Numerical Methods of Approximating Definite Integrals
|
|
- Silas Edwards
- 8 years ago
- Views:
Transcription
1 6 C H A P T E R Numericl Methods o Approimting Deinite Integrls 6. APPROXIMATING SUMS: L n, R n, T n, AND M n Introduction Not only cn we dierentite ll the bsic unctions we ve encountered, polynomils, eponentil nd logrithmic unctions, trigonometric nd inverse trigonometric unctions, nd rtionl unctions, but rmed with the Product nd Chin Rules, we cn hppily dierentite ny new unction constructed by multipliction, division, ddition, subtrction, or composition o these unctions. This gives us sense o competence nd stisction. Although t this point we cn integrte mny unctions, there re bsic unctions such s ln nd sec ) tht we hve not yet tckled. From the Chin Rule or dierentition we get the technique o substitution or ntidierentition; rom the Product Rule or dierentition we get technique o ntidierentition known s Integrtion by Prts. The ltter is something to look orwrd to lerning in Chpter 9.) Lerning more sophisticted methods o substitution nd lgebric mnipultion will enlrge the collection o unctions we cn ntidierentite. Tbles o integrls nd high-powered computer pckges cn provide ssistnce i we re in dire strits. Nevertheless, there re some very innocent-looking unctions tht cnnot be delt with esily. For instnce, ll the technicl skill in the world won t help us ind n ntiderivtive or e, or sin ), or sin ). Knowing tht there is no gurntee tht we cn ntidierentite cn be unnerving. This chpter will restore our sense o hving things under control when we re ced with deinite integrl. Suppose we re interested in evluting deinite integrl nd we hve not ound n ntiderivtive or the integrnd. In ny prcticl sitution we ll need to evlute the deinite ln d = ln + C veriy this or yoursel). This cn be ound either by some serious guess-work, methods given in Section 7., or using the technique clled Integrtion by Prts. With some work we cn integrte sec. sec d = sec +tn sec sec +tn d = sec +sec tn sec +tn d = ln sec + tn +C. Tht is, i we wnt to obtin n ntiderivtive tht is inite sum, product, or composition o the elementry unctions. 805
2 806 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls integrl with certin degree o ccurcy. Provided tht n pproimtion is stisctory, it is not necessry to be ble to ind n ntiderivtive o the integrnd. Insted, we return to the bsic ides tht led us to the limit deinition o the deinite integrl. The theoreticl underpinnings o clculus involve the method o successive pproimtions ollowed by limiting process. Dierentil Clculus Let be dierentible unction. We obtin numericl pproimtions o the slope o the tngent to the grph o t point P by looking t the slope o secnt lines through P nd Q, where Q is point on the grph o very close to P. By tking the limit s Q pproches P we determine the ect slope o the tngent line t P. y = ) Q P = = b Figure 6. Integrl Clculus Let be n integrble unction. We obtin numericl pproimtions o the signed re between the grph o nd the -is on [, b] by prtitioning the intervl [, b] into mny equl subintervls, treting s i it is constnt on ech tiny subintervl, nd pproimting the signed re with Riemnn sum. By tking the limit s the number o subintervls increses without bound we determine the deinite integrl. In this chpter we return to Riemnn sums to obtin pproimtions o deinite integrls. The numericl methods discussed here re oten used in prctice. Computers or progrmmble clcultors re idel or perorming the otherwise tedious clcultions. Approimting Sums: L n, R n, T n, nd M n In the contet o the ollowing emple we ll discuss let-hnd, right-hnd, midpoint, nd trpezoidl sums, sums tht cn be used to pproimte deinite integrl. In order to be ble to compre our pproimtions with the ctul vlue, we ll look t n integrl we cn evlute ectly. EXAMPLE 6. Approimte 5 d. Keep improving upon the pproimtion until you know the vlue to our deciml plces. SOLUTION i) To pproimte the integrl we chop up the intervl o integrtion into n equl subintervls, ii) pproimte the re under the curve on ech subintervl by the re o rectngle, nd iii) sum the res o these rectngles.
3 6. Approimting Sums: L n, R n, T n, nd M n 807 We ll construct three Riemnn sums: the let- hnd sum, denoted by L n ; the right-hnd sum, denoted by R n ; nd the midpoint sum, denoted by M n. We re lredy milir with the irst two.) We describe these sums s ollows. L n : The height o the rectngle on ech subintervl is given by the vlue o t the let-hnd endpoint o the subintervl. R n : The height o the rectngle on ech subintervl is given by the vlue o t the right-hnd endpoint o the subintervl. M n : The height o the rectngle on ech subintervl is given by the vlue o t the midpoint o the subintervl. On the i th intervl, [ i, i ], the height o the rectngle is c), where c is midwy between i nd i ; c = i + i. 4 Subdivisions Let n = 4; we chop [, 5] into 4 equl subintervls ech o length = 5 4 =. = 4 5 Below re the let- hnd sum, the right-hnd sum, nd the midpoint sum. 4 5 L 4 = = = R 4 = = = M 4 = = = Figure 6. By... we men there re more deciml plces but we hve stopped recording them.
4 808 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls The unction ) = is decresing; thereore the let-hnd sums provide n upper bound nd the right-hnd sums lower bound or 5 d = R 4 < d < L 4 = Suppose we tke the verge o R 4 nd L 4, L 4+R 4. This is closer to the vlue o the integrl thn either the right- or let-hnd sums. Geometriclly this verge is equivlent to pproimting the re on ech intervl by trpezoid insted o rectngle, s shown below in Figure 6.. E B F C A i ) D i 4 5 b) Figure 6. Averging the re o rectngles o ABCD nd AEFD in Figure 6.) gives the re o the trpezoid AECD. We reer to the verge o the let- nd right- hnd sums, L n+r n, s the trpezoidl sum or the Trpezoidl rule) nd denote it by T n. T 4 = = Becuse is concve up 4 on [, 5] we know tht on ech subintervl the re under the trpezoid is lrger thn tht under the curve. Thus T 4 gives n upper bound or the integrl, better upper bound thn tht provided by L 4. 5 R 4 < d < T 4 <L < d <.68...< At this point the mind o the criticl reder should be buzzing with questions. Perhps they include the ollowing. Wht re the conditions under which T n will be lrger thn the vlue o the integrl? Smller? Where does the midpoint sum it into the picture? Let s investigte the irst question by looking t some grphs. 4 t) = t, so t) = t nd t) = t. The ltter is positive on [, 5], so is concve up on the intervl under discussion.
5 6. Approimting Sums: L n, R n, T n, nd M n 809 Figure 6.4 I is concve up on [, b], then every secnt line joining two points on the grph o on [, b] lies bove the grph. I is concve down on [, b], then every secnt line joining two points on the grph o on [, b] lies below the grph. We conclude tht b t) dt < T n i is concve up on [, b], nd b T n < t) dt i is concve down on [, b]. Where does the midpoint sum it into the picture? It turns out tht T n nd M n re complementry pir. We will show tht b i is concve up on [, b], then M n < t) dt < T n b i is concve down on [, b], then T n < t) dt < M n. while To do this we ll give n lterntive grphicl interprettion o the midpoint sum. Figure 6.5i) illustrtes the midpoint pproimtion, pproimting the re under the grph o on [ i, i+ ] by the re o rectngle whose height is the vlue o t the midpoint o the intervl. In Figure 6.5ii) we pproimte the re under by the re o the trpezoid ormed by the tngent line to the grph o t the midpoint o the intervl. We clim tht these res re identicl; pivoting the line through A, B, nd C bout the midpoint C does not chnge the re bounded below. Look t Figure 6.5iii). The tringles ACD nd BCE re congruent. We rgue this s ollows. Angles CAD nd CBE re right ngles. Angles ACD nd BCE re equl. Thereore the two tringles in question re similr. But AC = CB becuse C is the midpoint o the intervl [ i, i+ ]. We conclude tht tringles ACD nd BCE re congruent nd hence hve the sme re. Thereore, rectngle i AB i+ nd trpezoid i DE i+ Figures 6.5i) nd ii), respectively) hve the sme re; we cn interpret the midpoint sum s the midpoint tngent sum.
6 80 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls A C B C E A C E B E i midpt. i + i) D i midpt. i + ii) D i midpt. i + iii) D A C B congruent tringles s long s the oblique line psses through midpoint C. Figure 6.5 Below is picture o M 4 using the midpoint tngent line interprettion. is concve up, so the tngent lines lie below the curve. We know tht on ech subintervl the re under the midpoint tngent line is less thn the re under the curve; thereore, M 4 gives lower bound or the integrl Figure 6.6 Where unction is concve up its tngent line lies below the curve; where unction is concve down its tngent line lies bove the curve. It ollows tht i is concve up on [, b], then M n < b t) dt; i is concve down on [, b], then b t) dt < M n. Figure 6.7 How do L n, R n, T n, nd M n improve s we increse n? 8 Subdivisions Let n = 8; we chop [, 5] into 8 equl pieces ech o length = 5 8 =.
7 6. Approimting Sums: L n, R n, T n, nd M n L 8 = = R 8 = = T 8 = L 8 + R 8 = midpoint rectngles OR midpoint tngent trpezoids M 8 = = We know tht.48...= R 8 <M 8 = < 5 d < T 8 =.68...<L 8 = ) = is decresing nd concve up on [, 5]. Thereore, we know tht or ny n 5 5 R n < d < L n nd M n < d < T n. I we re interested in more deciml plces, we cn simply choose lrger vlues o n. 50 Subdivisions Suppose n = 50; we chop [, 5] into 50 equl pieces ech o length = 5 50 = We don t ctully wnt to sum up 50 terms by hnd. Work like this is pinul to do by hnd but it s child s ply or progrmmble clcultor or computer. Get out your progrmmed clcultor or computer nd check the igures given below. 5 R 50 = M 50 = T 50 = L 50 = Subdivisions Suppose n = 400; we chop [, 5] into 400 equl pieces ech o length = = 0.0. We obtin R 400 = M 400 = T 400 = L 400 = Using T 400 s n upper bound nd M 400 s lower bound, we ve niled down the vlue o this integrl to 4 deciml plces. 5 You ll hve to enter the ollowing inormtion: the unction oten s Y ), the endpoints o integrtion, nd the number o pieces into which you d like to prtition the intervl. And, i you re using clcultor without L n, R n, T n, nd M n progrmmed, you ll hve to enter the progrm.)
8 8 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls REMARK M 50 nd T 50 give better pproimtions o 5 d thn do R 400 nd L 400. Summry o the Underlying Principles I is incresing on [, b], then L n < b t) dt < R n. L n R n I is decresing on [, b], then R n < b t) dt < L n. L n R n I is concve up on [, b], then M n < b t) dt < T n. T n M n I is concve down on [, b], then T n < b t) dt < M n. T n M n For ny given n, generlly the trpezoidl nd midpoint sums re much closer to the ctul vlue o the deinite integrl thn re the let- nd right-hnd sums. EXAMPLE 6. Approimte 0 e d by inding upper nd lower bounds diering by no more thn SOLUTION This is gret problem on which to prctice, becuse it is impossible to ind n ntiderivtive or ) = e unless we resort to n ininite sum o terms. Look t the grph o ). It is decresing on the intervl [0, ] nd ppers to hve point o inlection somewhere on this intervl. 7 ) = e Figure 6.8 Suppose we were plnning to use let- nd right-hnd sums only. As shown in Section., the dierence between the let- nd right-hnd sums is given by 6 The unction e k is o prcticl importnce becuse, or the pproprite k, its grph gives the bell-shped norml distribution curve. The re under the norml distribution curve over some given intervl is o vitl importnce to probbilists nd sttisticins. 7 How cn we be sure is decresing? ) =. As increses rom 0 to, increses, so e is positive nd incresing. e d Thereore, its reciprocl is decresing. Alterntively, d e = e 4) = 4 0 or 0. e
9 6. Approimting Sums: L n, R n, T n, nd M n 8 R n L n = b) ) = b) ) b n. Thereore, in this emple R n L n = e 8 e 0 0 n = e 8 n < n I we wnt R n L n < 0.00, then we cn solve.9994 n = 0.00 or n nd choose ny integer lrger thn this. n = = 999.4, so we cn choose n = 000. I your clcultor will ccept number this lrge, you re in good shpe; the let-hnd sum will be n upper bound nd the right-hnd sum will be lower bound, becuse is decresing on [0, ]. The computtion my tke some time or the mchine to perorm, depending upon its power. You should get L 000 = nd R 000 = REMARK Suppose is monotonic 8 over the intervl [, b], so the ctul vlue o b ) d is between L n nd R n. Although it is possible simply to try lrger nd lrger vlues o n until L n nd R n re within the desired distnce rom one nother, it is more eicient, i high degree o ccurcy is demnded, to use to ind n pproprite vlue o n. R n L n = b) ) Generlly, the midpoint nd trpezoidl sums give us much better bounds or prticulr n. I we ind the point o inlection, we cn use these sums to orm sndwich round the vlue o the integrl. ) = e ) = 4e ) = 4[e + 4e )] = 4e + 4 ) ) = 0, where + 4 = 0, tht is, where = 4. On [0, ] ) = 0 t =. Looking bck t ) = 4e + 4 ), we cn see tht chnges sign rom negtive to positive) t =, so = 0.5 is point o inlection. is concve down on [0, ] nd concve up on [, ]. Thereore, 8 is monotonic on [, b] i it is either lwys incresing on [, b] or is lwys decresing on [, b].
10 84 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls on [0, ], T n gives lower bound or the integrl nd M n gives n upper bound; on [, ], T n gives n upper bound or the integrl nd M n gives lower bound. ) = e Figure 6.9 The combined error on [0, ] nd [, ] must be no more thn We cn split up the llowble error however we wnt. We cn ply round with the clcultor or computer until the sum o T n M n on [0, ] nd T m M m on [, ] is less thn Here is the result o some plying round. I we try n = 50 on both intervls, we obtin the ollowing. On [0, ] T 50 = M 50 = On [, ] T 50 = M 50 = On ech intervl the dierence between T 50 nd M 50 is substntilly less thn To get the inl nswer we need to dd the lower bounds T 50 on [0, ] nd M 50 on [, ]) to obtin lower bound or the integrl, nd dd the upper bounds to obtin n upper bound. 0 e d =.5 0 e d + lower bound: = upper bound: = e d In the net section we will show n lterntive nd more eicient method o using the trpezoidl nd midpoint sums to pproimte this integrl. Numericl methods o integrtion, such s let- nd right-hnd sums nd trpezoidl nd midpoint sums, re very useul not only when we re trying to pproimte b ) d nd cn t ind n ntiderivtive or, but lso in situtions in which we don t even hve ormul or. A scientist my be tking periodic dt redings, or surveyor my be tking mesurements t preset intervls, nd these dt sets my constitute the only inormtion we hve bout. Numericl methods o integrtion cn be pplied directly to the dt sets. Eercise 6. below dels with inormtion rom dt set, nd the results o Eercise 6. nd Eercise 6. cn be esily pplied to dt sets in which dt hve been collected t eqully spced intervls. EXERCISE 6. Between noon nd 5:00 p.m. wter hs been leving reservoir t n incresing rte. We do not hve rte unction t our disposl, but we do hve some mesurements indicting the rte tht wter hs been leving t vrious times. The inormtion is given below. These mesurements were not tken t eqully spced time intervls.
11 6. Approimting Sums: L n, R n, T n, nd M n 85 Time noon :00 :0 :00 :00 :45 4:5 5:00 Rte out in gl/hr) EXERCISE 6. ) Find good upper nd lower bounds on the mount o wter tht hs let the reservoir between noon nd 5:00 p.m. b) Use trpezoidl sum to pproimte the mount o wter tht hs let the reservoir between noon nd 5:00 p.m. Answers re provided t the end o the section. Our im is to pproimte b ) d when we do not hve ormul or ). Our dt consist o collection o mesurements o ) tken t eqully spced intervls. Use trpezoidl sum to pproimte the deinite integrl. Prtition the intervl [, b] into n equl subintervls, ech o length = b n. 0,,,..., re s indicted. 0 = nd k = + k or k =,..., n. y k = k ) or k = 0,..., n. Show tht the trpezoidl sum cn be computed s ollows: ) T n = [y 0 + y + y + +y n + y n ]. y 0 y y n y n 0 = 4 n n Figure 6.0 = b Answers to Selected Eercises Eercise The nswer is provided t the end o the section. ) The rte is incresing, so the lower bound is given by the let-hnd sum nd the upper bound by the right-hnd sum. Lower bound: 40) ) ) + 00) ) ) ) = Upper bound: 60) ) ) + 50) ) ) ) = 0.5 Between nd 0.5 gllons o wter hve let the reservoir.
12 86 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls b) The trpezoidl pproimtion is the verge o the let- nd right-hnd sums given bove. The dierent lengths o the intervls do not chnge this bsic principle. Convince yoursel o this.) Trpezoidl sum = ) = 955 Eercise. T n = [y 0 + y + y y n ) + y + y + y y n ) ] = y 0 + y + y y n + y n ) P R O B L E M S F O R S E C T I O N 6. Some o the problems in this problem set require the use o progrmmble clcultor or computer. These re not highlighted in ny wy, but common sense should tell you tht summing couple o hundred terms is not good use o your time. On the other hnd, in order to mke sure tht you understnd the numericl methods, some problems eplicitly sk you to rerin rom using progrm to eecute the computtion.. Suppose we use right- nd let-hnd sums to pproimte b t) dt. We prtition the intervl [, b] into n equl pieces ech o length t. Let R n be the right-hnd sum using n subdivisions nd L n be the let-hnd sum using n subdivisions. ) Show tht R n = L n + b) t ) t. b) Conclude tht R n L n = [ b) )] t = [ b) )] b n.. ) Find t dt, where >0. b) ln cn be deined to be t dt. Sketch the grph o /t nd pproimte ln numericlly by prtitioning the intervl [, ] into 4 equl pieces nd computing L 4 nd R 4. Write out the sum long-hnd, without using clcultor progrm. c) Into how mny equl pieces would you hve to subdivide the intervl [, ] in order to pproimte ln so tht R n nd L n give upper nd lower bounds tht dier by no more thn 0.0? Cll this number p. d) Compute M p nd T p. e) Into how mny equl pieces would you hve to subdivide the intervl [, ] in order to pproimte ln so tht R n nd L n give upper nd lower bounds tht dier by no more thn 0.00? Tht dier by no more thn D?. Compute e ln d with error less thn Try, by eperimenttion, to see how mny subdivisions re required or M n nd T n to dier by less thn How mny re required or L n nd R n to dier by less thn 0.00?
13 6. Approimting Sums: L n, R n, T n, nd M n ) Let be the unction grphed below. is incresing nd concve down. b Let A = b ) d. Suppose tht estimtes o ) d re computed using the let, right, nd trpezoid rules, ech with the sme number o subintervls. We ll denote these estimtes by L n, R n, nd T n, respectively. Put the numbers A, L n, R n, nd T n in scending order. Justiy your nswer nd eplin why your nswer is independent o the vlue o n used. b) Answer the sme question s bove i the grph o is the one given below. is decresing nd concve up on [, b]. b 5. Consider 4 d. ) Find vlue o n or which L n b ) d 0.0. b) Use the vlue o n rom prt ) to ind L n nd R n. c) Is the verge o the let- nd right-hnd sums lrger thn the integrl, or smller? d) Compre your numericl pproimtions to the nswer you get using the Fundmentl Theorem o Clculus. 6. Approimte d with error less thn Give upper nd lower bounds or 0 0 d such tht the upper nd lower bounds + 5 dier by less thn Give upper nd lower bounds or ln d such tht the two bounds dier rom one nother by less thn Eplin how you know tht the upper bound is indeed n upper bound nd the lower bound is indeed lower bound. 9. Consider 9 4 d. ) Find vlue o n or which L n R n 0.0. b) Use tht vlue o n to ind L n nd R n.
14 88 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls c) I you tke the verge o the let- nd right-hnd sums, will your pproimtion be lrger thn the integrl, or smller? d) Compre your numericl pproimtions to the nswers you get using the Fundmentl Theorem o Clculus. 0. One o your riends is doing his mthemtics homework on the run. He mnged to squeeze in one problem between lunch nd his epository writing clss. He used his clcultor to ind L n, R n, T n, nd M n to pproimte deinite integrl nd jotted down the results on npkin. They were , 0.885, 0.489, nd He didn t write down the problem, nor did he record the vlue o n he used. But he did sketch the unction over the relevnt intervl) on the npkin; it looked like this. He wnts your help in lbeling the dt he recorded s let-hnd sum, right-hnd sum, midpoint sum, nd trpezoidl sum. Which is which? Eplin your nswer.. Mesurements o the width o pond re tken every 0 yrds long its length. The mesurements re: 0 yrds, 60 yrds, 50 yrds, 70 yrds, 50 yrds, nd 0 yrds. Approimte the surce re o the pond using the Trpezoidl rule Approimte ech o the ollowing integrls with error less thn /00. Note: I you look t ll the questions nd think bout strtegy in dvnce, you will only hve to compute two integrls in order to nswer ll our questions with the desired degree o ccurcy. There is no problem with being more ccurte thn is requested.) Briely eplin wht you hve done nd how mny subdivisions you used. i) 0 e ) d ii) e ) d iii) 0 e ) d iv) e ) d
15 6. Approimting Sums: L n, R n, T n, nd M n 89. Let ) =. Let nd b be positive constnts, 0 < < b. Below re two prtitions + o the intervl [, b]. One given with w s) prtitions [, b] into 8 equl subintervls, ech o length w; the other given with t s) prtitions the intervl into equl pieces ech o length t. w i = + i w, i = 0,,...,8 t i = + i t, i = 0,,..., Put the ollowing epressions in scending order, with < or = signs between them. Identiy the sum with the letter preceding it.) A = 7 w i ) w B = i=0 C = t i ) t D = E = i=0 b 8 w i ) w i= i= t i ) t w) dw F = ) b ) 4. Approimte rctn d using let- nd right-hnd sums to obtin n upper nd lower bound or the integrl with dierence less thn Sve time by grphing y = rctn nd using symmetry to simpliy the problem. 5. The unction ) is decresing nd concve down on the intervl [, 5]. Suppose tht you use right-hnd sum, R 00, let-hnd sum, L 00, trpezoidl sum, T 00, nd midpoint sum, M 00, ll with 00 subdivisions, to estimte 5 ) d. Select ll o the ollowing tht must be true. ) L 00 R 00 b) 5 ) d T 00 c) T 00 R 00 d) T 00 M 00 e) M 00 L 00 ) T 00 = L 00 + R 00 )/ 6. Suppose tht g is dierentible unction whose derivtive is g ) = +. Prtition [0, ] into n equl pieces ech o length nd let k = k, where k = 0,,..., n. Put the ollowing epressions in scending order with < or = signs between them). A = n n g i ) B = g i ) i= n C = lim g i ) n i=0 i=0 D = lim n n g i ) i=
16 80 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls 7. Suppose L 0 < b ) d < R 0 nd R 0 L 0 < 0.0. Which o the ollowing sttements must be true? Circle ll such sttements. ) R 0 b ) d < 0.0 b) c) T 0 b ) d < d) T 0 b ) d < 0.0 L 0 b ) d < Two o the ollowing three integrls you cn evlute ectly. One you cnnot, until lerning integrtion by prts. Identiy the one you cnnot evlute ectly nd pproimte it with n error under Find ect nswers or the other two integrls. ) e ln d b) 0 e d c) 0 e d 6. SIMPSON S RULE AND ERROR ESTIMATES Simpson s Rule We begn our discussion with let- nd right-hnd sums, sums tht provide bounds or b ) d i is monotonic on [, b]. Then we cme up with the clever ide o using the verge o L n nd R n to pproimte deinite integrl. This led us to the trpezoidl sum, which generlly gives much better pproimtion to deinite integrl thn either the let or right sums or the sme number o subdivisions. Encourged by our success, we might strt thinking bout verging the midpoint nd trpezoidl sums. An even better ide would be to tke weighted verge o these two sums. This is becuse the midpoint rule generlly gives us better it thn does the trpezoidl rule. In ct, nlysis o numericl dt indictes tht the error in using the trpezoidl rule is bout two times the size s tht rom using the midpoint rule. From the picture below this seems plusible. The weighting M n+m n +T n = M n+t n gives us method o pproimting deinite integrls known s Simpson s rule. i midpoint i+ = Error using trpezoid = Error using midpoint tngent trpezoid The shded re is bout double the blckened re. Weighted verge: weight midpoint twice s hevily s the trpezoidl sum. M n + T n Figure 6. = Simpson's rule Simpson s rule: S n = M n + T n
17 6. Simpson s Rule nd Error Estimtes 8 We cll it S n becuse we need n dt points or M n nd n + ) dierent dt points or T n. Sometimes it will be convenient to write S p = M n+t n where p = n. Simpson s rule generlly gives substntilly better numericl estimte thn either the midpoint or the trpezoidl sums. To get some pprecition or this, try it out on some integrls tht you cn compute ectly. Below we do this or the integrl 5 / d. EXAMPLE 6. Compre M n, T n, nd Simpson s rule the weighted verge) or 5 d, where n = 4, 8, 50, nd 00. Note tht 5 d = ln 5 = ln 5 ln = ln 5 = SOLUTION ) n = 4: M 4 = T 4 = Simpson s rule = M 4+T 4 = b) n = 8: M 8 = T 8 = Simpson s rule = M 8+T 8 = c) n = 50: M 50 = T 50 = Simpson s rule = M 50+T 50 = d) n = 00: M 00 = T 00 = Simpson s rule = M 00+T 00 = Like the trpezoidl sum, Simpson s rule is convenient tool or pproimting b ) d even when ormul or is not vilble. For instnce, the vlues o ) reerred to in the discussion below could be mesurements tken by surveyor estimting the re o plot o lnd or body o wter. Suppose our im is to pproimte b ) d. We cn collect vlues o ) t eqully spced intervls nd use Simpson s rule to pproimte the deinite integrl. Prtition the intervl [, b] into n equl subintervls, ech o length = b n. We need to use weighted verge o the midpoint nd trpezoidl sums; thereore, on ech subintervl we need not only the vlue o t ech endpoint or T n ) but lso the vlue o in the middle or M n ). Although we re chopping [, b] into n equl subintervls, we will use n + vlues o n + vlues or T n nd n vlues or M n ). Let y k = k ) or k = 0,...,n, where 0,,,..., n re s indicted below. ) k = + k or k = 0.,...,n 0 = 4... n n n or M n or M n or M n = b Figure 6. In the ollowing eercise you will show tht using this lbeling convention, Simpson s rule cn be used to pproimte the deinite integrl using the ollowing ormul.
18 8 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls Simpson s rule: ) S n = 6 [y 0 + 4y + y + 4y + y 4 + +y n + 4y n + y n ], where = b n. EXERCISE 6. In this eercise use the lbeling system described bove: Prtition [, b] into n equl pieces o length = b n. k = + k ) or k = 0,,...,n, where = b n Let y k = k ) or k = 0,...,n, where 0,,,..., n. ) Show tht using this lbeling convention T n = b) Show tht using this lbeling convention ) y 0 + y + y 4 + y 6 + +y n + y n ). M n = y + y + y 5 + y 7 + +y n ). c) Show tht using this lbeling convention Simpson s rule is given by 6) y 0 + 4y + y + 4y + y 4 + +y n + 4y n + y n ). The nswer to prt c) is given t the end o the section. Notice tht becuse Simpson s rule requires both the midpoint nd trpezoidl rules nd the relevnt dt points or ech o these two rules re dierent, Simpson s rule requires pproimtely double the number o dt points necessry or computing either M n or T n individully. REMARKS Simpson s rule cn be presented without reerence to the trpezoidl nd the midpoint sums. Insted, it cn be viewed s ollows. On ech subintervl [ i, i ] pproimte the re under by the re under the prbol pssing through the three points on the grph o corresponding to the endpoints o the intervl nd the midpoint, or i, i )), i, i )), nd i + i, i + i )).
19 6. Simpson s Rule nd Error Estimtes 8 A B C D E Find prbol through A, B, nd C. Find nother prbol through C, D, nd E. Sum the res under the prbols to pproimte the re under the curve. i i i+ Figure 6. When Simpson s rule is pplied to cubics, it will lwys give n ect nswer, even when using M +T. Error Bounds In Section 6. we used pproimting sums to provide upper nd lower bounds or the vlue o n integrl. When we use Simpson s rule we hve good estimte, but without getting bound on the error involved we cnnot be sure how good the estimte is. While we cnnot epect to know the ect error in using pproimting sums, it would be quite useul to get bound on the error. Consider, or instnce, Emple 6.. It would be useul to be ble to pproimte 0 e d using the trpezoidl rule or Simpson s rule, hving bound or the error but not worrying bout the point o inlection. Let s begin by thinking bout the error involved in using let- nd right-hnd sums. y y y The steeper the slope o the lrger the error in pproimting by rectngles. Figure 6.4 As illustrted in Figure 6.4, the lrger is, the lrger the mgnitude o the error involved in pproimting the re under the curve by the re o rectngle. When pproimting the re using the trpezoidl rule or the midpoint tngent trpezoid, the mgnitude o is not n issue; the rte tht is chnging, mesured by ), is ctor controlling the error. For L n, R n, T n, nd M n, the lrger n is, the smller the epected error. Put nother wy, the smller is, the smller the epected error. Using the Men Vlue Theorem, the ollowing error bounds cn be proven.
20 84 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls Let I = b ) d. Let L n, R n, T n, nd M n be let, right, trpezoidl, nd midpoint pproimtions o I, respectively. Then L n I M b ) n R n I M b ) n T n I M b ) n M n I M b ) 4n where M is the mimum vlue o on [, b]. where M is the mimum vlue o on [, b]. You might guess tht the error bound or Simpson s rule involves n n nd the third derivtive o. In ct, Simpson s rule is much more ccurte thn you might epect. It cn be shown tht S n I M 4b ) 5 80n) 4 I p is n even number, then where M 4 is the mimum vlue o 4) on [, b]. S p I M 4b ) 5 80p 4. EXAMPLE 6.4 Find upper bounds or the error involved in using T 0 nd M 0 to pproimte 0 e d. SOLUTION In order to use the error estimtes or T 0 nd M 0 we need to ind n upper bound or on [0, ]. In Emple 6. we computed ). I ) = e, then ) = 4e + 4 ) = ) e. We cn get crude upper bound or on [0, ] by noting tht the numertor is no more thn ) = 4 + 6) = 5 4 = 60. ) 60 = 60e e e on [ 0, ] so ) < 60. NOTE We hven t ound M, but we ound bound or M. T 0 I M b ) n where = 0, b =, n = 0, nd M < 60 T 0 I < M 0 I < = = 0.4 The error using T 0 is gurnteed to be less thn = 0.4 = 0. The error using M 0 is gurnteed to be less thn 0..
21 6. Simpson s Rule nd Error Estimtes 85 EXAMPLE 6.5 How big should n be to esily gurntee tht M n pproimtes 0 e d to within 0.00? SOLUTION From Emple 6.4 we know tht M n I < 480 = 0. We must ind n such tht 4n n 0 < We ll solve 0 = 0.00 nd pick the net higher integer. n n n 0.00) = 0 =>n = 0,000 Then n 4.4. So we choose n = 4. EXAMPLE 6.6 Get n upper bound or the error involved in using S 0 to pproimte ln = d. SOLUTION ) =. Compute,,, nd 4). Conirm tht 4) ) = 4 5. On [, ], 4 4. S p I M 4b ) 5 80p 4 where =, b =, p = 0, nd M 4 = 4. 4 ) < = Answers to Selected Eercises Eercise 6.c) Simpson s rule = M n+t n ) ) [y + y + y 5 + +y n ) + y 0 + y + y 4 + y 6 + +y n + y n ) ] ) = [y + y + y y n )] + ) ) [y 0 + y + y 4 + y 6 + +y n + y n )] ) = [4y + 4y + 4y y n )] 6 + [y 0 + y + y 4 + y 6 + +y n + y n )] 6 = [y 0 + 4y + y + 4y + y 4 + 4y 5 + y 6 + 4y 7 + +y n + 4y n + y n )] 6 ) )
22 86 CHAPTER 6 Numericl Methods o Approimting Deinite Integrls P R O B L E M S F O R S E C T I O N 6.. Approimte e ln d using Simpson s rule with p = 0. S 0 = M 5+T 5. Find n upper bound or the error. At certin point you ll use the ct tht e<.. Approimte the ollowing integrls using the trpezoidl rule, the midpoint rule, nd Simpson s rule or the speciied number o subdivisions nd compute error bounds using the ormuls given in this section. Compre your nswers to the ect nswer. ) 0 d n = 4 b) 0 + d n = 4. Approimte the ollowing integrl using the trpezoidl rule, the midpoint rule, nd Simpson s rule or the speciied number o subdivisions. Look t T n, M n, nd S n. Compute error bounds. ln d ) n = 4 b) n = 8 c) n = 0 4. ) Approimte 0 cos ) d using M 0. Find n upper bound or the error. b) Approimte 0 cos ) d using S 0. S 0 = M 0+T 0. Find n upper bound or the error. 5. A surveyor is mesuring the cross-sectionl re o 0-oot-wide river beneth bridge. He mesures the river s depth every 5 eet. The dt re given below. 5 t 5 t 5 t 5 t 5 t 5 t {}}{ {}}{ {}}{ {}}{ {}}{ {}}{ Depth in eet) ) Use the trpezoidl sum to pproimte the cross-sectionl re o the river. b) Use Simpson s rule to pproimte the cross-sectionl re o the river.
AREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More informationSPECIAL PRODUCTS AND FACTORIZATION
MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationIntroduction to Integration Part 2: The Definite Integral
Mthemtics Lerning Centre Introduction to Integrtion Prt : The Definite Integrl Mr Brnes c 999 Universit of Sdne Contents Introduction. Objectives...... Finding Ares 3 Ares Under Curves 4 3. Wht is the
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More information6.2 Volumes of Revolution: The Disk Method
mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationModule Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials
MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic
More informationCHAPTER 11 Numerical Differentiation and Integration
CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More informationExponential and Logarithmic Functions
Nme Chpter Eponentil nd Logrithmic Functions Section. Eponentil Functions nd Their Grphs Objective: In this lesson ou lerned how to recognize, evlute, nd grph eponentil functions. Importnt Vocbulr Define
More information15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style
The men vlue nd the root-men-squre vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time
More informationIntegration by Substitution
Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is
More information5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.
5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous rel-vlued
More informationwww.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)
www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More informationP.3 Polynomials and Factoring. P.3 an 1. Polynomial STUDY TIP. Example 1 Writing Polynomials in Standard Form. What you should learn
33337_0P03.qp 2/27/06 24 9:3 AM Chpter P Pge 24 Prerequisites P.3 Polynomils nd Fctoring Wht you should lern Polynomils An lgeric epression is collection of vriles nd rel numers. The most common type of
More informationMathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100
hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by
More informationPROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1
PROBLEMS - APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationLecture 3 Gaussian Probability Distribution
Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike
More informationWarm-up for Differential Calculus
Summer Assignment Wrm-up for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments - they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationThe Velocity Factor of an Insulated Two-Wire Transmission Line
The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the
More informationPure C4. Revision Notes
Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd
More informationSection 7-4 Translation of Axes
62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 7-4 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the
More information2005-06 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration
Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 25-6 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting
More informationAlgebra Review. How well do you remember your algebra?
Algebr Review How well do you remember your lgebr? 1 The Order of Opertions Wht do we men when we write + 4? If we multiply we get 6 nd dding 4 gives 10. But, if we dd + 4 = 7 first, then multiply by then
More informationThe Definite Integral
Chpter 4 The Definite Integrl 4. Determining distnce trveled from velocity Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: If we know
More information4.11 Inner Product Spaces
314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces
More informationAppendix D: Completing the Square and the Quadratic Formula. In Appendix A, two special cases of expanding brackets were considered:
Appendi D: Completing the Squre nd the Qudrtic Formul Fctoring qudrtic epressions such s: + 6 + 8 ws one of the topics introduced in Appendi C. Fctoring qudrtic epressions is useful skill tht cn help you
More informationEcon 4721 Money and Banking Problem Set 2 Answer Key
Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in
More informationSection 5-4 Trigonometric Functions
5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationReview Problems for the Final of Math 121, Fall 2014
Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since
More informationDistributions. (corresponding to the cumulative distribution function for the discrete case).
Distributions Recll tht n integrble function f : R [,] such tht R f()d = is clled probbility density function (pdf). The distribution function for the pdf is given by F() = (corresponding to the cumultive
More informationThinking out of the Box... Problem It s a richer problem than we ever imagined
From the Mthemtics Techer, Vol. 95, No. 8, pges 568-574 Wlter Dodge (not pictured) nd Steve Viktor Thinking out of the Bo... Problem It s richer problem thn we ever imgined The bo problem hs been stndrd
More informationExperiment 6: Friction
Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht
More informationApplications to Physics and Engineering
Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics
More informationDIFFERENTIATING UNDER THE INTEGRAL SIGN
DIFFEENTIATING UNDE THE INTEGAL SIGN KEITH CONAD I hd lerned to do integrls by vrious methods shown in book tht my high school physics techer Mr. Bder hd given me. [It] showed how to differentite prmeters
More informationBasic Analysis of Autarky and Free Trade Models
Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently
More information9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes
The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is so-clled becuse when the sclr product of two vectors
More informationBinary Representation of Numbers Autar Kaw
Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy
More informationAll pay auctions with certain and uncertain prizes a comment
CENTER FOR RESEARC IN ECONOMICS AND MANAGEMENT CREAM Publiction No. 1-2015 All py uctions with certin nd uncertin prizes comment Christin Riis All py uctions with certin nd uncertin prizes comment Christin
More informationBayesian Updating with Continuous Priors Class 13, 18.05, Spring 2014 Jeremy Orloff and Jonathan Bloom
Byesin Updting with Continuous Priors Clss 3, 8.05, Spring 04 Jeremy Orloff nd Jonthn Bloom Lerning Gols. Understnd prmeterized fmily of distriutions s representing continuous rnge of hypotheses for the
More informationHow To Understand The Theory Of Inequlities
Ostrowski Type Inequlities nd Applictions in Numericl Integrtion Edited By: Sever S Drgomir nd Themistocles M Rssis SS Drgomir) School nd Communictions nd Informtics, Victori University of Technology,
More informationHelicopter Theme and Variations
Helicopter Theme nd Vritions Or, Some Experimentl Designs Employing Pper Helicopters Some possible explntory vribles re: Who drops the helicopter The length of the rotor bldes The height from which the
More informationTreatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3.
The nlysis of vrince (ANOVA) Although the t-test is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the t-test cn be used to compre the mens of only
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nm-wide region t x
More informationMATH 150 HOMEWORK 4 SOLUTIONS
MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive
More informationg(y(a), y(b)) = o, B a y(a)+b b y(b)=c, Boundary Value Problems Lecture Notes to Accompany
Lecture Notes to Accompny Scientific Computing An Introductory Survey Second Edition by Michel T Heth Boundry Vlue Problems Side conditions prescribing solution or derivtive vlues t specified points required
More informationSmall Business Networking
Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology
More informationSmall Business Networking
Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology
More information4. DC MOTORS. Understand the basic principles of operation of a DC motor. Understand the operation and basic characteristics of simple DC motors.
4. DC MOTORS Almost every mechnicl movement tht we see round us is ccomplished by n electric motor. Electric mchines re mens o converting energy. Motors tke electricl energy nd produce mechnicl energy.
More informationBabylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity
Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University
More informationLECTURE #05. Learning Objective. To describe the geometry in and around a unit cell in terms of directions and planes.
LECTURE #05 Chpter 3: Lttice Positions, Directions nd Plnes Lerning Objective To describe the geometr in nd round unit cell in terms of directions nd plnes. 1 Relevnt Reding for this Lecture... Pges 64-83.
More informationReview guide for the final exam in Math 233
Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in point-direction nd twopoint
More informationDerivatives and Rates of Change
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk Derivtives nd Rtes of Cnge Te Tngent Problem EXAMPLE: Grp te prbol y = x 2 nd te tngent line t te point P(1,1). Solution: We ve: DEFINITION: Te
More information5.6 POSITIVE INTEGRAL EXPONENTS
54 (5 ) Chpter 5 Polynoils nd Eponents 5.6 POSITIVE INTEGRAL EXPONENTS In this section The product rule for positive integrl eponents ws presented in Section 5., nd the quotient rule ws presented in Section
More informationRIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS
RIGHT TRIANGLES AND THE PYTHAGOREAN TRIPLETS Known for over 500 yers is the fct tht the sum of the squres of the legs of right tringle equls the squre of the hypotenuse. Tht is +b c. A simple proof is
More informationModule 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur
Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives
More information3 The Utility Maximization Problem
3 The Utility Mxiiztion Proble We hve now discussed how to describe preferences in ters of utility functions nd how to forulte siple budget sets. The rtionl choice ssuption, tht consuers pick the best
More informationHow To Network A Smll Business
Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology
More informationUnit 6: Exponents and Radicals
Eponents nd Rdicls -: The Rel Numer Sstem Unit : Eponents nd Rdicls Pure Mth 0 Notes Nturl Numers (N): - counting numers. {,,,,, } Whole Numers (W): - counting numers with 0. {0,,,,,, } Integers (I): -
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS
CHAPTER ALGEBRAIC FRACTIONS,AND EQUATIONS AND INEQUALITIES INVOLVING FRACTIONS Although people tody re mking greter use of deciml frctions s they work with clcultors, computers, nd the metric system, common
More information10.6 Applications of Quadratic Equations
10.6 Applictions of Qudrtic Equtions In this section we wnt to look t the pplictions tht qudrtic equtions nd functions hve in the rel world. There re severl stndrd types: problems where the formul is given,
More informationSmall Business Cloud Services
Smll Business Cloud Services Summry. We re thick in the midst of historic se-chnge in computing. Like the emergence of personl computers, grphicl user interfces, nd mobile devices, the cloud is lredy profoundly
More informationChapter 2 The Number System (Integers and Rational Numbers)
Chpter 2 The Number System (Integers nd Rtionl Numbers) In this second chpter, students extend nd formlize their understnding of the number system, including negtive rtionl numbers. Students first develop
More informationSolving BAMO Problems
Solving BAMO Problems Tom Dvis tomrdvis@erthlink.net http://www.geometer.org/mthcircles Februry 20, 2000 Abstrct Strtegies for solving problems in the BAMO contest (the By Are Mthemticl Olympid). Only
More informationUnit 29: Inference for Two-Way Tables
Unit 29: Inference for Two-Wy Tbles Prerequisites Unit 13, Two-Wy Tbles is prerequisite for this unit. In ddition, students need some bckground in significnce tests, which ws introduced in Unit 25. Additionl
More informationThe Fundamental Theorem of Calculus
Section 5.4 Te Funmentl Teorem of Clculus Kiryl Tsiscnk Te Funmentl Teorem of Clculus EXAMPLE: If f is function wose grp is sown below n g() = f(t)t, fin te vlues of g(), g(), g(), g(3), g(4), n g(5).
More informationSmall Business Networking
Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd business. Introducing technology
More informationAAPT UNITED STATES PHYSICS TEAM AIP 2010
2010 F = m Exm 1 AAPT UNITED STATES PHYSICS TEAM AIP 2010 Enti non multiplicnd sunt preter necessittem 2010 F = m Contest 25 QUESTIONS - 75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD
More informationIs Competition Among Charities Bad?
Is Competition Among Chrities Bd? Inkyung Ch nd Willim Neilson Tes A&M University, College Sttion, TX 7783 December Abstrct This pper studies tht the eect o incresed competition mong chrities or dontions,
More informationExam 1 Study Guide. Differentiation and Anti-differentiation Rules from Calculus I
Exm Stuy Guie Mth 2020 - Clculus II, Winter 204 The following is list of importnt concepts from ech section tht will be teste on exm. This is not complete list of the mteril tht you shoul know for the
More informationSmall Business Networking
Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd business. Introducing technology
More informationCOMPONENTS: COMBINED LOADING
LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of
More informationThe Riemann Integral. Chapter 1
Chpter The Riemnn Integrl now of some universities in Englnd where the Lebesgue integrl is tught in the first yer of mthemtics degree insted of the Riemnn integrl, but now of no universities in Englnd
More informationMath 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More informationObject Semantics. 6.170 Lecture 2
Object Semntics 6.170 Lecture 2 The objectives of this lecture re to: to help you become fmilir with the bsic runtime mechnism common to ll object-oriented lnguges (but with prticulr focus on Jv): vribles,
More informationLecture 5. Inner Product
Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right
More informationSolution to Problem Set 1
CSE 5: Introduction to the Theory o Computtion, Winter A. Hevi nd J. Mo Solution to Prolem Set Jnury, Solution to Prolem Set.4 ). L = {w w egin with nd end with }. q q q q, d). L = {w w h length t let
More informationFirm Objectives. The Theory of the Firm II. Cost Minimization Mathematical Approach. First order conditions. Cost Minimization Graphical Approach
Pro. Jy Bhttchry Spring 200 The Theory o the Firm II st lecture we covered: production unctions Tody: Cost minimiztion Firm s supply under cost minimiztion Short vs. long run cost curves Firm Ojectives
More informationAP STATISTICS SUMMER MATH PACKET
AP STATISTICS SUMMER MATH PACKET This pcket is review of Algebr I, Algebr II, nd bsic probbility/counting. The problems re designed to help you review topics tht re importnt to your success in the clss.
More informationAnswer, Key Homework 10 David McIntyre 1
Answer, Key Homework 10 Dvid McIntyre 1 This print-out should hve 22 questions, check tht it is complete. Multiple-choice questions my continue on the next column or pge: find ll choices efore mking your
More informationLectures 8 and 9 1 Rectangular waveguides
1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves
More informationNumeracy across the Curriculum in Key Stages 3 and 4. Helpful advice and suggested resources from the Leicestershire Secondary Mathematics Team
Numercy cross the Curriculum in Key Stges 3 nd 4 Helpful dvice nd suggested resources from the Leicestershire Secondry Mthemtics Tem 1 Contents pge The development of whole school policy 3 A definition
More informationVector differentiation. Chapters 6, 7
Chpter 2 Vectors Courtesy NASA/JPL-Cltech Summry (see exmples in Hw 1, 2, 3) Circ 1900 A.D., J. Willird Gis invented useful comintion of mgnitude nd direction clled vectors nd their higher-dimensionl counterprts
More informationaddition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix.
APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The
More information