SOLVING LINEAR RECURSIONS OVER ALL FIELDS

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1 SOLVING LINEAR RECURSIONS OVER ALL FIELDS KEITH CONRAD 1. Itroductio A sequece {a } = a 0, a 1, a,... i a field K satisfies a liear recursio if there are c 1,..., c d K such that 1.1 a = c 1 a 1 + c a + + c d a d for all d. For example, the Fiboacci sequece {F } = 0, 1, 1,, 3, 5, 8,... is defied by the liear recursio F = F 1 +F with iitial values F 0 = 0 ad F 1 = 1. Ofte F 0 is igored, but the values F 1 = F = 1 ad the recursio force F 0 = 0. We will assume c d 0 ad the say the recursio has order d; this is aalogous to the degree of a polyomial. For istace, the recursio a = a 1 + a has order. The sequeces i K satisfyig a commo recursio 1.1 are a K-vector space uder termwise additio. The iitial terms a 0, a 1,..., a d 1 determie the rest ad if c d 0 the we ca set the iitial d terms arbitrarily 1, so solutios to 1.1 form a d-dimesioal vector space i K. We see a explicit basis for the solutios of 1.1 described by ice formulas. Example 1.1. Solutios to a = a 1 + a i R are a -dimesioal space. A power sequece with 0 satisfies it whe = 1 +, which is equivalet to = +1. That maes a root of x x 1, so = 1± 5. The sequeces 1+ 5 ad 1 5 are ot scalar multiples, so they are a basis: i R if a = a 1 + a for all the a = α β 1 5 for uique α ad β i R. The Fiboacci sequece {F } is the special case where a 0 = 0 ad a 1 = 1. Solvig α + β = 0 ad α β 1 5 = 1 we get α = 1/ 5 ad β = 1/ 5 whe a = F. The solutio with a 0 = 1 ad a 1 = 0 uses α = ad β = If we replace R by ay field K i which x x 1 has roots, eve a field of characteristic p, the computatios above still wor if we replace 1± 5 by the roots i K uless K has characteristic 5: there x x 1 = x + 4x + 4 = x + so 3 is the oly root ad the oly power sequece satisfyig a = a 1 + a is {3 } = 1, 3, 4,, 1, 3, 4,.... Example 1.. Let K have characteristic 5. What ice formula fits a = a 1 + a ad is liearly idepedet of {3 } = 1, 3, 4,...? The sequece a = 3 1 wors sice a 1 + a = = = = 3 1 = a. This starts out as 0, 1, 1,, 3, 0, 3, 3,..., so it is F mod 5 ad is ot a multiple of {3 }. 1 What if cd = 0? If a = a 1 the the first term determies the rest, so the solutio space is 1- dimesioal. Writig the recursio as a = a 1 + 0a does t mae the solutio space -dimesioal uless we isist the recursio is for rather tha for 1. We will ot address this optio. 1

2 KEITH CONRAD How is {3 1 } foud? If { } ad {µ } satisfy the same liear recursio, so does ay liear combiatio. If µ a liear combiatio is µ / µ, ad as µ this becomes { 1 }, which up to scalig is { }. This suggests that sice 3 is a double root of x x 1 i characteristic 5, {3 1 } should be a solutio ad we saw it really is. The formula F 3 1 mod 5 goes bac at least to Catala [3, p. 86] i Our goal is to prove the followig theorem about a basis for solutios to a liear recursio over a geeral field K. Theorem 1.3. Let a = c 1 a 1 + c a + + c d a d be a liear recursio of order d with c i K. Assume 1 c 1 x c x c d x d factors i K[x] over its reciprocal roots the such that 1/ is a root as 1 1 x e1 1 r x er, where the i are distict ad e i 1. A basis for the solutios of the liear recursio i K is give by the e i sequeces { i }, { i }, { i },..., { e i 1 i } for i = 1,..., r. Example 1.4. The simplest case of Theorem 1.3 is whe each reciprocal root has multiplicity 1: if 1 c 1 x c x c d x d = 1 1 x 1 d x has d distict reciprocal roots i the the solutios of 1.1 are uique liear combiatios of the i : a = α α d d. Example 1.5. The recursio a = 8a 1 4a +3a 3 16a 4 has order 4 if K does ot have characteristic. Sice 1 8x + 4x 3x x 4 = 1 x 4, the solutios have basis { }, { }, { }, ad { 3 }, so every solutio is b 0 + b 1 + b + b3 3 for uique b i K. The classical versio of Theorem 1.3 i C, or more geerally characteristic 0, is due to Lagrage ad says a basis of the solutios is the sequeces { i } for < e i ad i = 1,..., r. This wors i characteristic p if all e i p but breas dow if ay e i > p, so it s essetial to use { i } to have a result valid i all fields. While mod p has period p i, mod p has a loger period if p. For istace, i characteristic the sequece mod has period 4 with repeatig values 0, 0, 1, 1 whe rus through the oegative itegers. We will prove Theorem 1.3 i two ways: by geeratig fuctios ad by a aalogy with differetial equatios. Aa Medvedovsy brought this problem i characteristic p to my attetio ad the secod proof is a variatio o hers. After writig this up I foud a result equivalet to Theorem 1.3 i a paper of Fillmore ad Marx [4, Thm. 1, ] ad the case of fiite K i McEliece s Ph.D. thesis [5, p. 19]. The earliest paper I foud metioig the basis { i } i characteristic p is by Egstrom [, p. 15] i 1931 whe max e i = p, but { i } still wors i that case. I 1933, Mile-Thomso [7, p. 388] i characteristic 0 gave the basis { i } ad remared that the alterative { 1 i } is sometimes coveiet.. First Proof: Geeratig Fuctios It is easy to show the sequece { } fits 1.1 if is a reciprocal root of 1 c 1 x c d x d : = c c + + c d d for all 0 d = c 1 d 1 + c d + + c d d c 1 d 1 c d = 0, 1 c 1 c d d = 0. There is a aalogous result i differetial equatios: the solutio space to y t + ay t + byt = 0 has basis {e t, e µt } if ad µ are differet roots of x + ax + b. If x + ax + b has a double root the a basis of the solutio space is {e t, te t }. So e t ad te t. We ll retur to this aalogy i Sectio 3.

3 SOLVING LINEAR RECURSIONS OVER ALL FIELDS 3 It s more difficult to show for a 1 satisfies 1.1 if is a reciprocal root of 1 c 1 x c d x d with multiplicity greater tha. To do this, we will rely o the followig theorem characterizig liearly recursive sequeces i terms of their geeratig fuctios. Theorem.1. If the liear recursio 1.1 has c d 0 the a sequece {a } i K satisfies 1.1 if ad oly if the geeratig fuctio 0 a x is a ratioal fuctio of the form Nx/1 c 1 x c x c d x d where Nx = 0 or deg Nx < d. Proof. Set F x = 0 a x. Usig 1.1, F x = a 0 + a 1 x + + a d 1 x d 1 + d a x = a 0 + a 1 x + + a d 1 x d 1 + c 1 a 1 + c a + + c d a d x d d = a 0 + a 1 x + + a d 1 x d 1 + c i a i x = a 0 + a 1 x + + a d 1 x d 1 + = a 0 + a 1 x + + a d 1 x d 1 + = a 0 + a 1 x + + a d 1 x d 1 + d d c i x i a i x i d d c i x i d i a x d d i 1 c i x F i x a x. The term i the sum at i = d is just c d x d F x; the ier sum from = 0 to = 1 i this case is 0. Brigig d c ix i F x over to the left side, we ca solve for F x as a ratioal fuctio: Nx F x = 1 c 1 x c x c d x d where Nx, if ot idetically 0, is a polyomial of degree at most d 1. Coversely, assume for {a } i K that 0 a x = Nx/1 c 1 x c d x d where Nx = 0 or deg Nx < d. The =0 Nx = a x 1 c 1 x c x c d x d. 0 Equatig the coefficiet of x o both sides for d, 0 = a c 1 a 1 c a c d a d, which is the liear recursio 1.1. Corollary.. I the liear recursio 1.1 suppose c d 0. For K, if 1 x is a factor of 1 c 1 x c d x d with multiplicity e 1 the for 0 e 1 the sequece { } satisfies 1.1.

4 4 KEITH CONRAD Proof. Theorem.1 tells us that our tas is equivalet to showig the geeratig fuctio 0 x ca be writte i the form Nx/1 c 1 x c d x d where Nx = 0 or deg Nx < d. We ll do this with a Nx of degree d 1. I Z[[x]], differetiatig the geometric series formula 0 x = 1/1 x a total of times ad the dividig both sides by! gives us the formal power series idetity.1 0 x = 1 1 x +1. Sice Z has a uique homomorphism to ay commutative rig,.1 is true i K[[x]]. 3 Multiply both sides by x :. x x = 1 x +1. We chaged the sum o the left to ru over 0 istead of, which is oay sice = 0 for 0 1. Replacig x with x i.,.3 0 x = x 1 x +1. Sice e 1, 1 x +1 is a factor of 1 x e, which is a factor of 1 c 1 x c d x d. Set 1 c 1 x c d x d = 1 x e gx. If we multiply the top ad bottom of the right side of.3 by 1 x e +1 gx, which has degree d + 1 because c d 0, we get 0 x = where deg Nx = + d + 1 = d 1 < d. Nx 1 c 1 x c x c d x d We proved i Corollary. that the sequeces { i } for 1 i r ad 0 e i 1 satisfy 1.1 whe c d 0. The umber of these sequeces is 4 r e i = d, which is the dimesio of the solutio space, so to fiish the proof of Theorem 1.3 we will show these d sequeces are liearly idepedet: if b i K satisfy.4 r e i 1 =0 b i i = 0 for all 0 the we wat to show each b i is 0. The sequece { r geeratig fuctio b i i x = b i i x i i 0 0 ei 1 =0 b i i } for 0 has.3 = i b i i x 1 i x +1, 3 We ca t prove.1 i K[[x]] directly for all fields K usig repeated differetiatio, sice i fields of characteristic p the pth ad higher-order derivatives are idetically 0. It could be proved directly i K[[x]] if K has characteristic p by usig Hasse derivatives. 4 Here we require that the liear recursio has order d, or equivaletly that cd 0.

5 SOLVING LINEAR RECURSIONS OVER ALL FIELDS 5 so this double sum is 0, sice it s the geeratig fuctio of the zero sequece. For each i, the ier sum over is.5 b i0 1 i x + b i1 i x 1 i x + b i i x 1 i x b i e i 1 ei 1 i 1 i x e i x e i 1 Puttig these terms over a commo deomiator, the sum is q i x/1 i x e i for a polyomial q i x ad the vaishig geeratig fuctio for.4 becomes.6 q 1 x 1 1 x e + + q r x = 0. 1 er 1 r x By costructio, each q i x is 0 or deg q i x < e i. What ca we say about each q i x? Lemma.3. Let 1,..., r i K be distict such that.6 is satisfied, where e 1,..., e r are positive itegers ad q 1 x,..., q r x are i K[x] with q i x = 0 or deg q i x < e i for all i. The every q i x is 0. Proof. We argue by iductio o r. The case r = 1 is obvious. If r ad the result is true for r 1 the multiply.6 through by the product 1 1 x e1 1 r x er : r q i x1 1 x e1 1 i x e i 1 r x er = 0, where the hat idicates a omitted factor i the ith term, for every i. Each term i this sum is a polyomial, ad all the terms besides the oe for i = r have 1 r x er as a factor. Thus the term at i = r is divisible by 1 r x er. That term is q r x1 1 x e1 1 r 1 x e r 1. Sice 1,..., r 1 are distict from r, 1 r x er must divide q r x. But q r x, if ot 0, has degree less tha e r by hypothesis. Therefore q r x = 0, so the rth term i.6 is 0, which maes every other q i x equal to 0 by iductio. Remar.4. This lemma becomes obvious if a term i.6 is moved to the other side, say q r x/1 r x er. If q r x 0 the the right side blows up at x = 1/ r sice the umerator ca t completely cacel the deomiator because deg q r x < e r, but the left side without the term q r x/1 r x er has a fiite value at x = 1/ r. Thus q r x = 0. Theorem.5. For 1,..., r K ad positive itegers e 1,..., e r, the sequeces { i } for i = 1,..., r ad = 0,..., e i 1 are liearly idepedet over K. Proof. If the sequeces satisfy a K-liear relatio.4 the applyig Lemma.3 to.6 shows each q i x vaishes, so.5 vaishes for each i. Sice.5 is the geeratig fuctio of the sequece with th term e i 1 =0 b i i, we get.7 e i 1 =0 b i i = 0 for each i ad all 0. We passed from a liear relatio.4 ivolvig several i s to a liear relatio.7 that ivolves just a sigle i that is oe of the ier sums i.4. I.7 we ca cacel the commo ozero factor i : ei 1 =0 b i = 0 for all 0..

6 6 KEITH CONRAD Let s write this out as a system of liear equatios at = 0, 1,..., e i 1: b i b i b i = b i ei 1 The matrix is ivertible i K, so b i = 0 for all i ad. Remar.6. Liear recursios over fiite fields are of iterest to codig theorists because of their close relatio to cyclic codes, a special type of liear code. The importat costructios of cyclic codes, lie Reed Solomo ad BCH codes, are related to liear recursios whose characteristic polyomial 5 has distict roots. Cyclic codes where the characteristic polyomial has repeated roots have bee studied [1], ad for a umber of reasos they are ot competitive with the stadard distict root cyclic codes. 3. Iterlude: Aalogy with Differetial Equatios Liear recursios are aalogous to liear differetial equatios, ad our secod proof of Theorem 1.3 will be motivated by this aalogy, which we set up i this sectio. A sequece {a } satisfyig a = c 1 a 1 + c a + + c d a d ca be compared with a fuctio yt satisfyig 3.1 y d t = c 1 y d 1 t + c y d t + + c d yt, which is a dth-order liear ODE with costat coefficiets. The solutio space to such a ODE is d-dimesioal. How similar are solutios to the recursio ad the ODE? Example 3.1. A first-order liear recursio a = ca 1 has geeral solutio a = a 0 c, while a first-order ODE of the form y t = cyt has geeral solutio yt = y0e ct. The geometric progressio c is aalogous to the expoetial fuctio e ct. Example 3.. A secod-order liear recursio a = ba 1 + ca has a geeral solutio that depeds o whether or ot the factorizatio 1 bx cx = 1 x1 µx has distict or repeated reciprocal roots: { α + βµ, if µ, a = α + β, if = µ. A secod-order ODE of the form y t = by t + cyt has a geeral solutio that depeds o whether or ot the factorizatio x bx c = x x µ has distict or repeated roots: { αe t + βe µt, if µ, yt = αe t + βte t, if = µ. Lettig D = d/dt, the differetial equatio 3.1 ca be writte as 3. D d c 1 D d 1 c d yt = 0, so solutios of 3.1 are the ullspace of the differetial operator 3.3 D d c 1 D d 1 c d, 5 This is x d c 1x d 1 c d i our otatio.. 0

7 SOLVING LINEAR RECURSIONS OVER ALL FIELDS 7 which acts o the real vector space of smooth fuctios R R. O sequeces, the aalogue of D is the left-shift operator L: if a = a 0, a 1, a,... the La = a 1, a, a 3,..., or equivaletly L{a } = {a +1 }. This is a liear operator o the K-vector space SeqK of sequeces with coordiates i K. Here is the aalogue of 3. that ca serve as a characterizatio of sequeces satisfyig a liear recursio i place of Theorem.1. Theorem 3.3. A sequece a = {a } i SeqK satisfies the liear recursio 1.1 if ad oly if L d c 1 L d 1 c d Ia = 0, where I is the idetity operator o SeqK ad 0 = 0, 0, 0,.... Proof. For i 0, the sequece L i a has th compoet a +i, so the sequece c i L i a has th compoet c i a +i. The th compoet of L d c 1 L d 1 c d Ia is a +d c 1 a +d 1 c d a, which is 0 for all if ad oly if a satisfies 1.1. Example 3.4. If a sequece a satisfies a = a 1 + a the L L Ia has th compoet a + a +1 a, which is 0 for all, so L L Ia = 0. To solve the differetial equatio 3., factor the polyomial x d c 1 x d 1 c d over C to get a factorizatio of the differetial operator 3.3: x d c 1 x d 1 c d = x i e i = D d c 1 D d 1 c d = D i e i, where 1,..., r are distict ad e i 1. We have to allow i C, so for compatibility let D = d/dt act o the smooth fuctios R C fuctios whose real ad imagiary parts are ordiary smooth fuctios R R. The operators D i e i for i = 1,,..., r commute, so C-valued solutios yt to the differetial equatio D i e i yt = 0 are solutios to 3.1. This is eough to describe all solutios of 3.1: Theorem 3.5. Usig the above otatio, a C-basis of solutios to D i e i yt = 0 is e it, te it,..., t e i 1 e it, ad puttig these together for i = 1,..., r gives a C-basis of solutios to 3.1. We omit a proof of this theorem, which for us serves oly as motivatio. I the simplest case that each e i is 1, so x d c 1 x d 1 c d = d x i, a basis of the solutio space to 3.1 is e 1t,..., e dt, which is aalogous to Example 1.4. Remar 3.6. Sice the i i Theorem 3.5 are i C, the solutio space i the theorem is the complex-valued solutios of 3.1. If the coefficiets c i i 3.1 are all real, the eve if some i i Theorem 3.5 is ot real ad therefore some t e it is ot a real-valued fuctio, it ca be proved that the R-valued solutio space to 3.1 is d-dimesioal over R. 4. Secod Proof: Liear Operators We will apply the ideas from Sectio 3 to the liear operator L d c 1 L d 1 c d I i Theorem 3.3 to reprove Theorem 1.3 usig a argumet of Aa Medvedovsy [6, App. B]. Our first approach to provig Theorem 1.3 ivolved the polyomial 1 c 1 x c d x d ad its reciprocal roots ad their multiplicities. By aalogy with the method of solvig differetial equatios, we will ow use the polyomial x d c 1 x d 1 c d istead. These two polyomials are reciprocal i the sese that x d c 1 x d 1 c d = x d 1 c 1 x c d x d.

8 8 KEITH CONRAD Therefore x d c 1 x d 1 c d = x i e i 1 c 1 x c d x d = 1 i x e i, where i j for i j, so reciprocal roots of 1 c 1 x c x c d x d are ordiary roots of x d c 1 x d 1 c d, with matchig multiplicities. Sice c d 0, o i is 0. Theorem 3.3 tells us that a sequece a SeqK satisfies 1.1 precisely whe a is i the erel of L d c 1 L d 1 c d I. Sice 4.1 x d c 1 x d 1 c d = x i e i = L d c 1 L d 1 c d I = L i I e i ad the operators L i I e i for differet i commute, ay a i the erel of some L i I e i is a solutio of 1.1. Solutios to L i I e i a = 0 belog to the geeralized i -eigespace of L, which is the set of a illed by some positive iteger power of L i I. If e i = 1, such a form the i -eigespace of L: L i Ia = 0 if ad oly if La = i a. The i -eigevectors are the ozero vectors i the i -eigespace, ad the ozero vectors i a geeralized eigespace are called geeralized eigevectors. Our secod proof of Theorem 1.3, lie the first, is established i two steps by provig results lie Corollary. ad Theorem.5. Theorem 4.1. If K is a root of x d c 1 x d 1 c d with multiplicity e 1 the the sequece { } for = 0, 1,..., e 1 satisfies 1.1. Proof. Sice x e is a factor of x d c d 1 x d 1 c d, it suffices by Theorem 3.3 to show the sequece { } for 0 e 1 is illed by L I e to mae it satisfy 1.1. First we treat = 0. Sice L{ } = { +1 } = { }, we get L I{ } = 0. Therefore L I e { } = 0. Let 1. Applyig L I to { }, we get the sequece { } { } { } { } + 1 L I = L = For 1, +1 = 1 +, so By iductio, L I { } = { } L I i = { } { } +i i for 1 i. Thus L I { } = { + } = { }. The sequece { } is a -eigevector of L, ad hece is i the erel of L I, so applyig L I more tha times to the sequece { } ills it. Thus L I e { } = 0 for e >. Secod proof of Theorem.5. Suppose a K-liear combiatio of these sequeces vaishes, say r e i 1 { } 4. b i i = 0. =0

9 SOLVING LINEAR RECURSIONS OVER ALL FIELDS 9 with b i K. We wat to show each b i is 0. For each i, the sequeces { i } for 0 e i 1 are all illed by L i I e i by Theorem 4.1, so the ier sum v i := e i 1 =0 b i{ i } i 4. belogs to the geeralized i -eigespace of L. A stadard theorem i liear algebra says that eigevectors of a liear operator associated to differet eigevalues are liearly idepedet, ad this exteds to geeralized eigevectors of a liear operator associated to differet eigevalues; a proof of that is i the appedix ad serves as a aalogue of Lemma.3. Sice v 1,..., v r belog to geeralized eigespaces associated to differet eigevalues of L, ad v 1 + +v r = 0, each v i must be 0; if ay v i were ot 0 the the vaishig sum over the ozero v i would be a liear depedece relatio amog geeralized eigevectors associated to distict eigevalues. The equatio v i = 0 says e i 1 { } 4.3 b i i = 0. =0 The passage from 4. to 4.3 is a aalogue of the passage from.4 to.7, ad 4.3 for i = 1,..., r is exactly the same as.7, so we ca fiish off this proof i the same way that we did before: equatig the coordiates o both sides of 4.3 for = 0,..., e i 1 ad dividig by i leads to the matrix equatio.8 so all b i are 0. Appedix A. Liear Idepedece of Geeralized Eigevectors Theorem A.1. Let V be a K-vector space, A: V V be a liear operator, ad v 1,..., v r i V be geeralized eigevectors of A associated to distict respective eigevalues 1,..., r. The v 1,..., v r are liearly idepedet over K. Proof. Sice v i is a geeralized eigevector of A associated to the eigevalue i, v i 0 ad A i I e i v i = 0 for some e i 1. Suppose there is a liear relatio b 1 v b r v r = 0 for some b 1,..., b r K. We wat to prove each b i is 0, ad will argue by iductio o r. The result is clear if r = 1, sice v 1 0, so suppose r ad the lemma is proved for r 1 geeralized eigevectors associated to distict eigevalues. The operators A i I e i commute, so applyig the product A 1 I e1 A r 1 e r 1 to the liear relatio ills the first r 1 terms ad leaves us with b r A 1 I e1 A r 1 e r 1 v r = 0. If b r 0 the v r is illed by A 1 I e1 A r 1 e r 1. It is also illed by A r I er. I K[x] the polyomials x 1 e1 x r 1 e r 1 ad x r er are relatively prime sice r 1,..., r 1, so there s a polyomial idetity gxx 1 e1 x r 1 e r 1 + hxx r er = 1 for some gx ad hx i K[x]. Thus gaa 1 I e1 A r 1 I e r 1 + haa r I er = I, ad applyig both sides to v r implies 0 = v r, which is a cotradictio. Thus b r = 0. The liear relatio amog the v i simplifies to b 1 v b r 1 v r 1 = 0, so by iductio all the remaiig b i equal 0.

10 10 KEITH CONRAD Refereces [1] G. Castagoli, J. L. Massey, P. A. Schoeller, ad N. vo Seema, O Repeated-Root Cyclic Codes, IEEE Tras. Iform. Theory , [] H. T. Egstrom, O sequeces defied by liear recurrece relatios, Tras. AMS , Olie at S pdf. [3] E. Catala, Mauel des Cadidats à l École Polytechique, Tome 1, Olie at https: //boos.google.com/boos?id=zwditswmaamc&pg=pr1&lpg=pr1&dq=mauel+des+cadidats+a+l% 7ecole+polytechique&source=bl&ots=S07mvdvW8&sig=hgtPbUzXIQ_pM9i4fupwKQF c&hl=e& sa=x&ei=nujkvfsdceyogstjicoca&ved=0cciq6aewaq#v=oepage&q=mauel%0des%0cadidats%0a% 0l ecole%0polytechique&f=false. [4] J. P. Fillmore ad M. L. Marx, Liear Recursive Sequeces, SIAM Review , [5] R. J. McEliece, Liear Recurrig Sequeces over Fiite Fields, Ph.D. thesis, Caltech, Olie at [6] A. Medvedovsy, Lower bouds o dimesios of mod-p Hece algebras: The ilpotece method, Ph.D. thesis, Bradeis, 015. [7] L. M. Mile-Thomso, The Calculus of Fiite Differeces, MacMilla ad Co., Lodo, Olie at