exam3solns - Exam 3 Solutions 1. Using the bilinearity of...

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Unformatted text preview: Exam 3 Solutions 1. Using the bilinearity of the inner product we have k x + y k 2- k x- y k 2 = h x + y,x + y i - h x- y,x- y i = h x,x i + 2 h x,y i + h y,y i - ( h x,x i + 2 h x,- y i + h- y,- y i ) = h x,x i + 2 h x,y i + h y,y i - ( h x,x i - 2 h x,y i + h y,y i ) = 4 h x,y i . 2. For ( x,y ) 6 = (0 , 0) we have f = ( 1 2 ) ( g ( 2 1 + 2 2 + 2 2 ) , where g ( x ) = 1 x . Since x 2 + xy + y 2 = ( x + y ) 2- xy , it follows that ( 2 1 + 2 2 + 2 2 )( x,y ) 6 = 0 when ( x,y ) 6 = (0 , 0) and thus f is continuous at all such points. However, f is not continuous at (0 , 0) because f (0 , 1 k ) 0 and f ( 1 k , 1 k ) 1 3 as k . 3. (a) Take d f ( S ). Then there is c S such that f ( c ) = d . Since c S there is ( x k ) in S such that x k c . Since f ( x k ) f ( c ) = d by the continuity of f and f ( x k ) f ( S ) for each k 1, we have d = f ( c ) f ( S )....
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This note was uploaded on 04/09/2010 for the course MATH 312 taught by Professor Wynne during the Fall '09 term at Simons Rock.

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exam3solns - Exam 3 Solutions 1. Using the bilinearity of...

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