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m425b-THfinsoln-s08

# m425b-THfinsoln-s08 - MATH 425b TAKE-HOME FINAL EXAM...

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MATH 425b TAKE-HOME FINAL EXAM SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) Let γ ( t ) = p k + t ( p 0 k - p k ) , t [0 , 1], which traces a line from p k to p 0 k . Let σ t = [ p 0 · · · p k - 1 γ ( t )] be the corresponding simplex. Since these points are on opposite sides on H , there must be a value t 0 where γ ( t 0 ) H . The mapping σ 0 is given by σ 0 ( x ) = p 0 + k X m =1 x m ( p m - p 0 ) , so ( σ 0 ) i ∂x j = ∂x j k X m =1 x m ( p j - p 0 ) i = ( p j - p 0 ) i , which means the j th column of the derivative matrix σ 0 0 ( x ) is p j - p 0 , for all x . For σ t the derivative matrix is the same except the last column is γ ( t ) - p 0 instead of p k - p 0 . Let H 0 be the translate of H which passes through the origin; since p 0 , . . . , p k - 1 , γ ( t 0 ) all lie in H , the k vectors p 1 - p 0 , . . . , p k - 1 - p 0 , γ ( t 0 ) - p 0 all lie in the subspace H 0 . Since H 0 is ( k - 1)-dimensional, these k vectors (the columns of σ 0 t 0 ( x )) must be linearly dependent. This means σ 0 t 0 ( x ) has determinant 0, that is, J ( t 0 )( x ) = 0, for all x . Now each entry of the last column of σ 0 t ( x ) has the form a + bt with a, b R , so expanding the determinant around this last column shows that J ( t ) also has the form c + dt . Since σ 0 is invertible, we have J (0) 6 = 0, while J ( t 0 ) = 0, so we must have d 6 = 0. Since J ( t 0 ) = 0 this means J

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