Winter 2009 Quiz 1b

Winter 2009 Quiz 1b - MATH 239 Quiz — Winter 2009...

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Unformatted text preview: MATH 239 Quiz — Winter 2009 SOLUTIONS No calculators or other aids may be used. Total marks: 20 1. (a) Let m be a positive integer. Find an expression (you may use summation notation) for [ x m ](1- x 2 ) 37 (1- 3 x )- 3 . [5 marks] Solution [ x m ](1- x 2 ) 37 (1- 3 x )- 3 = [ x m ] 37 X j =0 37 j (- 1) j x 2 j ∞ X k =0 2 + k 2 3 k x k ! = [ x m ] 37 X j =0 ∞ X k =0 2 + k 2 3 k 37 j (- 1) j x 2 j + k . Since we get an x m term whenever 2 j + k = m , we can sum over the relevant values of j and set k = m- 2 j to get [ x m ] 37 X j =0 ∞ X k =0 2 + k 2 3 k 37 j (- 1) j x 2 j + k = 37 X j =0 2 + m- 2 j 2 3 m- 2 j 37 j (- 1) j (b) Let S be the set of all subsets of the positive integers { 1 , 2 , 3 , . . . } . Define the weight function ω as follows. For the empty set {} , we let ω ( {} ) = 0 and for any non-empty set σ we let ω ( σ ) equal the largest element of the subset σ . For example, if σ = { 1 , 2 , 4 } , then ω ( σ ) = 4, since 4 is the largest element in σ ....
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Winter 2009 Quiz 1b - MATH 239 Quiz — Winter 2009...

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