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Unformatted text preview: Modern Analysis, Homework 4 Possible Solutions 1. For h,k ∈ R not both zero, we have 1 √ h 2 + k 2 f ( x + h, + k ) f ( x,y ) 2 x + y x 1 h k =  h   h + k  √ h 2 + k 2 ≤  h   h  +  k  √ h 2 + k 2 (1) But  h  +  k  ≤ √ 2 √ h 2 + k 2 for any pair h,k ∈ R (square both sides, simplify and recognize a perfect square). Therefore equation 1 is bounded above by √ 2  h  which goes to 0 as h,k → 0. This proves that f is differentiable at any point ( x,y ) with differential matrix 2 x + y x 1 . 2. (a) For ( x,y ) 6 = (0 , 0), we have f x ( x,y ) = x 2 ( x 2 + 3 y 2 ) ( x 2 + y 2 ) 2 and f y ( x,y ) = 2 x 3 y ( x 2 + y 2 ) 2 . This implies that  f x ( x,y )  ≤ (3 x 2 +3 y 2 ) 2 ( x 2 + y 2 ) 2 = 9 (if ( x,y ) 6 = (0 , 0)). Similarly, since  2 xy  ≤ x 2 + y 2 , we get that  f y ( x,y )  ≤ ( x 2 + y 2 ) 2 / ( x 2 + y 2 ) = 1 (if ( x,y ) 6 = (0 , 0)). Notice that f x (0) and f y (0) both exist, and f x (0) = lim h → 1 h h 3 h 2 +0 = 1, f y (0) = lim h → 1 h · 0 = 0. Therefore f x and f y are bounded functions on R 2 . I claim that this implies that f is continuous at (0 , 0) (it is clearly continuous elsewhere): consider h,k ∈ R . Then by the mean value theorem (applied twice), we get  f ( h,k )  ≤  f ( h,k ) f (0 ,k )  +  f (0 ,k ) f (0 , 0)  ≤ (sup  f x  )  h  + (sup  f y  )  k  which goes to 0 as h,k → 0. This proves continuity. (b) Write ~u in coordinates: ~u = ( u 1 ,u 2 ). Then f ( h~u ) f (0 , 0) h = u 3 1 u 2 1 + u 2 2 = u 3 1 . This proves that ( D ~u f )(0 , 0) exists and is u 3 1 . Since ~u is a unit vector, this has absolute value at most 1. (c) Since f is clearly differentiable away from (0 , 0), by the chain rule g is differentiable at any point t ∈ R at which γ ( t ) 6 = (0 , 0). Suppose that γ ( t ) = (0 , 0). Writing γ in coordinates, γ ( t ) = ( γ 1 ( t ) ,γ 2 ( t )), we have g ( t + h ) g ( t ) h = 1 h γ 3 1 ( t + h ) γ 2 1 ( t + h ) + γ 2 2 ( t + h ) . (2) 1 Since γ 1 and γ 2 are differentiable at t , we have γ 3 1 ( t + h ) = ( γ 1 ( t )) 3 h 3 + o ( h 3 ) and γ 2 1 ( t + h ) + γ 2 1 ( t + h ) = ( γ 1 ( t )) 2 h 2 + ( γ 2 ( t )) 2 h 2 + o ( h 2 ). It follows that equation 2 can be written as ( γ 1 ( t )) 3 + ε 1 ( h ) k γ ( t ) k 2 + ε 2 ( h ) for some pair of functions ε 1 ( h ) ,ε 2 ( h ) satisfying ε 1 ( h ) ,ε 2 ( h ) → 0 as h → 0. Since k γ ( t ) k 6 = 0, it follows that g is differentiable at t with derivative ( γ 1 ( t )) 3 k γ ( t ) k 2 . (d) If f were differentiable at (0 , 0), then for a unit vector ~u = ( u 1 ,u 2 ) we would have ( D ~u f )(0 , 0) = f x (0 , 0) u 1 + f y (0 , 0) u 2 = u 1 but we saw that ( D ~u f )(0 , 0) = u 3 1 so f cannot be differentiable at the origin....
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This note was uploaded on 10/18/2011 for the course MATH S4062Q taught by Professor Staff during the Summer '11 term at Columbia.
 Summer '11
 Staff

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