m425b-hw7soln-s08

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

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Unformatted text preview: 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 (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 (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...
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This note was uploaded on 06/23/2008 for the course MATH 425B taught by Professor Alexander during the Fall '07 term at USC.

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

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