Discrete Mathematics with Graph Theory (3rd Edition) 144

Discrete - 142 Solutions to Exercises Exercises 5.3 1[BB The characteristic polynomial is x 2 X 6 the characteristic roots are 2,3 We obtain an = c

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142 Solutions to Exercises Exercises 5.3 1. [BB] The characteristic polynomial is x 2 - X - 6; the characteristic roots are -2,3. We obtain an = cI(-2)n + c2(3 n ). Since CI + C2 = 1, -2CI + 3C2 = 3, we find CI = 0, C2 = 1. The solution is an = 3 n . 2. The characteristic polynomial is x 2 + 6x - 7; the characteristic roots are -7, 1. We obtain an = cI(-7)n + c2(ln). Since CI + C2 = 32, -7CI + C2 = -17, we find CI = 49/8, C2 = 207/8. The solution is 3. [BB] The characteristic polynomial here, x 2 -6x+9, has the repeated root x = 3 and so Theorem 5.3.1 tells us that the solution is an = cI(3 n ) + c2n(3 n ). From the given values of ao and aI, we obtain cI(30) + c2(0)(30) = -5 (so CI = -5) and cI(3 1 ) + c2(1)(3 1 ) = 3; that is, 3CI + 3C2 = 3. SO CI + C2 = 1 and C2 = 1 - CI = 6. The solution is an = -5(3 n ) + 6n(3 n ). 4. To fit this problem to the notation of Theorem 5.3.1, we rewrite the given recurrence in the equivalent form an = 7an -1 -1Oan-2, n ;::: 3. The characteristic polynomial is
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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