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m425b-hw7soln-s08

# m425b-hw7soln-s08 - MATH 425b ASSIGNMENT 7 SOLUTIONS SPRING...

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MATH 425b ASSIGNMENT 7 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 9: (23) f ( x, y 1 , y 2 ) = x 2 y 1 + e x + y 2 . Clearly f (0 , 1 , - 1) = 0. We have ( D 1 f )( x, y 1 , y 2 ) = 2 xy 1 + e x so ( D 1 f )(0 , 1 , - 1) = 1 6 = 0. In the notation of the Implicit Function Theorem we have A x = ( D 1 f )( x, y 1 , y 2 ) (which is 1 × 1, i.e. a scalar.) Hence in a neighborhood of (0 , 1 , - 1) we can solve f ( x, y 1 , y 2 ) = 0 for x in terms of y = ( y 1 , y 2 ), obtaining x = g ( y 1 , y 2 ), so that f ( g ( y 1 , y 2 ) , y 1 , y 2 ) = 0. By the Implicit Function Theorem we have g 0 (1 , - 1) = - A - 1 x A y , where A x = ( D 1 f )(0 , 1 , - 1) = 1 and A y is the 1 × 2 matrix A y = [( D 2 f )(0 , 1 , - 1) ( D 3 f )(0 , 1 , - 1)] = [ x 2 1] = [0 1] , so g 0 (1 , - 1) = [0 - 1]. The entries of this matrix are ( D 1 g )(1 , - 1) = 0 , ( D 2 g )(1 , - 1) = - 1. (26) Let f : R 2 R 2 be given by f ( x, y ) = | x | . Then ( D 2 f )( x, y ) = 0 for all x, y , so ( D 12 f )( x, y ) = 0 for all x, y . But ( D 1 f )(0 , y ) would be the derivative of g ( x ) = | x | at x = 0, which doesn’t exist. (29) If we interchange i j and i j +1 for some j 1, we get D i 1 ...i j i j +1 ...i k f = D i 1 ...i j - 1 ( D i j i j +1 ( D i j +2 ...i k f )) . (1) Since f ∈ C k , we have D i j +2 ...i k f ∈ C j +1 ⊂ C 2 so D i j i j +1 ( D i j

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