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n-1 ( fei+29 + fei+1k Sei-1p+ foi-ah) foi+19 + foik fesg + fei-1k ) Tau-19+ fei-2k ) hII fep+ fei-1h %3D n-1 Joi+ap + foi+2h ( fei+2P + fei+zħ\ ( feis49+ fei+ak) fei+1P+ foih ) feie+39 + fes+2k ) ( foi+6p + foi+sh\ ( foi+29 + foi+ik + foirsh, n-1 Iên-2 = r n-1 PII fei+19 + foik n-1 (foi+4P+ foi+sh) ( Soreg + foirsk Soi+3p+ fei+2h) fei+59 + fei+ak) n-1 fei-sP+ feiezh) ( fei+s9 + foi+sk Jei+7P+ feisch) (Ffei+39 + fei+2k, 2p +h II Tên+1 = r P+h where z-4 = h, z-3 = k, r-2 = r, 1-1 = p, ro = q, {fm}- = {1,0,1,1,2,3, 5,8, ..}. Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n – 2. That is; kT foisap + fai-ah( Jaug + fei-ik Soi+3p+ fei+zh, n-2 Sei-19+ foi-ak) n-2 fei+2p + fei+zh) (feir49 + fesuak foi+1p + foih ) foi+39 + fei+zk ( fei+6P + fei+sh\ Sei+sp+ fei+ch i-0 n-2 fei+29+ fei+1k) fai+19 + Sosk n-2 ( fei+aP+ fei+sh ( feso9 + foi+sk = 9T Soi+ap+ fei+2h) Iên-6 Seins9 + fes+ek ) fei+sp+ fei+zh ( fei+sg + foi+sk fei+7p + Sos+sh n-2 2p+. Tên-5 = P+h Soi+39 + fes+2k,

n-1 ( fei+29 + fei+1k Sei-1p+ foi-ah) foi+19 + foik fesg + fei-1k ) Tau-19+ fei-2k ) hII fep+ fei-1h %3D n-1 Joi+ap + foi+2h ( fei+2P + fei+zħ\ ( feis49+ fei+ak) fei+1P+ foih ) feie+39 + fes+2k ) ( foi+6p + foi+sh\ ( foi+29 + foi+ik + foirsh, n-1 Iên-2 = r n-1 PII fei+19 + foik n-1 (foi+4P+ foi+sh) ( Soreg + foirsk Soi+3p+ fei+2h) fei+59 + fei+ak) n-1 fei-sP+ feiezh) ( fei+s9 + foi+sk Jei+7P+ feisch) (Ffei+39 + fei+2k, 2p +h II Tên+1 = r P+h where z-4 = h, z-3 = k, r-2 = r, 1-1 = p, ro = q, {fm}- = {1,0,1,1,2,3, 5,8, ..}. Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n – 2. That is; kT foisap + fai-ah( Jaug + fei-ik Soi+3p+ fei+zh, n-2 Sei-19+ foi-ak) n-2 fei+2p + fei+zh) (feir49 + fesuak foi+1p + foih ) foi+39 + fei+zk ( fei+6P + fei+sh\ Sei+sp+ fei+ch i-0 n-2 fei+29+ fei+1k) fai+19 + Sosk n-2 ( fei+aP+ fei+sh ( feso9 + foi+sk = 9T Soi+ap+ fei+2h) Iên-6 Seins9 + fes+ek ) fei+sp+ fei+zh ( fei+sg + foi+sk fei+7p + Sos+sh n-2 2p+. Tên-5 = P+h Soi+39 + fes+2k,

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 3E
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Br,-10n-2 YIn-1 + 8xn-4' In+1 = QI,n-2 + n = 0, 1,..., (1) 1 The following special case of Eq.(1) has been studied In-1Tn-2 In+1 = In-2 + (8) In-1 + In-4 where the initial conditions r-4, x-3, T-2, r-1,and ro are arbitrary non zero real numbers. Theorem 4. Let {rn}-4 be a solution of Eq.(8). Then for n = 0, 1,2, ... Ton-4 = hT foip + foi-1h foi-1p + foi-ah) ( fei+29 + Soi+2k foi+19 + ferk n-1 n-1 ( foisap + foi+ah foi+3P+ fei+2h) ( fei-19 + fei-2k) fe:9 + fei-1k I'ên-3 = kI i=0 T( foi-2p + foi+h) ( feir49 + fei+zk fei+39 + foi+2k) fei+29 + fei+1k). fei+19 + fesk Ten-2 = foi+1P + foih foi+ep + fei+sh Soi+sp + fei+sh , ( foi+ap + foi+ah ( foi+69 + foi+sk). Sei+ap + fei+2h) fei+59 + fei+ak ) (2p + h (fei+sP+ foi+zh (feit49 + fei+3k n-1 p+h) th II(a-p+farch) (Fars9 + Sa+zk ) where r-4 = h, x-3 = k, x-2 = r, -1 = P, xo = q, {fm}- = {1,0,1, 1,2, 3, 5, 8, ...}. Proof: For n = 0 the result holds. Now suppose that n >0 and that our assumption holds for n – 2. That is; kT (foisap + fei-3ah ( feg + fei-ik ) Soi+3p + foi+2h) (Toi-19 + foi-2k ) fei+2P + fei+1h ( fei+1p + feih ) \Foi+39 + fois+2k) fei+ep+ fei+sh ( fei+29 + foi+ik Sei+sp + fei+gh Ten-9 = i-0 Sei+49 + fei+3k n-2 foi+19 + Sauk ) (foisap + fo+3gh ( fei+69 + foi-sk) foi+ap + fei+zh ) Jei459 + foitak) I6n-6 = (2p +4) T(foi-ap + fes-zh) (foisaq + foi+sk Jei+7p+ fei+ch) Joi+39 + fei+2k ) I6n-5 = Now, it follows from Eq.(8) that Z6n-4 = T6n-7 + T6n-6 + 2ôn-9 11 foi+6P + fei+sh\ ( Soi+29 + fei+ik ( foi+6P + fei+sh PIIasp+ feirah ) foi+29 + foi+ik n-2 n-2 foi+6P + foi+sh ( foi+29 + fei+ik PIISP+ feigh) Tei+19 + feik ) n-2 paII (st fa) (2 fuk) n-2 Seseg+ fersk Joi-19+fer-ak im0 pa II ( (atl) = PTT (feisap + feirsh) (feis29 + feisak fei+sp + fei+sh, im0 n-2 Josreg+Sesk) (Tei-19+Se-ak i=0 -i )- ( foi+oP+ fsi+sh` Soi+sp + fei+ah) foi+29 + foi+1k fei+19 + feik =PII 12
Transcribed Image Text:Br,-10n-2 YIn-1 + 8xn-4' In+1 = QI,n-2 + n = 0, 1,..., (1) 1 The following special case of Eq.(1) has been studied In-1Tn-2 In+1 = In-2 + (8) In-1 + In-4 where the initial conditions r-4, x-3, T-2, r-1,and ro are arbitrary non zero real numbers. Theorem 4. Let {rn}-4 be a solution of Eq.(8). Then for n = 0, 1,2, ... Ton-4 = hT foip + foi-1h foi-1p + foi-ah) ( fei+29 + Soi+2k foi+19 + ferk n-1 n-1 ( foisap + foi+ah foi+3P+ fei+2h) ( fei-19 + fei-2k) fe:9 + fei-1k I'ên-3 = kI i=0 T( foi-2p + foi+h) ( feir49 + fei+zk fei+39 + foi+2k) fei+29 + fei+1k). fei+19 + fesk Ten-2 = foi+1P + foih foi+ep + fei+sh Soi+sp + fei+sh , ( foi+ap + foi+ah ( foi+69 + foi+sk). Sei+ap + fei+2h) fei+59 + fei+ak ) (2p + h (fei+sP+ foi+zh (feit49 + fei+3k n-1 p+h) th II(a-p+farch) (Fars9 + Sa+zk ) where r-4 = h, x-3 = k, x-2 = r, -1 = P, xo = q, {fm}- = {1,0,1, 1,2, 3, 5, 8, ...}. Proof: For n = 0 the result holds. Now suppose that n >0 and that our assumption holds for n – 2. That is; kT (foisap + fei-3ah ( feg + fei-ik ) Soi+3p + foi+2h) (Toi-19 + foi-2k ) fei+2P + fei+1h ( fei+1p + feih ) \Foi+39 + fois+2k) fei+ep+ fei+sh ( fei+29 + foi+ik Sei+sp + fei+gh Ten-9 = i-0 Sei+49 + fei+3k n-2 foi+19 + Sauk ) (foisap + fo+3gh ( fei+69 + foi-sk) foi+ap + fei+zh ) Jei459 + foitak) I6n-6 = (2p +4) T(foi-ap + fes-zh) (foisaq + foi+sk Jei+7p+ fei+ch) Joi+39 + fei+2k ) I6n-5 = Now, it follows from Eq.(8) that Z6n-4 = T6n-7 + T6n-6 + 2ôn-9 11 foi+6P + fei+sh\ ( Soi+29 + fei+ik ( foi+6P + fei+sh PIIasp+ feirah ) foi+29 + foi+ik n-2 n-2 foi+6P + foi+sh ( foi+29 + fei+ik PIISP+ feigh) Tei+19 + feik ) n-2 paII (st fa) (2 fuk) n-2 Seseg+ fersk Joi-19+fer-ak im0 pa II ( (atl) = PTT (feisap + feirsh) (feis29 + feisak fei+sp + fei+sh, im0 n-2 Josreg+Sesk) (Tei-19+Se-ak i=0 -i )- ( foi+oP+ fsi+sh` Soi+sp + fei+ah) foi+29 + foi+1k fei+19 + feik =PII 12
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