hw4-sol

# hw4-sol - Modern Analysis Homework 4 Possible Solutions 1...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

## This note was uploaded on 10/18/2011 for the course MATH S4062Q taught by Professor Staff during the Summer '11 term at Columbia.

### Page1 / 6

hw4-sol - Modern Analysis Homework 4 Possible Solutions 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online