Unformatted text preview: 4 n x n + ... ) 2 + 8x + 40x 2 + ... + [(1)4n + 2(2)4 n1 + 3(22)4n2 + ... +(n 1)(2n2)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 immediately 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 + 5an1 = 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 +...
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 Summer '10
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 Graph Theory, Trigraph, Characteristic polynomial, Recurrence relation, ur initial conditions

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