Discrete Mathematics with Graph Theory (3rd Edition) 158

Discrete Mathematics with Graph Theory (3rd Edition) 158 -...

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156 Solutions to Exercises (c) Let f(x) = Po + PIX + P2X2 + . .. be the generating function. Then f(x) 2xf(x) = x 2 f(x) = Po + PIX + P2X2 + P3 x3 + .. . 2pox + 2PIX 2 + 2P2X3 + .. . pox 2 + PIX 3 + . . . So (1 - 2x - x 2 )f(x) = Po = 1 and 1 1 f(x) = 1 - 2x - x2 1 - x(2 + x) = 1 + x(2 + x) + x 2 (2 + x)2 + x 3 (2 + x)3 + . .. = 1 + 2x + x 2 + x 2 (4 + 2x + x 2 ) + x 3 (8 + 12x + 6x 2 + x 3 ) + . .. = 1 + 2x + 5x 2 + 12x 3 + . .. 14. (a) f(x) ao + aIX + a2x2 + + anx n + xf(x) aox + aIx 2 + + an_IX n + x 2 f(x) aox 2 + + an_2 Xn + Therefore, f(x) - xf(x) - x 2 f(x) = ao + (al - ao)x + (a2 - al - ao)x 2 + . .. + (an - an-I - an_2)X n + . .. = 1, since ao = al = 1 and an = an-I + an-2 for n ~ 2. 1 Therefore, (1 - x - x 2 )f(x) = 1 and f(x) = 1 _ x _ x 2 · (b) The roots of 1 - x - x 2 = 0 are, evidently, the roots of 1 - ax
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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 -2-Thus, a 1 -2 1-J5 --~ ( -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 a-f3 J5 J5 13 131 1 B=-;A=-;J5 a =-J5f3 (d) From the previous parts, it follows that 1 A B f(x) = 1-x -x2 = 1-ax + 1-f3x = 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....
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