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show me steps of the determine blue and all information is here explain to me step by step i need every details

show me steps of the determine blue and all information is here explain to me step by step i need every details

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 27EQ
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show me steps of the determine blue and all information is here explain to me step by step i need every details

Let w2n-1 = Un, w2n–2 = Vn, z2n–1 = Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = µ, (11) Un+1 - ki Vn+1 - kị Un = µ, and Yn+1 – vo Xn - vo Yn = €, (12) Xn+1 - Vị Yn+1 – vị Xn = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko Un = H, (E – k)) Un – kị E V, = µ, (13) (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – k1) and ko, respectively, (E – k1) µ, (E – k1) (E – ko) Vn – (E – k1) ko Un ko (E – k1) Un – ko ki E V (15) ko H. (16) Thus, the combination of (15) and (16) is (E – (ko + k1+ ko k1) E + ko k1) Vn = (1 – kı + ko) µ, (17) (E² – O E + ko kı) Vn = (1 – kı + ko) µ, where o = ko + kı + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, 0 + Vø? – 4 ko ki 62 – 4 ko k1 (1- k + ko` V = A +B + (18) 2 2 1- kо — k1 since A and B are constants. By Substituting (18) in (13), we give 1 U, = A ko $ + Vø2 – 4 ko ki - 1 0 + Vø2 – 4 ko k, 2 2 – 4 ko ki - 1 $ - Vø2 – 4 ko ki (19) +B ko 2 * ((-) ( )- (1- k + ko 1- ko – ki - 1 ko H. 6. Therefore, 62 – 4 ko ki ø + Vo2 – 4 ko ki A 1- k + ko 1– ko – k. W2n-2 = + B 4, (20) 2 02 – 4 ko ki ² – 4 ko ki 1 A ko W2n–1 0 - Vø? – 4 ko k1 0 - Vo2 – 4 ko k1 +B ko - 1 2 1- kị + ko 1– ko – k1 1 u. ko + (21) For n = 0, 1,2,... Now, to calculate A and B, we solve the following system (1- kı + ko 1- ko – k1 w-2 = A+ B+ 62 – 4 ko ki (1– k + ko + 1- ko – k1, 0+ Vo2 – 4 ko ki wo = A + B which implies, (1- k - - ki + ko Vo2 – 4 ko ki 22 – 8 ko ki ) -) (v - ako ki – 4) + 2 (we w-2 - wo- ko - ki V – 4 ko ki 262 – 8 ko ki CG -ki + 1- ko-ki) ki + ko H- w-2) (V2 - 4 kg k1 + B +2 wo- - ko – ki Similarly, we apply the same previous process to get the solution of (12), which is given as + V2 - 4o v 2-4 vo V1 22n-2 = C + D 2 1 22n-1 = C + V2 – 4 vo v1 + V2 - 4 vo vi - V2 – 4 vo - V2 - 4 vo VI + D 2 2 • (÷-() (-)- (1- v + vo 1- v0 - V where C and D are - 4 vo vị 2 2 - 8 vo v1 1- v + vo - vị +o C = +2 - vo - v1 V2 – 4 vo Vi 1-vi + vo 1- v + o D +2 1 – 0 = --2) (V2 – 4 o vi + %3D 2 2 - 8 v0 Vı -0 - v The proof is complete.
Transcribed Image Text:Let w2n-1 = Un, w2n–2 = Vn, z2n–1 = Xn and z2n-2 = Yn, implies that Vn+1 – ko Un – ko Vn = µ, (11) Un+1 - ki Vn+1 - kị Un = µ, and Yn+1 – vo Xn - vo Yn = €, (12) Xn+1 - Vị Yn+1 – vị Xn = €. It is observe that, the equations (11) and (12) are system of linear non-homogeneous difference equations with constants coefficients. First, we solve the system in (11). Thus, consider (11) is (E – ko) Vn – ko Un = H, (E – k)) Un – kị E V, = µ, (13) (14) where E = A + 1, is the shift operator. Multiplying (13) and (14) by (E – k1) and ko, respectively, (E – k1) µ, (E – k1) (E – ko) Vn – (E – k1) ko Un ko (E – k1) Un – ko ki E V (15) ko H. (16) Thus, the combination of (15) and (16) is (E – (ko + k1+ ko k1) E + ko k1) Vn = (1 – kı + ko) µ, (17) (E² – O E + ko kı) Vn = (1 – kı + ko) µ, where o = ko + kı + ko k1. Clearly, (17) is linear difference equations from order two and it is easy to obtain the solution, 0 + Vø? – 4 ko ki 62 – 4 ko k1 (1- k + ko` V = A +B + (18) 2 2 1- kо — k1 since A and B are constants. By Substituting (18) in (13), we give 1 U, = A ko $ + Vø2 – 4 ko ki - 1 0 + Vø2 – 4 ko k, 2 2 – 4 ko ki - 1 $ - Vø2 – 4 ko ki (19) +B ko 2 * ((-) ( )- (1- k + ko 1- ko – ki - 1 ko H. 6. Therefore, 62 – 4 ko ki ø + Vo2 – 4 ko ki A 1- k + ko 1– ko – k. W2n-2 = + B 4, (20) 2 02 – 4 ko ki ² – 4 ko ki 1 A ko W2n–1 0 - Vø? – 4 ko k1 0 - Vo2 – 4 ko k1 +B ko - 1 2 1- kị + ko 1– ko – k1 1 u. ko + (21) For n = 0, 1,2,... Now, to calculate A and B, we solve the following system (1- kı + ko 1- ko – k1 w-2 = A+ B+ 62 – 4 ko ki (1– k + ko + 1- ko – k1, 0+ Vo2 – 4 ko ki wo = A + B which implies, (1- k - - ki + ko Vo2 – 4 ko ki 22 – 8 ko ki ) -) (v - ako ki – 4) + 2 (we w-2 - wo- ko - ki V – 4 ko ki 262 – 8 ko ki CG -ki + 1- ko-ki) ki + ko H- w-2) (V2 - 4 kg k1 + B +2 wo- - ko – ki Similarly, we apply the same previous process to get the solution of (12), which is given as + V2 - 4o v 2-4 vo V1 22n-2 = C + D 2 1 22n-1 = C + V2 – 4 vo v1 + V2 - 4 vo vi - V2 – 4 vo - V2 - 4 vo VI + D 2 2 • (÷-() (-)- (1- v + vo 1- v0 - V where C and D are - 4 vo vị 2 2 - 8 vo v1 1- v + vo - vị +o C = +2 - vo - v1 V2 – 4 vo Vi 1-vi + vo 1- v + o D +2 1 – 0 = --2) (V2 – 4 o vi + %3D 2 2 - 8 v0 Vı -0 - v The proof is complete.
In this paper, we solve and study the properties of the following system Wn-p 2 zn-h p=0 E wn-p zZn-h h=1 h=0 Wn+1 = p=1 + u and zn+1 = + €, (4) Zn - € Wn - u 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, z,} 2 be a solution of (4), then + V2 – 4 ko kı A - 4 ko ki - kị + ko w2n-2 +B ko ki + V – 4 ko kị $+ v2 – 4 ko ki - V – 4 ko kı 1 +B A ko w2n-1 1- ko + V2 – 4 vo VỊ V2 – 4 vo V - V + vo 22n-2 + D 0 - V1 1 (+ Ve2 – 4 v0 vi + V2 – 4 vo vi gb - V2 – 4 0 vỊ V2 – 4 vo v1 22n-1 +D 2 1- v + vo 1- vo - VI where A, B, C and D are constants defined as ko wo -4 ki = z-1 +z-2 $ = ko + ki + ko k1, w-1 +w-2 20 -e 20 -e w-1 + w-2 = vo+ v1 + o V, z-1 +2-2 wo -H V2 – 4 ko ki 262 – 8 ko ki 4) (V2 – 4 ko kı – 4) +2 A w-2- - 1- ko - ki V2 – 4 ko ki 2 62 - 8 ko ki - ko + k1 (1– ki + ko --) (v - sho ki +e) + 2 (wo - ( ) )]. B 4- w-2 - ko - ki 1- k - ko V2 – 4 vo vI 2 2 - 8 v0 v - vi + vo - vO - V , - v + o 1- vo - V +2 20 V2 – 4 v0 v1 1- v + vo 1-v + vo D E- 2-2 12 - 4 vO +2 2 2 since ko + ki #1 and vo + v 71 for n e 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 - 4 h=1 Zn+1 - € p=1 and (5) 2, - € Wn -H EWn-p 2 Zn-h h=0 p=0 Then, we assume that Wn+1 - u Wn - u kn+1 = 1 k, = Wn-p Wn-p p=0 p=1 (6) Zn+1 - € Zn - € Vn+1 = 1 Vn = 2 E zn-h E zn-h h=0 h=1 Substituting (6) in (5), we have 2 Zn-h h=1 kn+1 = 1 = -= kn-1, Vn (7) Wn-p p=1 1 = Vn-1. kn Vn+1 %3D Wn - u Hence, we see that 1 v1 = Vo wo - u 20 - € 1 ko = Vo = k = w-1 + w-2 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 V4 = Vo, at n = 4, kg = k1 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 = µ+ kn (wn-1 + Wn-2) and zn = €++ Vn (Zn-1 + zn-2). (9) Using (8) in (9), we have w2n = µ+ ko (w2n-1 + W2n-2), W2n+1 = H + k, (w2n + w2n-1), (10) 22n = € + vo (z2n-1+ 22n-2), Z2n+1 = €+ v1 (22n + z2n-1). 介
Transcribed Image Text:In this paper, we solve and study the properties of the following system Wn-p 2 zn-h p=0 E wn-p zZn-h h=1 h=0 Wn+1 = p=1 + u and zn+1 = + €, (4) Zn - € Wn - u 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, z,} 2 be a solution of (4), then + V2 – 4 ko kı A - 4 ko ki - kị + ko w2n-2 +B ko ki + V – 4 ko kị $+ v2 – 4 ko ki - V – 4 ko kı 1 +B A ko w2n-1 1- ko + V2 – 4 vo VỊ V2 – 4 vo V - V + vo 22n-2 + D 0 - V1 1 (+ Ve2 – 4 v0 vi + V2 – 4 vo vi gb - V2 – 4 0 vỊ V2 – 4 vo v1 22n-1 +D 2 1- v + vo 1- vo - VI where A, B, C and D are constants defined as ko wo -4 ki = z-1 +z-2 $ = ko + ki + ko k1, w-1 +w-2 20 -e 20 -e w-1 + w-2 = vo+ v1 + o V, z-1 +2-2 wo -H V2 – 4 ko ki 262 – 8 ko ki 4) (V2 – 4 ko kı – 4) +2 A w-2- - 1- ko - ki V2 – 4 ko ki 2 62 - 8 ko ki - ko + k1 (1– ki + ko --) (v - sho ki +e) + 2 (wo - ( ) )]. B 4- w-2 - ko - ki 1- k - ko V2 – 4 vo vI 2 2 - 8 v0 v - vi + vo - vO - V , - v + o 1- vo - V +2 20 V2 – 4 v0 v1 1- v + vo 1-v + vo D E- 2-2 12 - 4 vO +2 2 2 since ko + ki #1 and vo + v 71 for n e 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 - 4 h=1 Zn+1 - € p=1 and (5) 2, - € Wn -H EWn-p 2 Zn-h h=0 p=0 Then, we assume that Wn+1 - u Wn - u kn+1 = 1 k, = Wn-p Wn-p p=0 p=1 (6) Zn+1 - € Zn - € Vn+1 = 1 Vn = 2 E zn-h E zn-h h=0 h=1 Substituting (6) in (5), we have 2 Zn-h h=1 kn+1 = 1 = -= kn-1, Vn (7) Wn-p p=1 1 = Vn-1. kn Vn+1 %3D Wn - u Hence, we see that 1 v1 = Vo wo - u 20 - € 1 ko = Vo = k = w-1 + w-2 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 V4 = Vo, at n = 4, kg = k1 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 = µ+ kn (wn-1 + Wn-2) and zn = €++ Vn (Zn-1 + zn-2). (9) Using (8) in (9), we have w2n = µ+ ko (w2n-1 + W2n-2), W2n+1 = H + k, (w2n + w2n-1), (10) 22n = € + vo (z2n-1+ 22n-2), Z2n+1 = €+ v1 (22n + z2n-1). 介
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