m425b-hw6soln-s08

# m425b-hw6soln-s08 - MATH 425b ASSIGNMENT 6 SOLUTIONS SPRING...

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Unformatted text preview: MATH 425b ASSIGNMENT 6 SOLUTIONS SPRING 2007 Prof. Alexander Chapter 9: (7) Suppose E ⊂ R is open, f : E → R , and there exists M such that | ( D j f )( x ) | ≤ M for all x ∈ E and all j ≤ M . Let x ∈ E and let B be a neighborhood of x with B ⊂ E . Suppose x + h ∈ B for some h , and let v k = x + ∑ k i =1 h i e i , so v = x,v n = y , and all v k ∈ B . Fix some k ≤ n . Since D k f exists in B , the function g ( t ) = f ( v k- 1 + th k e k ) is differentiable in [0 , 1] with g ( t ) = ( D k f )( v k- 1 + th k e k ) h k . By the mean value theorem, there exists t ∈ (0 , 1) such that f ( v k )- f ( v k- 1 ) = g (1)- g (0) = g ( t )(1- 0) , so | f ( v k )- f ( v k- 1 ) | ≤ | g ( t ) | ≤ M | h k | . Summing, we get | f ( x + h )- f ( x ) | ≤ n X k =1 ( f ( v k )- f ( v k- 1 )) ≤ M n X k =1 | h k | , which approaches 0 as h → 0, so f is continuous. (8) If f has a local maximum, the for every i , the function g ( t ) = f ( x + te i ) has a local maximum at t = 0. By Theorem 5.8, g (0) = 0. But g (0) = ( D i f )( x ) so ( D i f ( x ) = 0 for all i , so all entries in the 1 × n matrix of f ( x ) are 0, that is, f ( x ) = 0. (16) f ( t ) = t +2 t 2 sin 1 /t . For t 6 = 0 , f ( t ) = 1+4 t sin 1 t- 2 cos 1 t , so for t ∈ (- 1 , 1) , | f ( t ) | ≤ 1 + 4 + 2 = 7. For t = 0, lim t → f ( t )- f (0) t- = lim t → 1 + 2 t sin 1 t = 1 , so f (0) = 1. Thus | f ( t ) | ≤ 7 for all t ∈ (- 1 , 1). Also | t | < 1 8 implies 4 t sin 1 t ≤ 1 2 . (1) For t = 1 /nπ , n even, with | t | < 1 / 8 we have 1- 2 cos 1 /t =- 1 so f ( t ) < 0. For t = 1 /nπ , n odd, with t < 1 / 8 we have 1- 2 cos 1 /t = 3 so f ( t ) > 0, using (1). Thus as t → 0, f alternates between intervals where f > 0 and f < 0 (i.e. where f increases and decreases) so f is not one-to-one....
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m425b-hw6soln-s08 - MATH 425b ASSIGNMENT 6 SOLUTIONS SPRING...

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