V. Adamchik 1. Recursions. Victor Adamchik Fall of 2005
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1 V. Adamchik Recursios Victor Adamchik Fall of 005 Pla. Solvig Liear Recurreces with Costat Coefficiets. Iteratios. Characteristic equatio.3 Geeral solutio.4 Higher order equatios.5 Multiple roots Solvig Secod Order Recurreces Iteratio Let us apply the method of iteratio to a secod order lier recurrece equatio. As a example we choose the Fiboacci sequece. Iteratig it oes, we obtai Iteratig it twice a a a a a a 3 a a 3 a a a a 3 a 4 a a a 3 a 4 a a 3 a 4 a 3 a 4 a 5 a 4 a 5 a a a 3 3a 4 3a 5 a 6 We see that iteratio does ot work i this case - the terms i the right-had side do ot cacel ad therefore the patter does ot emerge.
2 V. Adamchik -7: Cocepts of Mathematics Characteristic Equatio I this sectio we will develop a ew method of solvig liear recurreces. As a demo example, we choose the Fiboacci umbers F F F Let us observe that the Fiboacci sequece is growig F F F Therefore, if we formally replace F term by F, we get O the other had Therefore, We itroduce two ew sequeces such that F F F F F F F F F F F b b c c c F b Both equatios ca be easily solved by iteratio b c where ad are some costats. Thus, we proved that F What do we lear from this? We form a educated guess that the Fiboacci equatio has a expoetial solutio F where is to be determied. To fid, we plug this ito the origial equatio to obtai F F F
3 V. Adamchik 3 This leads a polyomial equatio i = The last equatio is called a characteristic equatio. The equatio has two roots Therefore, 5, 5 5 are solutios to the Fiboacci recurrece. Geeral Solutio ad 5 Let ad be two solutios. The their liear combiatio a c c where c ad c are arbitrary costats is also a solutio. Such solutio is called a geeral solutio. We prove that a geeral solutio is a solutio by substitutio a ito the Fiboacci equatio We obtai a a a 0 c c c c c c Collect terms by c ad c, yeilds c c Each term is zero sice ad are roots of the characteristic equatio. For the Fiboacci sequece the geeral solutio is a c 5 c 5
4 V. Adamchik -7: Cocepts of Mathematics How do you fid c ad c? We fid them from the iitail coditios. For the Fiboacci sequece they are This leads to a system of two equatios or The system ca be easily solved. We get Hece, I terms of the golde ratio a 0 0, a a 0 c 0 c 0 0 a c c c c 0 c 5 c 5 c 5, c 5 F 5 5 F Based o a closed form solutio, it's easy to prove lim Example. We solve the followig recurrece F F a 3 a 4a a 0 0, a Assume the solutio i the form a ad substitute it to the equatio to obtai The characteristic equatio
5 V. Adamchik 5 has two roots The geeral solutio is 3 40 ad 4 a c c 4 We fid c ad c from iitial coditios. This gives us the followig system Its solutio a 0 c c 0 a c 4c c 5, c 5 Fially, the solutio to the origial recurrece is Higher order equatios a The method of the charcteristic equatio works for ay order liear recurrece equatio with costat coefficiets. The latter meas that coefficiets by all a k terms i the equatio are free of. The followig are equatios with costat coefficiets a a a a a a 3 These are ot equatios with costat coefficiets Let us cosider the followig equatio The characteristic equatio has three roots a a a a a a a a a 3 a 0, a 3, a
6 V. Adamchik -7: Cocepts of Mathematics The geeral solutio,, 3 a c c c 3 We fid ukow coefficiets from the system Therefore, the solutio is Multiple roots Cosider the followig example. The characteristic equatio has two idetical roots The first solutio is a 0 c c c 3 0 a 0 c c c 3 3 a c c 4 c 3 9 a 3 a a a a 0, a 0 But what is the secod solutio? Geerally, what would be the solutio of the recurrece if all (or few) roots of the characteristic equatio are the same? To get a otio of the secod solutio we cosider a ew equatio b b b b 0, b It's easy to see that if 0 the the sequece b approaches a. The characteristic equatio for b sequece is It has two roots 0
7 V. Adamchik 7 Therefore, a geeral solutio is, b c c where c ad c are arbitrary. Let us choose them i the special form Now, cosider a geeral solutio whe 0 c, c... lim lim 0 0 This is the secod solutio for a sequece. The the geeral solutio is Takig ito accout iitial coditios we fid Hece, a c c a 0, a c, c a
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