+ au n+1 + bu n = 0.)

Transcription

1 Lecture 6 Recurreces - kth order: u +k + a u +k +... a k u k 0 where a... a k are give costats, u 0... u k are startig coditios. (Simple case: u + au + + bu 0.) How to solve explicitly - first, write characteristic polyomial (eg., T + at + b) the compute roots (λ ad µ, assume λ µ) The there are some costats α, β such that u αλ + βµ, determied with startig coditios. For example, ( ) ( ) ( ) α u0 λ µ β u Why is this the solutio? We ll see that ay sequece of form αλ + βµ satisfies. Sice startig coditios determie the etire sequece, the system of equatios above shows that with these values of α, β we have the uique solutio. so u u. u αλ + βµ satisfied the recursio u 0 u 0, u u u au bu 0 au bu 0 u, etc. So all we eed is to show that αλ + βµ satisfies liear recurrece. u + au + bu 0 If {u } satisfies recurrece, the for ay costat α {αu } also satisfies recurrece, if {v } also satisfies, the {u + v } also does solutios to recurrece are closed uder takig liear combiatios, so it s eough to show that if λ is root of T + at + b 0, the {λ } satisfies the liear recurrece (ad by symmetry µ, ad liear combiatio αλ + βµ also does). Plug i to get λ + aλ + bλ 0 which is by defiitio true A more illumiatio reaso - suppose {u } satisfies recurrece u + au +

2 bu 0. Look at geeratig fuctio with some variable z: U u z u 0 + u z + 0 u z azu au z + au 0 z + auz bz U 0 bu z + 0 bu z Add to get U + azu + bz U u 0 + (u + au 0 )z + } (u + au {{ + bu ) z } U( + az + bz ) u 0 + (u + au 0 )z u 0 + (u + au 0 )z U + az + bz 0 by recurrece defiitio liear(z) ( λz)( µz) α β + (partial fractios) λz µz u z α( + λz + λ z... ) + β( + µz + µ z... ) 0 u (αλ + βµ )z

3 Eg. Fiboacci F F + F, F 0 0, F T T + T T T ±, λ, µ F 0 α + β 0 F αλ + βµ α 5 β 5(( ) ( ) ) F 5 Eg. u 6u u + 6u 3 characteristic polyomial T 3 6T + T 6T (T )(T )(T 3) geeral solutio α + β + γ3 If d order with repeated root λ µ, the geeral solutio u αλ + βλ. Eg. u 3u 3u + u 3 (T ) 3 geeral solutio α + β + γ Above examples are all homogeous. A example of a o-homogeous liear recurrece: u 5 + 6u Idea - first solve homogeous ad fid geeral solutio, the fid particular solutio ad add the two solutios. homogeous: u g α + β3. 3

4 particular: try similar form u a + b + c, plug ito equatio u 5u + 6u a + (4a + b) + (9a 7b + c) 7 5 a, b, c. geeral: u α + β3 + ( ). Eg. Look-ad-say sequece ( Oly,, 3 appear, legths satisfy recurrece of degree 7. I ay such sequece, evetually splits ito sequece of atomic elemets which ever agai iteract with their eighbors. 9 of these cotaiig oly,, 3 Some coectios of recurreces to umber theory. Eg. F + F F () Ca prove by iductio, or explicitly λ µ F, λ, µ, λ µ λ + µ, λµ F + F F λ + µ + ) ( λ µ ) ( λ µ ) λ µ λ µ λ µ ( λ + µ λ + λ µ (λ λµ + µ ) (λ µ) (λµ) () µ λ + µ ( λ µ ) (λ µ) 4

5 I particular, F k+ F k F k + Eg. If prime p > 5, the p F p or p F p+ but ot both. First compute F p (( p p + ) ( ) ) 5 5 F p 5 ( + p 5) ( ( p ) ( ) ) p p Now p ( p ( ( ) ( ) ) 5) p p p p 5 ( p p + ) ( ) (( ) ( ) 5 5 p 5 p 3 p ) p 5 3 So we have (( ) ( ) ( ) ) p p p 5 p3 p Fp p p Wat to uderstad F p deomiator p So F p ±. The p umerator 5 by Fermat s Little Theorem p 5 ± F p F p+ F p F p F p+ 0 Ca t divide both because F p F p+ F p is ot 0. Eg. Sometimes work backwards to show umber theoretic properties. ( + 3) + is divisible by + Note that + 3 ad 3 are roots of T T. Easy to check that a ( + 3) + ( 3) satisfies a + a + a 0 ad is a iteger sequeces. If odd, the ( 3) is egative ad betwee 0, i absolute value, so a ( + 3) for odd. Set a 0, a, ow easy to show that a or a + divisible by + by iductio, a + (a + + a ). 5

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