quiz3solutions

# quiz3solutions - e x-1 2 e x We want the co-e ﬃ cient of...

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Quiz 2 Solutions Dilip Raghavan October 22, 2008 Problem 1. (a) The ordinary generating function is ( x 0 + x 2 + x 4 + x 6 + ··· ) 8 (b) We know that 1 1 - x = x 0 + x 1 + x 2 + x 3 + ··· . Now replacing x with x 2 in this, we get 1 1 - x 2 = x 0 + x 2 + x 4 + x 6 + ··· So we want the co-e cient of x 20 in 1 (1 - x 2 ) 8 . We know 1 (1 - x 2 ) 8 = ± 0 + 8 - 1 0 ! x 0 + ± 1 + 8 - 1 1 ! x 2 + ± 2 + 8 - 1 2 ! x 4 + ± 3 + 8 - 1 3 ! x 6 + ··· So the co-e cient of x 20 is ± 10 + 8 - 1 10 ! Problem 2. (a) The exponential generating function is ± x 1 1! + x 2 2! + x 3 3! + x 4 4! + ··· ! 2 ± x 0 0! + x 1 1! + x 2 2! + x 3 3! + ··· ! (b) The generating function is equal to (
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Unformatted text preview: e x-1) 2 e x . We want the co-e ﬃ cient of x r in this. We have ( e x-1) 2 e x = ( e 2 x-2 e x + 1) e x = e 3 x-2 e 2 x + e x . Using the Taylor series for e x this is equal to ∞ X r = 3 r x r r !-∞ X r = 2 r + 1 x r r ! + ∞ X r = x r r ! So the co-e ﬃ cient of x r r ! is 3 r-2 r + 1 + 1. 1...
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