solns2

solns2 - Mathematics 105 — Spring 2004 — M. Christ 1...

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Unformatted text preview: Mathematics 105 — Spring 2004 — M. Christ 1 Solutions to selecta from problem set #2, with problem set #3 2-5 . Let f ( x, y ) = x | y | / p x 2 + y 2 for ( x, y ) 6 = (0 , 0), and let f (0 , 0) = 0. Show that f is not differentiable at 0. (Note: It is customary to write 0 as shorthand for (0 , , ··· , 0) ∈ R n , just as the additive identity element of a vector space is customarily denoted by 0.) Solution. Suppose that f were differentiable at 0. We have f (0+ t, 0)- f (0 , 0) ≡ 0. Taking v = ( t, 0) = te 1 in the definition of Df (0) would give us 0 = lim | t |→ | f (0 + t, 0)- f (0 , 0)- T ( te 1 ) | | te 1 | = lim t → | T ( e 1 ) | 1 = | T ( e 1 ) | , so T ( e 1 ) = 0. Likewise f (0 , 0 + t ) = f (0 , 0) ≡ 0, so the same reasoning gives T ( e 2 ) = 0. Since e 1 , e 2 together form a basis for R 2 , this implies that T = 0, that is, T ( v ) = 0 for every v ∈ R 2 . Now consider v = ( t, t ). We are still assuming that Df (0) exists, and we have proved that if it does, it must equal 0. Therefore 0 = lim | t |→ | f ( t, t )- f (0 , 0)- T ( t, t ) | | t | √ 2 = lim | t |→ 2- 1 / 2 | f ( t, t ) | | t | = lim | t |→ 2- 1 / 2 t | t | , which is a contradiction. Therefore Df (0) cannot exist. 2-7 . Let f : R 2 → R satisfy | f ( x ) | ≤ | x | 2 . Show that f is differentiable at 0. Solution. This is one of those cases where it’s easier to prove more than is asked for: We’ll show that Df (0) = 0. (As is so often the case in vector calculus, the notation in this equation is imperfect; the zero on the left-hand side of the equation is the origin in R 2 , while the zero on the right-hand side refers to the linear transformation T : R 2 → R which satisfies T ( v ) = 0 for all v ∈ R 2 .) Letting T be the linear transformation 0, for any x 6 = 0 we have | f ( x )- f (0)- T ( x ) | | x- | = | f ( x ) | | x | ≤ | x | . This certainly tends to 0 as x → 0, so we’ve shown that Df (0) = 0. 2-15(a) . Define a function det : ( R n ) n → R by regarding an element x = ( x 1 , ··· , x n ) ∈ ( R n ) n as a matrix whose rows are the vectors x j , and taking its determinant. Show that det is a differentiable function, and show that D det( a )( x ) = n X i =1 det ( a 1 , a 2 , ··· , a i- 1 , x i , a i +1 , ··· , a n ) . (Here ( a 1 , a 2 , ··· , a i- 1 , x i , a i +1 , ··· ) ∈ ( R n ) n , so we can form its determinant. Note also that the notation (which is commonly used) is slightly misleading when i = 1 or n .) Comment. In this problem we systematically identify ( R n ) n with R n 2 via the corre- spondence (( x 1 1 , ··· , x n 1 ) , ··· , ( x 1 n , ··· , x n n )) ↔ ( x 1 1 , ··· , x n 1 , x 1 2 , ··· , x n 2 , x 1 3 , ··· ). And we also identify elements of ( R n ) n...
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This note was uploaded on 05/07/2008 for the course MATH 105 taught by Professor Michaelchrist during the Spring '04 term at Berkeley.

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solns2 - Mathematics 105 — Spring 2004 — M. Christ 1...

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