This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 146 Solutions to Exercises (b) 1,1,4,6,4, 4, 16, 24, 16, 16, .... The corresponding characteristic polynomial is x 2  2x + 2 whose roots are 1 i. By The orem 5.3.1, the solution is an = cl(I + i)n + c2(1  i)n. The initial conditions give ao = 1 = Cl + C2, al = 1 = cl(I + i) + c2(1 i), so Cl = !(I + 2i), C2 = !(I 2i) and an = !(I + 2i)(1 + i)n  !(I 2i)(1 2i)n. (c) 2,0,20, 40, 120,640, 80, 6240, 13280, 35840, .... The corresponding characteristic polynomial is x 2 + 2x + 10 whose roots are 1 3i. By Theorem 5.3.1, the solution is an = Cl (1 + 3i)n + C2( 1  3i)n. The initial conditions give ao = 2 = Cl + C2, al = = Cl( 1 + 3i) + C2( 1  3i), so Cl = k( 3 + i), C2 = k(3 + i) and an = k( 3 + i)( 1 + 3i)n  k(3 + i)( 1  3i)n. 21. [BB] In Problem 22, we showed that the nth term of the Fibonacci sequence is anl = Jg () n  ....!...(ly'S)n Thus la  ....!...(~)nl = 1....!...(ly'S)nl Now 1....!...(ly'S)nl < ....!... < 1 since y'S 2 . ,nl y'S 2 y'S 2...
View
Full
Document
 Summer '10
 any
 Graph Theory

Click to edit the document details