive of sin(x) on tt intervl re nd , respectively If we let n =, in wic cse = /6, our pproximtion is 3 sin xdx sin sin k sin k Subtrcting tis from te exct solution yields n error of Te minimum nd mximum vlues of te pproximtion of te error re nd , respectively, nd te error of our pproximtion flls between tese two If we now increse n = 0, in wic cse, = /60, our pproximtion is 3 9 sin xdx sin sin k sin k 0 0 Subtrcting tis from te exct solution yields n error of , nd s you cn see, tis is pproximtely oneundredt te previous error Of course, te bounds on te error would not cnge We cn now pply tis to Simpson s rule nd Simpson s 3 / 8 t rule Prctice questions Approximte te integrl 3 4 x dx using te composite trpezoidl rule wit n = 6 0 Find te exct error of your pproximtion nd find te bounds of te error using te given formuls 34

35 Composite Simpson s rule We will wnt to keep te number of function evlutions te sme Tus, we will ssume we re dividing te intervl into n even number of sub-intervls for Simpson s rule, nd into multiple of tree sub-intervls for Simpson s 3 / 8 t rule In te cse of Simpson s rule, we must ssume tt n is even, nd tus ec sub-intervl to wic we will pply Simpson s rule will be of widt : or b b f xdx f0 4 f f 4 f3 f4 fn 4 fn fn f n n b 4 b 4 f x dx f0 4 fk fk f n f 3 k k 80 If you wnted to be explicit wit respect to te function evlutions (s f f k b k 4 4 ), you could write tis s n n 4 b 4 f x dx f 4 f k f k f b f 3 k k 80 Wt is importnt, owever, is te weigts in front of te coefficients:, 4,, 4,, 4,, 4,, 4, You my note tt b, so tis is weigted verge of tese 3 n times evlutions of te function on te intervl [, b] multiplied by te widt of te intervl Viewing tis, we re finding interpolting nd integrting interpolting qudrtics over intervls of widt : Approximting our integrl of te sine function from to 3, we ve tt tis equls n n 3 sin xdx sin 4 sin k k sin k k 35

36 Subtrcting tis from te correct nswer yields n error of Te minimum nd mximum vlues of te pproximtion of te error re nd , respectively, nd te error of our pproximtion flls between tese two If we increse te number of intervls by fctor of 0 to 0, te pproximtion is now wit n error of ; gin, drop by lmost exctly fctor of 0 4 Prctice questions Approximte te integrl 3 4 x dx using te composite Simpson s rule wit n = 6 0 Find te exct error of your pproximtion nd find te bounds of te error using te given formuls 36

37 Composite Simpson s 3 / 8 t rule In te cse of Simpson s 3 / 8 t rule, we must ssume n is multiple of tree, nd tus ec sub-intervl will be of widt 3 or b b f xdx f0 3 f 3 f f3 3 f4 3 f5 f6 fn3 3 fn 3 fn fn 3 f n n n b b 4 f x dx f0 3 f3k 3 f3k f3k f n 8 f k k k 80 If you wnted to be explicit wit respect to te function evlutions (s f f k b k 4 4 ), you could write tis s n n n d b f x x f f k f k f k f b f 8 k k k 80 Wt is importnt, owever, is te weigts in front of te coefficients:, 3, 3,, 3, 3,,,, 3, 3, You my note tt b, so tis is weigted verge of tese 8 n times evlutions of te function on te intervl [, b] multiplied by te widt of te intervl Viewing tis, we re finding interpolting nd integrting interpolting cubic polynomils over intervls of widt 3: Agin, pplying tis out our integrl, we get n pproximtion n n n sin xdx sin 3 sin 3k 3 sin 3k 3k sin 3 8 k k k

38 Te error wen we subtrct tis from te ctul integrl is Our pproximtions for te limits of te error spn form to , nd we see tt tis error flls into tt intervl Agin, if we increse n by fctor of 0 to 0, we get te pproximtion wic s n error of wic is smller by fctor of 0 4 nd tis gin flls between te two end-points of te error estimtes You will notice tt te error is pproximtely twice tt of te pproximtion using Simpson s rule nd tt te intervls of possible error do not overlp Prctice questions Approximte te integrl 3 4 x dx using te composite Simpson s rule wit n = 6 0 Find te exct error of your pproximtion nd find te bounds of te error using te given formuls 3 Compre te nswers for te composite trpezoidl, Simpson s nd Simpson s 3 / 8 t rules 38

39 533 Integrtion rules implemented in C++ For tose keen to implement tese in C++, consider te following: #include <cssert> #include <iostrem> #include <cmt> int min(); double trpezoidl_rule( double f(double), double, double b, unsigned int n ); double simpsons_rule( double f(double), double, double b, unsigned int n ); double simpsons_rule( double f[], size_t cpcity, double ); double simpsons_3_8_rule( double f(double), double, double b, unsigned int n ); double trpezoidl_rule( double f(double), double, double b, unsigned int n ) { ssert( n > 0 ); double {(b - )/n; double sum{00; for ( unsigned int k{; k < n; ++k ) { sum += f( + k* ); return (05*)*(f() + 0*sum + f(b)); double simpsons_rule( double f(double), double, double b, unsigned int n ) { ssert( n > 0 ); ssert( (n % ) == 0 ); // 'n' must be even double {(b - )/n; double sum_{00; double sum_4{00; for ( unsigned int k{; k < n/; ++k ) { sum_4 += f( + (*k - )* ); sum_ += f( + *k * ); return (/30)*(f() + 40*(sum_4 + f( + (n - )* )) + 0*sum_ + f(b)); 39

40 double simpsons_rule( double f[], std::size_t cpcity, double ) { double sum{00; for ( std::size_t k{; k < cpcity - ; ++k ) { sum += f[k]; sum *= 0; sum += f[0] + f[cpcity - ]; return sum*/3; double simpsons_3_8_rule( double f(double), double, double b, unsigned int n ) { ssert( n > 0 ); ssert( (n % 3) == 0 ); // 'n' must be multiple of 3 double {(b - )/n; double sum_{00; double sum_3{00; for ( unsigned int k{; k < n/3; ++k ) { sum_3 += f( + (3*k - )* ) + f( + (3*k - )* ); sum_ += f( + 3*k * ); return (30*/80)*(f() + 30*(sum_3 + f( + (n-)* ) + f( + (n-)* )) + 0*sum_ + f(b)); int min() { std::coutprecision( 6 ); std::cout << trpezoidl_rule( std::sin,, 3, ) << std::endl; std::cout << trpezoidl_rule( std::sin,, 3, 0 ) << std::endl; std::cout << simpsons_rule( std::sin,, 3, ) << std::endl; std::cout << simpsons_rule( std::sin,, 3, 0 ) << std::endl; std::cout << simpsons_3_8_rule( std::sin,, 3, ) << std::endl; std::cout << simpsons_3_8_rule( std::sin,, 3, 0 ) << std::endl; std::cout << (std::cos(0) - std::cos(30)) << std::endl; return 0; Te output of tis progrm is

41 Appendix: Finding formul Finding interpolting polynomils requires you to find te polynomil tt interpoltes, for exmple, te points (t 0, y 0 ), (t 0, y ) nd (t 0, y ) We wnt to find polynomil t + bt + c tt psses troug ll tree of tese points; tt is, we require tt ll tree of tese re true: t bt c y t b t c y 0 0 t b t c y 0 0 Tis is system of liner equtions in, b nd c, nd tus we must solve t t y t t y t t y We must now perform Gussin elimintion nd bckwrd substitution, so we strt by dding Row onto Row, nd dding t 0 times Row onto Row 3: t0 t 0 times t0 t0 t0 y0 t0 t0 y0 t0 0 y t0 t0 t0 t0 4 t0 y0 t0 0 y t0 t0 t0 Now, we must dd 0 t0 t 0 t0 t0 t0 t times Row onto Row 3: t0 t0 y0 t0 t0 t0 0 y y 0 t0 t0 t0 t0 t0 0 0 y y y 0 t0 t0 t0 t 0 Te next step is to solve for c, ten b nd ten : 4

42 c t t t t t t y y y t0 t0 t0 t0 t0 t0 t0t0 y t t y t t y t t You cn now substitute tis into te second eqution to solve for b: 4 3 y t0 y t0 y0 t0 b You cn now substitute te vlues for bot b nd c into te first eqution to solve for : y0 y y Consequently, te best-fitting polynomil tt psses troug te tree points (t 0, y 0 ), (t 0, y ) nd (t 0, y ) is te polynomil 4 3 y0 y y t0 y t0 y 0 t0 y t0 t0 yt0 t0 y0 t0 t0 y t t You cn now, for exmple, differentite tis polynomil wit respect to t: nd evlute it t te point t = t 0 : Finding common denomintor, we get 4 3 y y y y t y t y t t y y y y t y t y t t y t 4y t y t y t y 4 t y 3 t nd expnding tis nd te collecting on t 0, we ve t y y y y y y y 4y 3y If you were to simplify te numertor, you would find tt , 3y0 4y y 4

43 We cn cncel out n in bot te numertor nd te denomintor to get 3y 4y y 0 Note lso if we differentite te interpolting polynomil twice, we get wic simplifies to y y y 0 y y y 0 wic you my recll is our pproximtion of te second derivtive: y t0 y t y t y t Finlly, let us integrte tis interpolting polynomil from t = t 0 to t = t 0 Note tt tis is simply polynomil in t, nd we ve tt, t0 3 t bt cdt t t t t b c 3 t0 Substituting te bove coefficients of te interpolting polynomil into tis expression, we get 3 y0 y y t0 t0 t0 t0 4 3 y t y t y t 3 y t0 t0 yt0 t0 y0 t0 t If you expnd nd simplify tis expression, you will get te very fmilir expression: y y y Exercise: Crefully expnd te bove expression nd demonstrte tt wen ll oter terms cncel, te only terms tt remin re tose in te formul for Simpson s rule It is esiest to note tt te bove expression s terms contining t 0, t 0, nd 3 t 0, nd if you were to collect terms, you would not ll of tese cncel, leving simply tose terms independent of t 0, wic would ten simplify to te expression bove 43

44 Acknowledgments Dviti Jignes Ptel Kinderdeep Sing Virdi Priynk Hrirn Deven Rjes Pttni Micel Georges Njrin Priynk Hrirn Adity Aror 44