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solnsA2a

# solnsA2a - C&O 330 SOLUTIONS#2 PROFESSOR D.M,JACKSON...

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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 λ ( π ) be the number of 1s in π , let μ ( π ) be the number of distinct parts in π ∈ P and let ω ( π ) = n where π turnstileleft n. Let A ( x,u ) = [( P ,ω ⊗ λ )] o ( x,u ) = ∑ n,k ≥ a n,k x n u k and let B ( x,u ) = [( P ,ω ⊗ μ )] 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 Ω: P ↔ × i ≥ 1 { i } star . Clearly, Ω is additively ω ⊗ λ-preserving, so A = productdisplay i ≥ 1 bracketleftbig( { i } star ) ,ω ⊗ λ bracketrightbig o ( x,u ) , by the Product Lemma. But bracketleftbig( { i } star ) ,ω ⊗ λ bracketrightbig o = summationdisplay j ≥ bracketleftbig( i j ,ω ⊗ λ )bracketrightbig o = summationdisplay j ≥ x ω ( i j ) u λ ( 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 ) ,ω ⊗ μ bracketrightbig o ( x,u ) where bracketleftbig( { i } star ) ,ω ⊗ μ bracketrightbig o ( x,u ) = summationdisplay j ≥ x ω ( i j ) u μ ( 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 D.M,JACKSON...

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