TITLE Curve Fitting Via The Criterion Of Least Squates . - Ed
DOCUMENT RESUMEEE(218 131SE 038 241.,AUTHORTITLEAlexander, John W., Jr.; Rosenberg, Nancy S.Curve Fitting via the Criterion of Least Squates.Applications of Algebra and Elementary Calculus toCurve Fitting. [and]%Linear Programming in TwoDimensions: I, Applications of High,SchoolAlgebra toOperations Research. Modules and Monographs inUndergraduate Mathematics and Its ApplicationsProject. UMAP Units 321, 453,.'Education Development Center, Inc.; Newton, Mass.,National Science Foundation, Washington, D.C.INSTITUTIONSPONSPAGENCY.,PUB DATE80GRANTNOTESEDr76-19615,7A02,EDRS PRICEMFOIPlus Postage. PC Not Available from EDRS.*Algebra; Answer Keys; *Calculus; *College"Mathematics; Computer Programs; Higher' - Education;Instructional Materials; Learning Modules; LinearPrograming; *Mathematical Applicatioh4;. Matrices;*problem Solving; Secondary-Education; *SecondarySchool Mathematics87/5:DESCRIPTORS.IDENTIFIERS*.Graphing (Mathematics)ABSTRACTThis document' consists ofitwo modules. The first ofthese views applications of algebra and elementary calculus- to, curvefitting. Theute'r is provided with information on how to: 1)gonstrUct,scatter:diagcamsj 2) choose .an appropriate. function to, fitspecific. data; 3) understnd SheShe underlying theory of least squares;4) use a'computei prograaktodesired curve fittimg; and 5) useI augmented matrix approacH to solv6 simultaneous equations;. The secondunit provides techniques to formulate siniple linear .programingprobleins and solve them graphically, Both modules contain exercisesand provide exams. Answers to all problems are supplied. ** *************#********************Reproductions supplied by EDRS athe best that cat. be madefrom the original *********************************,,-14
fzumap%.1CURVE FITTING k IA 111E CRITERION OF LLAS1 SQUARESUNIT 321zbyJohn h. lexander, Jr.Corporate \tuarial Depar tmentConn ecticut Mutual Life friur-ance CompanyU S DEPARWENT OF EDUCATION,Hartford, Connecticut 0 6115e,NATIONALUATITUTE OF EDUCATION'EOUCATiONAL,BESOURCES INFORMATIONCURVE FITTING VIA :1'HE CRITERIONCENTER (ERICAOF LEAST SOUAR ES,./T-hrs document ties been reproduCed asreceived from the person or organizationonTRemgqV14,), jhanges have been made to improveby John W. Alexander, Jr."PERMISSION TO REPRODUCE THISMATERIAL IN MICROFICHE ONLYHAGS BEEN GRANTED BY.reproduction qualdy/aPoints of Grew Or Opinios stated in thsdocu4ment do not necessarily represent offictfl NIETABLE Of CONTENTS TOTHE EDUCATIONAL RESOURCES.position or pokyINFORMATION CENTER (ERICF"1. INTRODUCTION2463.5. SCATTER HE LINE OF REGRESSIONe/COEFFICIENTIOF 'CORREIATIONr.?0REGRESSION FOR 10eARDTHMIC.SCATTERs*2461.0.6.REGRESSION FOR EXPONENTIAL SCATTERS7.POLYNOMIAL SCATTERScs2030.40ir7063.2Temperhture ( 4h,1.4t .1380C)8. MO" EXAM30:1'4APPLICATIONS OF ALGEBRA9.ANSWERS' TO MODEL EXAM31rAND ELEMENTARY CALCULUS 11 CURVE FITTING'10.ANSWERS TO EXERCISESL35isedc; urnape,st rv 11,17tor,, alas 0216()APPENDIXaez'42.e
.Intdinodukr Description Sheet:Title:Authoj:UMAP Unit 321MODULES AND MONOGRAPHS IN UNDERGRADUATECURVE FITTING VIA THE CRITERION OF J.EAST SQUARESMATHEMATICS AND ITS APPLICATIONS. PROJECT (UMAP)John 14,,Alexander, Jr.Corpornte Actuarial DepartmentConnecticut Mutual Life Insurance Company'Hartford, CT 06115The goal of UMAP is to develop, through a community of usersand developers, a system of instructional modules in undergraduate,mathematics and its applications which may be used to supplement .rexisting course, and from which complete courses may eventually .APPL ALG & ELEM CALC/CURVE FITTINGSuggested Support Materials:.The Project is guided by a National Steering Committee ofmathematicians, scientists, and educators.UMAP is funded by agrant from the National Science Foundation to Education DevelopmentCenter Inc, .0 a publicly supportea, nonprofit corporation engaged ineddcatiodal research in'the U.S. and abroad.A computer terminal on line to asystem with BASIC compiler (to beused for the appendix).Prerequisite Skills:1.2.3.4.Be able to do partial differentiation.Be able to ,malmize functions.Know how Co so ve simultaneous equations by elimination orsubstitution for 2 x 2 cases.Know how to .graph elementary, exponential, and logariththic, :Output Skills:' Director:tAssoctate Director/ConsortiumCoordinatorAssociate Director for Administration46ordinator for Materials ProductionAdministrative AssistantStaff AssistanttBarbara Kelczewsk4Paula M. Santillo.2achaiy 4,vitasL'-43.Toundeestand the underlying theory of the methodbf least5.Felicia Delray.Be able to construct scatter diagrams.Be able to choose an appropriate function to fit specific i.data.- 4.Ross L. FinneySolomon Garfunkel.-2.1.PROJECT STAFFfunctions.,-,squares.To be able to ulpe a. computer program to do desired curve.;fitting.Be able to use augmented matrix approach to solve simultaneousequations.4'4-NATIONAL STEERING tOMNITTEgW.Ty Martin.Steven,J. B'ij'amseltLlayron ClarksonErnest,J. HenleyWilliam:HoganDo'nald A. LarsonWilliam F. LucasR. Duncan LuceGeorge MillerFrederick MostellerWalter E. SearsGeorge Springer- Arnold A: StrassenburgAlfred B. Willcox'M.I.T. (Chair).New York UniversityTexas Southern UniversityUniversity of HoustonHarvard UniversitySUNY at BuffaloC9rnell UniversiiyHarvard UniversityNassau Community CollegeHarvard UniversityUniversity of Michigan PressIndiana UniversitySUNY at Stony BrookMathematical Association of America'The Project would.like to thank Thomas R. Knapp and RogerCarlson, member's of the UMAP Statistics Pariel, and LZd H. Minor,Nathan Simms, Jr., and Charles Votaw for their reviews, and allothers who assisted in the production of this unit.This material was prepared with the support of NationalScience Foundation Grant No. SED76-19615 A02. Recommeddati6nsekpressed are those of the author.and do not necessarily reflectthe views of the NSF, nor, of the National Steering Cbmmittee.EDC/ProjecE UMAP. -All rights reserved.
1.TABLE 1CURVE FITTING VIA THE CRITERION,OF LEAST SQUARESTemperature( C)INTRODUCTION.Length(mm)YIn'many instances, we wish to b.ovahle 'to predict theoutAme of certarn phenomena. For example: we mayto 'know which students in a graduating high school Crass44-2461.16 o2461.49tiwill do well in'their first year bf college.'One way to get a measure, or at least an indicatic4'would be to observe the high school grades inEngrish 12463,38.20 or so students who have gone to college.If we matchthe students' English grades with ,their grade.pointaverage after one semester, we would be able. to see ifWhen we draw a scatter diagram,letthng the horizontalaxis he the scale Lithe temperature and the verticalgood grades in English matched with high grade pointaverages.axis the :-cale for the length, we note that the plottedIf the "coxrelatton' is high, then, we might wish toassert that students who do well in high-school, Englishdo well in college.There may be exceptiong of course.ftpoints lie very close to a straight line,.Lt is, therefore,reasonable to make a quick and accurate estimate of thelength, of the rod for an):' temperature between 20.1 nay want to look at other indicators (e.g., mathand,t.-grades)chut., the point.is, 1,e'wish to look at two or2463.5.more statistics on the same individual, and we 'are.ihtereg'ted to know how these statistics relate.2463.0Ideas of the sort alluded to above are.the subjectof this module.2462.9 12462.54'2gSCATTER DIAGRAMS2462.0a:I IMany statistical prblems are concerned with morethan a single characterigtic of an indiVid,a1.0 For2461.5instance, the weight and height.of a number of people'2461,0412030405060could be recorded so Oat an examin'ation of Ihe rel'ation63.2shipjetweeuthe two measurements could be made. As a---further example, consider how the length of a copper rodrelates to its temperatUre.Temperature ( c)Figure 1.I.1'7080
r,7&.10:*tor example, if the temperature was 63.2 thedotted lines i.n Figure 1 indicate that the correspondingpoint on the line gives a length of approximately2462%9 mm.i4ex us explore another example that gives usla scatterdiagram where the-points are more scattered. Table 2gives us the weight in grams, x, and the length of therighthind foot in mtlltmetors, y, of a ,sampl,e of 14 adult23.5-O-23.0-isa6,Li' 22.5-field mice;''411,.TABLE 2-22.0 -L.,0GdWeight (g)Length (mm)xy21 5 1.822.44."21.0I15;1617II1819202122Weight (gm).'Figure 2.1)3.THE LINE OF REGRESSIONss.22.421.521.923.3The criterion traditionally used. to,Oefine a",best" fit dates back to the nineteenth century Frenchmathematician Adr7ien LegendreIt is-called thScriterion, or method, o'f least squaresa This criterionThe point in Figure 2 that is circled indicates wheretwo points of the data coincide% The points here are muchdata to minimize the sum of the squar.es of the veiticaZmore scattered than those of the previous set.deviations (distances) from the points'to theliate.requiries the line of regression which weit to our'It wouldbe.extreTely difficult, to deterpine-which strpIgh,t,line'In other words, the method requires the sum of-the.Pest fits this set of points.' In fact, if a number ofpeople were to attempt to fit a line to these points, theresquares of the distances represented by the sdlid linesegments of Figure 3 to be a.small as possible.is little doubt that each per'S'on would come up with a 6different line. What we need isa mathematiCal methodfor determining the line that comes "closest" to all ofthe points.From the figure, we 'See that the actual gradereceived for a student who studied 11 hours was 79.'Reading from the line of regression we predict a gradeof about 71.*We are hotitn a position to speculate about values outside ofthis range,(-).9,3
yTABLE 3-100-#- x80- 7970-[Difference]260*/-.2130b25 c'3046.b30 c50.51b50 c2028'b20 c,[28 - (20b c)]27048b70 c(48 - (70b ,c) ]b80 c[88 - (80b c)]9[51 - (50h c)12180X, 9175b91 --cP54652b46 c[52 - (46b c.) ]35'35b35 c(352528b25 c8095 88.:.3020105.1015;.[46 - (30b 0)2i"440E02[30 - (25b c)]50a-ro4''--,,.,71CDobxI- cy.2242(91b t c)L(35b c)]428 - (25b c))tb80 t[95.- (80b c)1/2'Z12t2'220Hours StudiedWe add up all,qf these squared di fferences,then be detervried what valueFigure 3.Line of regression fitted Co data on hours studiedand examination gilides.-.kIt 'must,of b and c thjst be used;orden tb have a line.such that the Sum of the verticaldistances from the line to the-date points is at a minimum,iThe Problem:Observe-that any",line can -be expressed:Find the values orb Pd ,c such'that'the sum indicattd below is a minimum.(1)y bx c-zXT,12"4 (30-25b-c) 2 (46-30b-c)2 (51-50b-c)2 (284-20b-c)2Or (48-70b-)2' (88-80b-c)x b'y c' ,(2).The symbol sigma can be emand, square the result,oyed on both sides, of.the equa-t-tpabove (i.e., ED2 iEl(yi-bxj-c)2).' Since,4.1D2 is a function of b and,e we'eda write.We 1 ake the difference in each caseConsider the values of x and y inTable '3.a10.n.If we consider Equation (1), knowing the valuesofgob and c will allow us to e mpare the actual values in they column with bx c.2,c's',are interpreted as the intercept of the axis.e104, (35-35b-c) 2 4 (28-25b-c).2 (95-80h-c)where the b's repreient the slope of the,sine and .the- (75-91b-c)2' (52-46C-e)5 f(h,c) X (y.-b;:-c).i 14To find our desired minimum we find the partialderivatives wi, h respect to band c and set the resultsequal to zero. We obtain twp equations in'two unknownsI."-6-1A ,'
e4which we solve simultaneously.This gives- us 'theThe normal equations for this line are obtained merelydesired values of b and cand thus our line of besi fit(the line of regression).by interchanging x and y in the original two equations,(3) and (4).'Trace through the actual development gi'ven below.nf(b,c) 46'9521161225625640055257634372xxyY1 2(y -bx.-c)(-4 0n1(-2y1.x.1 2bx. 2ck.1 ) .i 1finally,bx.f X-i 1X cx1 11 X x.y.1 1'To continue with the other 75013802550784230477 4562527041225 /8281-52900211626015603360,, facX 2(y.-bx. -c)(-1) -0i 1 As an e;cample, for x b'y c' we have:2 1 (-y. bx c) 11n0(3')b'y.2 c',X y.11 1 X y.x11.1andnnX bx.i 11.nl'c i 1(4')X yi.i 1b'nX yi nc'i 1Thus our two equations' which are traditionally nX x-From Table 4, our equations becdmq:called.normal equations are:b 1 x.(3)i 11nb 1 xi(4)i 12(3)n cx. andi 1i n n.552b llc 576.ii 1Thus,In order to solve these equations, we 'must calculatethe indicated,sums as is dime in Table 4.cn1llc 576-552b*e have alsoncludedthetableofly.21 s because we can use the sumiEly.34372b 552c 34382.1576 - 552b11to find.the line of regression x b'y c'.7Substitutingthe Value of c into (3) we get68
/0415521-34372b 552(5763438211378092b 317952'Y90304704b 37820273388b 60250801.70b 0.8209789andHence, Yc 11.165422.500.8209789x 11.165422.-Similarly,40-(3')35812b' 576c' 3438230(41,)576b' 11c'ilc'Ac' 55220 552 -'576b' 55210576b'eN1,1.0Substituting in '(3') we get358.12b' 576(552393932b' 317952.20Figlire n of y on x. 378202Y62156b' 6025090' 0.969335280and-c' -0:5760972x 0.9693352y - 0.5760972.6060-50-So, we now have the lines of best fit with respectto y and with respeFt to x. ,(Sep Figures 4a anh.-0b.)40We can use either one, depending on our Deeds.' Further4.30than that, having the two lines allows us to calculate whatAis called the coefficient Of correlation.20-014.10.-COEFFICIENT OF CORRELATION',010'20304050L60In order to get 4; mumerical indicatOr of how wellI70i80xII10090,the two sets or'scores compare, we take the geometric meanof the slopes of the two pines of regression (i.e., r ibb').*9Figure 4b.Regression of x on'y.10
.Q1The sign is chosen to. be negative if both slopes arenegative, and positive if bopt slopes are positive.'value, which ranges from'-1 toorcirrelation.ouble of finding t e lines of regiession.*x,yinyili 1bn,1x 12[1 1 1)x1 1the correlation is consideredObserve the distribution of points in the1 11 1n'If high values of one characteristic are associatednegative.alittlealgebra we can w ite:Poor correlation is indicated by a valuewith low values of the otherW ithis called the coefficientIf we have good cor relation the.value ris cl6se to 11near 0.1theThis /1andgraphs of Figure 5.nfx4Y4n1 14-Yii11 1.ni 1 xi]1n2) 2lilYijIExercise 1.:AhGiven the normal equations (3), (4), 067, and (4'), use algebrar F 0r -0.65.to obtain b and b' above.".Since r bb' we can write:42n X x.yrr - 0.4r 0.981 11/.An X x.1 1)(di(V(i 1nY14'1 11n.i lxi)21i 121(i dFigure 5.More simply:Using the data from the example in the previous0.section we have:1 1rr. bb' (0.8209789)(0.9693352)-1 1xilYi l'j(i 1. jy12x2Ini 1 ,/0.7958037"x.y.n,1t i 13i 1Tyi) 121ti 1 0.8921.As a check we substitute the indicated sums in thisnew formula:The value of 'r indicates a reasonably,goodcorrelation.We can determine b and b' directly from the twonormal, equations.12This allows us to calculate r without1174.6
6)2).,6-378202317952. 6025067540(73388)(621561445 0.89'20639 a 0.8921.C4.;This agrees.with the results obtained by employingthe explicit slopes, b and b' of the two lines ofregression. 3 215.REGRESSION FOR LOGXRITJJMIC SCATTERS631291521182427Age in YearsGonsider the graph' in Figure 6 of a,man's growthmeasured every three years after birth.Notice thatthere is a-great deal of growth" -between birth and 15 years.After that time., growth tapers off.Table 5 g *res theFigure 6.4-4data used' in plotting the graph.TABLE 6TABLE 5xAgein .17.xyY-.-0Heightin 21Using the techniques developed in Section 3, we caneasily fit a line to the data:See the carculationsbelow in Table 6.24Using bEx2bEx1.602.903.754.505.005.800.6.18,. 6.55772.596.106.156.17-t Exy cn Ey.13Therefore we can write:141ti
2565b 135c .772.59135b 46.5510c*oc46.55 - 135b10and we can further write,1351[46:552565b 13510772.594.25650b 135(46.5.5)(135)2b '7725.925650b 6284.2518225b 7725.i7425b 1441.651'441.65/7425.b 0.1942ap9c 2.0333121518212427Age in YearsFiiiireAnd we have our 'line of regression:7.Vy' 0.1942x ' 2,0333.In order to draw the line we need only'locate twopoints.For x 0, y 2.0333,y b log'e x cfor ,x 3, y 0.1942(3) 2.0333 2.6159.While this is not a bad fit, we c'an do better:It404turns el that the data will fit a logarithmic'curv.emuch betteran a straight line.In.general, logarithmiccurves lookWe can takethis example).log* of each of the x, values (age-inWe thene the same technique of leastsquakes to find a log dine .!.sest fit.are given below.The calculations-.Notice how mu,osetthis curve is to-the actual data.*Loge is also writtenthe general nature of logarithm'440ione shown in figure 8.are using loge here to emphaii0weWe can arbitrarily -use any base:)''.Figure 8.16
46.55cTABLE 722.6894(1.6529)2(1-ag x.-41.1w-y37.503341:00529. 9 7lagex46J54y 1.6529 logex 680 ---Letx i.then y 1.6529(1.0980) 1,p)52 2:821.x6,then y 1.0529(1.'918) 1.065.2' x 9,x X x,, x124.1198 then y 1.005Z 1.0529(2.19-2) then y 1.0529(2.4g49) 1.0052 then y 1.6529(2.7081) 1.0052 18, then 1:6529(2.8904) 1.0052 21, then y 1.6529(3.0445) 1.0052. 12,5.1125.IS,5.4814.y'45.7827.6.0375.x -.24, than-y 16529(3.1781) 1.0052 6.2583.x 27: then y 1.652,9(3.2958) 1.0052 '0.4528.We usebl(log x) tEiog x E(logex)yee7bElogex Ei."al6 -.Therefore we can write: I-4't5 -E4.3392b 22.6894c 124.119822.6894b1 4t 46.556cC22.6894b46.'SS39x2therefore1,61.3392b 22.6895(46'55-292.68.94h)124.1198-.1.I19(61.3392b) 22.6894(46.55)(22.6984)26 9(124.1198) A .03'6I-9I12115i18f21 '124i27'1.,552.0528b,t 1056es .1915515.2174b 1117.078236.8354b Age in Years60.8867Figure 9.-b 1.6529.17189:3
1000Exetcise 2.Fit a logarithmic-curve to the data given in the table below.--LI64212YAoo167153.90021-176.700., .r,Lo600c5006,REGRESSION FOR EXPONENTIAL SCATTERS4oc4400Consider now, an experiment where a large number of300corn seedlings were grown under fa.vorableconditions.200Eyery two weeks a few plants were weighed, and .theaverage of their weights-wasrecorded.We also give a graph in Figure 10.100(See Table .8.),It would be difficultto find a straight line that would fit very wen. The1026,4logarithmic curve does 410t fit so well either.Figure 10.aAge2468101214f6.AverageWeightin Grams 12858761704227068531820.924966.,Thisiset of data is prbbably best fit to anexponential curve.The-general shape of such curves(y ex) is given in Figure 11, Algebraically y excan be mYitten logey x.For a general exponential wecan write:y cebx With a little algebra, we can get a form that will allowus'to use the least squ'ares method.me t below.11012Age in WeeksTABLE 8in Weeksi8Analyze the develop-Figure 11.19.11'1416.18120
4y ceAnd furtherhxbx4- It g .bx.p4.15406e have: 110(52.959101540b 5825.567.logey / logecbx orlogey bx 1101662.5196121006 6625.196-.logec.3300b b We caLtt finA the "line" of best exponential fit bytaking the loge; of the y values and then proceeding withthe least squftestechnicill.logec 52.9597' - 26.65310799.6290.24232.63067(See Table 9.)logey 0.2423x 2.63067.TABLE 9For x logey.A0.2423(2)' 2.63067 3.11527 e 3.11527 22.5395.1X 4logeyYx2thereforex(logey)f.2.718283.11527. .20 00,144For x .4640256324400logey 5.0.2423(4) 2.6307y 36.5946.For x 6,y. 59.4122for x 8,52.95971540y 96.4573for\x 10 y 156.6008662.5196-.for x 12, y 254.24541.for.x 14, y 412.77391,1hping the equtionfor x 16, y '670.1'489bx logefor x -7 18, y 1068.0038cfor x 20, y 1766.4019.a:.in mind, we ,calculate'We could have solved for c when we obtained-q540b 110 logec 662.5196AS2 959752%9597 - 10 logb '110IfAog,e c 2.63067, then c 13.8831,therefdie'e,c.(r.6y 1.3.8831e 0.2423xItlog c 52.9597 - 110beOf'tiu.-10,,21If we substitute 2 for x we get a' value which is virtuallythe same as we got using the other form. That is, 62227
y 13.88310'2423(2)POLYNOMIAL SCATTERS7. 22.5395.rObserve the fitted curve in Figure 12.Si.A disc was rolled down an inclined plane and thedistance it travelled,was measured after 0, 2,seconds.117004,The results are organized in Table 10.1600-TABLE 101500-14001300-1200-Time (x)024681Q121416Distance (y)0135812172329We give a graphsof the data in Figure 13.' Notice,that it looks as if it could be fitted to an exponential.1100 --.However, this data fits closer to a second degree poly1000--nomial or a parabola, y ax2 bx c.960-800--30700--25600-20500-7. 15 ---F400--m300-- m10200-54100-701I24I1618I.10I- 12514161820Figure 13.,.4.1II151020Time In SecondsAge in WeeksFigure,12.In order, to fit a polynomial we must do a littlemore mathematics. Notice, that we now have three constantsto identify, namely a, b, and c.Exercise 3.Try to fiqthe data for the growth of the corn seedlingsWe mast consider minimizing the sumusing 15.as a 64q,e instead of 10.n23f) 8yi 1bx. c)]2.24
This means that we must calculate the partial derivdtivesTABLEr 1 IgikWe set theseof-this sum .with respect to a, b, and c.derivatives equal to 0 and come up with three equationsThat development isin three unknowns a, b, and c.xxlx.12bo-X.-c]x 1i 12.002441636646n8F(-2)(.2y 2ax.44-26)(.34- 2cx.2) 0.1012'i 11614419625672816142aEx.44-6Ex.3 cEx. 2 Exy.teZ(y11i 1000124641625631212964096100002073684166 5365'30644818051212002448512,,11216100x. y1given below:mna57 1-2(yi-aX12xy4Y.110017 827 440 632432298To solve this system, we can use an augmented matrix.*0(1)i 1 e first divide 'the top 'T.OWihrough by 140352 to obtainaEx2i bEx.1aEx.1 2 bEx. cE(1) 1 in the first row and first column:Ey.1 .nc Ey.10.07390.00580.1163103688167212188167291,Frbm these normal equations we, can obtaima best.'Ex.-3,,Exit,111i''give these1calculations below using the data from Table- 11.- -----a(140352) b(10368).a.(10368)1a(816).Next, we multiply the top row by -10168 And,add it to thesecond row.Then, we multiply the top row by -816 and addit to:the bottom row.The resulting matrix is givenWe Aust.find the indicated sums Ex.1 4parabolic fit.98.-1c(816) b(816) c(72) b(72) c(9) 16324.*For a more detailed discussion on'matrix manipulations seeElementary Differential Equations with Linear Algebra.byRoss. L.Finney and Donald R. Ostberg.98.,.2526
a 0.0739(0.0116) 12.2016041.07364.26723.0992 0.1163a 0.0008572 0.003566 0.1163'a 0.1119.Therefore the parabola of best fit is:continue we divide the second row by 423.0528 to obtain,If in the second row, second column.y 0.1119x21'a.10.07390041.07360.0116x 0.6148.We obtain the y values below:0.00580.14.63x 0,y 0.6148.0.02800.0288x y 0.1119(4) '0.0116(2) 0.61484.26723.0992\ 0.4475 00232 0.6148 1.0855.We now multiply the second row by -41.0736 and add it 631he thiid row.x14,y 1.66 3x 6,y 4.9684.y 7.8692.0.07390:00580.1163-x 010.02800.0288x 10,y 11.9208.00.3.11721.91,63x 12,y -16.806:x '14,y 22.1104.x 16,y 29.4468.1Ve divide the last row by 3.'1172 and obtain c from8,'.stem above:0.07390.0058When this data is graphed on.the original set of0:1163axes, we see that we .have a very close fit.10.02800.0288010.6148(SeeFigure 14.)1Exe cise 4.t the data to an exponential.Therefore,It should be convincing thatthe exponential does got .fit as well as the parbole.c ' \0.6148.Cb J.028(0.6148) 0.0288\ b 0.0172 0.0288There are sets of data-that Produce scatters thatfit higher order polynomials thdn 2.b a 0.0116.'227For example, the.
,regr.a-ii- uses the -.5,ame procedulws ,cadsKui you are spared the------.- ,-- It-shourd also be pointed out lhat an practice-,-.!-.the- ailloUnt -Of data coileCted Wertainore than Illely .be-more'7,.kigill.,,.We have also -kept x"e)Cterrs'ive.numbersumbers reasonably.co,,-- 1WithIter. pograiir-7.t,,o do the work r we can.-.efl-!fti- a. t-a, ro:Aumber ot-data and the nuriihers can be- ,.-77 ---:-or :very-sya 11.-.,.,-.-. '.-;e i-ther- xeiY2 la re,-.-""the'rear-e other functions such----.-----. :Ir.: :.-S ----.1.--powers and powers-,-:-Taised to powers .that can be employed, and data fitted5yto .4hem:"14-i .ey cxr1,1),Aprropri ateetc .):of the. data can be employed to handle these',10-in:Cent .The basi,c mathematics of the least square.method" can still he used.Hopefully, this material hasgiven enough background so that virtually any type ofsc-a-tter can he fitted.Figure 14.MODEL EXAM.8.corn seedling examyle in Section6 might he fit 'With acubic (i.e., y ax 3 bx 2 cx d)I1.Given the data in"the table below construct a scatterdiagram:However, this means 'that we- would have to solvefour equations in four unknowns (a, b, c,This isno small task. There are methods for finding ther, coefficients without going through all the work of 025232725'parttl' derivations, namely, the square' root .metheid andGauApo, method.* Z.Tre is still a great, dear of calculation 0,o dois the coefficientof correlation positive.- negative, or zero?even with these methods.In--fact, all curve fittingrequires a :ood Aeal of dalculation.Now that we havecomputers, weFor the data given in Question 1,Fit alane by eye through the points of the scatter diagramthAt iias constructed for Question 1.an write .programs to deal with any typeFit a linethrough the data using the least square technique.of scatter.We present, as an appendix, a BASIC program called"Super Fit."3.Liven the parabold y 2x2 3, let x take on thevalues 1, 2, 3, 4, 5, 6, and 7. Find the correspondingy values.Which type of function--logarithmic of.exponential will ,best fit the given paraboka?AfteN going through this unit the readershould be comfortable with using the program.The29IfIt i30.
xFit the data in Question 3 to either a' logarithmic4.or exponential curve depending on your choice from,Question 3.4.40'2.2289I.iANSWERS'TO MODEL 25253025301520252332352725253 24b 253c 5183'15 -:-253b ',1015c I/\c223.223 - 253b155;*6724b 253[2 23 - 253b1 - 51830510152025303515406724(15)b 1253(223) 100860b 564192532b 5183(15)- 64009b 7774536851b 21316 -.Positive.2.'b 0.5787.Since c223253b15we have c 223 - 253(0.5787)15thereforec 5.1059.tiThe line.of regression isy 0.5787x 5.1059.'y (5,11,21,35,53,75,1011An exponential would fit best.506'3.tf510t15t20i25I30i351x40 .3132
0e.4.xY---152113logeyx2xl.Igey11.6094'7)'899 943351614.22122519.85156A 4.61514925.9056'32.305723.5100140454x 5y 4.0139ex 6 4.0139(11.7165) 47.0289 4.0139(19.1672) ;76.9352 4.0139(31.3558)(--y 4.0139e0.4922(6)x 7y 4.0139e280.4922(5)0.4922(7)125.8491.107.8221140b 28logec 167.822128b 7log c 23.5100e130x- 23.51 - 28blog c e1207- 28b1140b 28(21'51.110107.82217,I1007(140)b 28(23.51) 7282b 7(107.8221)90980b 658.28- 784b 754.7547:80196b 96.474770b 0.4922.lo g c e23.51 - 28(0.4922)9.7284771.3896xtherefore.%.C 4 4.0139.407I-30-"17bx0.4922x.We have y, cewhich yields y 4.0139e.-* data.,.20--so ifx,x 161y 4.0139eM9 22(1). 4.0139(1.6359)' 16.566410--.-*11(4.Ix 20 .4922(2)0.y 4.0139e.X fitx1II3IiI456I170 4.0139(2.6762) 10.742xy 4.0139e0.4922(3 )A. 4.0139(4.378) 17.573x 40.4 22(4)y 4.0149e '4.0739(7.162) 26.7479-.44.34'%') 9.
,1.To solve for b' we do a similar procedure:ANSWERS TO EnRCISES10.GivenGivennnn(3')b'l cl(4')b' y. nc i 1'i 1. ,1n ,bLx 2 cIx i(3)n64x yi1 1i1 1n1 xab1 xi ncusing (4') w,e can write:i 1i 14nusing *(4) we can write:W1.y.nc 1 x.4nnc 11 1.i 11- b1 xi4ei 1i 1ni 1Cn-IIna,i 1n1 yi.,7 61 x;.1 1c Substitutecthe value of c into (3'):4 1Substitute this value of c lUitcr (3):rin41 y- .) :.n- b1 xii 1b1 x4 1.i'1 x. 1 x.y.i 111 1 1 1 i 1using the di'ltributive lay:Ari 121-in1 yif ntv4-.1/-i 1b'14x. X.x, nl'x.y.i 1 1i 1i 11 1ii 1n1 xi 1 y.ii 1nn.41 1nyi1 11n4xiyi -1 1'.1 11ii 1ni 1i 1yi[ 11 12.nxlogex(log x)'e(logex)yy.i 1n2z xi]i 1i 11 xiyi --nyixi.- .1 xi 1 yil1 yi 1 xi-n4i 1i 1yixinil xi i n1 yixiii.0.-41b'yli 1nnb'l y. 12yil1Further we can write:x12.-4niv iyxii 1nbq y.Mulftplyse equation obtained by n and remove the parenthesesnbi, xy.1 1 1inxi -i 38205510.006825.9166254630.5809,127.08871353LIU
UserbE(10geX)2CLIneX m s/(10geX)yb/lOgeXcn iy.Therefore we can write:25.91666 10.0068c 630.580910.00686 5c 254c -254 - 10.00686525.91666 10.0068:.(254 - 10.0068b5 l-63Q.58095(25.91666) 10.0068
these views applications of algebra and elementary calculus- to, curve. fitting. Theute'r is provided with information on how to: 1) gonstrUct,scatter:diagcamsj 2) choose .an appropriate. function to, fit. specific. data; 3) understnd She underlying theory of least. squares; 4) use a'computei prograakto She desired curve fittimg; and 5) use
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Part 5 - CURVE FITTING Describes techniques to fit curves (curve fitting) to discrete data to obtain intermediate estimates. There are two general approaches for curve fitting: Least Squares regression: Data exhibit a significant degree of scatter. The strategy is to derive a single curve that represents the general trend of the data .
