Unformatted text preview: = 0 and 1 f3x = 0; that is, 1 1 1 ± V( 1)2  4( 1)(1) 1 ± J5 1 ± J5 ;' ~ 2(1) 2 2 1 2 1+J5 ~(1 + J5) J5 1 2Thus, a 1 2 1J5 ~ ( 1 J5) J5 + 1 2 13 (c) Set 1 = _A_ + _B_ = (A+B) (Af3+ Ba )x. 1 x x 2 1 ax 1 f3x 1 x x 2 Thus, A + B = 1, Af3 + Ba = 0, so a a 1 A===a af3 J5 J5 13 131 1 B=;A=;J5 a =J5f3 (d) From the previous parts, it follows that 1 A B f(x) = 1x x2 = 1ax + 1f3x = A(l + ax + a 2 x 2 + . .. + anxn + . .. ) + B(l + f3x + f3 2 x 2 + . .. + f3 n x n + . .. ) and so an = A(a)n + B(f3)n = Jga n + 1 Jgf3 n + l , in agreement with the formula derived in Problem 22. The method of Section 5.3 seems preferable to the method of generating functions....
View
Full
Document
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.
 Summer '10
 any
 Graph Theory

Click to edit the document details