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Unformatted text preview: = 5a + 5b + 3, b = 5b. These equations give b = 0, a = !. Thus, Pn + qn = Cl (5)n +!. Using ao = 2, we get Cl = ~ and so an = ~(5)n + ! in agreement with our previous solution. 10. (a) f(x) ao + alX + a2x2 + + anx n + 4xf(x) = 4aox + 4alX2 + + 4an_lX n + 3x 2 f(x) = 3aox2 + + 3an_2Xn + Therefore, f(x) 4xf(x) + 3x 2 f(x) = ao + (al 4ao)x + (a2 4al + 3ao)x2 + ... + (an 4anl + 3an_2)X n + ... = 2 3x. 2 3x So, f(x)(1 4x + 3x 2 ) = 2 3x and f(x) = ( )( ). Now write Ix 13x 2 3x A B (A + B) + (3A B)x (lx)(13x) = Ix + 13x = (lx)(13x) Then A + B = 2, 3A B = 3; A = !, B = ~ and so = !(1 + x + x 2 + ... + xn + ... ) + ~(1 + 3x + 9x 2 + ... + 3 n xn + ... ) 2 + 5x + ... + (! + ~(3n))xn + ... and so an = ! + ~(3n). (b) f(x) ao + alX + a2x2 + lOxf(x) lOaox + lOalX2 + 25x 2 f(x) 25aox2 + + anx n + + lOan_lX n + + 25an_2Xn +...
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 Summer '10
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 Graph Theory

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