163 Discrete Mathematics Review 2 Use the following to answer ...

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$Math 171 Review for the Final Exam$

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10. 1 2 ,2 2 ,3 3 ,4 2 ,…. Solution. We see that 2 , 1,2, 2 2 an = n n= … Therefore a − n 1 a = n ( n + 1) − + n = 2n + 1 and the sequence can be described recursively as a = 1, a = a + 2n+ 1. 1 n+ 1 n Use the following to answer question 11: In the questions below solve the recurrence relation either by using the characteristic equation or by discovering a pattern formed by the terms. 11. a n = −10a n − 1 − 21a n − 2 , a 0 = 2, a 1 = 1. Solution. The characteristic equation of the above recurrence relation is 2 r + 10r+ 21 = 0 . The left part can be factored as ( r+ 3)( r+ 7) whence the solutions of the characteristic equation are r1 =− 3, r2 =−7 , and the general solution of the recurrence relation is n n an = α1( − 3) + α2( − 7) . From the initial conditions we find α1+ α2 = 2 3α1+ 7α2 =− 1 . By multiplying the first equation by 7 and subtracting from it the second equation we get 4α 1 = 15whence α = 15 1 4 and α = 2− α = 2− 15 =− 7 . Finally 2 1 4 4 15 n 7 a n = ( −3) − ( − 7) 4 4 n Page 4
12. What form does a particular solution of the linear nonhomogeneous recurrence relation a n = 4a n − 1 − 4a n − 2 + F(n) have when F(n) = (n 2 + 1)2 n Solution. The associated homogeneous recurrence relation is an = 4an− 1− 4an−2. Its 2 characteristic equation r − 4r+ 4= 0has solution 2 of multiplicity 2. By theorem 6 on page 469 there is a particular solution of our nonhomogeneous recurrence relation of the 2 n 2 n 4 3 2 form a = n 2( An + Bn+ C) = 2( An + Bn + Cn ). To find the values of the n coefficients A, B, and C let us introduce the polynomial Gn ( ) = + + 4 3 2 An Bn Cn n n n a = 2 G( n), a = 2⋅ 2 G( n+ 1), and a = 4⋅2 G( n+ 2). n n+ 1 n+ 2 . Then 2 Plugging in these expressions into the relation a = 4a − 4 a + [( n+ 2) + 1] ⋅2 n ⋅ 4 n+ 2 n+ 1 n and dividing both parts by 4⋅ 2 n we obtain the relation 2 Gn ( + 2) = 2 Gn ( + 1) − Gn ( ) + n + 4n + 5. Or 2 Gn ( + 2) − 2 Gn ( + 1) + Gn ( ) −n −4n− 5= 0. By expanding the expressions in the left part and combining the like terms we get 2 ( − 1+ 12 An ) + (6B+ 24A−4) n− 5 + 6B+ 2C+ 14A= 0 , 1 1 11 From here we find A = , B = , and C = . 12 3 12 Page 5
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