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# hw5sol - Math 146 Algebraic Combinatorics Homework 5...

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Unformatted text preview: Math 146: Algebraic Combinatorics Homework 5 Solutions A hint on problem 3.1, which is quite difficult otherwise: Given a partition on n into odd parts, you can center its Ferrers diagram. Then you can divide the centered diagram into L shapes, which is then a partition of the same n into distinct parts. For example, given 18 = 7 + 5 + 3 + 3, the centered diagram is to yield 18 = 7 + 6 + 4 + 1. You should describe more rigorously what the L shapes do, and prove that it is a bijection between the two types of partitions. 3.1 Give an explicit 1- 1 correspondence between partitions of n into distinct parts and partitions of n into odd parts. Solution. Note: This proof does not use the hint provided above. Let λ = ( λ 1 ,..., λ k ) be a partition of n so that n = λ 1 + ··· + λ k and each λ i is odd. Produce a partition μ from λ by reading λ left to right and each time you encounter a repeated part say λ i = λ i + 1 , add those two parts together. Repeat this process until you get (by construction) a partition μ of n with no repeated parts, ie a partition of n with distinct parts. To invert this procedure, let μ = μ 1 + μ 2 + ··· + μ ‘ be a partition with distinct parts. Read μ left to right and each time you see an even part divide it into two equal parts of half the size. Repeat as needed until you get (by construction) a partition with only odd parts. The two maps above display a bijection between partitions of n with odd parts and partitions of n with distinct parts. 3.21 In a country that has 1-cent, 2-cent, and 3-cent coins only, show that the number of ways of changing n cents is exactly the integer nearest to ( n + 3 ) 2 12 . Solution. Let f ( n ) be the number of ways of changing n with the given coins. Then f ( n ) is the coefficient of x n in the expansion of ( 1 + x + x 2 + x 3 + ··· )( 1 + x 2 + x 4 + x 6 + ··· )( 1 + x 3 + x 6 + x 9 + ··· ) . Thus we have the generating function ∑ n ≥ f ( n ) x n = ∑ n ≥ x n ! ∑ n ≥ x 2 n ! ∑ n ≥ x 3 n ! = 1 ( 1- x )( 1- x 2 )( 1- x 3 ) = 1 ( 1- x ) 3 ( 1 + x )( 1 + x + x 2 ) . Partial fractions yeilds (here ω can be determined using the quadratic formula on 1 + x + x 2 , and ¯ ω is its complex conjugate) 1 ( 1- x )( 1- x 2 )( 1- x 3 ) = 1 6 ( 1- x ) 3 + 1 4 ( 1- x ) 2 + 17 72 ( 1- x ) + 1 8 ( 1 + x ) + 1 9 ( 1- ω x ) + 1 9 ( 1- ¯ ω x ) . 1 Now if we expand each fraction on the right-hand-side as a power series and read off the coefficient of x n we get the formula f ( n ) = 1 6 n + 2 2 + 1 4 ( n + 1 )+ 17 72 + (- 1 ) n 8 + 2 9 cos ( 2 n π 3 ) which can be simplified as f ( n ) = ( n + 3 ) 2 12 +- 7 + 9 (- 1 ) n + 16cos ( 2 n π / 3 ) 72 where the second term is always less that 1 / 2 in absolute value. Thus f ( n ) is the unique integer whose distance from ( n + 3 ) 2 / 12 is less than 1 / 2 as required....
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hw5sol - Math 146 Algebraic Combinatorics Homework 5...

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