a6_soln

# a6_soln - x, y ) within 1 / 4 of (0 , π/ 4). Find an upper...

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Math 237 Assignment 6 Solutions 1. Let f ( x, y ) = e - 2 x + y . Use Taylor’s theorem to show that the error in the linear approximation L (1 , 1) ( x, y ) is at most 6 e [( x - 1) 2 + ( y - 1) 2 ] if 0 x 1 and 0 y 1. Solution: We have f x = - 2 e - 2 x + y f y = e - 2 x + y , f xx = 4 e - 2 x + y , f xy = - 2 e - 2 x + y , f yy = e - 2 x + y . Thus, on 0 x 1 and 0 y 1 we have | f xx | ≤ 4 e, | f xy | ≤ 2 e, | f yy | ≤ e. Since f C 2 , by Taylor’s Theorem there is a point c such that ± ± R 1 , (1 , 1) ( x, y ) ± ± = 1 2 ± ± f xx ( c )( x - 1) 2 + 2 f xy ( c )( x - 1)( y - 1) + f yy ( c )( y - 1) 2 ± ± 1 2 ² | f xx ( c ) | ( x - 1) 2 + 2 | f xy ( c ) || ( x - 1) || ( y - 1) | + | f yy ( c ) | ( y - 1) 2 ³ 2 e ( x - 1) 2 + 2 e | x - 1 || y - 1 | + 1 2 e ( y - 1) 2 2 e ( x - 1) 2 + e ( x - 1) 2 + e ( y - 1) 2 + 2 e ( y - 1) 2 = 3 e [( x - 1) 2 + ( y - 1) 2 ] 2. Consider f : R 2 R deﬁned by f ( x, y ) = 2 x 2 + 3 y 2 . Prove that for any ( a, b ) R 2 we have f ( x, y ) L ( a,b ) ( x, y ) for all ( x, y ) R 2 . Solution: We have f = (4 x, 6 y ) and so f xx = 4, f xy = 0 and f yy = 6. Hence, by Taylor’s theorem there exists a point ( c, d ) on the line segment joining ( a, b ) to ( x, y ) such that f ( x, y ) = L ( a,b ) ( x, y ) + 1 2 f xx ( c, d )( x - a ) 2 + f xy ( c, d )( x - a )( y - b ) + 1 2 f yy ( c, d )( y - b ) 2 Thus f ( x, y ) - L ( a,b ) ( x, y ) = 2( x - a ) 2 + 3( y - b ) 2 0 , as required.

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2 3. The magnitude of each second partial derivative of f is less than 2, for all (
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Unformatted text preview: x, y ) within 1 / 4 of (0 , π/ 4). Find an upper bound for the error in the linear approximation L (0 ,π/ 4) ( x, y ) of f for all ( x, y ) within 1 / 4 of (0 , π/ 4). Solution: Assuming f has continuous second partial derivatives, Taylor’s theorem gives that there exists a point c on the line segment joining ( x, y ) to (0 , π/ 4) such that | R 1 , (0 ,π/ 4) ( x, y ) | = ± ± ± ± 1 2 ² f xx ( c ) x 2 + 2 f xy ( c ) x ( y-π/ 4) + f yy ( c )( y-π/ 4) 2 ³ ± ± ± ± ≤ 1 2 ² | f xx ( c ) | x 2 + 2 | f xy ( c ) || x || y-π/ 4 | + | f yy ( c ) | ( y-π/ 4) 2 ³ ≤ 1 2 ² 2 x 2 + 4 | x || y-π/ 4 | + 2( y-π/ 4) 2 ³ ≤ x 2 + x 2 + ( y-π/ 4) 2 + ( y-π/ 4) 2 = 2 x 2 + 2( y-π/ 4) 2...
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## This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.

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a6_soln - x, y ) within 1 / 4 of (0 , π/ 4). Find an upper...

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