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# Math Biology I MATH 5110

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This 13 page Class Notes was uploaded by Miss Noel Mertz on Monday October 26, 2015. The Class Notes belongs to MATH 5110 at University of Utah taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/229939/math-5110-university-of-utah in Mathematics (M) at University of Utah.

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Date Created: 10/26/15

Introduction to Difference Equations Berton Earnshaw February 23 2005 1 The Difference Equation Aon ml The Take Home exercises are examples of dz erence equations As you might guess a difference equation is an equation that contains sequence differences We solve a difference equation by nding a sequence that satis es the equation and we call that sequence a solution of the equation The first three Take Home exercises ask for the solutions to difference equa tions of the form Aan nk where k is some natural number As the solutions hinted and you may have found for yourself these equations are very easy to solve if we can express nk in terms of falling factorialsi Recall that a falling factorial is de ned as nl nEWnn7ln72nik2nikli Thus n9l n7nniln2in n nniln72n373n22n Using these relationships we can write 1 n9 n nl n2 n27nnngnl n3n373n22n3n27nnn 3n2nl and if we had the time7 we could write every monomial nk as a linear combination of the falling factorials n97 n7 i i i 7 nil How does this help us nd a solution to Aan nk Simply because the difference equation Abn nE is really easy to solve Remember that Ankil h 1nE when k 2 0 Thus is a solution of Abn n5 So7 to solve Acn n7 for example7 we just remember that n n1 and solve Acn n1 which has 1 2 cu ne as a solution To solve Adn n27 we write n2 n2 nl and solve Adn n2 n which has d 1713 1712 n 7 3 2 as a solution To solve Aen n3 7 we write n3 nE 3n2 nl and solve Aen n 3n2 n1 which has 1 1 en 1n n in as a solution You may be wondering whether or not these are the only solutions to these difference equations In fact7 they are not To nd out what the other solutions are7 we need to a few results Lemma 1 The only solutions to the di erence equation Aan 0 are the constant sequences an c for some number c Proof Aan 0 means an1 7 an 07 or an1 an7 for all n Thus a0 a1 a2an El Theorem 1 Let an and bn be sequences such that Aan Abn Then an bn c for some number c Proof If Aan Ab then by the linearity of A 0 Aan 7 Abn Aan 7 127 So the difference of the sequence an 7 In is zero By Lemma 1 an 7 In is a constant sequence ie an 7 n c for some number or This implies an 12 or This theorem is really important and usefull It tells us that if we know just one solution of Aan 2n we actually know all of the solutions and those solutions are Pn C where pn is some particular solution that we know and c is any constantl Hence all the solutions of A0 n are 1 on n2 c all the solutions of Adn n2 are 1 1 dn n n2 c and all of the solutions of A6 n3 are 1 1 en 1n n n2c where c is any constantl 11 Exercises Find all the solutions for the following difference equations You may leave your solution in terms of falling factorialsl ll Aan n4 2 Abn2n27n4 12 Solutions 1 We begin by noting that n nn7 ln72n73 n4 7 6n3 Jrlln2 7 6n n4n476n3lln276n6n373n22n7n27nn n 6n 7n2nL Thus a solution of Aan n4 is 1 Q 3 i 7 3 1 2 an7gn n n ni Therefore all the solutions are represented by l 3 7 1 an 0 379 in n n2c Where c is any constanti E0 We Write 2n2 7 n4 2n2ni 7 nl4n9 2mg 7 nl 4mg Thus a solution of Abn 2n2 7 n 4 is 2 1 bn gm 7 n2 4n Therefore all the solutions are represented by 2 l bnc n 7 n24nic Where c is any constanti 2 The Difference Equation Aan on We now turn our attention to the last Take Home exercise It asks us to nd a solution of the following difference equation Ann ani We recall from the last lecture that Ac b 01 7 Db for any number I and constant c This is exactly what we want as long as b 7 1 17 that is7 b 2 Thus 02 is a solution to our difference equation7 for any constant c Are the solutions 02 the only solution to Ann an We analyze this question as follows The difference equation Ann an means an1 7 an an or an1 7 Zan 0 This is almost a difference equation Can we somehow manipulate this equation to make it a difference equation We can by dividing through the entire equation by 2n1z an1 7 Qan 0 a 1 a 21 7 27 0 d1V1de by 2n1 an A27 0 Look at that7 a difference equation By Lemma l7 anQ c for some number cl Therefore an 027K This represents all the solutions of Ann ani We state this as a theorem for convenience Theorem 2 The only solutions of the di erence equation Ann an are an 02 where c is a constant 21 Exercises 1 Find all the solutions of the difference equation Aan Aan Where A is some real number What happens to the solutions When A 71 2 Find a solution to the difference equation Abn bn 1 22 Solutions 1 The solutions of the difference equation Aan Aan are 2 a an c1 A When A a 71 here 5 is an arbitrary constant b the zero sequence an 0 When A 71 Proof Aan Aan is equivalent to the equation an1 7 1 Aan 0 If A f 71 then 1 A f 07 and we can divide through our equation by 1 A 17 giving us the difference equation a A7 0 1 A an 7 my By Lemma 1 c for some constant 57 and so an c1 A l If A 71 then we see that 0 an1 7 1 Aan an1 for all n D In 71 for example 3 The Difference Equation Aan on 1 Lets take a look at the difference equation of exercise 212 Ann an 1 Were you able to come up with a solution Our intuition tells us that the solutions of this equation should somehow be related to the solutions of Ann an7 namely 027K The next theorem tells us how they are related Theorem 3 Let pn be any solution of the di erence equation Ann an 1 If In is any other solution then In Pn 62 for some constant 5 Proof If pn and In be are both solutions of Ann an 1 then by the linearity of A AU n pnAbn APnbn1 pn1bn Pn Thus by Theorem 2 In 7 pn 02 for some constant c Adding pn to both sides of this equation gives bn 1 6273 D This theorem is really usefuli It tells us that if we know just one solution of the difference equation Ann an 7 we actually know them a i So how do we come up with a particular solution pn of Ann an 1 The theory of how to do this in general is a little too advanced at this point So what else can we do We could try a sequence and hope we get lucky Letls try a constant sequence pn d for some constant d We know Ad 07 so the difference equation yields 0 d 17 or d 71 Wow What luck The constant sequence pn 71 solves the difference equation By Theorem 37 we know that all of the solutions are of the form anpn02n52nili 31 Exercises 1 Find all the solutions of the difference equation Aan Aan 17 Where A is some real number What happens to the solutions When A 0 or 71 2 Find all the solutions of the difference equation Abn bn 2 32 Solutions 1 The solutions to the difference equation Aan Aan 1 are a an C1A 71 When A f 71 and A a 07 Where c is any constant b an n c When A 07 Where c is any constant c the constant sequence an 1 When A 71 Proof We already saw in Exercise 21111 that the solutions of Adn Adn are a 01 A When A f 717 for any constant 57 and b the zero sequence 0 When A 71 Notice that if A f 07 then 71 is a particular solution of our difference equation By a theorem similar to Theorem 37 51 A 7 l A represents all the solutions of Aan Aan 1 When A f 71 and A f 0 Of course7 When A 07 our difference equation reduces to Aan 17 Which we can solve by the method of falling fractions7 since 1 n91 Thus n 5 represents all solutions in this case When A 717 we see that 1 an1 7 1 Aan an1 for all n D E0 The constant sequence 72 is a particular solution of Abn bn 2 try it Therefore7 by a theorem similar to Theorem 37 02 7 2 represents all the solutions of Abn bn 2 Where c is any constanti 4 The Tower of Hanoi We are now in position to solve an old and interesting mathematical puzzle 7 The Tower of Hanoi The Tower of Hanoi sometimes referred to as the Tower of Brahma or the End of the World Puzzle was invented by the French mathematician Edouard Lucas in 1883 He was inspired by a legend that tells of a Hindu temple where the pyramid puzzle might have been used for the mental discipline of young priestsi Legend says that at the beginning of time the priests in the temple were given a stack of 64 gold disks each one a little smaller than the one beneath it Their assignment was to transfer the 64 disks from one of the three poles to another with one important proviso 7 a large disk could never be placed on top of a smaller one The priests worked very ef ciently day and night When they nished their work the myth said the temple would crumble into dust and the world would vanish 1 The mathematical puzzle is this 7 what is the least number of moves required to move n disks from the rst pole to the last pole according to the rules given in the last paragraph Let tn represent this number It would be great if we could come up with some recurrence relation for this sequence Notice that if we have n 1 disks we cannot move the bottom disk off the rst pole until we have moved all the others off of it onto the other polesi Let7s move the top n disks off to the second polei That takes tn movesi Then we can move the bottom disk to the third pole requiring one move We then move the n disks from the second pole onto the third pole requiring another tn movesi Thus tn tn1tn 2tn1i Subtracting tn from both sides of this equation gives Atn in 1 which is exactly the difference equation we just solvedl Hence tn 02 7 1 for some constant of We know that t1 1 by trying it ourselves duhl and we use this equation to compute c 1t1c21712071 or c 1 Therefore tn 2 7 1 Remember the legend said that the priests in the temple had 64 disks to work with The number of moves required to move these disks from the rst pole to the third pole is therefore 54 18446744073709551615 1httpwwwlawrencchallofscienceorgJavaTowerindexhtml 12 This is a huge number If the priests worked day and night7 making one move every second it would take slightly more than 580 billion years to accomplish the job This is an enormous amount of time7 considering that the generally accepted age of the Earth and the rest of our solar system is about 455 billion years i 2 http WWW V talkori gins V org faqs faqe age of earth html 13

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