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Unformatted text preview: Math 146: Algebraic Combinatorics Homework 3 Solutions 1.16 Find the coefficient of x n in the power series for f ( x ) = 1 ( 1 x 2 ) 2 , first by the method of partial fractions, and second, give a much simpler derivation by being sneaky. Solution. The partial fractions decomposition is 1 ( 1 x 2 ) 2 = 1 (( 1 x )( 1 + x )) 2 = 1 4 ( 1 x ) 2 + 1 4 ( 1 x ) + 1 4 ( 1 + x ) 2 + 1 4 ( 1 + x ) . Writing the power series expansion for each term on the righthandside above gives the coefficient of x n as n + 1 4 + 1 4 + ( 1 ) n ( n + 1 ) 4 + ( 1 ) n 4 which is 0 if n is odd and is n + 2 2 if n is even. A sneaky method would be to write down the series for 1 ( 1 u ) 2 and then replace u with x 2 . 1.17 Let b ( n , k ) be the number of permutations of n letters that have exactly k inversions. Find a simple formula for the generating function B n ( x ) = k b ( n , k ) x k . Make a table of values of b ( n , k ) for n 5. Solution. Fix j , 1 j n , and consider just those permutations of n letters that have ( j ) = n . Viewed in twoline notation, the entry n in the bottom row would be directly below the entry j in the top row. With ( j ) = n , then no inversions have j as the second member of the pair. This can be seen in twoline notation by noticing nothing to the left on n in the bottom row is greater than n . Also, exactly n j inversions have j as the first member of the pair. This can be seen in twoline notation by noticing that all elements to the right of n in the bottom row are smaller than n . Hence if we delete the entry n from the bottom row of (viewed in twoline notation) then we obtain a permutation of n 1 letters with n j fewer inversions. Thus b ( n , k ) = j b ( n 1 ) , k n + j ) . Multiply by x k and sum on k to obtain k b ( n , k ) x k = k j b ( n 1 , k n + j ) ! x k = k b ( n 1 , k n + 1 ) x k + + k b ( n 1 , k ) x k = ( x n 1 + x n 2 + + 1 ) k b ( n 1 , k ) x k ....
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 Spring '10
 GregKuperberg

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