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Then, we assume that Wn+1 - H Wn - u kin+1 kn 1 2 E Wn-p E Wn-p p=0 p=1 (6) Zn+1 - € Zn - E Vn+1 Vn = 2 E žn-h h=0 h=1 Substituting (6) in (5), we have E Zn-h 1 kn-11 h=1 kn+1 Zn Vn (7) 2 Wn-p 1 = Vn-1. Vn+1 Wn - l kn Hence, we see that 1 ki = 1 V1 = ko wo - H 20 - € ko W-1 + w_2 Vo = Z-1 + 2-2 and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, ks = k and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n+ 1, k2n+1 = k1 and V2n+1 = V1. Now, from the relations in (6) we get Wn = u + kn (Wn-1 + Wn-2) and zn = € + Vn (žn-1 + Zn-2). (9) Using (8) in (9), we have W2n = H + ko (W2n-1 + W2n-2), W2n+1 = l + kı (w2n + W2n-1), (10) 22n = € + vo (22n-1 + 22n-2), 22n+1 = € +vi (22n + 22n-1). 介

Then, we assume that Wn+1 - H Wn - u kin+1 kn 1 2 E Wn-p E Wn-p p=0 p=1 (6) Zn+1 - € Zn - E Vn+1 Vn = 2 E žn-h h=0 h=1 Substituting (6) in (5), we have E Zn-h 1 kn-11 h=1 kn+1 Zn Vn (7) 2 Wn-p 1 = Vn-1. Vn+1 Wn - l kn Hence, we see that 1 ki = 1 V1 = ko wo - H 20 - € ko W-1 + w_2 Vo = Z-1 + 2-2 and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, ks = k and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n+ 1, k2n+1 = k1 and V2n+1 = V1. Now, from the relations in (6) we get Wn = u + kn (Wn-1 + Wn-2) and zn = € + Vn (žn-1 + Zn-2). (9) Using (8) in (9), we have W2n = H + ko (W2n-1 + W2n-2), W2n+1 = l + kı (w2n + W2n-1), (10) 22n = € + vo (22n-1 + 22n-2), 22n+1 = € +vi (22n + 22n-1). 介

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 50E
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In this paper, we solve and study the properties of the following system E Wn-p > Zn-h Zn-h Wn-P p=0 h=1 p=1 h=0 Wn+1 +µ and zn+1 + €, (4) Zn - € Wn - H where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. Theorem 2.1. Let {wn, zn}- be a solution of (4), then n=-2 o+ V02 – 4 ko k1 $ - V62 – 4 ko k1 k1 + ko w2n-2 + B + 2 2 ko – k1 n $ + Vø2 – 4 ko k1 $+ Vø2 – 4 ko k1 0 - V02 – 4 ko k1 0 - Vo2 – 4 ko k1 1 A ko 1 + B ko w2n-1 - 1 - 1 2 2 + (-) (는)-) (1– ko + k1 1– k1 – ko 1 - 1 ko + V2 – 4 vo vi 1 – vị + vo + - V62 4 vo v1 22n-2 + D E, 2 2 - vo - v1 1 + V2 – 4 vọ v1 b + V2 – 4 vo v1 e;2 – 4 vo v1 al2 - 4 vo vị 22n-1 C + D - 1 2 2 2 +(G-) )-) . 1- vi + vo 1- vo - v1 E, where A, B, C and D are constants defined as wo - u z-1 + z-2 ko ki = O = ko + k1 + ko k1, w-1+ w-2 20 - E 20 - € w-1 + w-2 v1 = = vo + vi + vo v1, 2-1+ z-2 1 – Om Vo2 – 4 ko k1 2 62 – 8 ko ki 1– k1 + ko 1 – 1 – kị + ko` 1 — ко — k1 A = - 4 ko k1 – o) + 2 ( wo – w-2 - V62 – 4 ko ki 2 ф2 — 8 kо k1 1 – ko + k1 - ko – k1 (1 – kị + ko) µ)|, 1 – k1 – ko B u - w-2 62 – 4 ko k1 + ¢) +2 ( wo – = 4 vo v1 2 2 – 8 vo v1 1- vi + vo 1- vị + vo 2-2 - 1- vo - vi 1- vo - vi V2 – 4 vo vi 1 – vị + vo 1- v1 + vo E - z-2 2 – 4 vo v1 + +2 20 - 2 2 – 8 vo vi - vO - v1 1- vo - v1 since ko + ki 1 and vo +v1 71 for n € N. Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow form Zn-h Wn-p Wn+1 - H h=1 Zn+1 - € and (5) 1 1 Zn - € Wn - u E Wn-p > Zn-h p=0 h=0 4 || ||
Transcribed Image Text:In this paper, we solve and study the properties of the following system E Wn-p > Zn-h Zn-h Wn-P p=0 h=1 p=1 h=0 Wn+1 +µ and zn+1 + €, (4) Zn - € Wn - H where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. Theorem 2.1. Let {wn, zn}- be a solution of (4), then n=-2 o+ V02 – 4 ko k1 $ - V62 – 4 ko k1 k1 + ko w2n-2 + B + 2 2 ko – k1 n $ + Vø2 – 4 ko k1 $+ Vø2 – 4 ko k1 0 - V02 – 4 ko k1 0 - Vo2 – 4 ko k1 1 A ko 1 + B ko w2n-1 - 1 - 1 2 2 + (-) (는)-) (1– ko + k1 1– k1 – ko 1 - 1 ko + V2 – 4 vo vi 1 – vị + vo + - V62 4 vo v1 22n-2 + D E, 2 2 - vo - v1 1 + V2 – 4 vọ v1 b + V2 – 4 vo v1 e;2 – 4 vo v1 al2 - 4 vo vị 22n-1 C + D - 1 2 2 2 +(G-) )-) . 1- vi + vo 1- vo - v1 E, where A, B, C and D are constants defined as wo - u z-1 + z-2 ko ki = O = ko + k1 + ko k1, w-1+ w-2 20 - E 20 - € w-1 + w-2 v1 = = vo + vi + vo v1, 2-1+ z-2 1 – Om Vo2 – 4 ko k1 2 62 – 8 ko ki 1– k1 + ko 1 – 1 – kị + ko` 1 — ко — k1 A = - 4 ko k1 – o) + 2 ( wo – w-2 - V62 – 4 ko ki 2 ф2 — 8 kо k1 1 – ko + k1 - ko – k1 (1 – kị + ko) µ)|, 1 – k1 – ko B u - w-2 62 – 4 ko k1 + ¢) +2 ( wo – = 4 vo v1 2 2 – 8 vo v1 1- vi + vo 1- vị + vo 2-2 - 1- vo - vi 1- vo - vi V2 – 4 vo vi 1 – vị + vo 1- v1 + vo E - z-2 2 – 4 vo v1 + +2 20 - 2 2 – 8 vo vi - vO - v1 1- vo - v1 since ko + ki 1 and vo +v1 71 for n € N. Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow form Zn-h Wn-p Wn+1 - H h=1 Zn+1 - € and (5) 1 1 Zn - € Wn - u E Wn-p > Zn-h p=0 h=0 4 || ||
Then, we assume that Wn+1 Wn - l kin+1 kn 1 2 > Wn-p > Wn-p p=0 p=1 (6) Zn+1 - € Zn - € Vn+1 = .nע 2 E zn-h Zn-h h=0 h=1 Substituting (6) in (5), we have Zn-h 1 kn-1, h=1 kin+1 Zn - € Vn (7) > Wn-p p=1 1 Vn+1 Vn-1· Wn – l kn Hence, we see that 1 Vị Vo wo – µ 20 - € ko Vo W-1+ w_2 k1 Z_1 + 2-2 ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and VĄ = Vo, at n = 4, kz = k1 and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1= k1 and V2n+1 = Vi. Now, from the relations in (6) we get Wn = u + kn (Wn-1+Wn-2 and zn = € + Vn (žn–1 + Zn-2). (9) Using (8) in (9), we have W2n = µ + ko (W2n-1 + W2n-2), W2n+1 = µ + kı (w2n + W2n-1), (10) %2n = €+ vo (22n–1+ %2n-2), 22n+1 = € + v1 (22n + 22n–1). 5 || || || ..
Transcribed Image Text:Then, we assume that Wn+1 Wn - l kin+1 kn 1 2 > Wn-p > Wn-p p=0 p=1 (6) Zn+1 - € Zn - € Vn+1 = .nע 2 E zn-h Zn-h h=0 h=1 Substituting (6) in (5), we have Zn-h 1 kn-1, h=1 kin+1 Zn - € Vn (7) > Wn-p p=1 1 Vn+1 Vn-1· Wn – l kn Hence, we see that 1 Vị Vo wo – µ 20 - € ko Vo W-1+ w_2 k1 Z_1 + 2-2 ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and VĄ = Vo, at n = 4, kz = k1 and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1= k1 and V2n+1 = Vi. Now, from the relations in (6) we get Wn = u + kn (Wn-1+Wn-2 and zn = € + Vn (žn–1 + Zn-2). (9) Using (8) in (9), we have W2n = µ + ko (W2n-1 + W2n-2), W2n+1 = µ + kı (w2n + W2n-1), (10) %2n = €+ vo (22n–1+ %2n-2), 22n+1 = € + v1 (22n + 22n–1). 5 || || || ..
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