TANTON S TAKE ON FITTING EQUATIONS TO DATA CURRICULUM TIDBITS FOR THE MATHEMATICS CLASSROOM MAY 013 Sandard algebra courses have sudens fi linear and eponenial funcions o wo daa poins, and quadraic funcions o hree daa poins. Here are my houghs and approaches o hese maers. THREE DATA POINTS: This secion is aken from a new online course currenly under consrucion. For a sneak peak a EVERYTHING ABOUT QUADRATICS have a look a: www.gdaymah.org. (All is here bar he videos o go wih he e.) A ypical, dry ebook quesion: Find a quadraic funcion ha fis he daa: Tha is, find a quadraic funcion ha passes,10 and hrough he poins (,7 ), ( ) ( 7,3 ). The firs, uerly appropriae response, should be: WHY? Once (if?) sufficien cone and impeus has been provided hen my advice o sudens is: JUST WRITE DOWN THE ANSWER! (Don boher wih ha uninuiive, uneplained salis/quadres ugly calculaor business.) And here is he answer: ( )( 7) ( )( 7) ( 3)( ) ( 3) ( ) ( )( ) + 3 ( ) ( ) y= 7 + 10 Okay, I admi his looks scary. Bu undersanding and making sense of his crazy response is surprisingly sraighforward. Firs, if one were o epand his ou we d see ha i is indeed a quadraic funcion of he form y= a + b+ c. (And since he auhor did no specify in which form she waned he quadraic epression, here is no need o epand i! Why do work wihou a cone or purpose o ha work?) Ne To undersand his meay formula, simply plug in some relevan values. Le s sar wih he firs value lised: =. (Do i! Subsiue i in.) Noice ha he second and hird erms are designed o vanish a = leaving us wih only he firs erm o conend wih: ( )( 7) 7 3 And wih he value = insered, he numeraor and he denominaor mach (he denominaor was designed o do his) giving us: 7 1
which is he value 7 as desired by he able provided. In insering = only he middle erm 10 ( )( 7) 3 ( ) survives which, for =, gives 3 ( ) 10 = 10 as needed. 3 ( ) In he same way, for he value = 7 only he hird erm is non-vanishing and has value 3 = 3 Thus he quadraic EXAMPLE: Find a quadraic ha passes 10,π and hough he poins ( 3,87 ), ( ) ( 3, ). Answer: ( 10)( 3) ( 3)( 3) π ( 7)( 3) ( 7)( 8) ( 3)( 10) y= 87 + + 3 CHECK: Pu in = 3. Do you ge he oupu 87? Also, pu in = 10 and hen = 3. EXAMPLE: Find a quadraic ha fis he daa ( )( 7) ( )( 7) ( 3)( ) 3 ( ) ( )( ) y = 7 + 10 + 3 does indeed produce he values 7, 10 and 3 for he inpus,, and 7! A wee bi of easy simplificaion makes i a ad friendlier o read: 7 3 y = + 1 3 10 ( )( 7) ( )( 7) ( )( ) Answer: ( )( 3) ( 1)( 3) 1 y= a + b + c 1 1 1 1 ( ) a c = + ( )( 3) b( 1)( 3) ( 1)( ) PRACTICE: Find a quadraic ha fis he daa Do a iny bi of simplificaion o your answer. Despie he visual complicaion of he formula one can see ha is consrucion is relaively sraighforward: i. Wrie a series of numeraors ha vanish, in urn, a all bu one of he desired inpus. IF YOU NEED TO SIMPLIFY ALL THE WAY I can be done and acually isn as bad as you migh firs hink. For eample, consider he daa: ii. Creae denominaors ha cancel he numeraors when a specific inpu is enered. iii. Use he desired y-values as coefficiens. COMMENT: I akes bu a momen o ge he hang of his. All my sudens hus far in my career, afer a jus a wee bi of pracice, have masered i wih ease! A quadraic ha fis his daa is: ( )( ) ( )( ) + 1 3 3 + 1 y= 3 + 6 + 10 3 1 3 1 Side Quesion: Is i: A quadraic ha fis he daa is or The quadraic ha fis he daa is?
A iny bi of simplifying gives: 1 y= + + + + ( 1)( 3) ( )( 3) ( )( 1) To handle he fracions, i migh be easies o pu erms over a common denominaor: ( )( ) ( )( ) ( )( ) + 1 3 + 3 + + 1 y = Epanding each produc gives: ( ) ( ) ( ) 3 + + 6 + y = We now see: + 1+ The erms are: = The erms are: = 3 The consan erm is: 6 + 6 10 = 1 Thus he quadraic ha fis he daa is: y = 3 + 1. PRACTICE: Find a quadraic ha goes 3, 1,1 and hrough he poins ( ), ( ) ( 3, ). For fun, simplify your answer all he way! PRACTICE: Somehing ineresing happens if one ries o find a quadraic ha 6,1. fis he poins (,7 ), ( 3,9 ) and ( ) PRACTICE: Why sop a quadraics? Here is some daa and here is a polynomial formula ha fis i: ( 6)( 8)( 9) ( 3)( 8)( 9) ( 3)( )( 6) ( 3)( )( 3) ( 3)( 6)( 9) ( 3)( 6)( 8) 3 + 1 ( )( )( 1) ( 6)( 3)( 1) y= + 8 a) Wrie down a formula ha fis he daa: b) Here is a able of daa ha spells my name. (Do you undersand he connecion?) I wroe a formula ha fis he daa perfecly and hen had a compuer algebra sysem do he simplifying for me. Here s he polynomial ha spells my name: 83 331 3 167 67 p( ) = + + 10 1 1 [So puing in 0 puing in 1 = gives p ( ) = gives p ( 1) 1 3 0 = 10, and =, and so on.] a) Wrie down a quadraic ha seems o fi hese daa poins and simplify your answer. b) Wha happened and why? PRACTICE: Somehing goes wrong if one ries o find a quadraic ha fis he daa 1, 1, 1. ( 1,0 ), ( ) and ( ) a) Try o wrie a quadraic ha fis his daa. b) Wha goes wrong and why? c) Find an equaion of he form = ay + by+ c ha fis his daa! Wha is your personalized polynomial? CHALLENGE: Prove ha your polynomial has he propery ha, despie all he fracions, i is sure o give an ineger oupu for each and every ineger inpu! (Whoa!) PRACTICE: Eplain why 17 is a valid answer o his inelligence es quesion: Wha is he ne number in he sequence? 6 8
TWO DATA POINTS: The sandard curriculum wans sudens o do wo hings wih wo daa poins. Fi a sraigh line hrough hem and/or fi an eponenial funcion o hem. Two Daa Poins: Sraigh Line Fis Le s illusrae wo echniques hrough a specific eample. Find a linear funcion ha fis his daa: Approach 1: Jus follow he previous echnique and wrie down he answer! ( 8) ( 6) y= + ( ) ( ) = +. Tha is, y ( 8) ( 6) Approach : We like o believe ha sraigh lines have he same value for slope no maer how we compue i. We have wo 8,, on he line we poins, ( 6, ) and ( ) wan and we wan a formula ha mus be rue for he coordinaes of any oher general, y on he line. poin ( ) For insance, y 3 = is one possible 6 equaion. y y 8 = is anoher valid* 6 equaion. COMMENT* Well.. These equaions are valid a all locaions where he denominaors involved are no zero. Mos everyone likes o perform a lile algebra on hese equaions o avoid possible worries of y 3 dividing by zero. For eample, = 6 y = 3 6 - can be presened as ( ) ( ) and his holds even for = 6, y= - and y y 8 = as can be given as: 6 y 6 = 6 y 8, which also holds valid a he previous rouble spos. Two Daa Poins: Eponenial Fis An eponenial funcion is a funcion of he form f ( ) a b f 0 = a is he =. (Noe ( ) funcion s iniial value.) Since eponenial funcions arise in models of populaion growh, one migh be ineresed in fiing eponenial curves o daa. EXAMPLE: A scienis grows a yeas culure in a Peri dish. Afer 3 days he mass of he yeas was 8 grams. Afer 7 days, 13 grams. Assuming ha he yeas populaion grows eponenially, find an eponenial funcion ha fis his daa. Answer: We have he able: I can see hree ways o compue he slope: 3 = 8 6 or y 6 or y 8 and hey mus all be equal. Equaing any wo of hem gives an equaion for he line! where is ime (in days) and P( ) is he populaion of yeas measured in erms of weigh (grams). We seek a formula: P( ) = a b ha fis his daa. There are wo approaches.
DIFFICULT APPROACH Pu = 3 and = 7 ino his formula o gain wo equaions: 3 a b = 8 7 a b = 13 Solve for a in one equaion and subsiue ino he second OR divide he wo equaions o obain: 13 b =. 8 This gives: 1 b=. Now subsiue his value of b ino he firs equaion, say, o obain: 3 13 a = 8 giving Thus we have: 3 3 7 3 a= 8 8 13 = 8 13. 7 3 13 P( ) = a b = 8 13. SIMPLER APPROACH: JUST WRITE DOWN THE ANSWER! One has o be savvy, bu one can do i. Firs noe ha if he daa were simpler, hen i would be easier o read off a formula ha fis. For eample, for he able: we see ha he iniial value is 8 ( so a= 8 ) and afer 1 day, he daa has grown by a facor of 13 8 fis.. Thus he funcion 8 CHECK: Pu in = 0 and = 1 o see ha he values 8 and 13 do indeed appear. Now suppose he daa were slowed down by a facor of : Tha is, we are running hough he ime values four imes as slowly: Since ime has changed by a facor of four, he formula fiing he daa changes o: 8 CHECK: Pu in = 0 and = o see ha he values 8 and 13 do indeed appear. Bu our original daa is no only slow by a facor of four, bu also sars a = 3 raher han = 0. This means ha = 3 is behaving like zero for he ime values. We have ha: P( ) = 8 fis he daa perfecly! 3 CHECK: Pu in = 3 ino his formula o see wha I mean by 3 behaves like zero P 3 = 8. for he ime values. See we ge ( ) Also pu = 7 o check ha he value 13 does indeed appear. EXERCISE: Show ha P( ) = 8 agrees wih our previous answer of 7 3 13 P( ) = 8 13. 3
EXAMPLE: Find an eponenial funcion ha fis he daa: 6 Here ime is in weeks, and populaion is measured in couns of a housand. Answer: This funcion has iniial value 30, bu wih = behaving as zero. The daa grows by a facor of 0 =, bu over 30 3 a period of seven -values. (I is seven imes as slow as normal ). We mus have: 7 f ( ) = 30 3 CHECK: Pu in = and = 1 o see ha we are correc! EXAMPLE: Find an eponenial funcion ha fis he decreasing daa: Answer: This funcion has iniial value 18, bu wih = 9 as he new zero. I grows by a facor of, bu over a period 18 of eleven -values. We mus have: 9 11 f ( ) = 18 18 CHECK: Pu in = 9 and = 0 o see we are righ! PRACTICE: A silly preend real-world quesion A scienis is sudying he growh rae of yerbis. She has wo daa poins for a populaion model. (Why boher o have more han wo daa poins?) Having never sudied yerbis before, he scienis is no sure if populaion is growing linearly or eponenially. a) Fi a linear funcion o his daa. Wha does his model predic for he populaion of yerbis in week 10? b) Fi an eponenial funcion o his daa. Wha does his model predic for he populaion of yerbis in week 10? c) The scienis wais unil week 10 and couns.3 housand yerbis. Which model seems mos reasonable? USING LOGARITHMS FOR EXPONENTIAL FITS If one suspecs daa fis an equaion y= a b, hen one would epec log y= log b+ log a o hold, ha is, for and log y o have a linear relaionship wih slope log b and inercep log a. We could work wih and log y direcly and look a linear fi. [Did you ever plo daa on log-graph paper and look for linear paerns?] EXAMPLE: Find an eponenial fi o he daa using a linear fi on logarihmic values. Answer: Le s add of column of log y values.
A linear equaion ha fis and log y is: log y 0.769 1.07 0.769 = Tha is, log y= 0.16+ 0.77 giving: ( ) ( ) 0.77 0.16 y= 10 10 3 1.. [How does his answer o compare o he 16.13 3 answer.88?].88 Quesion : Find an eponenial funcion ha fis his daa. (??) Quesion 3: Consider he daa: 7 CHALLENGE EXERCISE: Find a b funcion of he form y= a ha fis he daa SOME MORE PRACTICE PROBLEMS: All his laer maerial comes from THINKING MATHEMATICS! Vol : Funcions and heir Graphs available a www.lulu.com. Quesion 1: For each of hese wo daa ses: find he equaion of a sraigh line y = m+ b ha passes hrough he wo daa poins and find he equaion of an eponenial funcion y = ab ha passes hrough he wo daa poins. (The second daa se is curious!) a) Wrie down an eponenial funcion ha fis his daa. b) Draw a able of -values and of log y values and find a linear funcion ha fis his new able. Use he linear formula o again find an eponenial funcion ha fis he original daa. Do you have he same funcion as he one you obained in a)? c) If possible, find a funcion of he form b y= a ha fis his daa. d) If possible, find a funcion of he form a y= + b ha fis his daa Quesion : a) Find a quadraic ha passes hrough he 10, and poins (,7 ), (, ) and ( ) simplify your answer. Wha do you noice? Eplain. b) Find a quadraic ha passes hrough he 7,. poins (,6 ), (,8 ) and ( ) Wha do you noice? c) Find a quadraic of he form = ay + by+ c ha passes hrough he poins (,6 ), (,8 ) and ( ) 7,. Skech i!
8 MORE DATA FITTING: There are wo more appearances of daafiing in he sandard curriculum. Each of hese will be discussed in laer curriculum leers. Bu for now, here are some places o learn abou hem. 1. Periodic Daa If you suspec your daa is cyclic - following some smooh up and down paern over a fied repeaing period - hen, ypically, in a pre-calculus course, sudens will be asked o find a rigonomeric funcion ha fis he daa. This work is also discussed in THINKING MATHEMATICS! Vol available a www.lulu.com.. Scaer Plos and Linear Regression Lines of bes fi and correlaion coefficiens measuring he goodness of fi, are discussed in saisics courses. These ideas are developed and eplained in THINKING MATHEMATICS! Vol 8: Beginning Probabiliy and Saisics. The shor video /?p=117 also delivers he basics on his. 013 James Tanon anon.mah@gmail.com