{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Discrete Mathematics with Graph Theory (3rd Edition) 154

# Discrete Mathematics with Graph Theory (3rd Edition) 154 -...

This preview shows page 1. Sign up to view the full content.

152 Solutions to Exercises 8. (a) First we look for a particular solution and try Pn = a(4n). Substituting, we obtain a(4n) = 4a(4 n - 1 )-4a(4n-2)+4n; that is, 4 n a = 4 n a-4 n - 1 a+4 n . This gives 4 n - 1 a = 4 n , so a = 4 and Pn = 4n+1. In Problem 21, the homogeneous recurrence an = 4an-1 -4an-2 was seen earlier to have general solution an = c1(2 n ) + c2n(2n), so Pn + qn = 4 n + 1 + c1(2 n ) + nC2(2n). Bringing in our initial conditions, we obtain 4 1 + C1 (2 0 ) + OC2(2 0 ) = 2 and 4 2 + C1 (21) + 1C2(21) = 8; that is, 4+C1 2 16 + 2C1 + 2C2 8. Solving, we obtain C1 = -2, C2 = -2, so (b) f(x) ao + a1X + a2x 2 + + anx n + 4xf(x) = 4aox + 4a1X2 + + 4an_1X n + 4x 2 f(x) = 4aox2 + + 4an_2 Xn + Therefore, f(x) - 4xf(x) + 4x 2 f(x) = ao + (a1 - 4ao)x + (a2 - 4a1 + 4ao)x2 + ... + (an - 4an-1 + 4an_2)X n + .... So f(x)(l - 4x + 4x 2 ) = 2 + (O)x + 4 2 x 2 + ... + 4 n x n + ... and f(x) 1 ( 2 nn) -:-----:-:=- 2 + 16x + ... + 4 x + ... (1- 2x)2 (1 + 2(2x) + 3(2x)2 + ... + (n + 1)(2x)n + ... )(2 + 16x 2 + ... +
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4 n x n + ... ) 2 + 8x + 40x 2 + ... + [(1)4n + 2(2)4 n-1 + 3(22)4n-2 + ... +(n -1)(2n-2)42 + (n + 1)2n(2)]xn + ... To find a formula for an, we must now sum the terms in the square brackets, but it is not imme-diately obvious how this can be done. An induction argument can be used to show that this sum equals 4 n + 1 - 2n+1(1 + n) as in (a). The authors prefer the method of the characteristic polynomial to generating functions in this instance. 9. (a) [BB] f(x) = ao + a1X + a2x2 + 5aox + 5a1X2 + (b) 5xf(x) = Adding, we obtain since ao = 2, an + 5an-1 = 0 for n ::::: 1. Thus, f(x) = _2_ = 2(1 -5x + 25x 2 + ... + (-5txn + " . ) 1 +5x and so an = 2( -5)n. The characteristic polynomial is x 2 + 5x with characteristic roots 0 and -5. We obtain an = C1 (-5)n. Since ao = 2, we have C1 (-5)0 = 2; that is, C1 = 2, so an = 2( -5)n as proven before. f(x) = 5xf(x) = ao + a1X + a2x2 + 5aox + 5a1X 2 +...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online