CURIOUS MATHEMATICS FOR FUN AND JOY

Transcription

1 WHOPPING COOL MATH! CURIOUS MATHEMATICS FOR FUN AND JOY APRIL 1 PROMOTIONAL CORNER: Have you a evet, a workshop, a website, some materials you would like to share with the world? Let me kow! If the work is about deep ad joyous ad real mathematical doig I would be delighted to metio it here. *** Lookig for Serious Mathematics Ifused with Levity? Check out a great summerschool program for math-ethusiastic studets. The time to apply for it is ow! Share this lik with all your high-school scholars. PUZZLER: (A mathy puzzle today!) A polyomial is a fuctio of the form: p( x) = a x + a x + + a x+ a. 1 1 I am thikig of a polyomial all of whose coefficiets are o-egative itegers. (Perhaps 56 ( ) for example, but ot ( ) p x = 9x + x + x+ 5, 1 = or p x x x p x x x 8x 1 ( ) = + +.) I wo t tell you what my polyomial is, but I will tell 1 9 p 1 = 1. you that p ( ) = ad ( ) What s p ( 5) for my polyomial?

2 SOLVING EQUATIONS A Too-Quick Historical Overview QUICK! Solve ax+ b=. I bet you aswered: x= b / a. But who said to solve for x? Actually Frech scholar Reè Descartes ( ) did! He proposed that mathematicias use the letters ear the ed of the alphabet, x, y ad z, to deote quatities that are assumed ukow (ukow ukows) ad letters early i the alphabet, a,b, c, for quatities assumed kow (ukow kows!). Mathematicias thought this a ifty idea ad have bee followig that covetio ever sice. LINEAR EQUATIONS, QUADRATIC EQUATIONS Scholars of very aciet times were able to solve liear equatios. The Egyptia Rhid Papyrus, dated ca.165 B.C.E., for istace, outlies a method of false positio for solvig equatios of the form ax= b. Clay tablets, datig ca. 17 B.C.E., show that Babyloia scholars were solvig liear equatios ad some quadratic equatios. Commet: I should poit out that throughout the ages all equatios ad their solutios were writte out i words. The idea of usig symbols to represet umbers was very log comig i the history of mathematics. Algebra, as we might recogize it today, was t fully developed util the 15s or so! The idea of completig the square to solve quadratic equatios was developed by Greek scholars of 5 B.C.E - via geometry of course. (Look at its ame!) The geeral formula we teach studets today: If ax bx c + + =, the b± b 4ac x=. a was t fully uderstood ad accepted util scholars were comfortable with egative umbers ad zero as a umber (thak you Brahmagupta, ca ), the full use of irratioal umbers as solutios (ca. 16), ad complex umbers as solutios (ca 175), ad, of course, the developmet of algebraic represetatio itself! (We expect a great deal of our ith ad teth graders!) See the followig videos o the mathematics of completig the square (literally!) ad derivig the quadratic formula, ad more For a overview of the history of algebra ad the stories of the people ad the mathematical cultures metioed here see the Facts o File ENCYCLOPEDIA OF MATHEMATICS (by yours truly). Aciet Greek scholars were also iterested i solvig cubic equatios, but struggled with the geometric thikig required to solve them. (They were locked ito the geometric midset.) Fidig a geeral method for solvig cubic equatios became a ifamous problem. CUBIC EQUATIONS, QUARTIC EQUATIONS ad BETRAYAL! James Tato 1 Mathematics was all the rage i 16 th -cetury Italy. Mathematicias were revered ad would hold public demostratios solvig mathematical problems ad challegig their peers with questios. Patros sposored these scholars ad mathematicias thus felt the eed to keep their problem-solvig methods secret. This way they could challege their peers with questios they themselves could solve but their peers could ot. Solvig cubic, quartic, quitic, ad higher-degree equatios become a favored theme. Commet: I was recetly asked: Why do we classify polyomials by their degree? Why the fuss about the highest power of the variable? The aswer is chiefly the history. Our huma story of solvig equatios is locked step-by-step with a icrease i the

3 James Tato 1 degree of the polyomial at had. It seems atural the we came to this classificatio. Girolamo Cardao ( ) was very iterested i geeral formulas for solvig polyomial equatios. Curiously, he ad his assistat, Lodovico Ferrara, discovered a method for solvig quartic equatios (degree 4) that relied o solvig a cubic equatio first - but either could solve the cubic equatio. Frustratig! I 159 Cardao caught wid that oe of his peers, Niccolò Tartaglia, was solvig cubic problems with ease. He visited Tartaglia ad urged him to reveal his clever methods. Surprisigly, Tartaglia coceded! He shared his work, but uder the strict proviso that Cardao ever reveal the methods. (After all, Tartaglia s fiacial support relied o his ability to cotiually impress his patro i public competitios.) The came betrayal Sometime later Cardao received the persoal otebook of recetly deceased Scipioe del Ferro ( ). Del Ferro apparetly dabbled i mathematics, recorded all his fidigs i his otebook, but ever shared his ideas. Del Ferro s so-i-law thought the famous Cardao might ejoy the mathematics i the book ad so gave it to him. Imagie Cardao s utter surprise upo opeig the tome to fid all the same methods ad techiques Tartaglia had devised for solvig cubic equatios spelled out i full glorious detail! After learig that all of Tartaglia s ideas had bee discovered by aother scholar idepedetly some years earlier, Cardao o loger felt obliged to hoor his promise to Tartaglia. He published all the results ad methods- the solutio to the cubic ad his solutio to a quartic - i his 1545 treatise Ars maga. Although Cardao properly credited Tartaglia, del Ferro, ad his assistat Ferrari i the work, Tartaglia was absolutely outraged by this act. His fiacial patroage was jeopardized ad Tartaglia took this as a act of betrayal. A public ad bitter dispute esued. BACK TO MATHEMATICS: SOME WEIRD SOLUTIONS! The cubic formula i Cardao s Ars maga ca lead to some very strage results. For example, i solvig equatio x = 6x 4 the method gives the solutio: x= (We ll show this i a momet.) Cardao would deem solutios like these as meaigless as they iclude the square roots of egative quatities. He would reject them. But Italia mathematicia Rafael Bombelli ( ) suggested ot beig so hasty! After all, every cubic curve must cross the x- axis somewhere ad so every cubic curve has at least oe real solutio. (Icludig the curve y= x 6x+ 4.) This gave Bombelli the audacity to maipulate a imagiary solutio like this to discover, i this case, that x= is actually the umber x= i disguise. Ad x= is a solutio to x = 6x 4! Exercise: Check this! Start by computig ( 1+ i) ad ( 1 i) roots of ± 4 are. to see what the cube Bombelli s observatio led mathematicias to chage their view of complex solutios ad ot dismiss them. Imagiary solutios might be real ad meaigful after all!

4 James Tato 1 THE CUBIC FORMULA: HERE IT IS There is a reaso why Cardao s cubic formula (or ay other versio of it) is ot taught i schools: it is mighty complicated ad there are o cute sogs for memorizig it! (Why would oe wat a cute sog i the first place?) Let me preset the formula here as a series of exercises. Here is the geeral cubic equatio: x Ax Bx C =. (Assume we have divided through by the ay coefficiets attached to the x term.) A STEP 1: Put x= z ito this equatio to show that it becomes a equatio of the form: z = Dz+ E for some ew costats D ad E. Thus, whe solvig cubic equatios, we ca just as well assume that o x term appears. This trick was well kow amog the Italia mathematicias of the 16 th cetury. But Tartaglia/del Ferro wet further with the followig truly-ispired idea. Istead of callig the costats D ad E, call them p ad q. This meas we eed to solve the equatio: z = pz+ q STEP : Show that if s ad t are two umbers that satisfy st = p ad s + t = q, the z= s+ t will be a solutio to the cubic. So our job ow is to fid two umbers s ad t satisfyig st = p ad s + t = q. STEP : Solve for t i the first equatio ad substitute the aswer ito the secod to obtais a quadratic equatio i s. Solve that quadratic equatio. Also write dow, ad solve, the quadratic equatio we would obtai for t if, istead, we solved for s first. THE FORMULA: Show that a solutio to the cubic equatio z = pz+ q is: z = q+ q p + q q p [Be careful about the choices of sigs here. Recall we must have s + t = q.] THAT S IT! That s the cubic formula! [Well we should utagle the meaig of p ad q, ad rewrite the formula i terms of x, A, B ad C. Feel free to do this o your ow!] EXAMPLE: Solve x + 6x 6x =. Aswer: Let s first divide through by the leadig coefficiet of to obtai: x + x x 11= A Step 1 says to put x= z = z 1. This gives: ( z ) ( z ) ( z ) z z z z z z z = = 6z 6= That is, we eed to solve: z = 6z+ 6 Thus p= 6 ad q= 6 givig: Thus: p= ad q= z = + + = 4+ givig the solutio: x= z 1=

5 James Tato 1 EXAMPLE: Solve x = 6x 4. Aswer: This is already of the correct form for step 1. So to solve it we simply use p= 6 ad q= 4 to get: p= ad q=. It yields the solutio: x= , which, as Bombelli would observe, is just x= i disguise! THE QUARTIC FORMULA: What ca I say? It s worse! For the vehemetly ethusiastic algebraists here is a brief outlie of a method for solvig quartics due to Descarte. (It differs slightly from Cardao s method). Dividig through by the leadig coefficiet we may assume we are workig with a quartic equatio of the form: 4 x + Bx + Cx + Dx+ E= B Substitutig x= y simplifies the 4 equatio further to oe without a cubic term: 4 y py qy r = Make the assumptio that this reduced quartic ca be factored as follows, for some appropriate choice of umber λ, m, ad : 4 y + py + qy+ r ( y λ y m)( y λ y ) = Solvig y + py + qy+ r= would the be equivalet to solvig ( y λ y m)( y λ y ) = which reduces to solvig two quadratic equatios (which we kow how to do). y + λ y+ m= λ + = y y So Are there umbers λ, m, ad for 4 which y + py + qy+ r factors as a pair of quadratics: ( y λ y m)( y λ y ) + + +? Expadig brackets ad equatig coefficiets gives the equatios: + m= p+ λ q m= λ m= r If we ca solve λ, m, ad, we ideed have the desired factorig. Summig the first two equatios gives q p+ λ + = λ ; subtractig them yields q p+ λ m= λ ; ad substitutig ito the third equatio yields, after some algebraic work, a cubic equatio solely i terms of λ : ( ) ( )( ) ( λ ) + p λ + p 4r λ q = Cardao s cubic formula ca ow be used to solve for λ, ad hece for λ ad the for m, ad. Now go back ad solve those two quadratic B equatios for y, ad the recall x= y. 4 EASY! (Hmm.) QUINTICS AND BEYOND! Durig the 16s ad 17s, there was great eageress to fid a similar formula for the solutio to the quitic (degree-5 equatio). The great Swiss mathematicia Leohard Euler ( ) attempted to fid such a formula, but failed. He suspected that the task might be impossible

6 James Tato 1 Commet: This would be mighty odd! Why would there be formulas for solvig degree 1 (liear), degree (quadratic), degree (cubic) ad degree 4 (quartic) equatios, but suddely ot for degree 5 equatios? I a series of papers published betwee the years 18 ad 181, Italia mathematicia Paolo Ruffii developed a umber of algebraic results that strogly suggested that there ca be o procedure for solvig a geeral fifth- or higher-degree equatio i a fiite umber of algebraic steps. Ad this surprisig thought was ideed prove correct a few years later by Norwegia mathematicia Niels Herik Abel (18-189). Thus although there is the quadratic formula for solvig degree- equatios, a formula for solvig degree- equatios, ad aother for degree-4 equatios there will ever be geeral formula for solvig all equatios of degree-5 equatios or all degree six equatios, or all degree seve equatios, ad so o! This was a surprisig ad shockig ed to a almost 4 year-log mathematical quest! Commet: Of course some specific degreefive equatios ca be solved algebraically with formulas. (Equatios of the form 5 x a=, for istace, have solutios 5 x= a.) I 181, Frech mathematicia Évariste Galois completely classified those equatios that ca be so solved with algebraic formulas. This work gave rise to a whole ew brach of mathematics today called group theory. RESEARCH CORNER: The opeig puzzler is itriguig! It is very surprisig that with just two pieces of iput-output iformatio oe ca completely determie a polyomial kow to have o-egative iteger coefficiets. For example, from p ( 1) 9 p ( 1) = 1, it must be that 4 p( x) = x + x + x+. = ad REASON: p x = a x + + a x+ a. Write ( ) 1 p( ) = a + a 1+ + a1 + a = 1 9 shows us that each a i is a digit betwee ad 9. p 1 = a 1 + a a 1+ a ( ) 1 = 1 shows us that ( ) p x = x + x + x+ sice there is oly oe way to write 1 i base 1! IN GENERAL: For a polyomial p( x ) with o-egative iteger coefficiets, explai why kowig the values p ( 1) ad p( p ( 1) 1) p( x ). + is eough to determie Research: Suppose I tell you that p( x ) has iteger coefficiets, oe of which is egative ad the rest are o-egative. Could you determie what the polyomial is from a fiite sequece of iput-output questios? How may do you eed? What if I told you that all but two of the iteger coefficiets were o-egative? 1 James Tato tato.math@gmail.com