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n-1 n- n-1 n-1 () %3D h - 1 h-p n-1 n-1 P-h = q h-p n-1 ()* (1 -4) n-1 %3D n-1 h - %3D

n-1 n- n-1 n-1 () %3D h - 1 h-p n-1 n-1 P-h = q h-p n-1 ()* (1 -4) n-1 %3D n-1 h - %3D

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.2: Properties Of Division
Problem 50E
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п-1 п-1 h p-h п-1 h п-1 h-p 1 h-p n-1 h p-h п-1 п-1 h п-1 h-p h-p п-1 п-1 1 п-1 h п-1 - h Therefore n р — h п-1 X6n-3 = q Also, we can prove the other relations by similar way, the proof is completed.
Transcribed Image Text:п-1 п-1 h p-h п-1 h п-1 h-p 1 h-p n-1 h p-h п-1 п-1 h п-1 h-p h-p п-1 п-1 1 п-1 h п-1 - h Therefore n р — h п-1 X6n-3 = q Also, we can prove the other relations by similar way, the proof is completed.
Brn-10n-2 YIn-1 + 8xn-4 In+1 = aIn-2 + n = 0, 1,., (1) 1. The following special case of Eq.(1) has been obtained is follows Tn-1Xn-2 In+1 = In-2 - (11) In-1 - In-4 where the initial conditions r4, 2-3, 2-2, x-1, To are arbitrary non zero real num- bers with r-A #1-1 and r-3 7 xo. Theorem 7. Let {In}-3 be a solution of Eq.(11). Then every solution of Eq.(11) is unbounded. Moreover {r,}-3 takes the form 100 k n-1 T6n-4 = X6n-3 = q h k I6n-2 T6n-1 = P %3D k – n+1 h X8n+1 =r %3D - . Proof: Forn=0 the result holds. Now suppose that holds for n – 1. That is; > 0 and that our assumption n-1 k n-2 p~hn-1 n-2 T6n-10 = p X6n-9 = q n-1 n-1 n-1 h q- k k n-1 X6n-8 X6n-7 = P n-1 n-1 n-1 h X6n-6 = q X6n-5 = r h - p Now, it follows from Eq.(11) that X6n-6X6n-7 X6n-4 = X6n-7 X6n-6 X6n-9 n-1 n-1 n-1 n-1 n-1 n-1 k-q = p n-1 п-1 p-h P-h n-1 k п-1 k-q п-1 = p (1) - g - 1 n-1 k n-1 k-q n-1 р k - %3D g-k k n-1 k n-1 %3D q - k n-1 k n-1 k Therefore k X6n-4 = P k - Also, from Eq.(11), we see that X6n-5X6n-6 X6n-3 = X6n-6 - X6n-5 - X6n-8 r(속)" (남)" "(뉴)"(무)"-" (속)""(부)" n n-1 п-1 h g-k p-h n-1 п-1 п-1 h-p = a n-1 g-k n-1 n-1 h h-p 22
Transcribed Image Text:Brn-10n-2 YIn-1 + 8xn-4 In+1 = aIn-2 + n = 0, 1,., (1) 1. The following special case of Eq.(1) has been obtained is follows Tn-1Xn-2 In+1 = In-2 - (11) In-1 - In-4 where the initial conditions r4, 2-3, 2-2, x-1, To are arbitrary non zero real num- bers with r-A #1-1 and r-3 7 xo. Theorem 7. Let {In}-3 be a solution of Eq.(11). Then every solution of Eq.(11) is unbounded. Moreover {r,}-3 takes the form 100 k n-1 T6n-4 = X6n-3 = q h k I6n-2 T6n-1 = P %3D k – n+1 h X8n+1 =r %3D - . Proof: Forn=0 the result holds. Now suppose that holds for n – 1. That is; > 0 and that our assumption n-1 k n-2 p~hn-1 n-2 T6n-10 = p X6n-9 = q n-1 n-1 n-1 h q- k k n-1 X6n-8 X6n-7 = P n-1 n-1 n-1 h X6n-6 = q X6n-5 = r h - p Now, it follows from Eq.(11) that X6n-6X6n-7 X6n-4 = X6n-7 X6n-6 X6n-9 n-1 n-1 n-1 n-1 n-1 n-1 k-q = p n-1 п-1 p-h P-h n-1 k п-1 k-q п-1 = p (1) - g - 1 n-1 k n-1 k-q n-1 р k - %3D g-k k n-1 k n-1 %3D q - k n-1 k n-1 k Therefore k X6n-4 = P k - Also, from Eq.(11), we see that X6n-5X6n-6 X6n-3 = X6n-6 - X6n-5 - X6n-8 r(속)" (남)" "(뉴)"(무)"-" (속)""(부)" n n-1 п-1 h g-k p-h n-1 п-1 п-1 h-p = a n-1 g-k n-1 n-1 h h-p 22
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