solnsA2a - C&O 330 - SOLUTIONS #2 PROFESSOR...

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Unformatted text preview: C&O 330 - SOLUTIONS #2 PROFESSOR D.M,JACKSON This assignment is about partitions. The questions are taken from the Exercises of Section 2.4 (page 38) of the Course Notes. A (15 points) Question 2. Solution: Let P be the set of all partitions, let ( ) be the number of 1s in , let ( ) be the number of distinct parts in P and let ( ) = n where turnstileleft n. Let A ( x,u ) = [( P , )] o ( x,u ) = n,k a n,k x n u k and let B ( x,u ) = [( P , )] o ( x,u ) = n,k b n,k x n u k . Note that clearly A,B C [ u ][[ x ]] . Then the total number of 1s in the set of all partitions of n is c n = n summationdisplay k =0 ka n,k = [ x n ] L u A u and the sum of the number of distinct parts in the set of all partitions of n is d n = n summationdisplay k =0 kb n,k = [ x n ] L u B u , where L u f ( x,u ) is defined to be f ( x, 1) for f C [ u ] [[ x ]] . Then c n = d n for all n L u A u = L u B u so, to establish the result, we demonstrate that the right hand side of the bi-implicant holds. A universal decomposition for P is : P i 1 { i } star . Clearly, is additively -preserving, so A = productdisplay i 1 bracketleftbig( { i } star ) , bracketrightbig o ( x,u ) , by the Product Lemma. But bracketleftbig( { i } star ) , bracketrightbig o = summationdisplay j bracketleftbig( i j , )bracketrightbig o = summationdisplay j x ( i j ) u ( i j ) = braceleftbigg 1 + ux + u 2 x 2 + if i = 1 , 1 + x i + x 2 i + if i negationslash = 1 , = braceleftbigg 1 1- ux if i = 1 , 1 1- x i if i negationslash = 1 , 1 2 PROFESSOR D.M,JACKSON so A = 1 1- ux productdisplay i 2 1 1- x i . Then L u A u = x 1- x productdisplay i 1 1 1- x i . Similarly, B = productdisplay i 1 bracketleftbig( { i } star ) , bracketrightbig o ( x,u ) where bracketleftbig( { i } star ) , bracketrightbig o ( x,u ) = summationdisplay j x ( i j ) u ( i j ) = 1 + u ( x i + x 2 i + ) = 1 + x i ( u- 1) 1- x i , so B = productdisplay i 1 1 + x i ( u- 1) 1- x i . Then log ( B ) = summationdisplay i 1 ( 1 + x i ( u- 1) )- summationdisplay i 1 ( 1- x i ) , and, differentiating partially with respect to u , we have L u 1 B B u = L u summationdisplay i 1 x i 1 + x i ( u- 1) = summationdisplay i 1 x i = x 1- x ....
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solnsA2a - C&O 330 - SOLUTIONS #2 PROFESSOR...

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