F2006Solns5 - k 2 ×-2 = 0 k 1 k 2 = 0 Since c 1 = 3 k 1 × 3 1 k 2 ×-2 1 = 3 3 k 1-2 k 2 = 3 Solving the system of equations in k 1 and k 2 we

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ECE 190 — Fall, 2006 Assignment 5 Solutions Section 8.1 2 1, 7, 16, 28 4 3, 9, 162, 78 732 6 -1, 2, 0, 4 10 LS = b k =4 k RS = 4 b k - 1 =4 · 4 k - 1 =4 k 14 LS = d k =3 k - 2 k RS = 5 d k - 1 - 6 d k - 2 = 5(3 k - 1 - 2 k - 1 ) - 6(3 k - 2 - 2 k - 2 ) =(5 · 3 - 6)3 k - 2 - (5 · 2 - 6)2 k - 2 =9 · 3 k - 2 - 4 · 2 k - 2 =3 2 · 3 k - 2 - 2 2 · 2 k - 2 =3 k - 2 k 28 We know by defnition that , For k 2 ,F k = F k - 1 + F k - 2 Equivalently, For k 1 ,F k +1 = F k + F k - 1 We want to show that, For k 1 ,F 2 k +1 - F 2 k - F 2 k - 1 =2 F k F k - 1 LS = F 2 k +1 - F 2 k - F 2 k - 1 =( F k + F k - 1 ) 2 - F 2 k - F 2 k - 1 = F 2 k +2 F k F k - 1 + F 2 k - 1 - F 2 k - F 2 k - 1 =2 F k F k - 1 =RS Section 8.3 2 b and F 6 We must show that b k = b k - 1 +6 b k - 2 LS = C · 3 n + D ( - 2) n RS = ± C · 3 n - 1 + D ( - 2) n - 1 ² +6 ± C · 3 n - 2 + D ( - 2) n - 2 ² = C (3 + 6)3 n - 2 + D ( - 2+6)2 n - 2 = C · 9 · 3 n - 2 + D · 4( - 2) n - 2 = C · 3 2 · 3 n - 2 + D ( - 2) 2 ( - 2) n - 2 = C · 3 n + D ( - 2) n =LS 1
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9 Characteristic equation is r 2 =7 r - 10 r 2 - 7 r +10=0 ( r - 2)( r - 5) = 0 r =2 r =5 Solution has the form b n = c 1 2 n + c 2 5 n Since b 0 =2 , c 1 × 2 0 + c 2 × 5 0 =2 c 1 + c 2 =2 Since b 1 =2 , c 1 × 2 1 + c 2 × 5 1 =2 2 c 1 +5 c 2 =2 Solving the system of equations in c 1 and c 2 , we get c 1 = 8 3 c 2 = - 2 3 Solution is f n = 8 3 2 n - 2 3 5 n 10 Characteristic equation is r 2 = r +6 r 2 - r - 6=0 ( r - 3)( r +2)=0 r =3 r = - 2 Solution has the form c n = k 1 3 n + k 2 ( - 2) n Since c 0 =0 , k 1 ×
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Unformatted text preview: k 2 × (-2) = 0 k 1 + k 2 = 0 Since c 1 = 3, k 1 × 3 1 + k 2 × (-2) 1 = 3 3 k 1-2 k 2 = 3 Solving the system of equations in k 1 and k 2 , we get k 1 = 3 5 k 2 =-3 5 Solution is c n = 3 5 · 3 n-3 5 (-2) n 2 • 13 See text. • 14 Characteristic equation is r 2 =-4 r-4 r 2 + 4 r + 4 = 0 ( r + 2) 2 = 0 r =-2 is a double root Solution has the form s n = c 1 (-2) n + c 2 n (-2) n Since s = 0, c 1 (-2) + c 2 (0)(-2) = 0 c 1 = 0 Since s 1 =-1, c 1 (-2) 1 + c 2 (1)(-2) 1 =-1-2 c 1-2 c 2 =-1 Solving the system of equations in c 1 and c 2 , we get c 1 = 0 c 2 = 1 2 Solution is s n = 1 2 n (-2) n 3...
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This homework help was uploaded on 04/19/2008 for the course ECE 190 taught by Professor Carter during the Fall '06 term at University of Toronto- Toronto.

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F2006Solns5 - k 2 ×-2 = 0 k 1 k 2 = 0 Since c 1 = 3 k 1 × 3 1 k 2 ×-2 1 = 3 3 k 1-2 k 2 = 3 Solving the system of equations in k 1 and k 2 we

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