The second difference is the sequence of differences of the first difference sequence, 2


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1 Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for a sequece that satisfies a equatio ivolvig differeces. These are useful i may applicatios where the idepedet variable is aturally discrete rather tha cotiuous. Differeces Give a sequece a, the first differece sequece of the a is the sequece δ = a+ a. The secod differece is the sequece of differeces of the first differece sequece, δ = δ δ + = ( a a ) ( a a = a+ a+ + a. We ca defie third differeces ad higher i a similar fashio. I should ote that these are all forward differeces (sice the th term of the differece is computed by goig forward i the sequece to the + st term, etc.). You ca also defie backward ad cetered differeces of differet orders. The distictios betwee these are useful whe usig differece equatios to set up umerical approximatios to partial differetial equatios, but we wo t get early that far here so we wo t worry about such poits ay further. Do ote that the differece operator is a liear operator. That meas that is we defie the differece operator Da { } = a + a = δ, the Da { + b} = Da { } + Db { } Dca { } = cda { } which you ca easily check is true. This will be importat i determiig what sort of solutios we should look for i the ext sectio. Differece Equatios A example of a secod order differece equatio is δ + 3δ + a = 0. We ca rewrite this i a more useful form by expadig the differece operators accordig to their defiitio. δ + 6δ + 8 a = ( a+ a+ + a) + 6( a+ a) + 8a = a + ( + 6) a + ( 6+ 8) a + + = a+ + 4a+ + 3a = 0. We ca solve for a + i the last equatio to rewrite the differece equatio i yet aother form, a = 4a 3a. I this form, it is clear that the iitial value problem, + + )
2 a a a a = 0, = α, a = β, 0 has a solutio for all values of α ad β, sice we ca just plug these values ito the recurrece a+ = 4a+ 3a ad the solve for a, a 3, ad so o. I geeral the forms that are writte solely i terms of the a are easier to work with tha the form that actually writes out the differece sequeces, ad so those are the forms we will use from ow o. The form with the differeces explicitly listed does help poit out the similarities betwee differece equatios ad differetial equatios, which ca be helpful i decidig how to approach fidig a solutio. I the ext sectio we will make those coectios explicit by givig some basic defiitios ad the relatig the theoretical ideas about liear equatios from sectio. of the text to the specific case of liear differece equatios. Defiitios ad Theoretical Cosideratios Just as for a differetial equatio, we will defie the order of a differece equatio to be the order of the highest differece sequece i the equatio. Of course, if the equatio is writte i oe of the forms that just uses a terms ad ot the differece sequeces, this defiitio ca be difficult to apply. Fortuately, we ca exted the defiitio to these forms easily eough. The order of a differece equatio ca also be defied as the largest differece betwee the idices of terms i the equatio. So our example of a+ + 4a+ + 3a = 0 is secod order because the highest term is of idex + while the lowest term is of idex. The geeral solutio of a th order equatio is a solutio with arbitrary coefficiets, just as the case for a differetial equatio. This strogly suggests that if we have two iitial coditios specified we should be able to solve for the two arbitrary coefficiets ad idetify a specific solutio, ad as oted above that is the case. A differece equatio is liear if it ca be writte i the form αk( a ) + k + αk ( a ) + k + + α0( a ) = f( ). Such a equatio ca also be writte as L{a } = f with L a liear differece operator, which meas the theoretical cosideratios of sectio. apply. I particular, to solve the homogeeous equatio where f() = 0, we () () ( k ) just eed to fid k liearly idepedet solutios a, a,, a (where the equatio () ( k ) has order k) ad the the geeral solutio is a = ca + + cka with arbitrary costats c,, c k. I the ihomogeeous case, we just eed to fid oe particular ( p) ( p) ( p) ( ) a αk( a ) + k + αk ( a ) + k + ( ) p solutio with + α ( ) 0 a = f ad the the geeral ( p) () ( k) solutio will be a = a + ca + + c a. k CostatCoefficiet Liear Homogeeous Differece Equatios A costatcoefficiet liear homogeous differece equatio is a equatio of the form αka+ k + αk a+ k + + α0a = 0 where all the α j are costats. Just as for differetial equatios, this equatio ca be solved usig algebraic techiques to factor the liear
3 operator. However, it is usually quicker ad easier to use guess ad check, particularly sice it is possible to establish a simple form for the solutio. I the case of a costatcoefficiet liear homogeeous differece equatio, the solutios will usually be of the form a = λ for the appropriate value of λ. Cosider the followig paradigm. Paradigm: Fid the geeral solutio of a + 4a + 3a =. Step : Guess a = λ ad plug this ito the equatio. + + This gives us λ + 4λ + 3λ =0. Step : Solve for λ. We divide by λ to get a quadratic equatio, λ + 4λ+ 3= 0. We ca factor this to get (λ + )( λ + 3) = 0, so the roots are λ = ad λ = 3. Step 3: The geeral solutio is a = cλ + cλ. So the geeral solutio to our paradigm is a = c ( ) + c ( 3). If we have a iitial value problem, we ca plug i the iitial values ad solve for the costats, just as we did for differetial equatios. Of course, you could also use the recurrece form to compute the terms of the sequece, but fidig the geeral solutio ad solvig for the costats gives you a formula for the terms of the sequece which you ca use to compute a directly without havig to compute all the terms leadig up to it. It is also ofte easier to deduce properties of the sequece from the formula. Example: Solve the iitial value problem a+ a+ + a = 0, a 0 = 3. a =. First we fid the geeral solutio. + + Step : λ λ + λ =0 Step : Dividig out by λ we get λ λ+ = 0 which factors as (λ )( λ ) = 0, so the roots are λ = / ad λ =. Step 3: The geeral solutio is a = c (/ ) + c. Secod, we plug i the iitial coditios ad solve for c ad c. We get the two equatios c + c = 3 (/ ) c+ c = We multiply the first equatio by ad subtract the secod equatio from it to get (3/ ) c =, so c = /3. The substitutig this back ito the first equatio we fid c = 7/3. So the solutio to our iitial value problem is ( / 3)(/ ) a = + (7 / 3).
4 Now just as for differetial equatios, the situatio is a little more complicated whe you have a double root. We eed two liearly idepedet solutios to solve a secod order equatio ad with a double root we oly get oe right away. But just as with differetial equatios, there is a quick trick to get a secod liearly idepedet solutio i the case of a double root. If you have a double root at λ = r, the the two liearly idepedet solutios are r ad r. It is possible to derive this from the algebra of differece operators just as we derived a similar result for differetial equatios whose characteristic equatio had a double root, but sice we have a very limited amout of time to sped o differece equatios we are t goig to bother with that i this class you ll just have to trust me o this (of course, you are always welcome to come to my office ad I ll go over the details with you). Example: Fid the geeral solutio of 4a+ 4a+ + a = Step : 4λ 4λ + λ =0. Step : Dividig through by λ we get 4λ 4λ+ = 0, which factors as ( λ ) = 0 so we have a double root at λ = /. c+ c Step 3: The geeral solutio is a = c(/ ) + c(/ ) =. Fially, it is possible that you will get complex roots for λ. Just as i the case of differetial equatios, if you have a complex solutio to a real differece equatio, the the real ad imagiary parts will each be solutios to the differece equatio. But i view of our limited time we wo t deal with this situatio i this class. Fiboacci Numbers As a applicatio of what we ve studied so far, we will deduce a geeral formula for the Fiboaci umbers ad demostrate a couple of properties about them. The Fiboacci sequece goes,,, 3,, 8, 3,, 34,, 89, 44,, where each term is the sum of the two precedig terms (so + =, + = 3, +3=, 3+ = 8, etc.). The sequece is amed after the mathematicia Leoardo Pisao. Leoardo icluded a problem givig rise to this sequece i his 0 work, Liber Abaci (The Book of Numbers). I the 9 th cetury, the umber theorist Edouard Lucas (who mistakely thought Leoardo wet by the ame of Fiboacci) deduced that the Fiboacci sequece had may iterestig properties. A Google search o Fiboacci turs up over 8 millio hits ad there is eve a etire joural, the Fiboacci Quarterly, devoted to articles o ideas arisig from this sequece. The defiitio of the Fiboacci sequece ca be writte as a secod order differece equatio, f+ = f+ + f, with the iitial coditios f 0 = 0, f =. So by solvig this differece equatio, we ca get a formula for the th Fiboacci umber, f. First, fid the geeral solutio. + + Step : Rewritig the equatio as f f f = 0, we get λ λ λ =
5 Step : Dividig by λ we get the quadratic equatio λ quadratic formula to get the roots are ±. Step 3: So the geeral solutio is c + + c. λ = 0. We use the Now we plug i the iitial values to solve for the costats. We get the system of equatios c + c = 0 + c + c = ad solvig this system gives c = ad c =. So our formula for the Fiboacci umbers becomes + f =. Oe strikig fact about this formula is that sice all the Fiboacci umbers are itegers, this messy formula with all the square roots of somehow is always equal to a iteger. Alog these lies, we otice that the secod term, < for all. Sice the f are always itegers, this meas we ca rewrite our formula as + f = roud. This is usually a quicker formula to compute. From our formula we ca also deduce the + limitig ratio of the Fiboacci umbers. To simplify the otatio, let ϕ = ad let ε =. Note that ϕ > while ε <. The + + ϕ ε f+ lim = lim f ϕ ε + ( ( / ) ) ϕ ε ϕ ϕ = lim (dividig top ad bottom by ) ( ε / ϕ) = ϕ
6 The umber ϕ is called the golde ratio ad shows up i a variety of applicatios. Amog other properties, rectagles where the ratio of the sides is ϕ are ofte cosidered the most aesthetically pleasig. The paiter George Seurat deliberately used golde rectagles i may of his paitigs. For more iformatio about the Fiboacci sequece you ca see the millios of web sites, or, if you are old fashioed like me, you might prefer the text Fiboacci ad Lucas Numbers with Applicatios by Koshy. You ca take Math 06, Elemetary Number Theory, to lear more about the Fiboacci sequece ad may other iterestig topics.
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