FL2009 Prelim2 solutions - Math 3210 1. second prelim...

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Math 3210 second prelim Wednesday 28 October 2009 10:10-11:00 am Solutions 1. (a) σ - 1 = σ ; (b) στ = ( n - 2, n , n - 1, n - 3, n - 4, . . . , 3, 2, 1 ) ; (c) τσ = ( n , n - 1, n - 2, n - 3, . . . , 6, 5, 4, 2, 1, 3 ) ; (d) στσ - 1 = ( 1, 2, 3, . . . , n - 4, n - 3, n - 1, n , n - 2 ) . 2. (a) φ i ( x 1 , x 2 ) = x n + 1 - i 1 x i - 1 2 , so φ * y i = x n + 1 - i 1 x i - 1 2 and φ * dy i = ( n + 1 - i ) x n - i 1 x i - 1 2 dx 1 + ( i - 1 ) x n + 1 - i 1 x i - 2 2 dx 2 . (b) φ * ( dy i dy j ) = φ * dy i φ * dy j = ( ( n + 1 - i ) x n - i 1 x i - 1 2 dx 1 + ( i - 1 ) x n + 1 - i 1 x i - 2 2 dx 2 ) × ( ( n + 1 - j ) x n - j 1 x j - 1 2 dx 1 + ( j - 1 ) x n + 1 - j 1 x j - 2 2 dx 2 ) = ( n + 1 - i )( j - 1 ) x 2 n + 1 - ( i + j ) 1 x i + j - 3 2 dx 1 dx 2 + ( i - 1 )( n + 1 - j ) x 2 n + 1 - ( i + j ) 1 x i + j - 3 2 dx 2 dx 1 = ( ( n + 1 - i )( j - 1 ) - ( i - 1 )( n + 1 - j ) ) x 2 n + 1 - ( i + j ) 1 x i + j - 3 2 dx 1 dx 2 = ( j - i ) nx 2 n + 1 - ( i + j ) 1 x i + j - 3 2 dx 1 dx 2 . 3.
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This note was uploaded on 04/14/2010 for the course MATH 3210 at Cornell University (Engineering School).

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